Sm Core Jacobs 3

Chapter 01 - Operations and Supply Chain Management

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CHAPTER 1OPERATIONS AND SUPPLY CHAIN MANAGEMENT

Review and Discussion Questions1. Look at the want ads at http://www.apics.org and evaluate the opportunities for an OSM major

with several years of experience.

There are pages and pages of these in the APICS Career Center. Here are some examples:

Global Active Ingredient Supply PlannerFMC CorporationUS - PA - PhiladelphiaThis is your opportunity to join the Agricultural Products Group (APG) and work on the teamresponsible for global active ingredient planning. You'll serve as a central Supply Plannerworking in tandem ...Sep-20-2011

Purchasing Manager (Buyer)Texas Dow Employees Credit Union (TDECU)US - TX - Lake Jackson - Nearest Metro area - South HoustonEducation Accredited university degree in Business or Marketing with certification in Purchasing,Inventory Management, or Logistics. Accredited Purchasing Practitioner (APP) or CertifiedPurchasing ...Sep-20-2011

Product Line ManagerCintas CorporationUS - OH - MasonHigh School diploma or GED required, 4 year degree preferred Knowledge of and experienceusing Internet and Microsoft Office (Word, Excel, PowerPoint, and Email) 3 years managementexperience preferred ...Sep-20-2011

Director PurchasingLegendary BakingUS - CO - DenverCERTIFICATIONS & LICENSES • Valid Driver’s License and car insurance. • CertifiedPurchasing Manager certification (C.P.M.) preferred. • Certified Food Purchasing Manager ...Sep-19-2011

Process Improvement ManagerARAMARK CorporationUS - TX - HoustonBachelor's Degree required. Technical Engineering discipline within Industrial, Mechanical,Chemical, or Food Operations strongly preferred. • Minimum 5 years Lean manufacturingexperience coupled ...Sep-19-2011


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2. What factors account for the resurgence of interest in OSCM today?

With companies facing competition on a global scale, and ever-advancing manufacturing andinformation technologies, firms realize the competitive advantage their OSCM functions canprovide if properly managed. Many have found that the same old way of doing business leavesthem unable to compete successfully. The 2011 tsunami in Japan has also brought to theforefront how important supply chains are, as well as the negative economic impact thatdisruptions in the supply chain can cause.

3. Using Exhibit 1.3 as a model, describe the source-make-deliver-return relationships in thefollowing systems:

a. An airlineSource: Aircraft manufacturer, in-flight food, repair parts, computer systemsMake: Aircraft and flight crew scheduling, ground services provided at airports, aircraftmaintenance and repairDeliver: Outbound and arriving passenger service, baggage handlingReturn: Resolve any post-service issues such as lost or damaged luggage

b. An automobile manufacturerSource: Suppliers of components and raw materialsMake: Manufacturing of vehicles and components or subassemblies to be sold as sparepartsDeliver: Delivery to and sales from dealerships, delivery of spare parts to the wholesalesystemReturn: Warranty and recall repairs, trade-ins

c. A hospitalSource: Medical supplies, cleaning services, disposal services, food services, qualifiedpersonnelMake: Inpatient rooms, outpatient clinics, emergency room, operating roomsDeliver: Scheduling patients, providing treatment, ambulance service, family counselingReturn: Billing errors, follow up visits

d. An insurance companySource: Supplies needed for the office, underwriters, legal authority to operateMake: Establish policy guidelines and pricing, field agent/representative and facilitynetwork, develop Internet service capabilities, establish preferred vehicle repair servicenetworkDeliver: Meet with and advise clients, write policies, process and pay claimsReturn: refund of overpayments

4. Define the service package of your college or university. What is its strongest element?What is its weakest one?

The categories with examples are:Supporting facility - location, buildings, labs, parkingFacilitating goods – class schedules, computers, books, chalkExplicit services – classes with qualified instructors, placement offices


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Implicit services – status and reputation (e.g., Ivy League schools)

At Indiana University and the University of Southern California, among their strongestelements are their business schools and their Operations Management programs (of course).Both also have very dedicated alumni networks. A weak element of Indiana University is itsweak football program; for USC, weak elements are on-campus parking and housing.

5. What service industry has impressed you the most with its innovativeness?

Our vote goes to cruise lines which have introduced such onboard innovations as wavemachines for belly boarding and rock climbing walls, as well as all sorts of other amenities tokeep cruisers involved. The industry is doing record business as well.

Some of the standout companies in less innovative industries are Bank of America (has aformalized research program to try out new customer services/amenities such as videoscreens in next to teller lines), Intuit (e.g., putting Quicken money management softwareonline), Ikea, JetBlue Airlines, and Progressive Insurance (discussed later in the book).

6. What is product-service bundling and what are the benefits to customers?

Product-service bundling is building Value-added services to a firm’s product offerings tocreate more value for the customer. This provides benefits in two areas. First, thisdifferentiates the organization from the competition. Secondly, these services tie customersto the organization in a positive way.

7. What is the difference between a service and a good?

A service is an intangible process (you can’t hold it in your hands), while a good is thephysical output of a process.

8. Recent outsourcing of parts and services that had previously been produced internally isaddressed by which current issue facing operations and supply management today?

The coordination of relationships between mutually supportive but separate organizations.

Chapter 02 - Strategy and Sustainability

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CHAPTER 2STRATEGY AND SUSTAINABILITY

Review and Discussion Questions1. Can a factory be fast, dependable, flexible, produce high-quality products, and still provide

poor service from a customer’s perspective?

Yes, if a customer’s needs are not considered and does not influence strategy development, anorganization could be delivering the wrong service or product. Even though the product is ofhigh quality, delivered quickly, and offers many options and features, overall service could berated “poor” by a customer who demands a different mix of features and attributes. It also couldbe a factor of how the product is delivered to the customer. Rude or inattentive salespersons,incompetent technical support, or difficulty in obtaining warranty service can all negativelyimpact the customer’s impression of a firm’s service, regardless of the ultimate quality of thegoods and the speed of delivery.

2. Why should a service organization worry about being world class if it does not competeoutside its own national border? What impact does the Internet have on this?

As the environment changes, firms can find themselves faced with competition from outside theirindustry, or from outside their home country. Even if they do not, the principles of a world-classfirm can be applied to any and all manufacturing and service concerns. Benchmarking or ratingyour firm’s performance to the best in your industry or class can provide future strategicdirections for improvements.

The Internet is global by its very nature. Retail stores must now compete with Internet stores.Local auction houses will be in competition with Internet auction sites such as eBay. Virtually allorganizations will be impacted in some form by the Internet. It is important that this impact beconsidered.

3. What are the major priorities associated with operations and supply chain strategy? How hastheir relationship to each other changed over the years?

The four major imperatives are cost, quality, delivery, and flexibility. In the sixties, these fourimperatives were viewed from a tradeoff’s perspective. For example, this meant that improvingquality would result in higher cost, and in many cases that was true. However, advances inmanufacturing and information technologies since then have reduced the size of those tradeoffs,allowing firms to improve on several or all of these imperatives simultaneously, gaining greatercompetitive advantage than was possible 50 years ago. The problem now becomes one ofprioritizing and managing towards orderly improvement.

4. Find examples where companies have used features related to environmental sustainability to“win” new customers.

Car companies use environmental concerns in marketing ads. The development of hybridand flex-fuel cars is one way they have operationalized those concerns. Consumer goodscompanies display the “made with recycled material” logo on the packaging. Bottled watermanufacturers are using and advertising bottles made with less plastic.


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5. For each of the different priorities in question 3, describe the unique characteristics of themarket niche with which it is most compatible.

Cost is most compatible with products that are commodities (i.e., highly standardized productswith many alternative suppliers). Quality provides companies a means of (1) differentiating aproduct and winning orders or (2) competing in a market and qualifying for orders. Quality isnow pervasive among all market niches in that customers now expect high quality. Speed andreliability of delivery are essential in those markets where there is a large degree ofcustomization. In addition, reliable delivery may be a competitive advantage in some regions ofthe world where delivery is difficult due to geographical or political reasons. Flexibility isimportant where customers demand low volume but wide varieties of products.

6. A few years ago the dollar showed relative weakness with respect to foreign currencies, suchas the yen, mark, and pound. This stimulated exports. Why would long-term reliance on alower valued dollar be at best a short-term solution to the competitiveness problem?

This approach is dependent on economic policies of other nations. This is a fragile dependency.A long-term approach is to increase manufacturing and service industry productivity in order toregain competitive advantage. At a national level, solutions appear to lie in reversing attitudes.At a firm level, competitive weapons are consistent quality, high performance, dependabledelivery, competitive pricing, and design flexibility.

7. In your opinion, do business schools have competitive priorities?

Yes. Their competitive priorities include:

Quality of professors and curriculum—consistent quality and high performance

Leader in development of new curriculum topics—design changes

Academic level of student attracted—consistent quality

Quantity and quality of research published—consistent quality

Quality of library resources—quality

What companies recruit at the school—after sales service

Success rate of graduates—consistent quality

Availability of financial aid—low price and after sales service

Cost of tuition—low price


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8. Why does the “proper” operations and supply chain strategy keep changing for companiesthat are world-class competitors?

The top three priorities have generally remained the same over time: make it good, make it fast,and deliver it on time. Others have changed. Part of this may be explained by realizing that worldclass organizations have achieved excellence in these three areas and are, therefore, focusingattention on some of the more minor areas to gain competitive advantage. The changes in theminor priorities may result from recognizing opportunities or from changes in customer desires orexpectations.

9. What is meant by the expressions order winners and order qualifiers? What was the orderwinner(s) for your last purchase of a product or service?

Order winners are dimensions that differentiate the product or service or services of one firmfrom another. Order qualifiers are dimensions that are used to screen a product or service as acandidate for purchase. Order qualifiers get a company’s “foot in the door.” Order winners arewhat make the sale. Obviously, answers will vary for the order winners from your last purchase.

10. Identify an operations and supply chain - related "disruption" that recently impacted acompany. What could the company have done to have minimized the impact of this type ofdisruption prior to it occurring?

The March 2011 tsunami that struck Japan was geographically concentrated but had globalimpact on multiple firms, many of which had no physical presence at all in the affected area.Examples include firms that had sole source agreements with suppliers in the affected area. Thetsunami left these companies scrambling to find new suppliers to feed into their supply chains.These firms could have reduced the impact of the tsunami by having a few high-quality,dependable suppliers located in different geographical regions. There are many other examplesthat could be taken from this one event. A simple Internet search will provide plenty of materialfor discussion.

11. What do we mean when we say productivity is a “relative” measure?

For productivity to be meaningful, it must be compared with something else. The comparisonscan be either intra-company as in the case of year-to-year comparisons of the same measure, orintercompany as in the case of benchmarking. Intercompany comparisons of single factorproductivity measures can be somewhat tenuous due to differences in accounting practices(especially when comparing with foreign competitors) and the balance of labor to capitalresources.. Total factor productivity measures are somewhat more robust for comparisonpurposes.


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Problems

1. As Operations Manager, you are concerned about being able to meet sales requirements in thecoming months. You have just been given the following production report.

JAN FEB MAR APR

Units Produced 2300 1800 2800 3000

Hours per Machine 325 200 400 320

Number of Machines 3 5 4 4

Find the average monthly productivity (units per machine hour).

To answer this we need to realize that the measure of hours given is per machine, so we have tomultiply that by the number of machines in each period to get the total machine hours in eachperiod. Those figures are used in the calculations below.

Average productivity: (2300/975 + 1800/1000 + 2800/1600 + 3000/1280)/4

Average productivity (2.36+1.80+1.75+2.34)/4= 2.06 units per machine hour

2. Sailmaster makes high-performance sails for competitive windsurfers. Below is informationabout the inputs and outputs for one model, the Windy 2000.

Units sold 1,217Sale price each $1,700Total labor hours 46,672Wage rate $12/hourTotal materials $60,000Total energy $4,000

Calculate the productivity in sales revenue/labor expense.We have to do some interim calculations here. Sales revenue is calculated by multiplying unitssold by the unit sales price. Labor expense is calculated by multiplying labor hours by the wagerate.

(1217*1700) / (46672*12) = 3.69


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3. Live Trap Corporation received the data below for its rodent cage production unit. Find the totalproductivity?

Output Input

50,000 cages Production time 620 labor hours

Sales price: $3.50 per unit Wages $7.50 per hour

Raw materials (total cost) $30,000

Component parts (total cost) $15,350

Total productivity could be expressed two ways here based on how you express output: in units sold, ordollars of sales.

Units sold:50,000 / ((620 * $7.50) + 30,000 + 15,350) = 1.00 units sold per dollar inputDollars of sales:(50000*3.5) / ((620 * $7.50) + 30,000 + 15,350) = 3.5 dollars in sales per dollar input

4. Two types of cars (Deluxe and Limited) were produced by a car manufacturer last year.Quantities sold, price per unit, and labor hours follow. What is the labor productivity for eachcar? Explain the problem(s) associated with the labor productivity.

QUANTITY $/UNITDeluxe car 4,000 units sold $8,000/carLimited car 6,000 units sold $9,500/carLabor, Deluxe 20,000 hours $12/hourLabor, Limited 30,000 hours $14/hour

Labor Productivity – units/hour

Model Outputin Units

Inputin Labor Hours

Productivity(Output/Input)

Deluxe Car 4,000 20,000 0.20

Limited Car 6,000 30,000 0.20

Labor Productivity – dollars

Model Outputin Dollars

Inputin Dollars


Deluxe Car 4,000($8,000)=$32,000,000

20,000($12.00)=$240,000

133.33

Limited Car 6,000($9,500)=$57,000,000

30,000($14.00)=$420,000

135.71

The labor productivity measure is a conventional measure of productivity. However, as a partialmeasure, it may not provide all of the necessary information that is needed. For example,increases in productivity could result from decreases in quality, and/or increases in material cost.


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5. A U.S. manufacturing company operating a subsidiary in an LDC (less-developed country)shows the following results:

U.S. LDCSales (units) 100,000 20,000Labor (hours) 20,000 15,000Raw materials (currency) $20,000 FC 20,000Capital equipment (hours) 60,000 5,000

a. Calculate partial labor and capital productivity figures for the parent and subsidiary. Do theresults seem misleading?

Labor Productivity

Country Outputin Units

Inputin Hours


U.S. 100,000 20,000 5.00

LDC 20,000 15,000 1.33

Capital Equipment Productivity


Inputin Hours


U.S. 100,000 60,000 1.67

LDC 20,000 5,000 4.00

Yes. You might expect the capital equipment productivity measure to be higher in the U.S. thanin a LDC. Also, the measures seem contradictory. Each plant appears to be far more productivethan the other on one measure, but much worse on the other.

b. Compute the multifactor productivity figures for labor and capital together. Do the resultsmake more sense?

b. Multifactor – Labor and Capital Equipment


Inputin Hours


U.S. 100,000 20,000 + 60,000=80,000

1.25

LDC 20,000 15,000 + 5,000=20,000

1.00

Yes, labor and equipment can be substituted for each other. Therefore, this multifactor measureis a better indicator of productivity in this instance.


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c. Calculate raw material productivity figures (units/$ where $1 = FC 10). Explain why thesefigures might be greater in the subsidiary.

Raw Material Productivity


Inputin Dollars


U.S. 100,000 $20,000 5.00

LDC 20,000 FC 20,000/$10 =$2,000

10.00

The raw material productivity measures might be greater in the LDC due to a reduced cost paidfor raw materials, which is typical of LDC’s, especially if there are local sources for the rawmaterials.

6. Various financial data for the past two years follow. Calculate the total productivity measure and thepartial measures for labor, capital, and raw materials for this company for both years. What do thesemeasures tell you about this company?

Last Year This YearOutput: Sales $200,000 $220,000Input: Labor 30,000 40,000

Raw materials 35,000 45,000Energy 5,000 6,000Capital 50,000 50,000Other 2,000 3,000

Total Productivity

Year Outputin Dollars

Inputin Dollars


Last Year $200,000 $30,000 + 35,000 +5,000 + 50,000 + 2,000= $122,000

1.64

This Year $220,000 $40,000 + 45,000 +6,000 + 50,000 +3,000= $144,000

1.53

Partial Measure – Labor


Inputin Dollars


Last Year $200,000 $30,000 6.67

This Year $220,000 $40,000 5.50


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Partial Measure – Raw Materials


Inputin Dollars


Last Year $200,000 $35,000 5.71

This Year $220,000 $45,000 4.89

Partial Measure – Capital


Inputin Dollars


Last Year $200,000 $50,000 4.00

This Year $220,000 $50,000 4.40

The overall productivity measure is declining, which indicates a possible problem. The partialmeasures can be used to indicate cause of the declining productivity. In this case, it is acombination of declines in both labor and raw material productivity, which were somewhat offsetby an increase in the capital productivity. Further investigation should be undertaken to explainthe drops in both labor and raw material productivity. An increase in the cost of both of thesemeasures, without an accompanying increase in the selling price might explain these measures.

7. An electronics company makes communications devices for military contracts. The companyjust completed two contracts. The navy contract was for 2,300 devices and took 25 workers twoweeks (40 hours per week) to complete. The army contract was for 5,500 devices that wereproduced by 35 workers in three weeks. On which contract were the workers more productive?

Contract Outputin Units

Inputin Hours


Navy 2300 25(2)40 = 2000 1.15

Army 5500 35(3)40 = 4200 1.31

The workers were more productive on the Army contract.

8. A retail store had sales of $45,000 in April and $56,000 in May. The store employs eight full-time workers who work a 40-hour week. In April the store also had seven part-time workers at10 hours per week, and in May the store had nine part-timers at 15 hours per week (assume fourweeks in each month). Using sales dollars as the measure of output, what is the percentagechange in productivity from April to May?


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Month Outputin Dollars

Inputin Hours


Percentage Change

April $45,000 (8(40)+7(10))*4 =1560

28.85

May $56,000 1820 30.77 (30.77-28.85)/28.85 = 6.66%

9. A parcel delivery company delivered 103,000 packages last year, when its averageemployment was 84 drivers. This year the firm handled 112,000 deliveries with 96 drivers. Whatwas the percentage change in productivity over the two years?

Year Outputin Packages

Inputin Drivers


Percentage Change

Last 103,000 84 1226.2

This 112,000 96 1166.7 (1166.7 -1226.2)/1226.2 = - 4.85%

10. A fast-food restaurant serves hamburgers, cheeseburgers, and chicken sandwiches. Therestaurant counts a cheeseburger as equivalent to 1.25 hamburgers and chicken sandwiches as 0.8hamburger. Current employment is five full-time employees who work a 40-hour week. If therestaurant sold 700 hamburgers, 900 cheeseburgers, and 500 chicken sandwiches in one week,what is its productivity? What would its productivity have been if it had sold the same number ofsandwiches (2,100), but the mix was 700 of each type?

Part Output inHamburgerEquivalents

Inputin Hours


700 Hamburgers900 Cheeseburgers (1.25)500 Chicken Sandwiches (.80)

2225 200 11.125

700 Hamburgers700 Cheeseburgers (1.25)700 Chicken Sandwiches (.80)

2135 200 10.675


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Case: Timbuk21 - Teaching Note

You can have a lot of fun with this case. Start off by logging on to the Timbk2 websiteand explore what is going on there. If you have a little money in a teaching account youmight actually order a custom bag and give it away or raffle it off in class, this will reallyget their attention. You make a big deal of it all when the back comes in and you give itto the lucky student. This also helps to reinforce the topic with the students.

1. Consider the two categories of products that Timbk2 makes and sells. For thecustom messenger bag, what are the key competitive dimensions that are driving sales?Are their competitive priorities different for the new laptop bags sourced in China?

This is one of the “other dimensions” and in this case it is the customization of the bag.Other than being able to get the colors they prefer, the customer also get pockets thatmeet the unique needs the customer has in mind. They can be successful withstandardizing the laptop bags since the purpose here is pretty well defined.

2. Compare the assembly line in China to that in San Francisco along the followingdimensions: (1) volume or rate of production, (2) required skill of the workers, (3) levelof automation, and (4) amount of raw materials and finished goods inventory.

Dimension China San Francisco

Volume/rate of production

Required skill of workers

Level of automation

Raw materials and finished goodinventory

High

Low

High

Low raw materials, butmay have finishedgoods

Low

High

Low

High raw materials, virtuallyno finished goods

3. Draw two diagrams, one depicting the supply chain for those products sourced inChina and the other depicting the bags produced in San Francisco. Show all the majorsteps including raw material, manufacturing, finished goods, distribution inventory, andtransportation. Other than manufacturing cost, what other costs should Timbuk2consider when making the sourcing decision?

1 Many thanks to Kyle Cattani for the idea behind this case. He does this regularly in his MBA class atIndiana University.


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The big cost other than manufacturing is the cost to transport material to the USA versusthe cost of transporting the completed bags to the USA. Here we assume that thematerial would be sourced in China. This is probably not a bad assumption.

RawMaterials(China)

Raw MaterialsInventory

Bag Fabricationand Assembly

(China)

Finished BagsInventory

(USA)

Transportto USA

RawMaterials(China)

Raw MaterialsInventory

Bag Fabricationand Assembly

(USA)

Transportto USA

Bag Fabrication and Assembly in China

Bag Fabrication and Assembly in USA

Chapter 03 - Forecasting

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CHAPTER 3FORECASTING

Review and Discussion Questions1. Examine Exhibit 3.3 and suggest which model you might use for (1) bathing suit demand, (2)

demand for new houses, (3) electrical power usage, (4) new plant expansion plans.

While any of the models in Exhibit 3.3 can potentially be used to forecast any of the items, thefollowing models are generally appropriate: (1) Bathing suit demand is highly seasonal. Of themodels in Exhibit 3.3, only the linear regression model can account for seasonality. Seasonaldecomposition methods as discussed in the chapter would also work well. (2) Demand for newhouses can be forecasted using linear regression. The time horizons are long, model complexityis medium high, model accuracy is medium high and data requirements are high. (3) Causal re-gression models might be used to forecast electrical power usage. The time horizon is long,model complexity is fairly high, model accuracy is high and data requirements are high. (4)New plant expansion plans can be forecast using qualitative forecasting techniques. This takesinto account non-quantifiable issues when planning plant expansion.

2. What is the logic in the least squares methods of linear regression analysis?

The least squares method tries to fit the line to the data that minimizes the sum of the squares ofthe vertical distance between each data point and its corresponding point on the line.

3. Explain the procedure to create a forecast using the decomposition methods of least squares re-gression.

Decomposition of a time series means finding the series’ basic components of trend, seasonality,and cyclical. The process is:

1. Decompose the time series into its components

a. Find seasonal componentsb. Deseasonalize the demandc. Find trend component

2. Forecast future values of each component.

a. Project trend component into the futureb. Multiply trend component by seasonal component

4. Give some very simple rules you might use to manage demand for a firm’s product. (An exam-ple is “limited to stock on hand.”)

Demand management can be in terms of: order control—“limited to stock on hand,” lead time—“allow six weeks for delivery,” need—“supply the parts to inoperative units first,” time open—“close up early every day,” or “closed on Saturdays,” plus others as mentioned in the text suchas price cuts, incentives, promotions, etc.


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5. What strategies are used by supermarkets, airlines, hospitals, banks, and cereal manufacturers toinfluence demand?

Supermarkets—several sales items, free giveaway items (such as a pound of butter or a loaf ofbread), an occasional “midnight madness” sale where the store is open late or even all night.

Airlines—excursion rates, age rates (senior citizens, children, youth fares), charter flights, off-season rates, exceptionally good meals (or no meals for further reduced prices), more flights, tie-in with hotels or auto rental, and tour agencies for “package tours,” free champagne, free stop-over at a third point, or new terminals.

Hospitals—patients generally go to the hospital recommended by their physician. Therefore,hospitals offer free office space, nursing assistance, lab equipment, staff positions, and patientbilling to physicians. Hospitals frequently advertise their occupancy rate and room rates, whichtend to influence demand. Also, they could become a preferred provider organization.

Banks—free gifts for new accounts, free checking, free safety deposit box, free financial advice,“club memberships,” and free use of “executive lounges” for depositors in various size accountranges, community rooms for club meetings.

Cereal manufacturers—TV advertising, sponsorship of some youth events, free prizes in cerealboxed, using prime display space.

6. All forecasting methods using exponential smoothing, adaptive smoothing, and exponentialsmoothing including trend require starting values to get the equations going. How would youselect the starting value for, say Ft-1?

Starting values can be simply an average of the early periods, or a guess. If the starting value istaken some period back (as opposed to starting to use the equations on very recent data) theequation will have a chance to adjust as it is carried forward to today.

7. From the choice of simple moving average, weighted moving average, exponential smoothing,and linear regression analysis, which forecasting technique would you consider the most accu-rate? Why?

The answer depends on the nature of the demand you are trying to forecast. If there is any trendto the demand, a simple moving average will always lag behind the actual demand. On the otherhand, if demand is relatively flat over the long run with only random variations from one periodto the next, the latter three methods will always be adjusting the forecast in response to randomvariation. In this case a simple moving average will be the best choice. If there are sesonal orcyclical components to demand, then none of thse methods will be particularly good as a fore-casting tool.

8. Give some examples that have a multiplicative seasonal trend relationship.

Multiplicative effects are the most common. Essentially all business sales are multiplicative.For example, in auto sales, as growth in total sales increases, the seasonal variation increases.

9. What is the main disadvantage of daily forecasting using regression analysis?


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The main disadvantage is the mathematical computation. There is very little time involved for acomputerized analysis, but hand computation is time consuming and prone to error.

10. What are the main problems with using adaptive exponential smoothing in forecasting?

Adaptive smoothing is computationally difficult; it requires solving a set of equations in se-quence. The user must select two constants for the equations; one is adjusted by the equations,but the other is not. It is not clear as to what the value should be. Lastly, there is some chal-lenge to the value of adaptive smoothing when compared to other methods.

11. How is a seasonal index computed from a regression line analysis?

Seasonal index is equal to the actual value (data point) divided by the value computed from theregression line. To lessen the effects of random errors, the indices may be averaged over severalyears for that same period.

12. Discuss the basic differences between the mean absolute deviation and mean absolute percenterror.

The mean absolute percent error can be used to compare forecasting accuracy when the averagedemand for each item is different. The mean absolute percent error (MAPE) is the expected er-ror measured as a fraction of demand, whereas the MAD is just the average error of the forecast.The MAPE is usually a better measure.

13. What implications do forecast errors have for the search for ultrasophisticated statistical fore-casting models?

The existence of unavoidable forecast errors seems to suggest that no matter what kind of modelone uses—simple or sophisticated—a perfect forecast is unattainable. Since forecasts are pre-dictions of the future based on present and past data, there is ample opportunity for very seriousforecast errors to be caused by changes in the conditions that generated the data. This could leadto an invalid forecast or at least one that contains added error. Therefore, one could be easilypersuaded to stop searching for ways to make more accurate forecasts and look instead for waysto quickly respond and adapt to demand changes.

14. Causal relationships are potentially useful for which component of a time series?

A time series creates an equation, such as y = a + bx, where a is the y-intercept. Therefore, ifthere is a relationship between y and x, it would show up as the slope b. If there is no relation-ship, b would be zero. There is some question as to the relationship being truly “causal,” sincemany relationships may depend on other factors outside of the analysis.


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Problems

1.

Three quarters ago (Oct, Nov, Dec last year) = 200 + 225 + 250 = 675Two quarters ago (Jan, Feb, Mar this year) = 125 + 135 + 135 = 395One quarter ago (Apr, May, Jun this year) = 190 + 200 + 190 = 580

For Jul Aug Sep, using a three-quarter average the forecast would be = (675 + 395 + 580)/3 = 550

2.

Third most recent quarter 275 + 375 +350 = 1000Second most recent quarter 425 +400 + 350 = 1175Most recent quarter 350 + 275 + 350 = 975

WMA = (.25*1000) + (.25 * 1175) + (.50 * 975) = 1031.25

3.

Use the actual demand from February through April to develop May’s forecast:

(45 + 81 + 90)/3 = 72

4.

a. Ft+1 = Ft + (At – Ft), = .20

Month Demand Forecast Absolute DeviationJanuary 100 80 20February 94 84 10March 106 86 20April 80 90 10May 68 88 20June 94 84 10

Total 90

b. MAD = 90/6 = 15


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5. a - c.

Month DemandExponentialsmoothing

Absolutedeviation Tt Ft FITt

Absolutedeviation

1 31 31.00 1.00 30.00 31.002 34 31.00 3.00 1.00 31.00 32.00 2.003 33 31.90 1.10 1.18 32.60 33.78 0.784 35 32.23 2.77 1.11 33.55 34.66 0.345 37 33.06 3.94 1.14 34.76 35.90 1.106 36 34.24 1.76 1.24 36.23 37.47 1.477 38 34.77 3.23 1.11 37.03 38.14 0.148 40 35.74 4.26 1.10 38.10 39.19 0.819 40 37.02 2.98 1.17 39.43 40.60 0.60

10 41 37.91 3.09 1.11 40.42 41.54 0.54

MAD 2.90 0.86

Based upon the MAD of each forecast, the exponential smoothing with trend is the betterforecasting model.

6.

a. F (this month) = (325 + 350 + 400)/3 = 358

b. F (next month) = (300 + 325 + 350)/3 = 325

c. F (two months ago) = 450 + .20(400 – 450) = 440

F (one month ago) = 440 + .20(350 – 440) = 422

F (this month) = 422 + .20(325 – 422) = 403

7.

6.3076.5302

6.54.28)308302(4.8)(

3026308)308288(3.308)(

3088300

288

40.30.

8300

111

11

1

ttt

tttt

tttt

ttt

t

tt

TFFIT

FITFTT

FITAFITF

TFFIT

A

TF


3-6

8.

Year Demand F(t)*2005 2702006 3562007 3982008 456 3412009 358 3642010 500 3632011 410 3902012 376 394

* Forecasts have been rounded to integer values, which may result in minor rounding differences.

F(2008) is calculated as average demand from years 2005-07. Later forecasts are based on

a simple exponential smoothing model with alpha = .20.

9. a. F5 = (700 + 600 + 400)/3 = 567

b. F4 = F3 + (A3 – F3) = 350 + .20(600 – 350) = 400

F5 = F4 + (A4 – F4) = 400 + .20(700 – 400) = 460

10. a. FSeptember = (170 + 180 + 140)/3 = 163.3

b. FSeptember = .50(170) + .30(180) + .20(140) = 167.0

a. FJuly = FJune + (AJune – FJune) = 130 + .3(140 - 130) = 133.00

FAugust = FJuly + (AJuly – FJuly) = 133.00 + .3(180 – 133.00) = 147.10

FSeptember = FAugust + (AAugust – FAugust) = 147.10 + .3(170 – 147.10) = 153.97


3-7

11. a. FOctober = (75 + 80 + 60 + 75)/4 = 72.5

b. FOctober = FSeptember + (ASeptember – FSeptember) = 65 + .2(75 – 65) = 67.0

c.

y = 405/6 = 67.5

t = 21/6 = 3.5

b =222 )5.3(691

5.67)5.3(61485

tnt

ytnty = 3.86

a = tby = 67.5 – 3.86(3.5) = 54.00

Y = a + bt = 54.0 + 3.86t

Using Excel, the intercept a can be found using the INTERCEPT() function. The slope b can befound with the SLOPE() function.

d. FOctober = 54.00 + 3.86(7) = 81.01

12.

t Y ty t2 y2 Y (y-Y)2

1 4200 4200 1 17640000 3958.97 58093.3602 4300 8600 4 18490000 4151.28 22117.0283 4000 12000 9 16000000 4343.59 118053.9124 4400 17600 16 19360000 4535.90 18468.1135 5000 25000 25 25000000 4728.21 73872.4526 4700 28200 36 22090000 4920.51 48625.9047 5300 37100 49 28090000 5112.82 35036.1608 4900 39200 64 24010000 5305.13 164128.8639 5400 48600 81 29160000 5497.44 9493.754

10 5700 57000 100 32490000 5689.74 105.19411 6300 69300 121 39690000 5882.05 174681.13112 6000 72000 144 36000000 6074.36 5529.257

Total 78 60200 418800 650 308020000 728205.128

t = 6.5

y = 5016.667

b = 22 tnt

ytnty

= 192.3077

a = tby = 3766.667

Using Excel, the intercept a can be found using the INTERCEPT() function. The slope b can befound with the SLOPE() function. You can also use the Regression tool from the Data Analysismenu. Either way, Excel’s results are identical.


3-8

Month Forecast

13 6266.6714 6458.9715 6651.2816 6843.5917 7035.9018 7228.2119 7420.5120 7612.8221 7805.1322 7997.4323 8189.7424 8382.05

b.212

128.728205

2

)( 2

1

n

YyS

i

n

ii

ty = 269.85

Therefore, 3 standard errors of the estimate would be 3(269.85) = 809.55 or 810

13.a. FJuly = .60(15) + .30(16) + .10(12) = 15.0

b. FJuly = (15 + 16 + 12) / 3 = 14.3

c. FJuly = FJune + (AJune – FJune) = 13 + .2(15-13) = 13.4

d.

t y ty t2

1 12 12 12 11 22 43 15 45 94 12 48 165 16 80 256 15 90 36

Total 21 81 297 91Average 3.5 13.5


3-9

t = 3.5

y = 13.5

b = 22 tnt

ytnty

= 0.77

a = tby = 10.8

Y = a + bt = 10.8 + .77t

e. FJuly, where July is the 7th month.

Y = a + bt = 10.8 + .77(7) = 16.2

14. TS 1

TS 1: Since there has been a rapid rise in the trend, the forecast will shortly be outside of the limits.Therefore, the forecasting model is poor.

TS 2


3-10

TS 2: This is within the limits. Therefore, the forecast is acceptable.

TS 3


3-11

TS 3: This series is rising rapidly, and is outside of the limits. Consequently, the model is poor.

15. a-c. For the exponential smoothing forecast we need a beginning forecast – this solution usesthe average of the first three months demand for the April forecast and the exponential smooth-ing model for the remaining forecasts. Other choices will produce different answers.

Month Demand 3-Mo. Absolute Exponential AbsoluteMA Deviation Smoothing Deviation

January 110February 130March 150April 170 130 40 130 40May 160 150 10 142 18June 180 160 20 147.4 32.6July 140 170 30 157.18 17.18August 130 160 30 152.03 22.03September 140 150 10 145.42 5.42

MAD 23.3 22.5

Based upon MAD, the exponential smoothing model appears to be slightly better.

16.

Month Forecast Actual Deviation RSFEAbsolutedeviation

Sum of ab-solute de-viations MAD TS

April 250 200 -50 -50 50 50 50.0 -1May 325 250 -75 -125 75 125 62.5 -2June 400 325 -75 -200 75 200 66.7 -3July 350 300 -50 -250 50 250 62.5 -4August 375 325 -50 -300 50 300 60.0 -5September 450 400 -50 -350 50 350 58.3 -6

-8

-6

-4

-2

01 2 3 4 5 6

TS

Period


3-12

For September, the MAD is 58.3 and the TS is –6. The model is performing poorly since the track-ing signal is –6 and moving in a downward direction. The model is consistently over-forecastingdemand.

17.


Sum of Absolutedeviations MAD TS

1 140 137 -3 -3 3 3 3.00 -1.002 140 133 -7 -10 7 10 5.00 -2.003 140 150 10 0 10 20 6.67 0.004 140 160 20 20 20 40 10.00 2.005 140 180 40 60 40 80 16.00 3.756 150 170 20 80 20 100 16.67 4.807 150 185 35 115 35 135 19.29 5.968 150 205 55 170 55 190 23.75 7.16

a. For month 8, the MAD is 23.75

b. The tracking signal for month 8 is 7.16

c. The tracking signal is too large, so the forecast should be considered poor. It is not effective-ly dealing with an apparent upward trend in demand.


3-13

18.

Week Forecast Actual Deviation RSFE Absolutedeviation

Sum of Abso-lute deviations

MAD TS

1 800 900 100 100 100 100 100 1.02 850 1000 150 250 150 250 125 2.03 950 1050 100 350 100 350 117 3.04 950 900 -50 300 50 400 100 3.05 1000 900 -100 200 100 500 100 2.06 975 1100 125 325 125 625 104 3.1

For week 6, the MAD is 104, and the tracking signal is 3.1. This is a fairly high value, which indi-cates the model is unacceptable.

19.

Period Forecast Actual Deviation RSFEAbsolutedeviation

Sum of Absolutedeviations MAD TS

1 1500 1550 50 50 50 50 50.0 1.002 1400 1500 100 150 100 150 75.0 2.003 1700 1600 -100 50 100 250 83.3 0.604 1750 1650 -100 -50 100 350 87.5 -0.575 1800 1700 -100 -150 100 450 90.0 -1.67


3-14

a. For period 5, the MAD = 90.00, and the TS = -1.67

b. Looking solely at the value of the TS, the model seems acceptable since the tracking signalis only 1.67 off the mean. However, the MAD has been increasing since the first period, andthe downward trend over the last several periods in the graph is cause for concern that theremay be some bias in the model.

20.

Month(t) Demand

3-mo.MA

Absolutedeviation

3-moWMA

Absolutedeviation Ft

Absolutedeviation Tt Ft FITt

Absolutedeviation

1 62 61.00 1.80 60.00 61.802 65 61.30 1.82 61.86 63.683 67 62.41 1.94 64.07 66.014 68 64.67 3.33 65.40 2.60 63.79 4.21 2.03 66.31 68.33 0.335 71 66.67 4.33 67.10 3.90 65.05 5.95 2.00 68.23 70.23 0.776 73 68.67 4.33 69.30 3.70 66.84 6.16 2.07 70.46 72.53 0.477 76 70.67 5.33 71.40 4.60 68.68 7.32 2.11 72.67 74.78 1.228 78 73.33 4.67 74.10 3.90 70.88 7.12 2.22 75.14 77.36 0.649 78 75.67 2.33 76.40 1.60 73.02 4.98 2.28 77.55 79.83 1.8310 80 77.33 2.67 77.60 2.40 74.51 5.49 2.11 79.28 81.39 1.3911 84 78.67 5.33 79.00 5.00 76.16 7.84 1.99 80.98 82.96 1.0412 85 80.67 4.33 81.60 3.40 78.51 6.49 2.08 83.27 85.35 0.35

MAD 4.07 3.46 6.17 0.89

Based upon MAD, the exponential smoothing with trend component appears to be the best method.This should not be a surprise given the upward trend that is apparent by inspecting the demand data.


3-15

21.


Sum ofabsolutedeviations MAD TS

May 450 500 50 50 50 50 50.00 1.00June 500 550 50 100 50 100 50.00 2.00July 550 400 -150 -50 150 250 83.33 -0.60

August 600 500 -100 -150 100 350 87.50 -1.71September 650 675 25 -125 25 375 75.00 -1.67

October 700 600 -100 -225 100 475 79.17 -2.84

The TS itself is acceptable. However, you would like to see the TS going back and forth betweenpositive and negative. It has been headed down since June. If this trend continues, the forecasts willbe unacceptable. This forecast should be closely monitored to see if the downward trend continues,or if this occurred by random chance.

22.

t yAverage fromsame quarter

Seasonalfactor

Deseasonalizeddemand t2

t*deseasonalizeddemand

1 4800 3833.33 1.23 3902.61 1 3902.612 3500 2766.67 0.89 3942.77 4 7885.543 4300 3500.00 1.12 3829.05 9 11487.144 3000 2366.67 0.76 3950.70 16 15802.825 3500 1.23 2845.65 25 14228.266 2700 0.89 3041.57 36 18249.407 3500 1.12 3116.67 49 21816.678 2400 0.76 3160.56 64 25284.519 3200 1.23 2601.74 81 23415.65


3-16

10 2100 0.89 2365.66 100 23656.6311 2700 1.12 2404.29 121 26447.1412 1700 0.76 2236.84 144 26842.08

Sum 78 37400 37400.00 650 219041.20

y = 3116.67

t = 6.5

b =22 tnt

ytnty dd

= -168.24

a = tby d = 4210.25

Period (t) Yd Seasonalfactor

Forecast(Yd*seasonal

factor)13 2023.08 1.23 2488.2814 1854.84 0.89 1646.5415 1686.60 1.12 1894.0416 1518.35 0.76 1152.97

Calculations were done in Excel. Hand calculations may result in some rounding differences.

23.


3-17

b - d.

t y

Average fromsame bi-monthly

periodSeasonal

factorDeseasonalized

demand t2 t*yt*deseasonalized

demand1 109 112.0 0.865 125.95 1 109 125.952 104 108.0 0.835 124.62 4 208 249.253 150 154.5 1.194 125.65 9 450 376.944 170 176.0 1.360 125.00 16 680 500.025 120 123.0 0.950 126.26 25 600 631.306 100 103.0 0.796 125.65 36 600 753.887 115 0.865 132.88 49 805 930.188 112 0.835 134.21 64 896 1073.689 159 1.194 133.19 81 1431 1198.68

10 182 1.360 133.83 100 1820 1338.2911 126 0.950 132.57 121 1386 1458.3112 106 0.796 133.19 144 1272 1598.23

Total 78 1553 1553.00 650 10257 10234.70

Deseasonalized Simple

y = 129.4167

t = 6.5

b=22 tnt

ytnty dd

= 0.9804 1.136364

a = tbyd = 123.04 122.0303

Period (t)Simple

Forecast Yd

Seasonalfactor

SeasonalForecast

13 136.80 135.79 0.865 117.514 137.94 136.77 0.835 114.115 139.08 137.75 1.194 164.416 140.21 138.73 1.360 188.717 141.35 139.71 0.950 132.818 142.48 140.69 0.796 112.0


3-18

24.

t y

Average fromsame quarterly

periodSeasonal


demand t2t*deseasonalized

demand

1 160 187.5 1.003 159.47 1 159.472 195 217.5 1.164 167.54 4 335.093 150 177.5 0.950 157.92 9 473.774 140 165.0 0.883 158.56 16 634.245 215 1.003 214.28 25 1071.426 240 1.164 206.21 36 1237.247 205 0.950 215.83 49 1510.798 190 0.883 215.18 64 1721.52

Total 36 1495 1495.00 204 7143.53

y = 186.875

t = 4.5

b = 22 tnt

ytnty dd

= 9.91

a = tbyd = 142.30

Period Yd Seasonal factor Forecast9 231.45 1.003 232

10 241.35 1.164 28111 251.26 0.950 23912 261.17 0.883 231

25. Within Excel, two methods can be used to compute the intercept and slope from a regressionmodel. We can use the Regression tool within Data Analysis in the Data menu to perform afull-blown regression analysis, or we can use the INTERCEPT() and SLOPE() functions to get aand b in a traditional regression model.

Using those functions we get:

a = 15.143 b = 1.024

We can now use decomposition to quantify the seasonal factors and forecast for the next fourquarters.


3-19

First procedure:

t yTrend

(a + tb)Actual/Trend Seasonal factor

1 12 16.17 0.742 0.7662 18 17.19 1.047 1.0873 26 18.21 1.427 1.3414 16 19.24 0.832 0.8025 16 20.26 0.7906 24 21.29 1.1287 28 22.31 1.2558 18 23.33 0.771

Period YSeasonal

factor Forecast9 24.36 0.766 18.66

10 25.38 1.087 27.6011 26.40 1.341 35.4212 27.43 0.802 21.99

Second procedure: We could also deseasonalize the data first, and perform a regression on thedeseasonalized data. Full calculations for this alternate method are shown below:

t yAverage from same

quarterly periodSeasonal


demand t2t*deseasonalized

demand1 12 14 0.71 16.93 1 16.932 18 21 1.06 16.93 4 33.863 26 27 1.37 19.02 9 57.064 16 17 0.86 18.59 16 74.355 16 0.71 22.57 25 112.866 24 1.06 22.57 36 135.437 28 1.37 20.48 49 143.378 18 0.86 20.91 64 167.29

Total 36 158 158.00 204 741.14

y = 19.75

t = 4.5

b = 22 tnt

ytnty dd

= 0.718

a = tbyd = 16.520


3-20

Period Yd

Seasonalfactor Forecast

9 22.98 0.71 16.2910 23.70 1.06 25.2011 24.42 1.37 33.3812 25.13 0.86 21.63

26.

t yAverage from same

quarterly periodSeasonal

factorDeseasonalize

d demand t2t*deseasonalized

demand1 205 340.0 0.736 278.48 1 278.482 140 207.5 0.449 311.63 4 623.253 375 530.0 1.147 326.80 9 980.404 575 770.0 1.667 344.91 16 1379.635 475 0.736 645.27 25 3226.336 275 0.449 612.12 36 3672.747 685 1.147 596.95 49 4178.668 965 1.667 578.84 64 4630.75

Total 36 3695 3695.00 204 18970.24

y = 461.88

t = 4.50

b = 22 tnt

ytnty dd

= 55.78

a = tbyd = 210.87

Period (t) Yd

Seasonalfactor


factor)9 712.88 0.736 525

10 768.66 0.449 34511 824.44 1.147 94612 880.22 1.667 1467


3-21

27. a.

Company A

Period EPSForecast = 0.10

Absolutedeviation

Forecast = 0.30

Absolutedeviation

2009-I 1.67 1.67 1.67II 2.35 1.67 0.68 1.67 0.68III 1.11 1.74 0.63 1.87 0.76IV 1.15 1.68 0.53 1.64 0.49

2010-I 1.56 1.62 0.06 1.50 0.06II 2.04 1.62 0.42 1.52 0.52III 1.14 1.66 0.52 1.67 0.53IV 0.38 1.61 1.23 1.51 1.13

2011-I 0.29 1.48 1.19 1.17 0.88II -0.18 1.36 1.54 0.91 1.09III -0.97 1.21 2.18 0.58 1.55IV 0.20 0.99 0.79 0.12 0.08

2012-I -1.54 0.91 2.45 0.14 1.68II 0.38 0.67 0.29 -0.36 0.74III 0.64 -0.14

MAD 0.96 0.79

Company B

Period EPSForecast = 0.10

Absolutedeviation

Forecast = 0.30

Absolutedeviation

2009-I 0.17 0.17 0.17II 0.24 0.17 0.07 0.17 0.07III 0.26 0.18 0.08 0.19 0.07IV 0.34 0.19 0.15 0.21 0.13

2010-I 0.25 0.20 0.05 0.25 0.00II 0.37 0.21 0.16 0.25 0.12III 0.36 0.22 0.14 0.29 0.07IV 0.44 0.24 0.20 0.31 0.13

2011-I 0.33 0.26 0.07 0.35 0.02II 0.40 0.26 0.14 0.34 0.06III 0.41 0.28 0.13 0.36 0.05IV 0.47 0.29 0.18 0.37 0.10

2012-I 0.30 0.31 0.01 0.40 0.10II 0.47 0.31 0.16 0.37 0.10III 0.32 0.40

MAD 0.12 0.08


3-22

b.

MAD

Company A Company B

= 0.10 .96 .12 = 0.30 .79 .08

Based upon MAD, an of .30 performs better than .10.

c.

Company A


Seasonalfactor



1 1.67 0.495 0.723 2.309 1 2.3092 2.35 1.148 1.677 1.401 4 2.8033 1.11 0.427 0.624 1.780 9 5.3414 1.15 0.577 0.843 1.365 16 5.4585 1.56 0.723 2.157 25 10.7836 2.04 1.677 1.217 36 7.2997 1.14 0.624 1.828 49 12.7988 0.38 0.843 0.451 64 3.6079 0.29 0.723 0.401 81 3.608

10 -0.18 1.677 -0.107 100 -1.07311 -0.97 0.624 -1.556 121 -17.11312 0.20 0.843 0.237 144 2.84813 -1.54 0.723 -2.129 169 -27.67614 0.38 1.677 0.227 196 3.172

Total 105 9.58 9.580 1015 14.165

y = 0.684

t = 7.5

b = 22 tnt

ytnty dd

= -0.254

a = tbyd = 2.5867


3-23

Period (x) Yd

Seasonalfactor


factor)15 -1.217 0.624 -0.7616 -1.470 0.843 -1.2417 -1.725 0.723 -1.2518 -1.978 1.677 -3.3219 -2.232 0.624 -1.3920 -2.485 0.843 -2.09

Company B


Seasonalfactor



1 0.17 0.263 0.764 0.223 1 0.2232 0.24 0.370 1.077 0.223 4 0.4463 0.26 0.343 0.999 0.260 9 0.7814 0.34 0.417 1.213 0.280 16 1.1215 0.25 0.764 0.327 25 1.6366 0.37 1.077 0.344 36 2.0617 0.36 0.999 0.360 49 2.5228 0.44 1.213 0.363 64 2.9029 0.33 0.764 0.432 81 3.887

10 0.40 1.077 0.371 100 3.71411 0.41 0.999 0.410 121 4.51312 0.47 1.213 0.388 144 4.65113 0.30 0.764 0.393 169 5.10414 0.47 1.077 0.436 196 6.110

Total 105 4.81 4.810 1015 39.672

y = 0.344

t = 7.5

b = 22 tnt

ytnty dd

= 0.016

a = tbyd = 0.225


3-24

Period (t) Yd

Seasonalfactor


factor)15 0.462 0.999 0.4616 0.478 1.213 0.5817 0.494 0.764 0.3818 0.510 1.077 0.5519 0.525 0.999 0.5320 0.541 1.213 0.66

c. The results indicate that Company A’s EPS is on a downward trend, while Company B’sEPS is growing.

28.

a. We can use the Regression tool or the LINEST() function within Excel. Answers usingLINEST function in Microsoft Excel follow.

Sales Price Advertising Fitted Values

400 280 600 451.72

700 215 835 977.21

900 211 1100 1090.98

1300 210 1400 1195.40

1150 215 1200 1095.85

1200 200 1300 1231.99

900 225 900 929.25

1100 207 1100 1118.62

980 220 700 898.79

1234 211 900 1025.98

925 227 700 850.42

800 245 690 722.80

Constant 2191.3374

Price -6.9094

Advertising 0.3250


3-25

212211 3250.9094.63374.2191 xxxbxbay

Wherea = y interceptx1 = priceb1 = slope of pricex2 = advertisingb2 = slope of advertising

b. Price has a larger effect on sales because it slope value is much higher (-6.9094 versus .3250).Price actually has a negative effect since raising price decreases sales.

c. Sales = 2191.3374 - 6.9094 (300) + .3250 (900)

Sales = 411.04

29.

QuarterI II III IV

Last year 23,000 27,000 18,000 9,000This year 19,000 24,000 15,000

Being left up to the student to develop a method, and given the limited amount of historical data,there could be several good answers to this problem. An inspection of the data makes it seemobvious that there is seasonality in demand. Also the year-to-year figures in each quarter sup-port an assumption of a negative long-term trend.

One simplistic solution would be to manually follow the patterns in the existing demand dataand forecast demand that is somewhat lower than fourth quarter last year, say 7,000.

Another more analytical approach would be to apply decomposition to the existing data, but wedo not have two full years’ worth of data, so computing a seasonal factor for quarter IV is a con-cern. We could develop seasonal factors based solely on last year’s data, or we could use a sub-stitute data point to complete this year’s data. The following solution uses last year’s quarter IVdata as this year’s quarter IV demand to build the seasonal factors. Regression is run on thedeseasonalized demand for the first seven quarters. Based on the results, this year’s quarter IVforecast is developed.


3-26

Same Qtr Seasonal Deasonalized Seasonal

Quarter t Demand Average Factor Demand Yd Forecast

Last Year I 1 23000 21000 1.167 19714

II 2 27000 25500 1.417 19059

III 3 18000 16500 0.917 19636

IV 4 9000 9000 0.500 18000

This Year I 5 19000 1.167 16286

II 6 24000 1.417 16941

III 7 15000 0.917 16364

IV 8 9000 0.500 15480.3 7740

Average 18000

Intercept = 20519.7 Found using INTERCEPT() function and deseasonalized data

Slope = -629.925 Found using SLOPE() function30. Since we our interested in forecasting the next four years, many of the procedures presented in

the text will not work. For example, moving average and exponential smoothing will only pro-ject one period into the future. Therefore, of the methods presented in the text, regression ap-pears to be the logical approach.

Examination of the graph of revenue over time suggests that there maybe a slight upward trend.Additionally, there may be a cyclical component, possibly 6 or 7 years. With the limited data, itis very difficult to determine the cycle. Consequently, simple regression appears to be the avail-able choice for the forecast.

t y

1 4865.92 5067.43 5515.64 5728.8


3-27

5 5497.76 5197.77 5094.48 5108.89 5550.6

10 5738.911 5860.0

Total 66 59225.8

Using LINEST():b = 55.62a = 5050.444

Period Forecast12 571813 577414 582915 5885

ANALYTICS EXERCISE: Forecasting Supply Chain Demand – Starbucks Corporation

A good first step in developing a forecasting model is to create a plot of the historical demand data.


3-28

There does not appear to be an obvious trend in demand at any of the DCs, though there is quite a bit of varia-tion. With no evident trend, both moving average and exponential smoothing models would be appropriate.

1, 2. Both the MA and ES models are shown on the following pages. A summary of their performancemeasures is shown in the table on the next page (best measures in each column are bolded). Performance ofthe two MA models is similar – neither one seems to be consistently better than the other. The same can besaid for the two ES models, though they do appear to perform slightly better than the MA models. The ESmodel with an alpha of .2 appears to be the best.

3. Aggregating demand at a single DC will result in better overall forecasting performance than forecasting forfive DCs separately. The variations in demand for the separate DCs will tend to cancel each other out some-what when aggregating demand and forecasting for a single DC. In each performance measure, the aggregatedforecast error was about half that of the sum or average measure for the separate DCs.

Other factors to consider would be the transportation costs which might be higher serving from a central DCrather than a regional one nearer the customers. We should also consider risk of natural disasters or othercauses that might disrupt transportation modes from the single DC and interrupt the supply chain for this item.

ATL BOS CHI DAL LAAvg

of DCs

3-w

eek

MA

MAD 11.44 10.08 19.85 10.64 9.87 12.37

MAPE 0.288 0.243 0.420 0.223 0.214 0.278

TS -0.117 1.224 -0.638 -1.222 1.992 N/A

5-w

eek

MA

MAD 11.17 10.63 18.45 12.37 9.58 12.44

MAPE 0.281 0.256 0.391 0.260 0.207 0.279

TS -0.322 1.750 -0.011 -0.808 1.982 N/A

ES,

=

.2

MAD 10.76 9.85 17.77 10.51 8.57 11.49MAPE 0.271 0.237 0.376 0.221 0.185 0.258

TS -0.512 1.225 -1.63 -0.65 1.52 N/A

ES,

=

.4

MAD 11.87 10.55 18.32 9.84 9.57 12.03

MAPE 0.299 0.254 0.388 0.207 0.207 0.271

TS -0.046 1.032 -0.241 0.024 1.510 N/A

Performance Measures of the Four Forecasting Models


3-29

Historical Demand 3-week MA

Week ATL BOS CHI DAL LA Total ATL BOS CHI DAL LA Total

-5 45 62 62 42 43 254

-4 38 18 22 35 40 153

-3 30 48 72 40 54 244

-2 58 40 44 64 46 252

-1 37 35 48 43 35 198

1 33 26 44 27 32 162 41.7 41.0 54.7 49.0 45.0 231.3

2 45 35 34 42 43 199 42.7 33.7 45.3 44.7 37.7 204.0

3 37 41 22 35 54 189 38.3 32.0 42.0 37.3 36.7 186.3

4 38 40 55 40 40 213 38.3 34.0 33.3 34.7 43.0 183.3

5 55 46 48 51 46 246 40.0 38.7 37.0 39.0 45.7 200.3

6 30 48 72 64 74 288 43.3 42.3 41.7 42.0 46.7 216.0

7 18 55 62 70 40 245 41.0 44.7 58.3 51.7 53.3 249.0

8 58 18 28 65 35 204 34.3 49.7 60.7 61.7 53.3 259.7

9 47 62 27 55 45 236 35.3 40.3 54.0 66.3 49.7 245.7

10 37 44 95 43 38 257 41.0 45.0 39.0 63.3 40.0 228.3

11 23 30 35 38 48 174 47.3 41.3 50.0 54.3 39.3 232.3

12 55 45 45 47 56 248 35.7 45.3 52.3 45.3 43.7 222.3

13 40 50 47 42 50 229 38.3 39.7 58.3 42.7 47.3 226.313-weekAverage: 39.69 41.54 47.23 47.62 46.23 222.31

Absolute Deviation RSFE


1 8.67 15.00 10.67 22.00 13.00 69.33 -8.7 -15.0 -10.7 -22.0 -13.0 -69.3

2 2.33 1.33 11.33 2.67 5.33 5.00 -6.3 -13.7 -22.0 -24.7 -7.7 -74.3

3 1.33 9.00 20.00 2.33 17.33 2.67 -7.7 -4.7 -42.0 -27.0 9.7 -71.7

4 0.33 6.00 21.67 5.33 3.00 29.67 -8.0 1.3 -20.3 -21.7 6.7 -42.0

5 15.00 7.33 11.00 12.00 0.33 45.67 7.0 8.7 -9.3 -9.7 7.0 3.7

6 13.33 5.67 30.33 22.00 27.33 72.00 -6.3 14.3 21.0 12.3 34.3 75.7

7 23.00 10.33 3.67 18.33 13.33 4.00 -29.3 24.7 24.7 30.7 21.0 71.7

8 23.67 31.67 32.67 3.33 18.33 55.67 -5.7 -7.0 -8.0 34.0 2.7 16.0

9 11.67 21.67 27.00 11.33 4.67 9.67 6.0 14.7 -35.0 22.7 -2.0 6.3

10 4.00 1.00 56.00 20.33 2.00 28.67 2.0 13.7 21.0 2.3 -4.0 35.0

11 24.33 11.33 15.00 16.33 8.67 58.33 -22.3 2.3 6.0 -14.0 4.7 -23.3

12 19.33 0.33 7.33 1.67 12.33 25.67 -3.0 2.0 -1.3 -12.3 17.0 2.3

13 1.67 10.33 11.33 0.67 2.67 2.67 -1.3 12.3 -12.7 -13.0 19.7 5.0MAD: 11.44 10.08 19.85 10.64 9.87 31.46

Sum of DC MAD: 61.87

MAPE: 0.288 0.243 0.420 0.223 0.214 0.142

Average DC MAPE: 0.278

TS: -0.117 1.224 -0.638 -1.222 1.992 0.159


3-30

Historical Demand 5-week MA


-5 45 62 62 42 43 254

-4 38 18 22 35 40 153

-3 30 48 72 40 54 244

-2 58 40 44 64 46 252

-1 37 35 48 43 35 198

1 33 26 44 27 32 162 41.6 40.6 49.6 44.8 43.6 220.2

2 45 35 34 42 43 199 39.2 33.4 46.0 41.8 41.4 201.8

3 37 41 22 35 54 189 40.6 36.8 48.4 43.2 42.0 211.0

4 38 40 55 40 40 213 42.0 35.4 38.4 42.2 42.0 200.0

5 55 46 48 51 46 246 38.0 35.4 40.6 37.4 40.8 192.2

6 30 48 72 64 74 288 41.6 37.6 40.6 39.0 43.0 201.8

7 18 55 62 70 40 245 41.0 42.0 46.2 46.4 51.4 227.0

8 58 18 28 65 35 204 35.6 46.0 51.8 52.0 50.8 236.2

9 47 62 27 55 45 236 39.8 41.4 53.0 58.0 47.0 239.2

10 37 44 95 43 38 257 41.6 45.8 47.4 61.0 48.0 243.8

11 23 30 35 38 48 174 38.0 45.4 56.8 59.4 46.4 246.0

12 55 45 45 47 56 248 36.6 41.8 49.4 54.2 41.2 223.2

13 40 50 47 42 50 229 44.0 39.8 46.0 49.6 44.4 223.8

13-weekAverage: 39.69 41.54 47.23 47.62 46.23 222.31



1 8.60 14.60 5.60 17.80 11.60 58.20 -8.6 -14.6 -5.6 -17.8 -11.6 -58.2

2 5.80 1.60 12.00 0.20 1.60 2.80 -2.8 -13.0 -17.6 -17.6 -10.0 -61.0

3 3.60 4.20 26.40 8.20 12.00 22.00 -6.4 -8.8 -44.0 -25.8 2.0 -83.0

4 4.00 4.60 16.60 2.20 2.00 13.00 -10.4 -4.2 -27.4 -28.0 0.0 -70.0

5 17.00 10.60 7.40 13.60 5.20 53.80 6.6 6.4 -20.0 -14.4 5.2 -16.2

6 11.60 10.40 31.40 25.00 31.00 86.20 -5.0 16.8 11.4 10.6 36.2 70.0

7 23.00 13.00 15.80 23.60 11.40 18.00 -28.0 29.8 27.2 34.2 24.8 88.0

8 22.40 28.00 23.80 13.00 15.80 32.20 -5.6 1.8 3.4 47.2 9.0 55.8

9 7.20 20.60 26.00 3.00 2.00 3.20 1.6 22.4 -22.6 44.2 7.0 52.6

10 4.60 1.80 47.60 18.00 10.00 13.20 -3.0 20.6 25.0 26.2 -3.0 65.8

11 15.00 15.40 21.80 21.40 1.60 72.00 -18.0 5.2 3.2 4.8 -1.4 -6.2

12 18.40 3.20 4.40 7.20 14.80 24.80 0.4 8.4 -1.2 -2.4 13.4 18.6

13 4.00 10.20 1.00 7.60 5.60 5.20 -3.6 18.6 -0.2 -10.0 19.0 23.8

MAD: 11.17 10.63 18.45 12.37 9.58 31.12


MAPE: 0.281 0.256 0.391 0.260 0.207 0.140

Average DC MAPE: 0.278

TS: -0.322 1.750 -0.011 -0.808 1.982 0.765


3-31

Historical Demand ES, = 2


-5 45 62 62 42 43 254

-4 38 18 22 35 40 153

-3 30 48 72 40 54 244

-2 58 40 44 64 46 252

-1 37 35 48 43 35 198

1 33 26 44 27 32 162 41.7 41.0 54.7 49.0 45.0 231.3

2 45 35 34 42 43 199 39.9 38.0 52.5 44.6 42.4 217.5

3 37 41 22 35 54 189 40.9 37.4 48.8 44.1 42.5 213.8

4 38 40 55 40 40 213 40.2 38.1 43.5 42.3 44.8 208.8

5 55 46 48 51 46 246 39.7 38.5 45.8 41.8 43.9 209.7

6 30 48 72 64 74 288 42.8 40.0 46.2 43.6 44.3 216.9

7 18 55 62 70 40 245 40.2 41.6 51.4 47.7 50.2 231.1

8 58 18 28 65 35 204 35.8 44.3 53.5 52.2 48.2 233.9

9 47 62 27 55 45 236 40.2 39.0 48.4 54.7 45.5 227.9

10 37 44 95 43 38 257 41.6 43.6 44.1 54.8 45.4 229.5

11 23 30 35 38 48 174 40.7 43.7 54.3 52.4 43.9 235.0

12 55 45 45 47 56 248 37.1 41.0 50.4 49.5 44.8 222.8

13 40 50 47 42 50 229 40.7 41.8 49.3 49.0 47.0 227.913-weekAverage: 39.69 41.54 47.23 47.62 46.23 222.31



1 8.67 15.00 10.67 22.00 13.00 69.33 -8.7 -15.0 -10.7 -22.0 -13.0 -69.3

2 5.07 3.00 18.53 2.60 0.60 18.47 -3.6 -18.0 -29.2 -24.6 -12.4 -87.8

3 3.95 3.60 26.83 9.08 11.48 24.77 -7.5 -14.4 -56.0 -33.7 -0.9 -112.6

4 2.16 1.88 11.54 2.26 4.82 4.18 -9.7 -12.5 -44.5 -35.9 -5.7 -108.4

5 15.27 7.50 2.23 9.19 2.15 36.35 5.6 -5.0 -42.3 -26.8 -3.6 -72.0

6 12.78 8.00 25.78 20.35 29.72 71.08 -7.2 3.0 -16.5 -6.4 26.1 -1.0

7 22.22 13.40 10.63 22.28 10.23 13.86 -29.4 16.4 -5.8 15.9 15.9 12.9

8 22.22 26.28 25.50 12.82 13.18 29.91 -7.2 -9.9 -31.3 28.7 2.7 -17.0

9 6.78 22.98 21.40 0.26 0.54 8.07 -0.4 13.1 -52.7 29.0 2.2 -9.0

10 4.58 0.38 50.88 11.79 7.44 27.46 -5.0 13.5 -1.9 17.2 -5.3 18.5

11 17.66 13.69 19.29 14.43 4.05 61.03 -22.7 -0.2 -21.2 2.7 -1.2 -42.5

12 17.87 4.04 5.44 2.55 11.24 25.17 -4.8 3.8 -26.6 0.2 10.0 -17.4

13 0.70 8.24 2.35 7.04 2.99 1.14 -5.5 12.1 -28.9 -6.8 13.0 -16.2

MAD: 10.76 9.85 17.77 10.51 8.57 30.06Sum of DC MAD: 57.47

MAPE: 0.271 0.237 0.376 0.221 0.185 0.135

Average DC MAPE: 0.258TS: -0.512 1.225 -1.628 -0.652 1.520 -0.539


3-32

Historical Demand ES, = 4


-5 45 62 62 42 43 254

-4 38 18 22 35 40 153

-3 30 48 72 40 54 244

-2 58 40 44 64 46 252

-1 37 35 48 43 35 198

1 33 26 44 27 32 162 41.6 40.6 49.6 44.8 43.6 220.2

2 45 35 34 42 43 199 38.2 34.8 47.4 37.7 39.0 196.9

3 37 41 22 35 54 189 40.9 34.9 42.0 39.4 40.6 197.8

4 38 40 55 40 40 213 39.3 37.3 34.0 37.6 45.9 194.3

5 55 46 48 51 46 246 38.8 38.4 42.4 38.6 43.6 201.8

6 30 48 72 64 74 288 45.3 41.4 44.6 43.6 44.5 219.5

7 18 55 62 70 40 245 39.2 44.1 55.6 51.7 56.3 246.9

8 58 18 28 65 35 204 30.7 48.4 58.2 59.0 49.8 246.1

9 47 62 27 55 45 236 41.6 36.3 46.1 61.4 43.9 229.3

10 37 44 95 43 38 257 43.8 46.6 38.5 58.9 44.3 232.0

11 23 30 35 38 48 174 41.1 45.5 61.1 52.5 41.8 242.0

12 55 45 45 47 56 248 33.8 39.3 50.6 46.7 44.3 214.8

13 40 50 47 42 50 229 42.3 41.6 48.4 46.8 49.0 228.113-weekAverage: 39.69 41.54 47.23 47.62 46.23 222.31



1 8.60 14.60 5.60 17.80 11.60 58.20 -8.6 -14.6 -5.6 -17.8 -11.6 -58.2

2 6.84 0.24 13.36 4.32 4.04 2.08 -1.8 -14.4 -19.0 -13.5 -7.6 -56.1

3 3.90 6.14 20.02 4.41 13.42 8.75 -5.7 -8.2 -39.0 -17.9 5.9 -64.9

4 1.34 2.69 20.99 2.36 5.95 18.75 -7.0 -5.5 -18.0 -15.5 -0.1 -46.1

5 16.20 7.61 5.59 12.41 2.43 44.25 9.2 2.1 -12.4 -3.1 2.4 -1.9

6 15.28 6.57 27.36 20.45 29.46 68.55 -6.1 8.6 15.0 17.3 31.8 66.7

7 21.17 10.94 6.41 18.27 16.32 1.87 -27.2 19.6 21.4 35.6 15.5 64.8

8 27.30 30.44 30.15 5.96 14.79 42.12 0.1 -10.8 -8.8 41.6 0.7 22.7

9 5.38 25.74 19.09 6.42 1.12 6.73 5.4 14.9 -27.9 35.1 1.8 29.4

10 6.77 2.56 56.55 15.85 6.33 25.04 -1.3 12.3 28.7 19.3 -4.5 54.4

11 18.06 15.53 26.07 14.51 6.20 67.98 -19.4 -3.2 2.6 4.8 1.7 -13.5

12 21.16 5.68 5.64 0.29 11.72 33.21 1.8 2.5 -3.0 5.1 13.4 19.7

13 2.30 8.41 1.39 4.82 1.03 0.93 -0.5 10.9 -4.4 0.2 14.4 20.6

MAD: 11.87 10.55 18.32 9.84 9.57 29.11


MAPE: 0.299 0.254 0.388 0.207 0.207 0.131

Average DC MAPE: 0.271TS: -0.046 1.032 -0.241 0.024 1.510 0.708

Chapter 04 - Strategic Capacity Management

4-1

CHAPTER 4STRATEGIC CAPACITY MANAGEMENT

Review and Discussion Questions1. What capacity problems are encountered when a new drug is introduced to the market??

The primary concerns come from uncertain demand for the drug and the high capital investmenttypically needed for modern drug production. Being a new drug, there are no historical sales data onwhich to base forecasts of future demand. If forecasts are too high, significant capital resources willbe underutilized. If forecasts are too low, there may be insufficient capital resources to meet theactual demand, resulting in lost sales when the price for the new drug is typically highest.

2. List some practical limits to economies of scale; that is, when should a plant stop growing?

The obvious answer is that a plant should stop growing when its long-run average cost curve hits theinflection point and starts increasing. Factors leading to this situation include difficultiescoordinating and managing a facility of that size, demand variations that can lead to regular periods oflow capacity utilization, and capacity imbalance within the facility.

3. What are some capacity balance problems faced by the following organizations or facilities?

a. An airline terminal.

Congested flight arrival/departure scheduling typically leads to problems throughout the system,including waiting areas, distances from boarding gates, ground crew requirements, runways, baggagehandling, etc.

b. A university computing lab.

The number of computer workstations, the size of each workstation (room for student papers, etc.),the mix of different computer types (Mac or PC), the number of printers, the capacity of the networkaccess, study space for students waiting. These problems are exacerbated by surges in demand duringcertain points in the semester (e.g. finals week).

c. A clothing manufacturer.

Many manufacturers now use highly decentralized shops to make clothes. This means that capacityof multiple sites must be accounted for in planning production.

4. What are some major capacity considerations in a hospital? How do they differ from those of afactory?


4-2

Some capacity considerations are size and composition of nursing staff (RNs vs. LPNs), balancebetween operating room and intensive care units, emergency rooms, etc., and, of course, how manybeds are to be available. One of the differences in capacity considerations between a hospital and afactory is that a hospital can add capacity rather quickly in the short run, through “simply” addingmore staff and more beds. A factory is usually technologically limited, and, therefore, must plan wellin advance to add major chunks of capacity. On the other hand, though, the general uncertaintywhich surrounds the demand for hospital services on any given day is much greater than would befaced by a factory. Additionally, factory management generally has the ability to backlog demand insuch a way as to achieve more efficient levels of capacity utilization than does a hospital. Sick andinjured patients cannot be put on a shelf and made to wait during periods of peak demand.

5. Management may choose to build up capacity in anticipation of demand or in response to developingdemand. Cite the advantages and disadvantages of both approaches.

The strategy of building up capacity ahead of demand is a risk-taking stance. Investment is based onprojections. This investment involves costs for new facilities, equipment, human resources, andoverhead. If the demand materializes, the investment is worthwhile since the firm may capture alarge amount of market share. If it does not materialize, the firm must redirect the invested resources.This strategy is most appropriate in high growth areas.

If the demand materializes, but the capacity planning strategy is risk averse, i.e., building capacityonly as demand develops, then most likely market share will be lost. The growth in demand willencourage new entrants, resulting in more competition. The risk averse strategy may be mostappropriate for small firms who cannot afford to invest in unproven prospects. To prevent potentialloss of market share, firms may choose to incrementally increase capacity to match the increase indemand.

6. What is capacity balance? Why is it hard to achieve? What methods are used to deal with capacityimbalances?

In a perfectly balanced plant, the output of each stage provides the exact input requirement for thesubsequent stage. This continues throughout the entire operation. This condition is difficult toachieve because the best operating levels for each stage generally differ. Variability in productdemand and the processes may also lead to imbalance, in the short run.

There are various ways of dealing with capacity imbalances. One is to add capacity to those stagesthat are the bottlenecks. This can be achieved by temporary measures such as overtime, leasingequipment, or subcontracting. Another approach is to use buffer inventories so that interdependencebetween two departments can be loosened. A third approach involves duplicating the facilities of onedepartment upon which another is dependent.

7. What are some reasons for a plant to maintain a capacity cushion? How about a negative capacitycushion?

A plant may choose to maintain a capacity cushion for a number of reasons. If the demand is highlyunstable, maintaining cushion capacity will ensure capacity availability at all times. Also, capacitycushions can be useful if high service quality levels are established. Some organizations choose to usecapacity cushions as a competitive weapon to create barriers to entry for competitors.


4-3

Negative capacity cushions may be maintained when demand is expected to decrease rapidly andcapacity investment is high enough to discourage short run capacity acquisitions. It may also makesense where capital investment needed to achieve a capacity cushion is extremely expensive, andcapacity can be easily increased in the short run by methods such as overtime or subcontracting.

8. At first glance, the concepts of the focused factory and capacity flexibility may seem to contradicteach other. Do they really?

This is not necessarily true. This will depend on the available technology of the facility and on the typeof industry it competes in. An FMS plant may, for example, use flexible processes to enlarge the varietyof products produced and delivered in a very short time. Therefore, it can choose to compete on fastdelivery of customized products rather than on cost. The PWP concept can capitalize on the overallfacility economies of scale while maintaining focus within each individual PWP.

Problems1.

Bronze Year 1 Year 2 Year 3 Year 4

Demand for bronze sprinklers 21 24 29 34

Percentage of capacity used 58.3% 66.7% 80.6% 94.4%

Machine requirements 1.75 2.00 2.42 2.83

Labor requirements 3.50 4.00 4.84 5.66

There is sufficient capacity to meet expected demand over the 4-year planning horizon. The only concernmight be year 4 on the bronze line. Capacity is approaching 100% in that year, and forecast error mightlead to an over-capacity situation. It is probably not a large concern at this point in time, but managementshould pay special attention to that point in time as forecasts are updated in the future.

Plastic Year 1 Year 2 Year 3 Year 4

Demand for plastic sprinklers 97 115 136 141


Machine requirements .485 .575 .680 .705



4-4

2. Requirements for plastic remain unchanged.




Machine requirements 2.67 3 3.42 4.33

Labor requirements 5.33 6 6.84 8.67

It is obvious that not enough capacity is available after year two to meet the increased demand.AlwaysRain will have to consider purchasing additional machines for the bronze operations.

3.




Machine requirements 2.67 3.00 3.42 4.33


No. An additional machine will provide enough capacity cushion until the third year. AlwaysRainmust consider additional ways of meeting the fourth year demand. This can include purchasing orleasing an additional machine, or outsourcing some of the demand.

4.

Year 1 Year 2 Year 3 Year 4

Labor requirements-bronze 5.33 6.00 6.84 8.67

Labor requirements-plastic 1.94 2.30 2.72 2.82

Total labor requirements 7.27 8.30 9.58 11.49

AlwaysRain will face a problem of not having enough trained personnel for running the equipmentafter the third year. At that time, they will need to either hire new trained employees or initiate atraining program for existing employees from other workstations who can be utilized at the bronze orplastic molding machines.


4-5

5.

For the small facility,

NPV = .40 ($12 Million) + .60 ($10 Million) - $6 Million = $4.8 Million

Do nothing,

NPV = $0

For the large facility

NPV = .40($14 Million) + .60($10 Million) - $9 Million = $2.6 Million

Therefore, build the small facility.

Build Small FactoryEV = $10.8 – 6.0 millionEV = $4.8 million

$11.6 million

Build Large FactoryEV = $11.6 – 9.0 millionEV = $2.6 million

Do Nothing, EV = $0

$10.8 million

$10.0 million

$12.0 million

Low growthP = .60

High growthP = .40

$10.0 million

$14.0 million

Low growthP = .60

High growthP = .40


4-6

6.

The “Do Nothing” option is included here for completeness.

Rezoned shopping center (includes $1.0 rezoning costs):

Point 1: Expected value = .70($4 Million) + .30($5 Million) - $1.0 million = $3.3 Million

Rezoned apartments:

Point 2: Expected value = .60($4.5 Million) + .40($3 Million) - $1.0 million = $2.9 Million

Since a shopping center has more value, prune the apartment choice. In other words, if rezoned, build ashopping center with arevenue of $4.3 Million - $1 Million = $3.3 Million. (The purchase cost could beincluded here if desired, but would need to be included in the calculations for all development options.This solution shows it at the leftmost part of the tree.)

If not rezoned the revenue will be $2.4 million from building homes:

Point 3: Expected value of developing the land is .6*($3.3 million) + .4*($2.4 million) = $2.94million.

Expected profit of buying and developing the land is $2.94 million - $2 million purchase cost =$940,000. Since this is a positive expected value, prune the option of doing nothing.

EV = $2.94 million

Not RezonedP = .40

RezonedP = .60

Sell to insurance co. @ $5.0 million

1,500 apts. @ $2,000 = $3.0 million

Build 600 homes @ $4,000 = $2.4 million

EV = $3.9 - $1.0 million = $2.9 million

Buy/Develop PropertyEV = $2.94 – 2.0 millionEV = $940,000

Do NothingEV = $0

EV = $4.3 – 1.0 million = $3.3 million

Sell to dept. chain @ $4.0 million

P = .30

P = .70

1

1,500 apts. @ $3,000 = $4.5 million

Low priceP = .40

High priceP = .60

3

2


4-7

Case: Shouldice Hospital - A Cut Above - Teaching Note

1. Mon. - Fri. Operations with 90 beds (30 patients per day)

Beds Required

Monday Tuesday Wednesday Thursday Friday Saturday Sunday

Monday 30 30 30

Tuesday 30 30 30

Wednesday 30 30 30

Thursday 30 30 30

Friday

Saturday

Sunday 30 30 30

Total 60 90 90 90 60 30 30 450

Utilization 66.7% 100.0% 100.0% 100.0% 66.7% 33.3% 33.3% 71.4%

2. Mon. - Sat. Operations with 90 beds (30 patients per day)

Beds Required


Monday 30 30 30

Tuesday 30 30 30

Wednesday 30 30 30

Thursday 30 30 30

Friday 30 30 30

Saturday

Sunday 30 30 30

Total 60 90 90 90 90 60 60 540

Utilization 66.7% 100.0% 100.0% 100.0% 100.0% 66.7% 66.7% 85.7%

Check-inOn

Check-inOn


4-8

3. Mon. - Fri. Operations with 135 beds (minimum)

Beds Required


Monday 45 45 45

Tuesday 45 45 45

Wednesday 45 45 45

Thursday 45 45 45

Friday

Saturday

Sunday 45 45 45

Total 90 135 135 135 90 45 45 675

Utilization 66.7% 100.0% 100.0% 100.0% 66.7% 33.3% 33.3% 71.4%

Can the capacity of the rest of Shouldice keep up?

One operating room can handle about 1 patient every hour. Since there are five operatingrooms, each must be able to handle 45/5 or 9 patients per day. This means they must beoperated 9 hours a day. In order to finish operating early enough for all patients to recoverby the evening, Shouldice would probably have to add operating room capacity although itmight be easy to just start earlier in the day. With 45 patients each day the total number ofoperations each week is 225. The 12 surgeons would need to do between 18 and 19 eachweek or between 3 and 4 a day. This should be feasible and even if it were not Shouldicecould hire some additional surgeons. These guys would be making over $450,000/year (3ops/day x 5 days/week x 50 weeks/yr x $600 = $450,000)!

Check-inOn


4-9

4. Using the financial data given in the fourth discussion question it is easy to justify theexpansion to 135 beds. The following is the analysis as presented in the spreadsheet. Basedon average costs and full capacity utilization, the hospital would pay back its investment inabout 86 weeks, or 1.72 years.

Beds RequiredMon Tues Wed Thurs Fri Sat Sun

Mon 45 45 45Tues 45 45 45Wed 45 45 45

Check-in day Thurs 45 45 45FriSat

Sun 45 45 45

Total Beds Total 90 135 135 135 90 45 45 675135 Utilization 66.7% 100.0% 100.0% 100.0% 66.7% 33.3% 33.3% 71.4%

Operating Rooms Operations 455 Oper/Room 9

Surgeons12 Oper/Surg 3.75

Cost of expansion Beds 45Cost/Bed $100,000Total $4,500,000

IncrementalRevenue

Rev/Oper $1,300

Surgeon $600Incr Rev $700

Additional Oper/Week 75Rev/Week $52,500Payback 85.7 Weeks

Appendix 04A - Learning Curves

4A-1

APPENDIX 4ALEARNING CURVES

1. Ratio for 50th unit from Exhibit 4A.5 is .5518; standard time for 50th unit is .20. Divide the standardtime for the 50th unit by .5518 to get an equivalent time for the first unit. Then you can apply the ratiofor the desired units from Exhibit 4A.5 to that first-unit time directly.

Unit Ratio (Exhibit 4A.5) Time

100th .49665518.

20.(.4966) = .18 hours

200th .44695518.

20.(.4469) = .16 hours

400th .40225518.

20.(.4022) = .15 hours

2. a.

Unit Ratio (Exhibit 4A.5) Cost1st ten 1.0000 $2,500.002nd ten .7000 $2,500 x .7000 = $1,750.003rd ten .5682 $2,500 x .5682 = $1,420.504th ten .4900 $2,500 x .4900 = $1,225.005th ten .4368 $2,500 x .4368 = $1,092.00

b. The simple answer is as high a price as we can get. If we are being “fair” and considering theeffect of the learning curve, we should ask for the sum of the prices listed above for the last twolots of 10, or $2317.00 for 20 units = $115.85 per unit.


4A-2

3.

For labor the following learning was experienced:

Unit 1 to 2 = 1500/2000 = 75%, from units 2 to 4 = 1275/1500 = 85%

Based on this, we estimate an average labor learning rate of 80%

For cost the following learning was experienced:

Unit 1 to 2 = 37050/39000 = 95%, from units 2 to 4 = 31492/37050 = 85%

Based on this, we estimate an average cost of parts learning rate at 90%

Labor for 12 more units:

From Exhibit 4A.6 16 units = 8.920

-4 units = 3.142

5.778

Therefore, Labor for 12 more units = 2,000(5.778) = 11,556 hours

Cost for 12 more units:


-4 units = 3.556

8.484

Therefore, Cost for 12 more units = 39,000(8.484) = $330,876

4. a. Labor:

LR = 3500/5000 = 70%


-2 units = 1.700

3.801

Therefore, Labor cost for 10 more units = 5,000(3.801)(30) = $570,150

Material:

LR = 200000/250000 = 80%


4A-3


-2 units = 1.800

5.427

Therefore, Labor cost for 10 more units = 250,000(5.427) = $1,356,750

Total Cost is $570,150 + $1,356,750 = $1,926,900

b. The minimum cost would be as calculated in part a, $1,926,900. This would assume noforgetting. However, the worst case would be total forgetting, which would imply thatthere was no benefit to having produced units 1 and 2. This cost would be as follows.

Complete forgetting:

Labor: 4.931(5,000)(30) = $ 739,650

Material: 6.315($250,000) = $1,578,750

Total = $2,318,400

Therefore, the range is from $1,926,900 to $2,318,400.

5.

a.

Units 1 to 2 = 640/970 = 65.98%Units 2 to 4 = 380/640 = 59.37%Units 4 to 8 = 207/380 = 54.47%

Average learning rate = 59.94% Round this to 60% to use the tables.

b. From Exhibit 4A.6 200 units = 12.090

-10 units = 3.813

8.277

Therefore, time for 190 more units = 970(8.277) = 8,029 hours

c. For 1,000th unit from Exhibit 4A.5:

.0062(970) = 6.0 hours


4A-4

6.

a.

For units 1 to 2 = 10/12 = 83.33%For units 2 to 4 = 6.5/10 = 65 %For units 4 to 8 = 3.6/6.5 = 55.38%For units 8 to 16 = 2.7/3.6 = 75%

The average learning rate is = 69.67 Round this to 70%.


-20 units = 7.407

12.163

Therefore, cost for 100 more units = 12 million (12.163) = $145,956,000. LTI should add theappropriate profit margin to this expected cost.

c. From Exhibit 4A.5: Cost for 120th unit is $12 million(.0851) = $1,021,200

7. LR = 1,800/2,000 = 90%


-3 units = 2.746

2.355

Therefore, time for 3 more units = 2,000(2.355) = 4,710 hours


4A-5

8. Southwest Honda’s LR = 6.3/9.0 = 70%

The breakeven point is where the AVERAGE labor hours on all cars serviced by the mechanicequals 3 hours. According to the following table, that doesn't occur until after car 25. Unless eachmechanic is going to service over 25 cars, it's not a good deal. Since the dealer expects toperform the service on 300 vehicles with 6 mechanics (approximately 50 per mechanic), itappears that Honda’s rate is more than fair.

LearningRate

70%

Unit Unit Time Average Time1 9.0 9.002 6.3 7.653 5.1 6.804 4.4 6.215 3.9 5.756 3.6 5.397 3.3 5.098 3.1 4.849 2.9 4.6310 2.8 4.4411 2.6 4.2712 2.5 4.1313 2.4 3.9914 2.3 3.8715 2.2 3.7616 2.2 3.6617 2.1 3.5718 2.0 3.4919 2.0 3.4120 1.9 3.3321 1.9 3.2622 1.8 3.2023 1.8 3.1424 1.8 3.0825 1.7 3.0326 1.7 2.9727 1.7 2.9228 1.6 2.8829 1.6 2.8330 1.6 2.79


4A-6

9. Labor LR = 3,200/4,000 = 80%

Material LR = 21,000/30,000 = 70%

a. For the 22nd unit:

Labor 4,000(.3697)($18 per hour) = $26,618.40

Material $30,000(.2038) = $ 6,114.00

Total = $32,732.40


-2 units = 1.800

9.430

Labor time for 20 more units = 4,000(9.430) = 37,720 hours

Average time = 37,720/20 = 1,886 hours

c. From Exhibit 4A.6 22 units = 7.819

-2 units = 1.700

6.119

Therefore, material cost for 20 more units = $30,000(6.119) = $183,570

Average material cost = $183,570/20 = $9,178.50

Average labor cost = 1,886($18 ) = $33,948.00

Average total cost = $9,178.50 + $33,948.00 = $43,126.50

Chapter 05 - Projects

5-1

CHAPTER 5PROJECTS

Review and Discussion Questions1. What was the most complex project that you have been involved in? Give examples of the following

as they pertain to the project: the work breakdown structure, tasks, subtasks, and work package.Were you on the critical path? Did it have a good project manager?

Obviously, the answer will vary. Remember that the project could be in a non-profit environment as well.School plays (especially musicals) are a good example, because there are many major tasks that need tobe broken down and scheduled in parallel, but all be completed by the time opening night comes. Thiswould include selecting the play and obtaining the rights, auditions, rehearsals of the actors, rehearsals ofthe musicians, construction of the sets, setting the lighting, printing tickets and programs, staffing thetheater, advertising and fund raising.

2. What are some reasons project scheduling is not done well?

Several problems with project scheduling are discussed at the end of the chapter. The uncertaintiesinherent in the activities comprising the network of any project make it necessary to update the scheduleon a regular basis. Maintaining accurate time and cost estimates is often difficult and frustrating.Managing this evolving process requires a discipline that is not always available.

3. Discuss the graphic presentations in Exhibit 5.4. Are there any other graphic outputs you would liketo see if you were project manager?

The various graphs and charts presented are typical of the graphical techniques for presenting thenecessary data. Most are adaptable to computer programming. The major requirements in the graphicspackage include planned activities related to time, a milestone chart to show major achievements, abreakdown to show how funds were spent plus a plot of actual completion versus planned.

4. Which characteristics must a project have for critical path scheduling to be applicable? What types ofprojects have been subjected to critical path analysis?

Project characteristics necessary for critical path scheduling to be applicable are:

a. Well-defined jobs whose completion marks the end of the project.

b. The jobs of tasks are independent in that they may be started, stopped, and conductedseparately within a given sequence.

c. The jobs or tasks are ordered in that they must follow each other in a given sequence.

d. An activity once started is allowed to continue without interruption until it is completed.

A wide variety of projects have used critical path analysis. Some industries that more commonly use thisapproach include aerospace, construction, and computer software.

5. What are the underlying assumptions of minimum-cost scheduling? Are they equally realistic?


5-2

The underlying assumptions of minimum cost scheduling are that it costs money to expedite a projectactivity and it costs money to sustain or lengthen the completion time of the project.

While both assumptions are generally realistic, it often happens that there are little or no out-of-pocketcosts associated with sustaining a project. Personnel are often shifted between projects, and in the shortrun there may be no incentive to compete a project in “normal time.”

6. “Project control should always focus on the critical path.” Comment.

In many project situations, it is not the activities on the critical path which cause problems, but rathernoncritical activities, which, for various reasons, become critical. In the context of PERT, it may turn outthat the activities on the critical path have small variances associated with them and can be treated as nearcertain. At the same time, activities off the critical path may have extremely large variances and, in fact,if not closely monitored, may delay the project. Thus, while project control must keep track of criticalpath activities, it may be more useful to focus on those activities which are not on the critical path but, forone reason or another, have a high degree of uncertainty associated with them.

Along these lines, some authors have suggested that the critical path approach should be replaced by acritical activity approach in which simulation is used to estimate which activities are likely to becomesources of project delay. These activities, rather than critical path, would become the focus of managerialcontrol.

Additionally, the critical path focuses on the time or schedule aspects of the projects. Certain activitiescould be "critical" because of cost or quality considerations.

7. Why would subcontractors for a government project want their activities on the critical path? Underwhat conditions would they try to avoid being on the critical path?

A subcontractor might want his activities on the critical path in situations where cost incentives areprovided for early project completion. Since the critical path ultimately determines project length, itstands to reason that activities on the path would be the ones that would draw additional funds to expeditecompletion. A subcontractor might want his activities off the critical path because of some error on hispart or because he doesn’t want to be bothered by the close monitoring of progress which often goes withcritical path activities.


5-3

Problems1.

ActivityExpectedRevenue

PlannedDuration

PlannedStart Date

PlannedComp. Date

Expected% Complete

Actual %Complete

Actual Rev.to Date

1. Solicit $25,000 25 0 25 100% 90%2. Donations $150,000 30 5 35 83.3% 50%3. Matching Funds $50,000 10 30 40 0% 0%Total $225,000 $175,000

BCWS

BCWP

Scheduled Variance = 97,500 - 149,500 = -52,000Scheduled Performance Index = 97,500/149,500 = .652

Cost Variance = 97,500 - 175,000 = -77,500Cost Performance = 97,500/175,000 = .557

Because we are working with revenues instead of costs, we have to invert the evaluation ruleslisted in the text. Our performance measures here are actually good. Although we are behindschedule on completing tasks 1 and 2, we have brought in more money than expected for theamount of work we have completed.

Activity 1 100% of $25,000 = 22,500Activity 2 83% of $150,000 = 124,500Activity 3 0 % of 50,000 = 0

$149,500

Activity 1 90% of 25,000 = 22,500Activity 2 50% of 150,000 = 75,000Activity 3 0% of $50,000 = 0

97,500


5-4

2.

ActivityExpected

CostExpected

% CompleteActual %

CompleteActual Cost to

Date1. Site preparation $1,420,000 100% 100% $1,300,0002. Pour concrete $10,500,000 100% 100% $9,000,0003. Construction $8,500,000 50% 60% $5,000,000Total $20,420,000 $15,300,000

BCWS

Activity 1 100% of $1,420,000 = $1,420,000Activity 2 100% of $10,500,000 = $10,500,000Activity 3 50 % of $8,500,000 = $4,250,000

$16,170,000

BCWP

Activity 1 100% of $1,420,000 = $1,420,000Activity 2 100% of $10,500,000 = $10,500,000Activity 3 60 % of $8,500,000 = $5,100,000

$17,020,000

Scheduled Variance = 17,020,000-16,170,000 = 850,000Scheduled Performance Index = 17,020,000/16,170,000 = 1.053

Cost Performance = 17,020,000/15,300,000 = 1.11

Ahead of schedule and under budget.


5-5

3.

a.

b. A-C-D-E-G, also shown in the network above as the bold path.

c. 6+7+2+4+7 = 26 weeks.

d. Activity B has 6 weeks of slack – the difference between its early and late start times.

4. a.

b. A-B-D-E-H, also shown in the network above as the bold path.

c. 15 weeks, 1+4+2+5+3.

d. C, 3 weeks; F, 1 week; and G, 1 week.

B (3)

G (7)

D (2)

E (4)

C (7)

6 13

6 9 15 19

13 15

19 26

2619

15

19151512

136 13

A (6)

0 6

0 6

F (3)

15 18

1916

C (3)

G (2)

H (3)

D (2)

E (5)

B (4)

1 5

1 4 7 12

5 7 9

12 15

11

15

1210

12

7

12774

51 5

A (1)

0 1

0 1

F (2)

7 9

108


5-6

5. a.

Note that G has both D and F as immediate predecessors. However, D is redundant because F also has Das an immediate predecessor.

b. A-C-F-G-I, and A-D-F-G-I at 18 weeks.

c. B is not on a critical path and has slack of 4; therefore, do not shorten as it will not change theproject completion time. Shorten C, D, and G one week each. C and D are on parallel criticalpaths, reducing them both will only reduce project completion time by 1 week.

d. A-C-F-G-I; and A-D-F-G-I remain the critical paths. Project completion time is reduced from 18to 16 weeks.

C (4) G (2)

H (3)D (4)

E (6)B (2)

3 5

3 7

5 11

3 7

13

7 10

15

15

1513

127

159

73

97

3

A (3)

0 3

0 3

F (6)

7 13

137

I (3)

15 18

1815


5-7

6. a.

TIMES (DAYS)

ACTIVITYIMMEDIATE

PREDECESSORS a m b ET A — 1 3 5 3 0.4444B — 1 2 3 2 0.1111C A 1 2 3 2 0.1111D A 2 3 4 3 0.1111E B 3 4 11 5 1.7778F C, D 3 4 5 4 0.1111G D, E 1 4 6 3.833 0.6944H F, G 2 4 5 3.833 0.2500

b. B-E-G-H

c. 2.00+5.00+3.83+3.831 = 14.66

d. Variance of project completion time is found by adding the variances of activities on the criticalpath. .1111 + 1.7778 + .6944 + .2500 = 2.833

833.2

)67.1416( Z = .79

P(T<16) = P(Z<.79) = .7852 (From Excel’s NORMSDIST() function)

B (2)

G (3.83)

H (3.83)

F (4)

E (5)

C (2)

1 5

1 4 7 12

5 7

9

12 15

1115

1210

12

7

12774

51

5A (3)

0 1

0 1

D (3)

1 4

74


5-8

7.

ActivityMost

optimisticMostlikely

Mostpessimistic

ExpectedTime

Variance

ABCDE

23164

53387

1135

1010

5.53387

2.250

.444

.4441

First find the value of Z that results in a probability of .82. Using Excel’s NORMSINV(.82) = .915.Then find the critical path (ABD) and the variance on the critical path: 2.25+ 0 + .444 = 2.694. Finally,use equation 5.3 to solve for D.

Paths

ABD 16.5694.2

5.16915.

DD = 18

ACE 15.5


5-9

8. a.

Job No. a m b ET 2

1 2 3 4 3.00 .1112 1 2 3 2.00 .1113 4 5 12 6.00 1.784 3 4 11 5.00 1.785 1 3 5 3.00 .4446 1 2 3 2.00 .1117 1 8 9 7.00 1.788 2 4 6 4.00 .4449 2 4 12 5.00 2.78

10 3 4 5 4.00 .11111 5 7 8 6.83 .25

b. 1-3-6-8-9-11.

c. 3.00+6.00+2.00+4.00+5.00+6.83 = 26.83

d. None of these. Job 5 is not on the critical path; therefore, reducing its time by two days will notreduce project completion time. If you reduce job 3 by two days, then path 1-4-7-10-11 becomescritical and the project length is 25.83 days. You’ve saved $1,000 but paid $1,500. Task 7 is noton the critical path, so reducing it alone will not shorten the project. You could reduce both 3 and7 to reduce the project length by two days, but would only save $2,000 while spending $3,000.

e. First you need to compute the variance of the C.P.: .111+1.78+.111+.444+2.78+.25 = 5.47. Thenuse equation 5.3 to find the correct Z and look its value up in Appendix E.

91.355.147.5

83.2630

PZ


5-10

9.a.

b. 1-3-6-9-10, length = 26 weeks.

c. The most logical option would be to cut activity 3 by 3 weeks, and then reduce activity 6 or 9by one week. This is the lowest cost option, and does not create an additional critical path.Another option would be to reduce activities 6 and 9 by a total of 3 weeks, and then reduceactivity 3 by one week. This also has the same cost, but creates an additional critical path (1-3-5-8-10).

10.


5-11

a. A-E-G-C-D.

b. 26 weeks.

c. No difference in completion date. Neither B nor F is on the critical path.

11.

a. A-C-D-F-G

b.

Activity NormalTime (NT)

Crash Time(CT)

Normal Cost(NC)

Crash Cost(CC)

NT-CT Cost/day toexpedite

A 7 6 $7,000 $8,000 1 $1,000B 3 2 5,000 7,000 1 2,000C 4 3 9,000 10,200 1 1,200D 5 4 3,000 4,500 1 1,500E 2 1 2,000 3,000 1 1,000F 4 2 4,000 7,000 2 1,500G 5 4 5,000 8,000 1 3,000

First, lowest cost activities to crash are A and E at $1,000 per day. E is not on the critical path, thereforeselect A. Critical path remains the same. Second, lowest cost activity on the critical path is C. Crashactivity C. Now two paths have become critical. Third, D and F are next lowest cost activities on thecritical paths. Both have a cost of $1,500 per day. Select D then F or reverse the order (F then D). Fcannot be reduced by two days because it would cause E to become part of a critical path.


5-12

Summary of steps to reduce project by four days:

Step Activity to crash Cost to crash Days saved

1 A $1,000 1

2 C 1,200 1

3 D (or F) 1,500 1

4 F (or D) 1,500 1

12.

a. 100 Hours.

b. Activities b and d are not on the critical path. Their start can be delayed without delaying thestart of any subsequent activities. Activity b can be delayed by 10 hours and d can be delayed by30 hours without affecting the project completion date.

c (20)

g (20) h (0)

d (30)

e (20)

b (10)

0 10

0 20 20 40

20 50

80 100 100100

10010080 100

80

4020200

2010 50

a (0)

0 0

0 0

f (40)

40 80

8040


5-13

13.

a. A-B-D-G, 5+10+6+4 = 25 weeks.

b.


Normal Cost(NC)

Crash Time(CT)

Crash Cost(CC)

NT-CT Cost/weekto expedite

A 5 $7,000 3 $13,000 2 $3,000B 10 12,000 7 18,000 3 2,000C 8 5,000 7 7,000 1 2,000D 6 4,000 5 5,000 1 1,000E 7 3,000 6 6,000 1 3,000F 4 6,000 3 7,000 1 1,000G 4 7,000 3 9,000 1 2,000

First, reduce D (lowest cost activity on the critical path) by one week. This adds an additional criticalpath with activities C and E in it. Second, crash activity G by one week. Critical paths remain the same.Third, crash activity A by one week at a cost of $3,000, which is the least expensive.

Summary of activities crashed:

Step Activity Cost to crash Weeks reduced

1 D $1,000 1

2 G 2,000 1

3 A 3,000 1

Total cost $6,000


5-14

14.

a. A-C-D-F-G.

b. 7+4+5+4+5 = 25 weeks.

c. B, 2weeks; E, 2 weeks.


Crash Time(CT)

Normal Cost(NC)

Crash Cost(CC)


A 7 6 $7,000 $8,000 1 1,000

B 2 1 5,000 7,000 1 2,000

C 4 3 9,000 10,200 1 1,200

D 5 4 3,000 4,500 1 1,500

E 2 1 2,000 3,000 1 1,000

F 4 2 4,000 7,000 2 1,500

G 5 4 5,000 8,000 1 3,000

First, shorten activity A by one week at a cost of $1,000. This is the lowest cost/week activity on the criticalpath. Second, shorten activity C by one week at a cost of $1,200. This is the next lowest cost/week activity onthe critical path. The total cost is $2,200 and the critical path remains unchanged.


5-15

15. a.

b. and c.

Path Normal Length Crashed LengthABDF 24 15ACDF 31 20ACEF 36 23

d. We would only crash the project until 29 weeks since the cost of crashing C is $4000 which is greaterthan the $3500 in additional profit.

LENGTH

ABDF - 24 24 24 24 24 23 22 21 21

ACDF - 31 31 31 31 31 30 29 28 27

ACEF - 36 35 34 33 32 31 30 29 28

Activity Crashed E E E E A A A C

Crash cost 2500 2500 2500 2500 3000 3000 3000 4000

Cumulative Cost 2500 5000 7500 10000 13000 16000 19000 23000

A (9)

B (8)

C (15) E (10)

D (5)

F (2)


5-16

16.


Crash Time(CT)

Normal Cost(NC)

Crash Cost(CC)


A 2 1 $50 $70 1 20

B 4 2 80 160 2 40

C 8 4 70 110 4 10

D 6 5 60 80 1 20

E 7 6 100 130 1 30

F 4 3 40 100 1 60

G 5 4 100 150 1 50

a.

b. The critical path is A-B-E-G, with a length of 2+4+7+5 = 18c. The normal time project cost is $500 at 18 weeks. Minimum cost crashing to 13 weeks is shown

below.

LENGTH

ABEG - 18 17 16 15 14 13

ACG - 15 14 14 13 13 13

ADFG - 17 16 16 15 14 13

Activity Crashed A E G B+D B+F

Crash cost 20 30 50 60 100

Cum. Crash Cost 20 50 100 160 260

Total Direct Cost 520 550 600 660 760

D (6)

G (5)

E (7)

F (4)

C (8)

B (4)

2 6

2 8

2 10

8 12

13

6 13

18

13

18

13

6

13

13

5

93

62

9

A (2)

0 2

0 2


5-17

d. Computations build off the earlier table and are shown below. The total project cost in normal timeis $500 + $400 = $900

LENGTH

ABEG - 18 17 16 15 14 13

ACG - 15 14 14 13 13 13

ADFG - 17 16 16 15 14 13

Activity Crashed A E G B+D B+F

Crash cost 20 30 50 60 100

Cum. Crash Cost 20 50 100 160 260

Total Direct Cost 520 550 600 660 760

Indirect Cost 350 300 250 200 150

Total Project Cost 870 850 850 860 910

We can achieve minimum cost by reducing the project to either 16 or 15 weeks. Both durations have thesame total cost of $850.

Analytics Exercise: Product Design Project

1, 2.

The project will take 37 weeks to complete.The critical path is P1-P2-P3-P4-S1-D2-I2-I3-I4-V2.

P1 (2)

0 2

0 2

P4 (5)

11 16

11 16

P2 (4)

2 6

2 6

P3 (5)

6 11

6 11

S2 (6)

16 22

19 25

S1 (5)

16 21

16 21

V1 (10)

22 32

25 35

D1 (1)

21 22

23 24

D3 (1)

21 22

23 24

D2 (2)

21 23

21 23

D4 (4)

22 26

24 28

I1 (3)

26 29

28 31

I2 (4)

23 27

23 27

I3 (4)

26 29

28 31

I4 (4)

31 35

31 35

V2 (2)

35 37

35 37


5-18

3. Slack for each activity is listed in the following table.

Major Subprojects/ActivitiesActivity

ID DependenciesDuration(Weeks) ES, LS Slack

Project Specifications (P) 16Market research P1 -- 2 0, 0 --Overall product specification P2 P1 4 2, 2 --Hardware P3 P2 5 6, 6 --Software P4 P3 5 11, 11 --

Supplier specifications (S) 6Hardware S1 P4 5 16, 16 --Software S2 P4 6 16, 19 3

Product design (D) 6Battery D1 S1 1 21, 23 2Display D2 S1 2 21, 21 --Camera D3 S1 1 21, 23 2Outer cover D4 D1, D2, D3 4 22, 24 2

Product integration (I) 12Hardware I1 D4 3 26, 28 2User interface I2 D2 4 23, 23 --Software coding I3 I2 4 27, 27 --Prototype testing I4 I1, I3 4 31, 31 --

Subcontracting (V) 12Supplier selection V1 S1, S2 10 22, 25 3Contract negotiation V2 I4, V1 2 35, 35 --

4. Assume that the activity lengths remain the same, but the precedence relationships withinsubprojects no longer apply as all activities in a subproject will be worked on in parallel.Therefore, the length of each subproject will now be equal to the length of the longest activityin the subproject. The subprojects are outlined in dashed lines in the network drawing above.

Subproject P will take 5 weeks to complete; subproject S, 6 weeks; D, 4 weeks; I, 4 weeks;and V, 10 weeks. Since all of these subprojects will completed in series now, the length ofthe project is the sum of the subproject times: 5+6+4+4+10 = 29 weeks, a decrease of 8weeks.

5. The revised network diagram is shown below. Adding P5 extends the time of subproject P to12 weeks while subprojects D and I remain at 4 weeks each. By eliminating subprojects Sand V, the project length is now down to 20 weeks – 17 weeks less than the original schedule.


5-19

6. Having the team focus on a single subproject at a time will allow more collaboration on eachsubproject as opposed to the team being split across several subprojects. This might result in higherquality of work as the entire project team is focused on a single subproject at a time and you will havemore input and a wider variety of experiences working on each subproject than you would otherwise.

By combining subprojects S and V with P, Nokia can perhaps take advantage of supplier expertise indesigning the new phone. Nokia would better understand the technological capabilities of theirsuppliers and include them in the phone design from the beginning.

One concern might be the feasibility of eliminating the original precedence relationships whenchanging to the new project structures in parts 4 and 5. Assuming there were good reasons for thoserelationships originally, eliminating them might cause problems in the project if Nokia does not fullyaddress those reasons in the new project structure. For example, if they select suppliers and negotiatecontracts before the product specifications are complete, and they do not include their suppliers in theproduct specification process, they might end up with a supplier that cannot supply the neededmaterials nor do so at the proper level of quality.

Assuming the technical/managerial precedence issues are properly addressed, the new projectstructures make sense. In addition to reducing project time, there are other possible benefits to begained from the increased collaboration the new structures would bring.

P1 (2)

0 2

10 12

P4 (5)

0 5

7 12

P2 (4)

0 4

8 12

P3 (5)

0 5

7 12

P5 (12)

0 12

0 12

D1 (1)

12 13

155

16

D3 (1)

12 13

15 16

D2 (2)

12 144

14 16

D4 (4)

12 16

12 16

I1 (3)

16 19

17 20

I2 (4)

16 20

16 20

I3 (4)

16 20

16 20

I4 (4)

16 20

16 20

12 weeks 4 weeks4 weeks

Chapter 06 - Manufacturing and Service Processes

6-1

CHAPTER 6MANUFACTURING AND SERVICE PROCESSES

Review and Discussion Questions1. What kind of layout is used in a physical fitness center?

Workcenter (job shop) layout—similar equipment or functions are grouped together, such as rowingmachines in one area, and weight machines in another. The exercise enthusiasts move through thefitness center, following an established sequence of operations.

2. What is the objective of assembly-line balancing? How would you deal with the situation whereone worker, although trying hard, is 20 percent slower than the other 10 people on a line?

The objective is to create an efficient balance between the tasks and workstations to minimize idletime. If the employee is deemed valuable, training may enhance his/her speed. It is also possible toplace him/her in the “choice” job, i.e., that workstation which has most idle time to adjust for theslowness. Also, faster workers may assist the slowpoke if the balance and physical features of theline permit.

3. How do you determine the idle-time percentage from a given assembly-line balance?

Idle-time percentage is given as “balance delay” in the chapter. It is simply one minus efficiency,where efficiency is equal to the sum of the task times divided by the number of workstations times thecycle time.

4. What is the essential requirement for mixed-model lines to be practical?

A company needs to develop a cycle mix that minimizes inventory build-up while keeping cycle timeconstant.

5. Why might it be difficult to develop a manufacturing cell?

a. Distinct parts families must exist. This requires developing and maintaining a computerized partsclassification and coding system. This can be a major expense.

b. Several of each type of machine must be available. This could be an expensive proposition, giventhe cost of purchasing and maintaining duplicate sets of machinery.

c. Taking a machine out of a cluster should not rob a cluster of all of its capacity.

d. There may be parts that cannot be associated with a family and specialized machinery that cannotbe placed in a cell because of its general use.

e. Training personnel to perform multiple types of tasks may be initially difficult. Unionregulations and interpersonal problems within a group working in a cell must be resolved beforethe cell is implemented.


6-2

6. How would you characterize the most important difference for the following issues whencomparing a facility organized with workcenters versus a continuous process?

Workcenters would require more frequent changeovers, but they could be done independently, whilea continuous process would try to minimize changeovers due to the significant coordination requiredto setup the line for a new product. Workcenters would typically have more labor content, while acontinuous process would have more specialized machines that remove labor from the process.Workcenter arrangements provide more flexibility in the kinds of different products produced at anypoint in time, and are also more flexible when introducing new products. To achieve similarflexibility with a continuous process requires significant planning and investment.

7. A certain custom engraving shop has traditionally had orders for between 1 and 50 units ofwhatever a customer orders. A large company has contacted this shop about engraving “ reward”plaques (which are essentially identical to each other). It wants the shop to place a bid for thisorder. The expected volume is 12,000 units per year and will most likely last four years. Tosuccessfully bid (low enough price) for such an order, what will the shop likely have to do??

Determine capacity analysis, look at moving from job shop to continuous flow production, price lowand make up profits in volume, analyze production deadlines in order to determine labor needs (i.e.can the shop get by with a part time employee to meet the extra capacity).

8. The product–process matrix is a convenient way of characterizing the relationship betweenproduct volumes (one-of-a-kind to continuous) and the processing system employed by a firm at aparticular location. In the boxes presented below, describe the nature of the intersection betweenthe type of shop (column) and process dimension (row).

Workcenters Continuous Process

Engineering Emphasis Product variety and improvements Process improvements

General Workforce Skill Skilled workforce Lower skill levels

Statistical Process Control Less important More important

Facility Layout Process, functional Product, line, flow

WIP Inventory Level Depends on the product Lower

9. For each of the following variables, explain the differences (in general) as one moves from aworkcenter process to a continuous process environment.

a. Throughput time decreases as you move from a job shop to flow shop environment.

b. Capital intensity increases as you move to a flow shop (need for more machinery). Laborintensity (assuming you mean # of workers) would probably increase as well (depends on thelevel of automation). Ratio would increase in a flow shop because capital requirements are somuch greater.

c. Bottlenecks would probably decrease in a continuous flow shop.


6-3

Problems

1. Wait Time = 50 Minutes

a. 10 Cars * 5 minutes per car

2. Using Little’s Law we know that Inventory = Throughput * FlowtimeBefore change: Inventory = 1 per hour * 15 hours = 15 * $1500 = $22,500After change: Inventory = 1 per hour * 10 hours = 10 * $1500 = $15,000Reduction in WIP = $22,500 - $15,000 = $7,500

3. a. The throughput rate is 30 buses per hour. The throughput time is 20/60 =1/3 hours.Inventory = Throughput * Flowtime = 30 * 1/3 = 10 buses are traveling to and from the airport atany one time.

b. This reduces the average waiting time, but has no effect on the average traveling time. Theaverage traveling time does not change and will remain 20 min. This would however result in 40buses being in transit between the airport and the office at any given time. Does Avis have thatmany buses?

4. TR = 15 planes per hour. TT = (0.75 +4.25)/60 = 1/12 of an hour.WIP = TT * TR = 15 * 1/12 = 1.25 airplanes in line or taking off.

5. a. TR = 60 per week, TT = 2/7 weeks.WIP = TT *TR = 60*2/7 = 120/7 = 17.1On average there are 17.1 new mothers in the Children’s Hospital.

b. TR = 210 per week, TT = 2/7 weeks.WIP = TT * TR = 210*2/7 = 60.On average there are 60 new mothers in Swedish Hospital.

c. TR = 270 per week, TT = 2/7 weeks.WIP = TT * TR = 270*2/7 = 540/7 = 77.1On average there will be 77.1 new mothers in the unified hospital. No, it will not decrease thetotal number of mothers in the hospital.

6. a. Throughput Rate = 50 people/month. WIP = 1000 people.Average working time = Throughput Time = WIP/TR = 1000/50 = 20 months.

b.Current annual cost for hiring and training:Throughput Rate = 50 people/month = 600 people/yearAnnual hiring and training cost is 600*1000=$600,000

New annual cost for hiring and training:Average working time = Throughput Time = 24 months = 2 yearsThroughput Rate = WIP/TT = 1000 people/ 2 years = 500 people/ yearAnnual hiring and training cost is 500*1000=$500,000.Annual saving on hiring and training cost is $100,000


6-4

7. a. What is the capacity of the washing stage?10 * 60/30 = 20 orders / hour

b. What is the capacity of the drying stage?15 * 60/40 = 22.5 orders/ hour

c. Identify the bottleneck(s). Briefly explain.Washers

d. What is the capacity of Money laundry? Briefly explain.20 orders/hour

e. If Money laundry would like to increase the capacity by buying one more machine, should theybuy a washer or a dryer? Why?

Washer as that is the bottleneck.

The manager, Mr. Money, decided not to buy a machine. He still has 10 washers and 15 dryers.The manager estimates that on average Money Laundry receives 8 orders every hour. Themanager also finds that on average there are 5 orders in the washing stage and 7 orders in thedrying stage.

f. What is the utilization of washers, on average?8 orders/hour / 20 orders/hour = 40%

g. What is the utilization of dryers, on average?8 orders/hour / 22.5 orders/hour = 35.56%

h. On average, how long does it take an order to finish washing process, from the time the order isreceived?

WIP = 5, TR = 8 orders/hour. TT = WIP / TR = 5/8 = 0.625 hour or 37.5 minutes.

i. On average, how long does it take an order to finish drying process, from the time the orderfinishes the washing process?

WIP = 7, TR = 8 order/hour. TT = WIP / TR = 7/8 = 0.875 hour or 52.5 minutes.

j. On average, how long does an order stay in Money laundry?0.625+0.875 = 1.5 hour.

Double check: WIP = 5+7=12, TR = 8 order/hour. TT = 12/8 = 1.5 hour.


6-5

8 a.

b. C = production time per day/required output per day =

seconds)75(orminutes25.1360

450C

c.

Work station Task Task time Idle time

IACE

303015 0

II F 65 10

IIIBG

3540 0

IVDH

3525 15

d. Efficiency = 91.7%or917.)75(4

275

CN

T

a

A

B D

C

E

F

G30

35

65

30

H35

15

40

25


6-6

9. a.

itseconds/un271000

60*60*5.7

DayperOutputRequired

(seconds)DayperTimeProductionC

b.

A

B

D

C

E

F

G

15

24

6

H

14

11

18

I

J

K

L

7

15

10

12

7

9

Work station Task Task time Idle timeI A

C156 6

II B 24 3III E

F187 2

IV ID

1412 1

V GHJ

1197 0

VI KL

1510

∑ 1482

c. Efficiency =CN

T

a

=)27(6

148= .914 or 91.4%

d. Reduce cycle time to 25 ((7.5*3600)/1100 = 24.54 seconds), which requires rebalancing theline, or work overtime: (100 units) 27 seconds per unit = 2700 seconds or 45 minutes ofovertime.


6-7

10. a. itseconds/un90300

60*60*5.7DayperOutputRequired(seconds)DayperTimeProduction

C

b. nsworkstatio556.490

410

C

TN t

c.

a

b

d

c

e

f

g

70

40

10

h

60

15

30

i

j

k

l

25

25

45 20

50

20

d.

Work station Task Task time Idle timeI a

d7010 10

II gj

6025 5

III cb

4540 5

IV eh

3050 10

V fIkl

20152025

∑ 41010


6-8

e. Efficiency =CN

T

a

=)90(5

410= .911 or 91.1%

f. Reduce cycle time to 81 seconds per unit. This produces (7.5 hours)(3600 seconds perhour)/81 seconds per units = 333.3 units. Another option is to work 45 minutes overtime (7.5x 10% = .75 hour or 45 minutes). There are many other options possible that arecombinations of these two options.

11. a. itseconds/un6.33750

60*60*7DayperOutputRequired(seconds)DayperTimeProduction

C

b. ons workstati451.36.33

118

C

TN t

c.

A B

D

C E

F

G

20 7

20

H

15

10

16 8

22

d.

Work station Task Task time Idle timeI A

B207 6.6

II DF

2210 1.6

III C 20 13.6

IV EG

1516 2.6

V H 8∑ 118

25.6


6-9

e. 70.2%or702.6.335

118Efficiency

CN

T

a

f. Reduce cycle time to 32 which can be done with the current balance. New production level is (7hours/day)*(3600 seconds per hour)/32 seconds per unit) = 787.5 units per day. Therefore, they are800 – 787.5 = 12.5 units short. Work (12.5 units)*(32 seconds per unit) = 400 seconds or 6.67minutes overtime.

g. 1000 – 787.5 = 212.5 units short, work (212.5 units)(32 seconds per unit) = 6800 seconds or 113.3minutes or 1.89 hours of overtime. May want to consider rebalancing the line.

12. a. itseconds/un30900


C

b. 100%or00.1304

120Efficiency

CN

T

a

This balance is 100 percent efficient.

c. To make 900 units per shift, station 3 will need to be duplicated.

Efficiency formula = 135 / (5(30)) = .9 This assumes that each of the other operatorsstill take 30 seconds to complete. The system becomes less efficient due to the idle timeat the parallel workstations 3.


6-10

13. a.

b. (60 minutes/hr)/(15 units/hr) = 4 minutes per unit. C = 4.

c.Work station Task Task time Idle time

IAC

13 0

II E 3 1

IIIBF

22 0

IVDG

13 0

d. 93.75%or9375.44

15Efficiency

CN

T

a

A

B

D

C E

1

2

1

3

G

F

3 3

2


6-11

ANALYTICS EXERCISE: Designing a Manufacturing Process

1. What is the daily capacity of the assembly line designed by the engineers?

The line operates for 7.5 hours per day. Workstation 9 is the bottleneck in the initial line balance, limitingthe cycle time to 2 minutes, so output is limited to 30 units/ hours * 7.5 hours = 225 units per day.

2. When it is running at maximum capacity, what is the efficiency of the line relative to its use of labor?Assume that the “supporter” is not included in efficiency calculations.

For this efficiency calculation, only consider the tasks that are performed at workstations usinglabor, not the 310 seconds for the software load. Therefore the sum of the task times for thiscalculation is 583 seconds.

3. How should the line be redesigned to operate at the initial 250 units per day target, assuming that noovertime will be used? What is the efficiency of your new design?

Without using overtime, the cycle time will have to be reduced. The maximum cycle time thatwill meet this production rate is:

itseconds/un108250


C

All current stations are under that cycle time except for station 6 (position 9). Because of theprecedence relationships for tasks 16 and 17, they must be split across two stations to meet thenew cycle time. A simple way to meet this cycle time is to just put task 17 into position 10 andadd an additional worker. The efficiency of this solution is:

4. What about running the line at 300 units per day? If overtime were used with the engineer’s initialdesign, how much time would need to be run each day?

With the original design, output is 30 units per hour. To reach output of 300 units would require2.5 hours of overtime per day.

80.97%or8097.1206

583Efficiency

CN

T

a

77.12%or7712.1087

583Efficiency

CN

T

a


6-12

5. Can the assembly line produce 300 units per day without using overtime?

The cycle time to meet this production rate without overtime is:

itseconds/un90300

60*60*5.7

DayperOutputRequired

(seconds)DayperTimeProductionC

1 (75)

3 (24)

2 (61)

4 (36)

5 (22)

7 (32)

6 (39)

8 (44)

9 (29)

12 (7)

11 (52)

10 (26)

13 (5) 14(11) 16 (60) 17 (60)15 (310)


6-13

This may be possible with a redesigned line, but we might not have enough Line Positions toaccommodate the new design. The following is a design that was constructed using the “longesttask” priority run. This is actually a pretty good design and it would be possible to make 300units per day. The labor efficiency of this line is 583/(8(90)) = .809 or 81%.

Station Task Task time LaborIdle time

Stationidle time

1 1 75 15 15

223

6124

5 5

348

3644

10 10

467

3932

15 15

595

10

292226

13 15

6

1112131415

52751115

15 0

7 15 90 0 0

8 15 90 0 0

9 15 90 0 0

101516

2560

30 5

11 17 60 30 30

6. What other issues might Toshihiro consider when bringing the new assembly line up to speed?

The total costs of the various options should be considered. For example, is it less expensive towork overtime to meet increased demand or to add another workstation requiring another full-time worker? Also, the quality of demand forecasts should be assessed. Redesigning the line toachieve higher output will be expensive, and should not be done unless there is strong confidencein demand forecasts. Finally, the cost of redesigning the line once it is operational and resultantdowntime should be considered. A thorough analysis of demand and line design options shouldbe performed to minimize the risk of having to redesign the line once operational in the nearfuture. Perhaps the best option would be the balance from part 3, allowing maximum productionof about 273 per day with no overtime, and the ability to reach 300 units per day with less thanone hour overtime per day.

Appendix 06A – Break-Even Analysis

6A-1

APPENDIX 6ABREAK-EVEN ANALYSIS

Problems

1.)( VCP

FCBEP

(FC = fixed cost, P = price, VC = variable cost, BEP is break-even point)

a. books000,20)823(

000,300

BEP

b. Higherc. Lower

2. units500,7))2545($90($

000,150$

)(

VCP

FCBEP

3. units900)50.4$50.5($

900$)(

VCP

FCBEP

a. Break-even = 900 units.

b. With a BEP of 900 units, every unit over 900 will bring in $1 profit ($5.50-4.50). Tomake a profit of $500 will require selling 500 units above the BEP = 1400 units.

c. Profit per unit = ((P-VC)*V – FC))/V$.25 = (($5.50 - $4.50)*V - $900)/V.25V = V - 900.75V = 900V = 1,200 units.

Profit per unit = ((P-VC)*V - FC)/V$.50 = (($5.50 -$4.50)*V - $900)/V.50V = V - $900.50V = 900V = 1,800 units.

The unit profit margin is only $1 ($5.50-4.50). It is impossible to average a unit profit of$1.50

4. miles9500)144.36($.

2052$)(

VCP

FCBEP

Appendix 06A – Break-Even Analysis

6A-2

5. FC = (Pc - VCc) * Vc + (Ps –VCs) * Vs

whereFC = fixed costPc = price for chairs, Ps = price for bar stoolsVCc, VCs = variable cost for each productVc, Vs = volume for each product

a. V = Vc = Vs

$20,000 = ($50 - $25) * V + ($50 - $20) * V20,000 = 25V + 30 V20,000 = 55VV = 364 units each of chairs and bar stools.Break-even in dollars = $50(364 + 364) = $36,400 in sales.

b. V = Vc = .25Vs → 4V = Vs

$20,000 = ($50 - $25) * V + ($50 - $20) *4V20,000 = 25V + 120V20,000 = 145VV = 138 units of chairs and (4 * 138) = 552 bar stools.Break-even in dollars = $50(138 + 552) = $34,500 in sales.

Chapter 07 - Service Processes

7-1

CHAPTER 7SERVICE PROCESSES

Review and Discussion Questions1) Evaluating a bank home office manager vs. a bank branch manager.

Bank home office would not have the amount of contact with the outside customer as a bank branch.The bank branch manager would need to be evaluated on how fast people were moved through thedrive through windows and the line inside the branch. There is a high level of face-to face contactand that contact should be fairly standardized as most customers coming to the branch do so todeposit or withdrawal money. The manager of the home office would have more customizationinvolved in dealing with the customer.

2) Identify the high-contact and low-contact operations of the following services:

Service High contact Low contact

Dental Office Cleaning teeth, taking x-rays,filling teeth…etc

Reminder postcards, phone callsto make appointments

Airline Pilot and flight attendantsinteraction with passengers

Online ticket purchase, baggagepick-up

Accounting Office Interaction with customer,custom forms

Customer website, standardizedforms

Automobile Agency Face-to face meeting withcustomer at the dealer

Agency website, online ornewspaper advertisements

Amazon.com Internet suggested purchases forregistered customer

Online order system, delivery oforder

3) Secondlife.com

a) Answers will vary – companies and even educators can use the website site to create customizedlearning opportunities for participants.

4) Student answers will vary widely. Assuming they model a simple process, it may very well be easilyanalyzed with queuing models in the chapter


7-2

Problems

1.

The answers will vary based on personal experience. With patient interaction, maintenance ofmedical records is very low. With an established process, the efficiency will be very high.Similarly, billing is not going to require a high level of patient interaction. The efficiency will bevery high, but since it may depend on outside factors such as insurance claim procedures, it may notbe as high as other processes. Admissions will be a structured process but will depend on moreinteraction with the patient.

When we get to the Lab Tests and Diagnostic Test, these will become a little more unique and willhave a high level of interaction. At the top of the chart, you will see Nurses and Doctors having totreat each patient case by case. Efficiency is going to be much lower.

2. Responses will vary

High Low

Low High

Buffered Core (none) Permeable Systems (Some) Reactive System (Much)

DoctorNurse

Billing

MedicalRecords

Lab Tests

Admissions

DiagnosticTests

Level of Patient Interaction

Efficiency


7-3

3. a. What are the important aspects of the service package?

The service package at McDonalds is the fast and friendly service along with the food beingprepared to your expectations. Depending on your personal situation you may be looking foractivities for kids such as a playground. Cleanliness is another factor that may be considered.

b. Which skills and attitudes are needed by the service personnel?

Being customer focused and willing to listen to the customers are some of the attitudes thatare needed by service personnel. Being able to ensure people are provided the condimentsthat are available is an example of being customer focused, e.g. by offering “Ranch”,“Italian” or “Vinaigrette” dressing when someone orders a salad.

c. How can customer demand be altered?

Customer demand can be altered by offering something else. For instance, if someone ordersa salad and requests French dressing, demand can be altered by suggesting the customer canhave “Ranch”, “Italian” or “Vinaigrette” dressing.

d, e, and f. answers will vary

4. Worker Skill

a. Low CCS – Worker would be able to offer minimal assistance with the bulk of the assistancecoming towards the end of the process. (Example – A worker at a grocery store maybe ableto direct a customer towards the baking isle.)

b. High CCS – Work would be able to offer a high level of knowledge and assistance throughthe process. (Example – A car salesperson may be able to help a person make a purchasebased on the individual needs of a car buyer.)

Capacity Utilization

a. Low CCS – Capacity utilization will be high. The physical area will be smaller but all theoptions will be made available in this area. (Example – A grocery store will have a definedarea in which the groceries are kept.)

b. High CCS – Capacity utilization will be low.

Level of Automation

a. Low CCS – Automation will be high, designed for efficient processing of routine tasksaccording to narrow specifications

b. High CCS – Automation will be low except for tasks that do not vary from customer tocustomer.


7-4

5. Use model 1.

4/hour 6/hour

a. 3333.6

4111

or 33.33%

b.)46(6

4

)(

22

qL =1.33,4

33.1

q

q

LW = 1/3 hour or 20 minutes

c.)46(6

4

)(

22

qL = 1.33 students

d. At least one other student waiting in line is the same as at least two in the system. Thisprobability is 1-(P0+P1).

n

nP

1

0

0 6

4

6

41

P = .3333

1

1 6

4

6

41

P = .2222

Probability of at least one in line is 1-(.3333 + .2222) = .4444

6. a. Use model 1.

4/hour 10/hour

)410(10

4

)(

22

qL = .267 students

Waiting cost = average number in line * goodwill loss per hour * number of hours per day.= .267(10)8 = $21.33 per day

Total cost = waiting cost + additional service cost= $21.33 + $99.50 = $120.83 per day


7-5

b. Use model 3.

M=2,6

4

=.667

Interpolating from Exhibit 7.12, qL = .0837

Waiting cost = number in line * goodwill loss per hour * of hour per day

= .0837(10)8 = $6.70 per day

Total cost = waiting cost + additional service cost

= $6.70 + $75.00 = $81.70 per day

Consequently, it is better to hire another clerk.

7. Use model 1.

10/hour 15/hour

a. 1- =15

101 = .3333 or 33.33%

b. P(both clerks busy) = P(one clerk busy) * P(one clerk busy)

= 2 =2

15

10

= .4444 or 44.44

c. P(both clerks idle) = P(one clerk idle) * P(one clerk idle)= (.3333)(.3333) = .1111 or 11.11%

d.)1015(15

10

)(

22

qL = 1.33 customers

e. 21015

10

sL

2.10

2

s

s

LW or 12 minutes


7-6

8. Use model 3

20/hour 15/hour

a. S =2 , and 3333.115

20

, from Exhibit 7.12, qL = 1.122

1

S

LP qw 1.122((2(15)/20) - 1) = 56%

b. 1.122 customers

c. qs LL =1.122 + 20/15 = 2.455

123.20

455.2

s

s

LW hours or 7.4 minutes

d. Yes, fewer customers in line and shorter time in system.

9. Use model 2.

72 per hour 80 per hour

minutes.

cars05.4)7280)(80(2

)72(

)(2

22

qL

qs LL = 4.05 + (72)/(80) = 4.95 cars

minutes375.3hours05625.)72(

05.4

q

q

LW

minutes4.125hours06875.)72(

95.4

s

s

LW


7-7

10. Use Model 1.


a.100120

100

sL = 5

100

5

s

s

LW = .05 hours or 3 minutes

b. Now, 180 per hour

100180

100

sL = 1.25

100

25.1

s

s

LW = .0125 hours or .75 minutes or 45 seconds

c. Using model 3, 100 per hour 120 per hour

S =2 , and, from Exhibit 7.12, qL = .1756

120

1001756.

qs LL = 1.01

100

01.1

s

s

LW = .0101 hours or .605 minutes or 36.3 seconds

11. Use model 2.


a.)1012)(12(2

10

)(2

22

qL = 2.083 people

b. qs LL = 2.083 + 10/12 = 2.917 people

c. 2083.10

083.2

q

q

LW hours.

d.10

917.2

s

s

LW = .2917 hours

e. It will cause it to increase, at 12 per hour,

)1212)(12(2

12

)(2

22

qL


7-8

12. Use model 1

3 per minute 4 per minute

a.34

3

sL = 3 customers

b.3

3

s

s

LW = 1 minute

c.4

3

= .75 or 75%

d. Probability of 3 or more is equal to 1 – probability of 0, 1, 20

0 4

3

4

31

P = .2500,

1

1 4

3

4

31

P = .1875,

2

2 4

3

4

31

P = .1406

Total of P0 + P1 + P2 = (.2500 + .1875 + .1406) = .5781

Therefore, the probability of three or more is 1 - .5781 = .4219

e. If an automatic vendor is installed, use model 2.(a. revisited)

customers1.125=)34)(4(2

3

)(2

22

qL

qs LL = 1.125 + ¾ =1.875

(b. revisited)

minutes375.3

125.1

q

q

LW

minutes625.3

875.1

s

s

LW

By converting to constant service time, the number in line is reduce from 3 to 1.875 people (areduction of 1.125, and time in system is reduced from 1 minute to .625 minutes (a reduction of.375 minutes or 22.5 seconds).


7-9

13. Use model 1. 20 per hour 30 per hour

a.2030

20

sL = 2 people in the system

b.20

2

s

s


c. Probability of 3 or more is equal to 1 – (probability of 0, 1 or 2)n

nP

1

0

0 30

20

30

201

P = .3333

1

1 30

20

30

201

P = .2222

2

2 30

20

30

201

P = .1481

Total of P0 + P1 + P2 = (.3333 + .2222 + .1481) = .7036


d. %7.66667.30

20

= .67 or 67%

e. Use model 3.

30

20

=.6667

From Exhibit 7.12, qL = .0093

qs LL = .0093 + 20/30 = .676

20

676.

s

s

LW = .0338 hours or 2.03 minutes


7-10

14. Use model 1


a.56

5

sL = 5 people in the system

5

5

s

s

LW = 1 hour

b.)56(6

5

)(

22

qL = 4.17 people on average

c. It would be busy6

5

=.833 or 83.3% of the time, for a 12 hour day .833(12) = 10 hours

d. Utilization is .833 so idle probability is 1 - .833 = .167. Double-check with formula:00

0 6

5

6

511

P = .167 or 16.7%

e. sW .75 minutes, = 5 per hour,

75.35

75. sss

s LLL

W

33.65

575.3

sL , therefore, the service rate must be increased to at least 6.33

customers per hour, which works out to an average service time of just under 9.5 minutes.

15. Use model 1.


a.)23(3

2

)(

22

qL = 1.333 customers waiting

b.2

333.1

q

q


c.2

2,2

23

2

s

ss

LWL = 1 hour

d.3

2

= .67 or 67% of the time


7-11

16. Use model 3.


a.3

2

, from Exhibit 7.12, qL = .0837

b. qq LW = .0837/2 = .0418 hours or 2.51 minutes

c. 2/7504./,7504.3/20837. ssqs LWLL = .3751 hours or 22.51 minutes

17. Use model 1.


a.610

6

sL = 1.5 people

6

5.1

s

s


b.10

6

= .60 or 60%

c. Probability of more than 2 people is equal to 1 – (probability of 0, 1 or 2)

n

nP

1

0

0 10

6

10

61

P = .4000

1

1 10

6

10

61

P = .2400

2

2 10

6

10

61

P = .1440

Total of P0 + P1 + P2 = (.4000 + .2400 + .1440) = .7840

Therefore, the probability of more than two customers is 1 - .7840 = .2160


7-12

d. Use model 3.

10

6

= .60, S = 2, from Exhibit 7.12, qL = .0593

6593.10/60593. qs LL

ss LW = .6593/6 = .1099 hours or 6.6 minutes

18. Use model 1.

60/55 per minute 60/50 per minute

a.55/6050/60

55/60

sL = 10.00 people

)55/60(

00.10

s

s

LW = 9.167 minutes (includes waiting + service time)

b.55/6050/60

55/60

sL = 10.00 people (waiting and being served)

c. Probability of 3 or more is equal to 1 – (probability of 0, 1, or 2)n

nP

1

0

0 50/60

55/60

50/60

55/601

P = .0909

1

1 50/60

55/60

50/60

55/601

P = .0826

2

2 50/60

55/60

50/60

55/601

P = .0751

Total of P0 + P1 + P2 = (.0909 + .0826 + .0751) = .2487


d. 90.91%or.9091=50/60

55/60


7-13

e. Use model 2.

)55/6050/60)(50/60(2

55/60

)(2

22

qL = 4.545

qs LL = 4.545 + (60/55)/(60/50) = 5.455 minutes

ss LW = 5.455/(60/55) = 5.0 minutes

19. Use model 1.


a.30

25

= .833 or 83.3%

b.2530

25

sL = 5.00 documents in the system

c.25

00.5

s

s


d. Probability of 4 or more is equal to 1 – (probability of 0, 1, 2, or 3)

n

nP

1

0

0 30

25

30

251

P = .1667

1

1 30

25

30

251

P = .1389

2

2 30

25

30

251

P = .1157

3

3 30

25

30

251

P = .0965

Total of P0 + P1 + P2 + P3 = (.1667 + .1389 + .1157 + .0965) = .5178

Therefore, the probability of three or more is 1 - .5178 = .4822 or 48.22%

e.

)3030(30

30

)(

22

qL


7-14

20. Use model 1.


a.6

4

= .667 or 66.7%

b.24

4

sL = 2.00 students in the system

c.4

00.2

s

s


d. Probability of 4 or more is equal to 1 – (probability of 0, 1, 2, or 3)

n

nP

1

0

0 6

4

6

41

P = .3333

1

1 6

4

6

41

P = .2222

2

2 6

4

6

41

P = .1481

3

3 6

4

6

41

P = .0988

Total of P0 + P1 + P2 + P3 = (.3333 + .2222 + .1481 + .0988) = .8024


e.

)66(6

6

)(

22

qL


7-15


a.)1012(12

10

)(

22

qL = 4.17 vehicles

b.1012

10

sL = 5,10

5

s

s


c.12

10

= .833 or 83.3%

d. Probability of 3 or more is equal to 1 – (probability of 0, 1, or 2)

n

nP

1

0

0 12

10

12

101

P = .1667

1

1 12

10

12

101

P = .1389

2

2 12

10

12

101

P = .1157

Total of P0 + P1 + P2 + P3 = (.1667 + .1389 + .1157) = .4213



a.12

10

= .8333, S = 2, from Exhibit 7.12, qL = .175 cars

qs LL = .175 + .8333 = 1.008

b.10

008.1

s

s



7-16

When adding a second lane, treat it as two separate models 1, and cut in half.

c.)512(12

5

)(

22

qL = .298

Since there are two lines, the total number in line is 2 times .298, which is .596 cars

d.512

5

sL = .7143

5

7143.

s

s


23. Use model 1.

2 per hour

With one repair person: 2

22

2

sL

With two repair people: 3

223

2

sL cars

With three repair people: 4

124

2

sL car

Number ofrepair personnel

Service rateper hour

()

sn Cost of waitingper hour1

Cost ofservice per

hour2

Totalcost per

hour

1 2 $ $ 20 $

2 3 2 80 40 120

3 4 1 40 60 100

Note: 1 = cost of waiting is number in system times downtime cost of $40 per hour.2 = cost of service is number of repair personnel times wage rate ($20 per hour).Should use three repair persons.


7-17

24. Use model 2.


a. ss LW = 2.92/750 = .003889 hours or .2333 minutes or 14 seconds (Ls computed below)

b.)750900)(900(2

750

)(2

22

qL = 2.083 cars

qs LL = 2.083 + 750/900 = 2.92 cars

25. = 52/hour = 20/hour / = 2.6 Use Model 3

a. How many agents should Global staff?

By inspection we see that Global must use at least 3 agents; otherwise the arrival rate wouldexceed the total service rate.

Consider 3 agents: From Exhibit 7.12, with S = 3, Lq = 4.9322. Wq = Lq/ = .0949 hours = 5.69minutes. The total cost is the cost of waiting plus the cost of service. The hourly cost of waitingis the average wait in minute * the number of customers per hour * $1 = 5.69*52 = $295.88. Thecost of service is the hourly wage * number of agents = $20*3 = $60. The total cost per hourwith 3 agents is $295.88 + 60.00 =$355.88.

Consider 4 agents: From Exhibit 7.12, with S = 4, Lq = 0.6581. Wq = Lq/ = .79 minutes. Thetotal cost of this option is (.79)(52)+(20)(4) = $121.08



Comparing the costs across these options, we see that the optimal number of agents is five.

b. What is the average waiting time in line (with 5 servers)?From a. above we see that this is .186 minutes.

c. What is the average number of customers in the system?

customers76.220/521609. qs LL

d. What is the average number of customers that are in service?

52.)20(5

52nUtilizatio

S

Each agent is busy 52% of the time. With 5 agents the number of customers in service is.52(5) = 2.6


7-18

26. = 12/hour cashier = 15/hour, prep = 20/hour, consider it as 2 serial models 2.

a. What is the average length of the “place order” line? What is the average length of the “pick-up”line?

Place order: customers6.1)1215(15)2(

12

)(2

22

qL

Pick-up: This can be tricky. The key here is the constant service times. Every orderplacement takes exactly 4 minutes after which the customer spends 3 minutes waiting for thefood to be prepared (in service at food prep). Since every order placement takes longer thanevery food prep, there will never be anyone “in line” at food prep. How many will be in thefood prep system? With 12 customers per hour, each spending no time in line and 3 minutesin service at food prep, there will be (12*3)/60 = .6 customers on average in the food prepsystem.

b. What is the average number of customers in the drive through area?

You can use the formulas or reason it out based on information we know. The cashier has autilization of 12/15 = 0.8, so there will be 0.8 on average in service at the cashier. We figuredabove that there will be 1.6 customers on average in line at the cashier, and 0.6 customers inthe food prep service. So the total number in the drive through system is 0.8 + 1.6 + 0.6 = 3.0customers.

c. The manager feels that the cashier has slowed down the process and send him to a strict trainingclass. After the class, the cashier can take an order in exactly 2 minutes. What is the capacity ofthe system now?

It takes the cashier 2 minutes to serve one customer, so the cashier can serve 30customers/hour. It takes the food preparer 3 minutes to serve one customer, so the foodpreparer can serve 20 customers/hour. We see the bottleneck shifting and the new capacity ofthe system is 20 customers per hour.

d. After the training, what is the average length of the “place order” line?

customers1333.0)1230(30)2(

12

)(2

22

qL

27. = 36/hour = 6/hour, = 6.0, S = 7, use Model 3.

a. Utilization = /S = 36/(7*6) = 85.7%

b. From Exhibit 7.12, Lq = 3.6878


7-19

28. = 18/hour = 3/hour, = 6.0, S = 10, use Model 3.

a. How long, on average, does a customer have to wait (before a representative picks up her call)?

From Exhibit 7.12, Lq = 0.1518. Wq = Lq/ = 0.1518/18 = .00843 hours = .5 minutes

b. The Northern Emergency Computer Support is going to merge with the Southern EmergencyComputer Support, which also has a call center with 10 customer service representatives andfaces the exact same customer demand pattern. While the managers want to maintain the samewaiting time criteria --- average waiting time to be less than 1 minute, a consultant points out thatafter the merger, the company can in fact fire 3 representatives and retain only 17 customerservice representatives. Is the consultant correct?

We cannot solve this using Exhibit 7.12 in the text because S = 17. You can solve it usingthe queuing spreadsheet or any “queuing calculator” on the web. Using the latter, athttp://www.wlu.ca/sbe/ttucker/si/QueueCalc.html: Wq = .00844 hours = .507 minutes. Thewait time is virtually unchanged from before, but they will save the cost of three agents. Yesthe consultant is correct.

If they go down to 16 agents the wait time will increase to 1.023 minutes.

29. = 8/hour = 10/hour, = 0.8, use Model 1.

a. What is the utilization rate? = 0.8

b. What is the average waiting time in line?

minutes24hours4.0)810(10

8

)(

q

q

LW

30. Independent Model 1 systems, = 2/hour = 3/hour, = 0.6667

a. What is the utilization rate of a clinician?Utilization = = 0.6667

b. What is the average waiting time in line?

minutes40hours6667.0)13(3

2

)(

q

q

LW

To improve patients’ experience, Wilshire Clinic decides to divide twenty clinicians into five teams,each of which consists of 4 clinicians. A patient’s information is shared within the team, so a cliniciancan serve a patient who comes to any clinician on the same team.

Now independent Model 3 systems, S=4, = 8/hour, = 3/hour, = 2.6667

c. What is the arrival rate for each team?2*4 = 8 patients/hour


7-20

d. What is the utilization rate of a clinician?

Utilization = /S = 8/(4*3) = .6667, unchanged.

e. What is the average waiting time in line?

From Exhibit 7.12, Lq = 0.7721. Wq = Lq/ = 0.7721/8 = .0965 hours = 5.79 minutes

f. What is the average number of patients waiting in line for each team?

From Exhibit 7.12, Lq = 0.7721. (This answer depends on linear interpretation of Exhibit7.12. Using an online calculator to check yields an answer of .7568.)

ANALYTICS EXERCISE1. Draw a diagram of the process using the format in Exhibit 7.3.

2. Consider a base case where a customer arrives every 40 seconds and the Customer Service Championcan handle 120 customers per hour. There are two Food Champions each capable of handling 100orders per hour. How long should it take to be served by the restaurant (from the time a customerenters the kiosk queue until her food is delivered)? Use queuing models to estimate this.

Customer arrives andenters order queue

Greetcustomer

Customerplaces order

Confirm orderand price

Customer enters pickupwindow queue

Food Champprepares order

Service Champprepares drinks

Service Champcollects payment

Service Champserves food

Customer departs


7-21

= 90, order = 120, prep = 100.

For the Customer Service Champion, the average service time is .0083 hours (40 seconds) and theaverage time in line is .0250 hours (from queuing formulas) for a total of .0333 hrs (120 seconds).This is calculated using Model 1. For preparing the food for the two Food Champions, the averageservice time is .01111 hours (40 seconds). Using model 3, and with lambda/mu = .9, Lq from Exhibit7.12 is 0.2285 customers. The time a customer spends in the system for food to be prepared is .01254hours or 45.1 seconds. The total time is 120 + 45.1 = 165.1 seconds or 2.75 minutes.

3. On average, how busy are the Service Champion and the two Food Champions?

The Service Champion is busy = 90/120 = 75% of the time and the Food Champions are busy

= 90/200 = 45% of the time.

4. On average, how many cars do you expect to have in the drive-thru line? (Include those waiting toplace order and waiting for food.)

Lq due to the order taking = λ2/µ(µ - λ)

= 8100/3600 = 2.25 customers

Lq due to the food preparation = .2285 customers

Average total customers waiting in line = 2.4785 customers

Ls due to the order taking = 3 customers

Ls due to the food preparation = 1.1285 customers

Average number of customers in the system = 4.1285

5. If the restaurant runs a sale and the customer arrival rate increases by 20%, how would this changethe total time expected to serve a customer? How would this change the average number of cars inthe drive-thru line?

The current customer arrival rate is 90 per hour, this would go up to 90(1.2) = 108 per hour.

Ls due to order taking = (λ/µ – λ) = 108/(120-108) = 8.1 cars

Ws = Ls/λ = 8.1/108 = .075 hours = 4.5 minutes

Ls due to order preparation = Lq + λ/µ = .445 + 108/100 = 1.525 cars(note λ/µ = 108/100 = 1.08, Lq = .445 from the spreadsheet)

Ws = Ls/λ = 1.525/108 = .01412 hours = .8472 minutes


7-22

Average total number of cars in the drive thru = 9.625 carsAverage total time in the system = 5.3472 minutes

6. Currently, relatively few customers (less than ½ percent) order the Crunchwrap Supreme. Whatwould happen if we ran the sale and demand jumped on the Crunchwrap Supreme and 30% of ourorders were for this item? Take a quantitative approach to answering this question. Assume that thetwo processes remain independent.

Just to make this interesting, let’s assume that the customer arrival rate went to 108 customers perhour.

Assume the order taking rate stays the same at 120/hour.

For the order preparation, assume that 70% of the orders still take (60*60)/100 = 36 seconds onaverage to prepare, but 30% of them now take 72 seconds.

The average time to make an order would be .7(36) + .3(72) = 46.8 seconds

So they could prepare (60*60)/46.8 = 76.92 orders per hour (each Food Champ).

Ls due to order taking = (λ/µ – λ) = 108/(120-108) = 8.1 cars

Ws = Ls/λ = 8.1/108 = .075 hours = 4.5 minutes this is the same as before.

Ls due to order preparation = Lq + λ/µ = 1.3449 + 108/76.92 = 2.749 cars(note λ/µ = 108/76.92 = 1.4, Lq = 1.3449 from the table 7.12)

Ws = Ls/λ = 2.749/108 = .2545 hours = 1.527 minutes

Average total number of cars in the drive thru = 10.849 carsAverage total time in the system = 6.027 minutes Not much of an impact.

7. For the type of analysis done in this case, what are the key assumptions? What would be the impacton our analysis if these assumptions were not true?

A big assumption is that there is no interference between the two processes. This would occur ifone process was causing a delay in the other process for some reason. It does not appear that thiswould be true in this process. We are also assuming the Food Champs are working independent,not as a team. Other major assumptions are the distribution associated with arrival and servicerates, and that these rates are valid during the peak noon period.

8. Could this type of analysis be used for other service type businesses? Give examples to support youranswer.

Yes, many examples can be given.

Chapter 08 - Sales and Operations Planning

8-1

CHAPTER 8SALES AND OPERATIONS PLANNING

Review and Discussion Questions1. What are the major differences between aggregate planning in manufacturing and aggregate

planning in services?

There are two main differences. One is that services typically need to be provided when de-manded – there are not many opportunities for backorders in a service firm. When demandcannot be met, the typical result is lost sales. The second difference compounds that problem:services cannot be inventoried during slow periods to satisfy demand during peak periods.Capacity in excess of demand in any period is almost always wasted capacity, unlike in manu-facturing.

2. What are the basic controllable variables of a production planning problem? What are the fourmajor costs?

Basic controllable variables: production rate, work force levels, and inventories.

Major costs: production costs (fixed and variable), production rate change costs, inventoryholding costs, and backlog costs.

3. Distinguish between pure and mixed strategies in production planning.

Pure strategies use only one variable to absorb demand fluctuations. Mixed strategies com-bine variables from two or more pure strategies.

4. Define yield management. How does it differ from the pure strategies in production planning.

With yield management, we allocate capacity to customers at the right price and time to max-imize revenue. Rather than trying to match demand by adjusting our workforce, inventory orproduction we adjust demand so that it better matches our production capability. The two ap-proaches compliment one another.

5. How does forecast accuracy relate, in general, to the practical application of the aggregateplanning models discussed in the chapter?

A highly accurate forecast encourages the use of deterministic techniques such as linear pro-gramming which in turn permits the development of near optimal plans. Clearly, though, anyreduction in uncertainty enhances the likely accuracy of any production planning method.


8-2

6. In which way does the time horizon chosen for an aggregate plan determine whether it is thebest plan for the firm?

Many factors affect the selection of an appropriate time horizon. Perhaps, the most importantis what the firm intends to plan during that time period. An aggregate plan implies a period ofup to 18 months wherein the firm takes its forecast and plans production using inventory,work force size, overtime and under time, subcontracting, and backlogging orders to achieve areasonable schedule at reasonable costs. A very stable firm in a very stable environment witha very stable demand really doesn’t need to go out very far with its aggregate plan. However,when there is variation, especially when this variation is considerable, then a longer aggregateplan will show the need to find subcontractors, new workforce availability, etc. Planning forthese can start early.

7. Review the opening vignette, how does sales and operations planning help resolve productshortage problems?

Sales and Operations planning helps reduce shortages by getting all the key players (sales, fi-nance, operations and product development) to work together to help balance supply and de-mand. When a firm does a good job of sales and operations planning it is less likely to havedemand and supply so far out of balance that there are product shortages. The opening vi-gnette shows that better communication between the executives may have averted the problemthey are discussing.

8. How would you apply yield management concepts to a barbershop? A soft drink vending ma-chine?

The first step would be to determine when peak and off-peak times existed. For the barber-shop, lower prices could be given during off-peak times. For example, price discounts couldbe given during days of the week, or times of the day when demand is low. Another approachwould be to offer a discount and an appointment to people that walk-in during peak times, thustransferring them to an off-peak time.

Hopefully, lack of capacity would not be a problem for a vending machine, so reallocatingpeak demand should not be an issue. But, trying to increase usage during non-peak times isdifficult because most vending machine can charge only one price. However, new technologycould allow the prices to be changed based on time of day, or even the day of the week.Therefore, during off-peak times, a lower price could be charged to stimulate sales.


8-3

Problems

1. Answers will vary, but Production Plan 1 is very difficult to beat in terms of total cost. Thehire/layoff costs are low relative to the inventory and shortage costs, which naturally favors achase strategy.

2.

Fall Winter Spring SummerForecast 10000 8000 7000 12000Beginning inventory 500 -2300 0 200Production required 9500 10300 7000 11800Production hours required 19000 20600 14000 23600Production hours available1 14400 14400 14400 14400Overtime hours 6200Temp workers2 20Temp worker hours available 9600Total hours available 14400 20600 14400 24000Actual production 7200 10300 7200 12000Ending inventory -2300 0 200 200Workers hired 20Workers laid off 20

Straight time $72,000 $72,000 $72,000 $120,000Overtime 0 49600 0 0Inventory $1,000 $1,000Backorder $23,000Hiring $2,000Layoff $4,000Total $95,000 $121,600 $73,000 $127,000

$416,600130 workers*8*602Temp workers to be hired = (23,600-14400)/(8*60) = 19.17 workers


8-4

3.

February March April MayForecast 80,000 64,000 100,000 40,000Beginning inventory - - - (16,000)Production required 80,000 64,000 100,000 56,000Production hours required 20,000 16,000 25,000 14,000Regular workforce 125 100 100 100Regular production 80,000 64,000 64,000 64,000Overtime hours - - 5,000Overtime production - - 20,000 -Total production 80,000 64,000 84,000 64,000Ending inventory - - - 8,000Ending backorders - - 16,000 -Workers hired 25Workers laid off - 25

Straight time $200,000 $160,000 $160,000 $160,000Overtime - - $ 75,000 -Inventory - - - 80,000Backorder $0 $0 $320,000 $0Hiring 1,250 - - -Layoff - 1,750 - -Total $201,250 $161,750 $555,000 $240,000

$1,158,000


8-5

4.

Spring Summer Fall WinterForecast 20,000 10,000 15,000 18,000Beginning inventory 1,000 - - -Production required 19,000 10,000 15,000 18,000Production hours required 38,000 20,000 30,000 36,000Regular workforce 70 50 75 75Regular production 14,000 10,000 15,000 15,000Overtime hours 10,000 - -Overtime production 5,000 - - -Total production 19,000 10,000 15,000 15,000Ending inventory - - - -Ending backorders - - - 3,000Workers hired - 25Workers laid off - 20

Straight time $280,000 $200,000 $300,000 $300,000Overtime 150,000 - - -Inventory - - - -Backorder $0 $0 $0 $24,000Hiring - - 2,500 -Layoff - 4,000 - -Total $430,000 $204,000 $302,500 $324,000

$1,260,500


8-6

5.

Jan. Feb. March April May June July Aug. Sept. Oct. Nov. Dec. Avg.

Forecast 2500 3000 4000 3500 3500 3000 3000 4000 4000 4000 3000 3000Beginning inventory 500 1250 1500 2000 1750 1750 1500 1500 2000 2000 2000 1500Production requirements 3250 3250 4500 3250 3500 2750 3000 4500 4000 4000 2500 3000 3458.3Ending inventory 1250 1500 2000 1750 1750 1500 1500 2000 2000 2000 1500 1500

Total CostForecast 2500 3000 4000 3500 3500 3000 3000 4000 4000 4000 3000 3000 40500Beginning inventory 500 1360 1720 1080 940 800 1160 1520 880 240 -400 -40Production plan 3360 3360 3360 3360 3360 3360 3360 3360 3360 3360 3360 3360 40320 $403,200Ending inventory 1360 1720 1080 940 800 1160 1520 880 240 -400 -40 320Safety stock 1250 1500 2000 1750 1750 1500 1500 2000 2000 2000 1500 1500Excess inventory 110 220 20 350 $1,750Back order 400 40 440 $8,800

Total $413,750

This plan uses a workforce of 21 workers. Assumptions include no carrying cost for inventoryused to satisfy safety stock, nor any cost for not having enough safety stock to satisfy companypolicy. Costs would vary under different assumptions.

Next, try increasing or decreasing the number of workers by one, and recalculate the total cost. Abetter solution may be found.


8-7

6. There is more than one solution. The following solution assumes no backordered work at theend of the plan.

January February March April May June

Forecast work hours 5,000 4,000 6,000 6,000 5000 4,000

Beginning inventory (work done earlier) 200 1,400 600 (200) -

Work hours required 5,000 3,800 4,600 5,400 5,200 4,000

Regular work hours available 4,000 4,000 4,000 4,000 4,000 4,000

Overtime hours 1,200 1,200 1,200 1,200 1200 -

Total planned hours 5,200 5,200 5,200 5,200 5,200 4,000

Ending inventory (early work completed) 200 1,400 600 -

Ending backorders (work to be done later) - - - 200 - -

Straight time $120,000 $120,000 $120,000 $120,000 $120,000 $120,000

Overtime 54,000 54,000 54,000 54,000 54,000 -

Inventory 1,000 7,000 3,000 - - -

Backorder $0 $0 $0 $2,000 $0 $0

Total $175,000 $181,000 $177,000 $176,000 $174,000 $120,000

$1,003,000

Allowing backordered work at the end of the plan can reduce the cost but will leave work to bedone in the second half of the year. Following allows up to 500 hours backordered work.

January February March April May June

Forecast work hours 5,000 4,000 6,000 6,000 5000 4,000

Beginning inventory (work done earlier) (0) 1,200 400 (400) (500)

Work hours required 5,000 4,000 4,800 5,600 5,400 4,500

Regular work hours available 4,000 4,000 4,000 4,000 4,000 4,000

Overtime hours 1,000 1,200 1,200 1,200 900 -

Total planned hours 5,000 5,200 5,200 5,200 4,900 4,000

Ending inventory (early work completed) - 1,200 400 -

Ending backorders (work to be done later) 0 - - 400 500 500

Straight time $120,000 $120,000 $120,000 $120,000 $120,000 $120,000

Overtime 45,000 54,000 54,000 54,000 40,500 -

Inventory - 6,000 2,000 - - -

Backorder $0 $0 $0 $4,000 $5,000 $5,000

Total $165,000 $180,000 $176,000 $178,000 $165,500 $125,000

$989,500


8-8

7. The decision variables are how many regular and OT hours to assign to production of eachproduct each month. The constraints are the limits of total regular and OT hours each month, andno backorders. The costs are a combination of production and inventory carrying costs.

APRIL MAY JUNE JULYDemand A 800 600 800 1200Demand B 600 700 900 1100Demand C 700 500 700 850

Total Demand 2100 1800 2400 3150Regular hours Available 1500 1300 1800 2000

Overtime Available 700 650 900 1000 CostsRegular Hours A 200 100 200 50 4Regular Hours B 600 700 900 1100 5Regular Hours C 700 500 700 850 6

Total Regular Hours 1500 1300 1800 2000OT Hours A 600 500 750 1000 6OT Hours B 0 0 0 0 7.5OT Hours C 0 0 0 0 9

Total OT Hours 600 500 750 1000Total Hours A 800 600 950 1050Total Hours B 600 700 900 1100Total Hours C 700 500 700 850

Excess Hours A 0 0 150 0 3Excess Hours B 0 0 0 0 4Excess Hours C 0 0 0 0 5

Production Costs 11600 9900 14000 16800Inventory Costs 0 0 450 0

TOTAL COST: 52750

Objective value = $52,750. There may be alternative optimal solutions.


8-9

8.

Number of workers = (6700-200)10/(249*8) = 32.6 or 33 workers

Monthly production (except July) = 22(8)33/10 = 580 units/month

Jan. Feb. March April May June July August Sept. Oct. Nov. Dec. TotalForecast 600 800 900 600 400 300 200 200 300 700 800 900 6700Beginning inventory 200 180 -40 -360 -380 -200 80 64 444 724 604 384Available Production 580 580 580 580 580 580 184 580 580 580 580 580 6564Ending inventory 180 -40 -360 -380 -200 80 64 444 724 604 384 64

Costs TotalLost Sales 0 800 7200 7600 4000 0 0 0 0 0 0 0 19600Inventory 900 0 0 0 0 400 320 2220 3620 3020 1920 320 12720Total 900 800 7200 7600 4000 400 320 2220 3620 3020 1920 320 32320

9. The following solution assumes no backorders, and includes safety stock in inventory cost cal-culations.

January February March

Forecast 1,000 1,500 1,200

Safety stock 500 750 600

Beginning inventory 500 503 751

Net production required 1,000 1,747 1,049

Workers required 57 115 63

Hired 7 58

Laid off 52

Actual production 1,003 1,748 1,058

Ending inventory 503 751 609

Labor $60,192 $104,880 $63,504

Inventory $ 1,509 $ 2,253 $ 1,827

Hiring $ 1,400 $ 11,600 $ -

Layoff $0 $0 $15,600

Total $63,101 $118,733 $80,931

Total: $262,765


8-10

10. The following plan assumes no backorders. The only cost data provided is for inventory car-rying costs. The 24% per year works out to 2% per month based on the $40 cost per unit, or$0.80 per unit per month.

May June July August

Forecast 3200 2,800 3,100 3,000Beginning inventory 403 158 148 3Production required 2,797 2,642 2,952 2,997Regular workforce 12 12 12 12Regular production 2,460 2,460 2,460 2,460Temp workforce 3 2 3 4Temp production 495 330 495 660Total production 2,955 2,790 2,955 3,120Ending inventory 158 148 3 123

Inventory Cost $126.40 $118.40 $2.40 $98.40$345.60

ANALYTICS EXERCISE: Bradford Manufacturing

This exercise can be left as a homework exercise or used as a teaching case. A solution to theproblem is shown in the plan below. Afterwards, teaching notes for use as a case are presented.

Aggregate Plan Quarter (Week Numbers)

1st (1-13) 2nd (14-26) 3rd (27-39) 4th (40-52)

Lines run 10 10 12 11

Overtime hours per day 0 0 0 0

Beginning Inventory 200.0 393.8 387.5 520.0

Production 2,193.8 2,193.8 2,632.5 2,413.1

Expected Demand 2,000.0 2,200.0 2,500.0 2,650.0

Ending Inventory 393.8 387.5 520.0 283.1

Deviation from Inventory Target 55.3 2.9 112.3 -55.3

Employees 60 60 72 66

Cost of Plan

Labor Regular Time $624,000 $624,000 $748,800 $686,400

Labor Overtime $0 $0 $0 $0

Hiring and Training $0 $0 $60,000 $0

Layoff $0 $0 $0 $18,000

Inventory Carry Cost $13,822 $721 $28,077 $0

Stockout Cost $0 $0 $0 $33,202

Quarter Budget $637,822 $624,721 $836,877 $737,602


8-11

Total Cost of Plan $2,837,022

The plan above is from the Excel spreadsheet at the book website, and is just one possible solu-tion. It is based on the following assumptions:

- Inventory carrying costs are based on inventory in excess of safety stock.- Backorder costs are incurred on negative deviation from planned safety stock, even

though total inventory may be positive.- Overtime is planned in hours per day across an entire quarter. A more reasonable ap-

proach might be to plan on overtime-weeks in a quarter (integer constraint, <= 26).

A discussion of the students’ approach to the problem, including any assumptions made would bea worthwhile exercise.

Teaching Note

This is a case that is designed to give the student experience with developing an aggregate plan.A follow up in-class simulation exercise can also be done with the students. The simulation in-volves the operation of the plant over the first 13 to 20 weeks of the year. The simulation allowsstudents to experience the problems associated with implementing an aggregate plan.

Assign the case as a homework assignment. The student should be instructed to develop an ag-gregate plan. Remind them to use the spreadsheet named “Bradford Manufacturing” from theCD. You might want to take 10 minutes in the class prior to the day when you plan to do thesimulation exercise to quickly familiarize students with the spreadsheet.

Remind students to bring a printout of their aggregate plan to class and to bring their notebookcomputer, if they have one.

Start the class by asking about their aggregate plans. Generate a range of costs that students ob-tained on the board.

Next, ask students to describe how they obtained their solution to the problem. Try to character-ize the different approaches. Some likely categories would be “Trial and Error”, “A simple heu-ristic”, and “Excel Solver”.

Following this, the spreadsheet can be brought up and some of the better solutions displayed.You can also run the Solver if you like at this time. You may need to “unprotect” the spreadsheetto run the Solver (Tools > Protection > Unprotect). Finish this section by putting a solution in theAggregate Plan portion of the spreadsheet that seems to be a good one.

Now move to the Simulation Worksheet part of the spreadsheet. Here the plan has been reor-ganized into a weekly master schedule with the data from the Aggregate Plan initially seeding theschedule. The idea is to now work through the weekly schedule by putting in what actually hap-pened in terms of sales and production rates. After seeing the data each week, students should begiven the opportunity to change next week’s schedule. You should do this for at least the first 13weeks. Then you can click on the Actual Costs worksheet and compare the budgeted cost to theactual cost of running the plant.


8-12

To make the simulation interesting use actual demand that demonstrates the old “hockey stick”phenomenon. Sales should be real slow at the beginning of the quarter and then surge at the end.Remember there is a sale at the end of the 1st quarter. Try to be real straight when you go fromweek to week and don’t hint at the fact that demand will take off at the end. This can be a goodlesson for the student.

The following are a set of production rates and demand that work well:

Week ProductionRate

Demand Week ProductionRate

Demand

1 423 140 11 465 1122 455 120 12 450 2003 430 100 13 455 4504 435 125 14 450 1605 435 125 15 430 1656 460 105 16 450 1607 465 115 17 455 1458 470 120 18 470 1509 455 105 19 460 15510 460 110 20 455 160

You can complete the exercise by discussing the following items:

- Why did demand vary the way it did during the first quarter?

- Why is it important for manufacturing and marketing to coordinate plans?

- What types of things can marketing do to make it easier on manufacturing? (Separate thedeals from the deliveries. Everyday low pricing, etc.)

- Do you think that management should change their inventory target?


8-13

Teaching Plan for a class using Bradford Manufacturing

Explain how Aggregate Planning fits into the overall process of Planning and Control – showchart.

What is Aggregate Planning?- Setting workforce levels- Aggregate inventory levels- Production rate- 6-18 month horizon- Product groups – rather than individual products

A strategy for how demand will be met, given current resource constraints.

Why is Aggregate Planning important?- Key interface to the capital budgeting process

10 minutes into the classBradford Manufacturing

- What are the key drivers of this plan?Forecast -> Marketing/market ResearchEnding Inventory Target -> ManagementTechnical Parameters – define current resource constraints and costs.

- Evaluate the costs associated with the current plan.Develop a solver plan. Rationalize the plan. – Integerize

30 minutes into class- Two basic strategies – chase demand or level demand (use inventory)- Put a high hiring and firing cost into the solution and generate a level plan. Use hiring

and training cost of $15,000 and layoff cost of $5,000.

35 Minutes into class- Explain the relationship between the Aggregate Plan and the Master Schedule

Run the simulation (takes about 40 minutes)Conditions

Inventory target – 1 weekHiring/training = $5,000Layoff cost = $3,000

Initial inventory = 200(000) units

Offer prize!

First, each student (or pair of students) needs to finalize an Aggregate Plan, and thenmove to the simulation worksheet. Make sure initial inventory is set correctly. Show actual costworksheet. Run simulation per the previous instructions.

Chapter 09 - Material Requirements Planning

9-1

CHAPTER 9MATERIAL REQUIREMENTS PLANNING

Review and Discussion Questions1. Discuss the meaning of MRP terms such as planned order release and scheduled order receipts.

A planned order release is an order currently planned to be released. It has not been released. Consequently,the planned order release can be changed based upon changes in demand as one example.

A scheduled order receipt, on the other hand, reflects an order that has already been released. The scheduledorder receipt indicates the anticipated arrival of the released order. Due to variations in delivery times, it maynot arrive exactly at the planned arrival time.

2. Many practitioners currently update MRP weekly or biweekly. Would it be more valuable if it were updateddaily? Discuss.

The performance of any operation will naturally vary from day to day. When the observed time period inwhich performance is measured is a week or two, the daily variations are smoothed; that is, the variations inperformance are averaged. For example, below-average performance in one day may be offset by a higher-than-average performance the next day. Daily MRP runs monitor performance too closely and may even createan exception report calling a normal variation an abnormal deviation from expected output.

3. Should safety stock be necessary in an MRP system with dependent demand? If so, why? If not, why do firmscarry it anyway?

In most systems, there are some reasons to carry safety stock, even for items that only have dependent demand.Most reasons stem from uncertainty in the system and its environment. Some of these include:a. Possible short-notice increase in the production order quantity for the parent independent demand item.b. Rush orders for the parent independent demand item.c. Potential quality or yield issues with the MRP ordersd. Lead time variation with the MRP orderse. Scrapped items in the MRP system

4. Contrast the significance of the term lead time in the traditional EOQ context and in an MRP system.

In the traditional context, lead time is fixed—either as a discrete time or as a probability distribution. Such leadtime constancy or variation is outside of the inventory model.

Lead time in an MRP system is assumed to be a variable. While specific lead times are stated for planningpurposes, these times may be speeded up or delayed as conditions warrant. Indeed, it is this ability to detectneeded changes in lead times—either by expediting or de-expediting—that many users cite as one of the mostvaluable features of MRP.

5. Discuss the importance of the master production schedule in an MRP system.

The master production schedule “drives” the system. It states the planned due dates for end items. Materialrequirements planning computer runs, however, involve an iterative process. The master production schedule“proposes” or “hypothesizes” a tentative schedule. After the MRP run with this schedule, the shop schedulerexamines the MRP plan for impractical loads on the productive system—either by stating excessive demandson personnel or equipment, or in excessive idle time. Then the master production schedule is revised and theprogram is run again.


9-2

Because the entire MRP system is geared to satisfying the master production schedule, it is critical that themaster production schedule be correct at the start of the first MRP run. The production scheduler then knowswhat effects any changes he makes on the schedule will have on the original MRP schedule. He can then takeappropriate action as necessary, such as requesting that customers be contacted to try to extend promised datesif they are too close, or to arrange for early delivery or additional storage space if products will be completedprior to the promised delivery date.

6. “MRP just prepares shopping lists. It does not do the shopping or cook the dinner.” Comment.

An MRP system generates schedules to meet material needs. It starts with the master schedule and develops atime phased schedule which specifies what, when, and how many units of each material are required. Whetherthis schedule is adhered to, depends first on the master scheduler who may change the schedule. Then aninventory control personnel may choose to change order quantities or timing. Then the purchasing departmentmay make further modifications to a purchase order, and finally the production scheduler may actually releasethe work to production—(which may be at some time other than that called for in the MRP schedule).

7. What are the sources of demand in an MRP system? Are these dependent or independent, and how are theyused as inputs to the system?

An MRP system has both dependent and independent item demands. The major demands on the system occurthrough the master production schedule (these are usually of independent origin). From here on throughout thesystem, the demands are then dependent on the master production schedule.

Orders for spare parts and repair parts normally do not go through the master production schedule unless theiramounts are large enough to place a significant load on the productive system. These demands (which areusually independent) are fed into the inventory records file by-passing the master production schedule. Oncethere, they are then exploded into the required parts and materials needed during the normal course of the MRPrun. The parts and materials needed to make the spares and repair parts are, therefore, dependent demand.

8. State the types of data that would be carried in the bill of materials file and the inventory record file.

The Bill of Materials file contains information about the product, including a listing of parts numbers, quantitiesneeded per unit or product, and the assembly or process flow stipulating how the unit is structured. Engineeringdesign changes that affect the product structure are placed into the Bill of Materials file. Also, parts or materialchanges that occur through a change of vendors or material composition are also added to update the file.

The Inventory Record file contains a great deal of information about each inventory item. At a minimum, thefile would contain the number of units on hand and on order, the number reserved for prior commitments, thecost of the item, the name and address of the vendor, the lead time needed to obtain a shipment, and anyshipment size restrictions. Additional information may be added as desired, such as that contained in Exhibit14.8.

9. Why is the MRP process referred to as an “explosion?”

Dependent demand for items managed in an MRP system is driven by production orders for the independentdemand end item. As end item orders are entered, the MRP system evaluates the impact on demand for all ofthe dependent demand items. The system calculates the dependent demand by examining the end item productstructure tree one level at a time. For complex items, a small order for a single end item could result ininventory calculations and planned orders for hundreds of component parts and assemblies. One small order“explodes” into a very complex series of orders.


9-3

Problems

1.

a.

b.

X A B C D E F G105 404 150 295 270 -10 180 490

(Sample MRP schedule worksheet)

Period: 1 2 3 4 5 6 7 8 9 10Item: Gross requirements

OH: Scheduled receipts

LT: Projected available balance

SS: Net requirements

Q: Planned order receiptsPlanned order releases

Item: Gross requirements










Item: Gross requirementsOH: Scheduled receiptsLT: Projected available balanceSS: Net requirementsQ: Planned order receipts

Planned order releases





Q: Planned order receipts

X

A(4) C(3)

F(4) G(2)

B(2)

D(3) E(1)


9-4


2.Period: 1 2 3 4 5

Item: J Gross requirements 75 50 70OH: 40 Scheduled receiptsLT: 1 Projected available balance 40 0 0 0 0SS: 0 Net requirements 35 50 70Q: L4L Planned order receipts 35 50 70

Planned order releases 35 50 70

3.

Period: 1 2 3 4 5 6 7 8 9 10Item: X Gross requirements 100OH: 20 Scheduled receipts

LT: 1 Projected available balance 20 20 20 20 20 20 20 20 20SS: 0 Net requirements 80Q: L4L Planned order receipts 80

Planned order releases 80Item: Y Gross requirements 160OH: 40 Scheduled receipts


Planned order releases 120Item: Z Gross requirements 240OH: 30 Scheduled receipts


Planned order releases 210Item: A Gross requirements 420 120OH: 50 Scheduled receiptsLT: 2 Projected available balance 50 50 50 50 50 0 0 0 0SS: 0 Net requirements 370 120Q: L4L Planned order receipts 370 120

Planned order releases 370 120Item: B Gross requirements 240OH: 100 Scheduled receipts


Planned order releases 140Item: C Gross requirements 840OH: 900 Scheduled receipts

LT: 3 Projected available balance 900 900 900 900 900 60 60 60 60SS: 0 Net requirements

Q: L4L Planned order receiptsPlanned order releases


9-5

4.

Z

A(2) B(4)

C(3) D(4)

E(2)

Level

0

1

2

3

Period: 1 2 3 4 5 6 7 8 9 10Item: Z Gross requirements 50OH: 0 Scheduled receipts

LT: 2 Projected available balance 0 0 0 0 0 0 0 0 0 0SS: 0 Net requirements 50Q: L4L Planned order receipts 50

Planned order releases 50Item: A Gross requirements 100OH: 0 Scheduled receipts


Planned order releases 100Item: B Gross requirements 200OH: 0 Scheduled receipts


Planned order releases 200Item: C Gross requirements 300OH: 0 Scheduled receiptsLT: 1 Projected available balance 0 0 0 0 0 0 0 0 0 0SS: 0 Net requirements 300Q: L4L Planned order receipts 300

Planned order releases 300Item: D Gross requirements 400OH: 0 Scheduled receipts


Planned order releases 400Item: E Gross requirements 800OH: 0 Scheduled receipts


Planned order releases 800


9-6

5.

A

B(3)

C

E(2) D

Level

0

1

2

3D(2)

B

E(2)

FF

E(2) D

F 4

Period: 1 2 3 4 5 6 7 8 9 10Item: A Gross requirements 30OH: 0 Scheduled receipts 10LT: 2 Projected available balance 0 10 10 10 10 10 10 0SS: 0 Net requirements 20Q: L4L Planned order receipts 20


LT: 1 Projected available balance 10 10 10 10 10 40 40 40SS: 0 Net requirements 10Q: 50 Planned order receipts 50

Planned order releases 50Item: B Gross requirements 50 60OH: 0 Scheduled receiptsLT: 1 Projected available balance 0 0 0 0 0 0 0 0SS: 0 Net requirements 50 60Q: L4L Planned order receipts 50 60

Planned order releases 50 60Item: D Gross requirements 50 60 40OH: 0 Scheduled receipts

LT: 2 Projected available balance 0 0 0 0 40 0 0 0SS: 0 Net requirements 50 60 0Q: 50 Planned order receipts 50 100

Planned order releases 50 100Item: E Gross requirements 100 220OH: 50 Scheduled receipts 50LT: 1 Projected available balance 100 100 100 0 180 180 180 180SS: 0 Net requirements 0 220Q: 200 Planned order receipts 400

Planned order releases 400Item: F Gross requirements 400OH: 150 Scheduled receipts 50LT: 1 Projected available balance 200 200 200 0 0 0 0 0SS: 0 Net requirements 200Q: L4L Planned order receipts 200



9-7

6. Product structure tree

A

B(2) C(3)

E F(2)

Level

0

1

2

3

D(2)

F(2)

D(2)

D

Low-level coded product structure tree

A

B(2) C(3)

E F(2)

Level

0

1

2

3D(2)

F(2)

D(2) D

Indented bill of materials Single level bill of materials

A A

B(2) B(2)

E C(3)

D(2) D(2)

F(2) B

C(3) E

F(2) F(2)

D C

D(2) D

F(2)

E

D(2)


9-8

Period: 1 2 3 4 5 6 7 8 9 10Item: A Gross requirements 20OH: 0 Scheduled receiptsLT: 1 Projected available balance 0 0 0 0 0 0 0 0SS: 0 Net requirements 20Q: L4L Planned order receipts 20


LT: 2 Projected available balance 0 0 0 0 0 0 0 0SS: 0 Net requirements 40Q: L4L Planned order receipts 40


LT: 1 Projected available balance 15 15 15 15 15 15 0 0SS: 0 Net requirements 45Q: L4L Planned order receipts 45

Planned order releases 45Item: E Gross requirements 40OH: 0 Scheduled receipts 20LT: 2 Projected available balance 0 20 20 20 30 30 30 30SS: 0 Net requirements 20Q: 50 Planned order receipts 50

Planned order releases 50Item: F Gross requirements 80 90OH: 0 Scheduled receipts

LT: 1 Projected available balance 0 0 0 0 100 10 10 10SS: 0 Net requirements 80Q: 180 Planned order receipts 180

Planned order releases 180Item: D Gross requirements 100 45 40OH: 50 Scheduled receipts

LT: 1 Projected available balance 50 50 0 0 0 0 0 0SS: 0 Net requirements 50 45 40Q: L4L Planned order receipts 50 45 40



9-9

7.

A

B

C(4)

D(2) E E

F(3) F(3) F

C

D(2) E

F(3)

Level0

1

2

3

4

Period: 1 2 3 4 5 6 7 8 9 10Item: A Gross requirements 50OH: 20 Scheduled receipts 10LT: 2 Projected available balance 30 30 30 30 30 30 30 30 30 0SS: 0 Net requirements 20Q: L4L Planned order receipts 20


LT: 2 Projected available balance 0 0 0 0 0 0 0 30 30 30SS: 0 Net requirements 20Q: 50 Planned order receipts 50

Planned order releases 50Item: C Gross requirements 200 20OH: 50 Scheduled receipts 100LT: 1 Projected available balance 150 150 150 150 150 50 50 30 30 30SS: 0 Net requirements 50Q: 100 Planned order receipts 100

Planned order releases 100Item: D Gross requirements 200OH: 100 Scheduled receipts 100LT: 3 Projected available balance 100 100 200 200 0 0 0 0 0 0SS: 0 Net requirements 0Q: L4L Planned order receipts


Item: E Gross requirements 100 50OH: 10 Scheduled receiptsLT: 2 Projected available balance 10 10 10 10 0 0 0 0 0 0SS: 0 Net requirements 90 50Q: L4L Planned order receipts 90 50

Planned order releases 90 50Item: F Gross requirements 270 150 50OH: 0 Scheduled receipts

LT: 2 Projected available balance 0 0 30 30 30 30 30 30 30 30SS: 0 Net requirements 270 120 20Q: 50 Planned order receipts 300 150 50



9-10

8.

A

B(2)

C

D(3) F

Level

0

1

2

3D

B(3)

E E(4)

D(3) F

E E 4

Period: 1 2 3 4 5 6 7 8 9 10Item: A Gross requirements 20OH: 5 Scheduled receipts 10LT: 2 Projected available balance 5 5 15 15 15 15 15 15 15 15SS: 0 Net requirements 5Q: 20 Planned order receipts 20



Planned order releases 20Item: B Gross requirements 60 40OH: 10 Scheduled receipts 20LT: 2 Projected available balance 10 10 10 10 10 10 10 10 10 10SS: 0 Net requirements 30 30Q: 40 Planned order receipts 40 40

Planned order releases 40 40Item: D Gross requirements 120 120 20OH: 100 Scheduled receiptsLT: 3 Projected available balance 100 100 100 100 140 20 0 0 0 0SS: 0 Net requirements 20 0Q: 160 Planned order receipts 160

Planned order releases 160Item: F Gross requirements 40 40OH: 0 Scheduled receipts 40LT: 2 Projected available balance 0 0 0 0 0 0 0 0 0 0SS: 0 Net requirements 0 40Q: L4L Planned order receipts 40

Planned order releases 40Item: E Gross requirements 160 80OH: 100 Scheduled receipts 60LT: 2 Projected available balance 100 0 0 0 0 0 0 0 0 0SS: 0 Net requirements 0 80Q: L4L Planned order receipts 80



9-11

9.

Period: 1 2 3 4 5 6 7 8 9 10Item: A Gross requirements 20 20 60 50OH: 20 Scheduled receipts

LT: 1 Projected available balance 0 0 0 0 0 0 0 0SS: 0 Net requirements 0 20 60 50Q: L4L Planned order receipts 20 60 50

Planned order releases 20 60 50Item: B Gross requirements 40 120 100OH: 50 Scheduled receipts 30LT: 1 Projected available balance 40 40 40 40 0 0 0 0SS: 0 Net requirements 80 100Q: L4L Planned order receipts 80 100

Planned order releases 80 100Item: C Gross requirements 60 180 150OH: 60 Scheduled receipts

LT: 2 Projected available balance 0 0 0 0 0 0 0 0SS: 0 Net requirements 0 180 150Q: L4L Planned order receipts 180 150

Planned order releases 180 150Item: D Gross requirements 180 150OH: 25 Scheduled receipts

LT: 1 Projected available balance 25 25 45 45 45 45 45 45SS: 0 Net requirements 155 105Q: 50 Planned order receipts 200 150

Planned order releases 200 150Item: E Gross requirements 180 150OH: 0 Scheduled receiptsLT: 2 Projected available balance 0 0 20 20 70 70 70 70SS: 0 Net requirements 180 130Q: 100 Planned order receipts 200 200

Planned order releases 200 200Item: F Gross requirements 360 80 300 100OH: 0 Scheduled receipts

LT: 2 Projected available balance 0 0 40 60 60 60 60 60SS: 0 Net requirements 360 40 240 40Q: 100 Planned order receipts 400 100 300 100

Planned order releases 400 100 300 100


9-12

10.

Period: 1 2 3 4 5 6 7 8 9 10Item: A Gross requirements 30 30 40OH: 20 Scheduled receiptsLT: 1 Projected available balance 20 0 0 0 0 0 0 0SS: 0 Net requirements 10 30 40Q: L4L Planned order receipts 10 30 40

Planned order releases 10 30 40Item: B Gross requirements 10 30 40OH: 0 Scheduled receipts 10LT: 1 Projected available balance 0 0 0 0 0 0 0 0SS: 0 Net requirements 0 30 40Q: L4L Planned order receipts 30 40

Planned order releases 30 40Item: C Gross requirements 20 60 80OH: 10 Scheduled receipts 50LT: 1 Projected available balance 40 40 40 30 30 30 0 0SS: 0 Net requirements 20 50Q: 50 Planned order receipts 50 50

Planned order releases 50 50Item: D Gross requirements 10 100 30 100 40OH: 20 Scheduled receipts

LT: 2 Projected available balance 10 10 10 80 80 80 40 40SS: 0 Net requirements 90 20 20Q: 100 Planned order receipts 100 100 100

Planned order releases 100 100 100Item: E Gross requirements 150 150OH: 10 Scheduled receipts

LT: 2 Projected available balance 10 10 10 10 10 10 10 10SS: 0 Net requirements 140 140Q: 50 Planned order receipts 150 150

Planned order releases 150 150


9-13

11.

Least Total Cost

Period 1 2 3 4 5 6 7 8 9 10Gross Requirements 30 50 10 20 70 80 20 60 200 50Projected available balance 60 10 0 230 160 80 60 0 50 0Net requirements 20 200Planned order receipts 250 250


Least Unit CostPeriod 1 2 3 4 5 6 7 8 9 10Gross Requirements 30 50 10 20 70 80 20 60 200 50Projected available balance 60 10 0 430 360 280 260 200 0 0Net requirements 20 50Planned order receipts 450 50Planned order releases 450 50

CalculationsWeeks Quantity

orderedCarrying

costOrdercost Total cost Unit cost

4 20 $0.00 $10.00 $10.00 $0.5004 to 5 90 0.70 10.00 10.70 0.1194 to 6 170 2.30 10.00 12.30 0.0724 to 7 190 2.90 10.00 12.90 0.0684 to 8 250 5.30 10.00 15.30 0.0614 to 9 450 15.30 10.00 25.30 0.056

4 to 10 500 18.30 10.00 28.30 0.0579 200 0.00 10.00 10.00

9 to 10 250 2.50 10.00 12.50

For Least Total Cost, order for periods 4 through 8, since carrying cost is the closest to ordering cost. For Least UnitCost, order for periods 4 through 9, since this has the lowest unit cost.


9-14

12. Assume all items ordered L4L.

A

B(2)

D(3)

Level

0

1

2

C(4)

E(2) F(2) E(2)

Period: 1 2 3 4 5 6 7 8 9 10Item: A Gross requirements 100OH: 0 Scheduled receipts




Planned order releases 200Item: C Gross requirements 400OH: 0 Scheduled receiptsLT: 2 Projected available balance 0 0 0 0 0 0 0 0 0 0SS: 0 Net requirements 400Q: L4L Planned order receipts 400

Planned order releases 400Item: D Gross requirements 600OH: 0 Scheduled receipts


Planned order releases 600Item: E Gross requirements 1200OH: 0 Scheduled receipts


Planned order releases 1200Item: F Gross requirements 800OH: 0 Scheduled receipts




9-15

13. a. Product Structure Tree

A

B(2) C(3)

E(4) F(3)

Level

0

1

2

3

D(3)H(2)

D

E(5) G(2)

b. Low-level Coded Product Structure Tree

A

B(2) C(3)

E(4)

F(3)

Level

0

1

2

3

D(3)H(2) D

E(5) G(2)

c.Indented bill of materials Single level bill of materials

A A

B(2) B(2)

E(4) C(3)

F(3) D

C(3) B

D(3) E(4)

H(2) F(3)

E(5) C

G(2) D(3)

D H(2)

H

E(5)

G(2)


9-16

d. Level 0 100 units of ALevel 1 200 units of B

300 units of CLevel 2 600 units of F

600 units of H1000 units of D (3x3x100 + 1x100)

Level 3 3800 units of E (4x2x100 + 5x2x3x100)1200 units of G

14.

Period 1 2 3 4 5 6 7 8 9 10

Gross Requirements 20 10 15 45 10 30 100 20 40 150

Projected available balance 50 40 25 160 150 120 20 0 150 0

Net requirements 20 0 40 0

Planned order receipts 180 190


WeeksQuantityordered

Carryingcost

Ordercost Total cost Unit cost

4 20 $0.00 $9.00 $9.00 $0.450

4 to 5 30 0.20 9.00 9.20 0.307

4 to 6 60 1.40 9.00 10.40 0.173

4 to 7 160 7.40 9.00 16.40 0.103

4 to 8 180 9.00 9.00 18.00 0.100

4 to 9 220 13.00 9.00 22.00 0.100

4 to 10 370 31.00 9.00 40.00 0.108

Least Total Cost method indicates that 180 units should be ordered to cover the needs for periods 4 through 8, sincethe carrying cost is equal to the order cost ($9). Least Unit Cost it tied at $.100 for ordering for periods 4 through 8and 4 through 9. Therefore, order either 180 or 220 units in period 2.


9-17

15. Assume all items are ordered L4L.

Period: 1 2 3 4 5 6 7 8 9 10Standard Model Demand: 300 400

Sports Model Demand: 200 100Radio/CD Gross requirements 300 200 500

OH: 50 Scheduled receiptsLT: 2 Projected available balance 50 50 50 0 0 0 0 0SS: 0 Net requirements 250 200 500Q: L4L Planned order receipts 250 200 500

Planned order releases 250 200 500Standard Trim Gross requirements 300 400

OH: 0 Scheduled receipts

LT: 2 Projected available balance 0 0 0 0 0 0 0 0SS: 0 Net requirements 300 400Q: L4L Planned order receipts 300 400

Planned order releases 300 400Standard HW Gross requirements 300 400OH: 0 Scheduled receipts


Planned order releases 300 400Sport Trim Gross requirements 200 100OH: 0 Scheduled receipts


Planned order releases 200 100Sport HW Gross requirements 200 100

OH: 0 Scheduled receipts




9-18

ANALYTICS EXERCISE: An MRP Explosion – Brunswick Motors

Recently, Phil Harris, the Production Control Manager at Brunswick, read an article on Time-Phased RequirementsPlanning. He was curious about how this technique might work in scheduling Brunswick’s engine assemblyoperations, and decided to prepare an example to illustrate the use to Time-Phased Requirements Planning.

Phil’s first step was to prepare a master schedule for one of the engine types produced by Brunswick - the Model1000 engine. This schedule indicates the number of units of the Model 1000 engine to be assembled each weekduring the past twelve weeks, and is shown below. Next, Phil decided to simplify his requirements planningexample by considering only two of the many components that are needed to complete the assembly of the Model1000 engine. These two components, the Gear Box and the Input Shaft, are shown in the Product Structure Diagramshown below. Phil noted that the Gear Box is assembled by the Sub-Assembly Department, and is subsequentlysent to the main engine assembly line. The Input Shaft is one of several component parts manufactured byBrunswick that are needed to produce a Gear Box sub-assembly. Thus, levels 0, 1, and 2 are included in the ProductStructure Diagram to indicate the three manufacturing stages that are involved in producing an engine: the EngineAssembly Department, the Sub-Assembly Department, and the Machine Shop.

The manufacturing lead times required to produce the Gear Box and Input Shaft components are also indicated inthe Product Structure Diagram. Note that two weeks are required to produce a batch of Gear Boxes, and that all ofthe Gear Boxes must be delivered to the assembly line parts stockroom before Monday morning of the week inwhich they are to be used. Likewise, it takes three weeks to produce a lot of Input Shafts, and all of the shafts thatare needed for the production of Gear Boxes in a given week must be delivered to the Sub-Assembly Departmentstockroom before Monday morning of the week.

In preparing the MRP example Phil planned to use the worksheets shown on the next page and make the followingassumptions:

1. Seventeen gear boxes are on hand at the beginning of week 1, and five gear boxes are currently on order tobe delivered at the start of week 2.

2. Forty input shafts are on hand at the start of week 1, and 22 are scheduled for delivery at the beginning ofweek 2.


9-19

Solution/Teaching Note

This is a simple case that can be used as an in-class exercise or a group assignment. To start the class, quicklyexplain the basics of calculating net requirements. Then, distribute this case, have the students read the case, andgive a brief explanation or what they are expected to do. It is probably best to have students work in pairs or smallteams for this exercise.

1. Initially, assume that Phil wants to minimize his inventory requirements. Assume that each order will be onlyfor what is required for a single period.

Engine Assembly Master Schedule

Week 1 2 3 4 5 6 7 8 9 10 11 12Demand 15 5 7 10 15 20 10 8 2 16

Gear Box RequirementsWeek: 1 2 3 4 5 6 7 8 9 10 11 12

Gross Requirements 15 5 7 10 0 15 20 10 0 8 2 16Scheduled Receipts 5Projected Available Balance 2 2 0 0 0 0 0 0 0 0 0 0Net Requirements 5 10 0 15 20 10 0 8 2 16Planned Order Receipt 5 10 15 20 10 8 2 16Planned Order Release 5 10 15 20 10 8 2 16

Input Shaft RequirementsWeek: 1 2 3 4 5 6 7 8 9 10 11 12

Gross Requirements 10 20 30 40 20 16 4 32Scheduled Receipts 22Projected Available Balance 30 32 32 2 0 0 0 0 0 0 0 0Net Requirements 38 20 0 16 4 32 0 0Planned Order Receipt 38 20 16 4 32Planned Order Release 38 20 16 4 32

2. Phil would like to consider the costs that his accountants are currently using for inventory carrying and setup forthe gearbox and input shafts. These costs are as follows:

Part CostGear Box Setup = $90/order

Inventory Carrying Cost = $2/unit/period

Input Shaft Setup = $45/orderInventory Carrying Cost = $1/unit/period

Gear Box Input ShaftSetup Cost = 8 orders x $90 = $720 Setup Cost = 5 orders x $45 = $225Inventory = (2+2) x 2 = $8 Inventory = (30+32+32+2) x 1 = $96Total = $728 Total = $321

Total Cost = $1,049


9-20

3. Calculate a schedule using least-total-cost lot sizing. What are the savings with this new schedule?

Engine Assembly Master Schedule

Week 1 2 3 4 5 6 7 8 9 10 11 12Demand 15 5 7 10 15 20 10 8 2 16

Gear Box RequirementsWeek: 1 2 3 4 5 6 7 8 9 10 11 12

Gross Requirements 15 5 7 10 0 15 20 10 0 8 2 16Scheduled Receipts 5Projected Available Balance 2 2 10 0 0 30 10 0 0 18 16 0Net Requirements 5 0 0 15 0 0 8 0Planned Order Receipt 15 45 0 26Planned Order Release 15 45 26

Input Shaft RequirementsWeek: 1 2 3 4 5 6 7 8 9 10 11 12

Gross Requirements 30 90 52Scheduled Receipts 22Projected Available Balance 10 32 32 0 0 0 0 0 0 0 0 0Net Requirements 58 52Planned Order Receipt 58 52Planned Order Release 58 52

Gear BoxSetup Cost = 3 orders x $90 = $270Inventory = 88 x $2 = $176Total = $446

Input ShaftSetup Cost = 2 orders x $45 = $90Inventory = 74 x 1 = $74Total = $164

Total Cost = $610

Total cost of this solution is $439 less than the initial solution, a reduction of about 42%.

Note: Inventory carrying costs in parts 2 and 3 are figured based on ending inventory levels (projected availablebalance).

Chapter 10 - Quality Management and Six Sigma

10-1

CHAPTER 10QUALITY MANAGEMENT AND SIX SIGMA

Review and Discussion Questions1. The capability index allows for some drifting of the process mean. Discuss what this means

in terms of product quality output.

When Cpk is larger then 1.33 or 1.5, this means that the mean of the process can drift (up to alimit) while still producing virtually all output within specifications. This allows the process toproduce acceptable output until monitoring systems are able to notice the change in processoutput and prompt management to take corrective action. This is what is implied by the phrase “acapable process."

2. Discuss the purposes and differences between the p-charts and X-bar and R charts.

p-charts are used to monitor the process for attribute data. These are typically binomial “go, no-go” data. An example of a p-chart is percent of pieces nonconforming. X-bar charts are used forcharting population values for continuous measurement. X-bar charts operate effectively withsmaller sample sizes than p-charts, but it is more involved to analyze the sample for an X-barchart since a measurement must be taken. A rule of thumb for the sample size of a p-chart is tohave on average at least two defective items in each sample. This can require a relatively largesample size in some cases. If the process is slow, an X-bar chart will generally be a better choicesince it functions with smaller sample sizes. An example of an X-bar chart is average time tocomplete a mile run for one person. R charts are used to compute process ranges for variabledata, and are generally used in concert with X-bar charts.

3. In an agreement between a supplier and a customer, the supplier must ensure that all parts arewithin tolerance before shipment to the customer. What would be the effect on the cost ofquality to the customer?

Before the agreement was made, the customer probably inspected each part to protect against off-spec supplies. This agreement (ideally) eliminates the need for this inspection. Appraisal costs,such as materials and supplies inspection and reliability testing, will be reduced since theagreement would ensure that the supplies are totally within tolerance. This allows the customerto focus attention on quality improvement within his or her own processes, requiring an increasein prevention cost. Scrap and rework costs will initially drop because of the improvement in thequality of the part supply. Once prevention programs are in force, scrap, repair, rework, anddowntime costs will drop even further because of improvements in the internal process. Externalfailure costs will drop because of improvement in the product and the process. Ultimately wewould expect this agreement to lower the total costs of quality for the customer.


10-2

4. In the situation described in Question 3, what would be the effect on the cost of quality to thesupplier?

If operating under the traditional definition for the quality control function, appraisal costs willincrease since all parts, not just a sample, must be inspected before shipment. However, the fullinspection policy should reduce or eliminate external failure costs, mitigating the increase inappraisal costs.

5. Discuss the trade-off between achieving a zero AQL (acceptable quality level) and a positiveAQL (e.g., an AQL of 2 percent).

The tradeoff involves a cost/precision tradeoff. This is analogous to the service level/costtradeoff. From a classical economic point of view, if the cost of defects is very high, an AQL ofzero is economical. If defect costs are nominal, the cost of achieving near perfect quality can beprohibitive. This assumes that conformance is asymptotic to the cost axis.

Problems

1. Not so good. Each of the 1500 units either passes or fails, so has one opportunity for errors. Sohe has

333,15000,000,11500

23DPMO

This works out to a 1.533% defective rate. Compare this to the ideal six-sigma rate of 2 parts perbillion and you see we are far from six-sigma ideals. In fact this is about a 2.4-sigma process.

2. Defective average = .04, inspection rate = 50 per hour, cost of inspector = $9 per hour, andrepair cost is $10 each.

a.

Calculation Cost per hourNo inspection .04 * (50) * $10 $20Inspection 9

Therefore, it is cheaper to inspect in this case.

b. Cost per unit for inspection = $9/50 = $.18

c. Benefit form the current inspection process is

Hourly: cost of no inspection – cost of inspection ($20 - $9 = $11)

Per unit: average cost of quality – cost of inspection ((.04)$10 - $.18 = $.22)


10-3

3.

a. 333.1,889.min)003(.399.002.1

,)003(.3002.101.1

min3

,3

min

LTLXXUTL

Cpk

= .889

b. The process could be capable of producing the desired quality, but at present it is not capabledue to the process center. The process center should be adjusted.

4.

a. Ten defectives were found in 10 samples of size 15.

)15(10

10p = .067

15

)067.1(067.)1(

n

ppS p = .0645

UCL = p + 1.96 S p = .067 + 1.96(.0645) = .194

LCL = p - 1.96 S p = .067 - 1.96(.0645) = -.060 zero


10-4

b. Stop the process and look for the special cause, it appears to be out of statistical control.

5.

a. Defective average = .02, inspection rate = 20 per hour, cost of inspector = $8 per hour, andreplacement cost is $25 each.

Calculation Cost perhour

No inspection .02(20)$25 $10Inspection 8


b. Cost per unit for inspection = $8/20 = $.40

Benefit from the current inspection process per unit is: average cost of quality – cost ofinspection ((.02)$25 - $.40 = $.10)

6. Defective average = .03, inspection rate = 30 per hour, cost of inspector = $8 per hour, andcorrection cost is $10 each.

Calculation Cost perhour

Cost per unit

No inspection .03(30)$10 $9 $.300Inspection 8 .267


7.

a. Irregularities1 32 53 24 65 56 47 68 39 410 5Total 43

Average Number of Defects 3.4c


10-5

07.2 csp

16.0)07.2(23.4

44.8)07.2(23.4

czcLCL

czcUCL

b.

The process is under control. The UCL is 8.44

8.

Sample 1 2 3 4 mean range1 1010 991 985 986 993.00 252 995 996 1009 994 998.50 153 990 1003 1015 1008 1004.00 254 1015 1020 1009 998 1010.50 225 1013 1019 1005 993 1007.50 266 994 1001 994 1005 998.50 117 989 992 982 1020 995.75 388 1001 986 996 996 994.75 159 1006 989 1005 1007 1001.75 18

10 992 1007 1006 979 996.00 2811 996 1006 997 989 997.00 1712 1019 996 991 1011 1004.25 2813 981 991 989 1003 991.00 2214 999 993 988 984 991.00 1515 1013 1002 1005 992 1003.00 21


10-6

X 999.1, R = 21.733

Control limits for X-bar chart:

UCL, LCL = RAX 2 = 999.1 ± .73(21.733) = 1014.965, 983.235

Control limits for R chart:

UCL = RD4 = 2.28(21.7333) = 49.551

LCL = RD3 = 0(21.7333) = 0.00

The process is in statistical control.

9.

a. AQL = .03, LTPD = .10

LTPD/AQL = .10/.03 = 3.333

From Exhibit 10.16, c = 5.

Also from this Exhibit, n (AQL) = 2.613

nAQL

2 613 2 613

03

. .

.= 87.1, round up to 88

b. Allow up to 5 defective components


10-7

10.

Date

NumberUnsatisfactoryMeals

SampleSize p

1-Dec 74 1000 0.0742-Dec 42 1000 0.0423-Dec 64 1000 0.0644-Dec 80 1000 0.085-Dec 40 1000 0.046-Dec 50 1000 0.057-Dec 65 1000 0.0658-Dec 70 1000 0.079-Dec 40 1000 0.0410-Dec 75 1000 0.075Sum: 600 10000

a.)1000(10

600p = .06

1000

)06.1(06.)1(

n

ppS p = .0075

UCL = p + 2 S p = .06 + 2(.0075) = .075

LCL = p - 2 S p = .06 - 2(.0075) = .045

b. The chart indicates that the process is out of control. The administrator should investigate thequality of the patient meals.


10-8

11.

a. AQL = .15, LTPD = .40

LTPD/AQL = .40/.15 = 2.667

From Exhibit 10.16, c = 8.

Also from this Exhibit, n(AQL) = 4.695

nAQL

4 695 4 695

15

. .

.= 31.3, round up to 32

b. Randomly sample 32 LSI, reject the lot if more than 8 defective

12.

Month CrimesSampleSize

CrimeRate

January 7 1000 0.007February 9 1000 0.009March 7 1000 0.007April 7 1000 0.007May 7 1000 0.007June 9 1000 0.009July 7 1000 0.007August 10 1000 0.01September 8 1000 0.008October 11 1000 0.011November 10 1000 0.01December 8 1000 0.008SUM: 100 12000


10-9

)1000(12

100p = .008333

1000

)008333.1(008333.)1(

n

ppS p = .00287

UCL = p + 1.96 S p = .008333 + 1.96(.00287) = .01396

LCL = p - 1.96 S p = .008333 - 1.96(.00287) = .0027

The process appears in control. Therefore, it can be stated that the crime rate has not increased.However, there appears to be a gradual increasing trend in the crime rate in later data, including thenew Jan – Mar figures given in the problem. That may warrant investigation into the cause, or at theleast should be watched to confirm if there is indeed a trend that continues in future periods.

13.

Area CrimesSampleSize

CrimeRate

1 14 1000 0.0142 3 1000 0.0033 19 1000 0.0194 18 1000 0.0185 14 1000 0.0146 28 1000 0.0287 10 1000 0.018 18 1000 0.0189 12 1000 0.01210 3 1000 0.00311 20 1000 0.0212 15 1000 0.015


10-10

13 12 1000 0.01214 14 1000 0.01415 10 1000 0.0116 30 1000 0.0317 4 1000 0.00418 20 1000 0.0219 6 1000 0.00620 30 1000 0.03SUM: 300 20000

)1000(20

300p = .015

1000

)015.1(015.)1(

n

ppS p = .00384

UCL = p + 1.96 S p = .015 + 1.96(.00384) = .0225

LCL = p - 1.96 S p = .015 - 1.96(.00384) = .0075

Process is out of statistical control. Specifically, three areas outside of UCL warrant furtherinvestigation into the excessive crime rate, and four areas below LCL warrant investigation for causesof the low crime rates.

14.Sample number 1 2 3 4 5 Mean Range

1 .486 .499 .493 .511 .481 .494 .0302 .499 .506 .516 .494 .529 .509 .0353 .496 .500 .515 .488 .521 .504 .0334 .495 .506 .483 .487 .489 .492 .0235 .472 .502 .526 .469 .481 .490 .0576 .473 .495 .507 .493 .506 .495 .0347 .495 .512 .490 .471 .504 .494 .0418 .525 .501 .498 .474 .485 .497 .0519 .497 .501 .517 .506 .516 .507 .02010 .495 .505 .516 .511 .497 .505 .02111 .495 .482 .468 .492 .492 .486 .027


10-11

12 .483 .459 .526 .506 .522 .499 .06713 .521 .512 .493 .525 .510 .512 .03214 .487 .521 .507 .501 .500 .503 .03415 .493 .516 .499 .511 .513 .506 .02316 .473 .506 .479 .480 .523 .492 .05017 .477 .485 .513 .484 .496 .491 .03618 .515 .493 .493 .485 .475 .492 .04019 .511 .536 .486 .497 .491 .504 .05020 .509 .490 .470 .504 .512 .497 .042

X .499, R = .037

Control limits for X-bar chart:

UCL, LCL = RAX 2 , = ..499 + .58(.037) = .520, .478

Control limits for R chart:

UCL = RD4 = 2.11(.037) = .078 LCL = RD3 = 0(.037) = 0.00

Process appears to be in statistical control, though there is a run of five below the center line in the X-barchart.

15.

a.

667,.333.min)002(.3

997.3001.4,

)002(.3

001.4003.4min

3,

3min

LTLXXUTL

C pk

= .333


10-12

b. No, the machine is not capable of producing the part at the desired quality level, even if theprocess were centered on the target.

16.

a. Assuming the process is centered on the target specification, it is not capable:

8333.,8333.min)4(3

90100,

)4(3

100110min

3,

3min

LTLXXUTL

Cpk

= .8333

b. Assuming a shift in the process mean, things get worse:

1667.,5000.1min)4(3

9092,

)4(3

92110min

3,

3min

LTLXXUTL

Cpk

= .1667

c. Many defects will be produced. Assuming a normal distribution, the left tail is z = (92 - 90)/4= .50, which corresponds to a probability of .1915. The right tail is z = (110-92)/4 = 4.5,which approximately corresponds to a probability of .5000. Therefore, .6915 (.1915 + .5000)are inside the specifications, while 1- .6915 = .3085 or approximately 31% are outside thespecification limits.

ANALYTICS EXERCISE: Quality Management – Toyota

Part A: Toyota – Under the Radar Recall Responses

1. Develop a diagram that summarizes what Toyota has done in response to its recent quality recallproblems. Focus on the changes by functional area (i.e. Management, Product Design, Quality,and Manufacturing).


10-13

ImmediateManagement Changes

ProductDesign/Engineering Quality Management Manufacturing

1. 20% pay cut forsenior officials.

2. Lowered thresholdfor effecting recalls.

1. Managing director tooversee safety-relatedissues.

2. Hired 1,000engineers to spot checkquality.

3. Improved perceptionof quality – e.g. reducevibration in steeringwheel.

1. TAQIC – ToyotaAdvanced QualityInformation Center.Global computer database to track vehiclerepairs and trackcustomer complaints.

2. SMART – SwiftMarket AnalysisResponse Teams.Rapid response teamsto determine causes ofaccidents – beyond USand Japan to Chinaand Europe.

1. No major changesother than an increasedawareness for theworkers.

2. Evaluate the statement in the case made by Toru Sakuragi that “… Toyota has been caughtbetween a need to cut costs to overcome the strong yen and the need to improve quality to preventrecalls.” And that “They are now pursuing both strategies but they are essentially at odds withone another.” Is this a realistic strategy? Do you have suggestions for how the strategy might beimproved?

It appears that Toyota has spent a significant amount of money to improve quality. Hiring the1,000 engineers in product design, and appointing a director to oversee safety-related issues aresignificant steps. Clearly, Toyota sees this as very much a product design issue and not related tomanufacturing or build quality. Also, the implementation of TAQIC and SMART are intended tospeed the time that information related to quality problem become apparent. These are fairlyexpensive moves and Toyota is investing in infrastructure to solve their quality problems.

There is nothing in the case to indicate that their problems are related to cost cutting initiatives,but they need to be careful that they don’t do things related to this initiative that impacts thesafety of their cars. Their problems at this point seem to be largely the consumer’s perception ofToyota quality due to the recalls and lawsuits since no defects have actually been found in thecars.

Many things can be done to improve the safety associated with how the cars are used such asconsumer awareness of the floor mat issues and the programming of the throttle control computer(i.e. such as not allowing brakes and the accelerator to be applied heavily at the same time).


10-14

3. Suggest improvements that you feel could be made to Toyota’s quality program. Also, what might Toyotado to improve its image to the consumer relative to quality?

Here are some thoughts, but many other ideas are possible. It was very important that Toyota focus on twoaspects of its operations, the design of the cars, and its response to issues as they arise. First, relative todesign, anything that can be done to make the cars more foolproof should be pursued. For example, thebrake and accelerator pedal system should be designed so that they cannot be hung up by items lying on thefloor. These types of changes should be driven by the information they learn through their TAQIC andSMART initiatives.

A lot of Toyota problems now are image and consumer perception. Toyota will need to aggressivelyadvertise the changes it makes to its vehicles relative to safety improvements. They need to develop astrategy that directly addresses the improvement in their image relative to safety.

Part B: Quality Control Analytics at Toyota

a. If the specification is such that no washer should be greater than 2.4 millimeters, assuming that thethicknesses are distributed normally, what fraction of the output is expected to be greater than thisthickness?

The average thickness in the sample is 1.9625 and the standard deviation is .209624. The probability thatthe thickness is greater than 2.4 is Z = (2.4 – 1.9625)/.209624 = 2.087068 1 - NORMSDIST(2.087068) =.018441 fraction defective, so 1.8441 percent of the washers are expected to have a thickness greater than2.4.

b. What if there is an upper and lower specification, where the upper thickness limit is 2.4 and the lowerthickness limit 1.4, what fraction of the output is expected to be out of tolerance?

The upper limit is given in a. The lower limit is 1.4 so Z = (1.4 – 1.9625)/.209624 = -2.68337.NORMSDIST(-2.68337) = .003644 fraction defective, so .3644 percent of the washers are expected to havea thickness lower than 1.4. The total expected fraction defective would be .018441 + .003644 = .022085 orabout 2.2085 percent of the washers would be expected to be out of tolerance

c. What is the Cpk for the process?

6957.06957.,8944.min6289.4375.

,6289.5625.

min3

,3

min

LTLXXUTL

Cpk

d. What would be the Cpk for the process if it were centered (assume the process standard deviation is thesame)?

The center of the specification limits is 1.9, which is used for X-bar in the following:

795.0795.,795..min6289.

5.,

6289.5.

min3

,3

min

LTLXXUTL

Cpk

e. What percentage of output would be expected to be out of tolerance if the process were centered?

Z = (2.4 – 1.9)/.209624 = 2.385221Fraction defective would be 2 x (1-NORMSDIST(2.385221)) = 2 x .008534 = .017069, about 1.7 percent.


10-15

f. If the process could be improved so that the standard deviation was only about .10 millimeters, what wouldbe the best that could be expected with the processes relative to fraction defective?

The best that could be done is for the process to be centered at 1.9, and given a standard deviation of .10,then Z = (2.4-1.9)/.1 = 5. The fraction defective = 2 x (1 – NORMSDIST(5)) = 5.73303E-07 which wouldonly be about 573 defects per billion washers.

g. Setup X-bar and Range control charts for the current process. Assume the operators will take samples of10 washers at a time.

Observation

Sample 1 2 3 4 5 6 7 8 9 10 X-bar R

1 1.9 2 1.9 1.8 2.2 1.7 2 1.9 1.7 1.8 1.89 0.5

2 1.8 2.2 2.1 2.2 1.9 1.8 2.1 1.6 1.8 1.6 1.91 0.6

3 2.1 2.4 2.2 2.1 2.1 2 1.8 1.7 1.9 1.9 2.02 0.7

4 2.1 2 2.4 1.7 2.2 2 1.6 2 2.1 2.2 2.03 0.8

Mean: 1.9625 0.65

From Exhibit 10.13, with sample size of 10, A2 = .31, D3 = .22 and D4 = 1.78

The upper control limit for the X-bar chart = 1.9625 + .31 x .65 = 2.164The lower control limit for the X-bar chart = 1.9625 - .31 x .65 = 1.761The upper control limit for the Range chart = 1.78 x .65 = 1.157The lower control limit for the Range chart = .22 x .65 = .143

h. Plot the data on your control charts. Does the current process appear to be in control?

With respect to the control limits, the process appears to be in control, though it should be noted that it isdifficult to have confidence in that conclusion based on just four samples. Students should also point out thatthere appears to be a positive trend, both in process mean and variability. Again, it would be difficult to say sowith much confidence based on just four samples.

Chapter 11 - Inventory Management

11-1

CHAPTER 11INVENTORY MANAGEMENT

Review and Discussion Questions1. Distinguish between dependent and independent demand in a McDonald’s, in an integrated manufacturer of

personal copiers, and in a pharmaceutical supply house.

The key to the answer here is to consider what must be forecasted (independent demand), and, given the forecast,what demands are thereby created for items to meet the forecasts (dependent demand).

In a McDonald’s, independent demand is the demand for various items offered for sale—Big Macs, fries, etc.The demand for Egg McMuffins, for example, needs to be forecasted. Given the forecast, then, the demand forthe number of eggs, cheese, Canadian bacon, muffins, and containers can then be computed based on the amountneeded for each Egg McMuffin.

The manufacturer of copiers is integrated, i.e., the parts, components, etc. are produced internally. The demandfor the number of copiers is independent (must be forecasted). Given the forecast, the Bill of Materials isexploded to determine the amounts of raw materials, components, parts, etc. that are needed (more on the BOMin chapter 16).

The pharmaceutical supply company is an extreme case where only end items are carried and nothing isproduced internally. The bill of materials is the end item and, therefore, the independent demand (forecastedfrom customers) is the same as the dependent demand. One might attempt to consider that when the demand foritems occurs together, that this is similar to a bill of materials. However, this is not a bill of materials, but rathera causal relationship making it easier to forecast.

2. Distinguish between in-process inventory, safety stock inventory, and seasonal inventory.

In-process inventory consists of those items of materials components and partially completed units that arecurrently in the production process.

Safety-stock inventory is set so that inventory is maintained to satisfy some maximum level of demand. It couldbe stated that safety stock is that level of inventory between the minimum expected demand and the desired levelof demand satisfaction.

Seasonal inventory is that inventory accumulated to meet some periodic increase in demand.

3. Discuss the nature of the costs that affect inventory size.

There are three main categories of costs: purchase cost, ordering costs, and holding costs. The purchase costmay affect inventory levels if quantity discounts are offered. Suppliers will offer a discount for placing largerorders, which might provide an incentive for carrying the resultant larger inventory levels. Ordering costsdirectly influence the optimum order quantity. As ordering costs increase, the effect is to order less often but inhigher quantities, thus increasing inventory levels. Holding costs have an inverse effect on inventory levels. Asholding costs increase, there is an incentive to reduce order quantities resulting in lower average inventorylevels.

a. How does shrinkage (stolen stock) contribute to the cost of carrying inventory? How can this costbe reduced?

Stock cannot be stolen unless it is on hand, and it is reasonable to assume that shrinkage willincrease as on-hand inventory levels increase. Shrinkage costs can be reduced through increased


11-2

security measures (security-related workforce, electronic tracking tags) and/or reducing theamount of inventory on hand at any one point in time.

b. How does obsolescence contribute to the cost of carrying inventory? How can this cost bereduced?

Obsolescence costs are relevant primarily in hi-tech industries. As inventory ages, marketinfluences and advancements in technology drive the value of that inventory lower. The cost ofmaking the item in inventory has likely decreased, and the “latest and greatest” new innovationlowers the value of the older technology. Obsolescence costs can be reduced by producing insmaller quantities as the product matures. That however will increase total setup costs.

4. Under which conditions would a plant manager elect to use a fixed-order quantity model as opposed to a fixed-time period model? What are the disadvantages of using a fixed-time period ordering system?

Fixed-order quantity models–when holding costs are high (usually expensive items or high deprecation rates), orwhen items are ordered from different sources.

Fixed-time period models—when holding costs are low (i.e., associated with low-cost items, low-cost storage),or when several items are ordered from the same source (saves on order placement and delivery charges).

The main disadvantage of a fixed-time period inventory system is that inventory levels must be higher to offerthe same protection against stockout as a fixed-order quantity system. It also requires a periodic count and closersurveillance than a fixed-order quantity system. A fixed-order quantity system can operate with a perpetualcount (keeping a running log of every time a unit is withdrawn or replaced) or through a simple two-bin or flagarrangement wherein a reorder is placed when the safety stock is reached. This latter method requires very littleattention.

5. What two basic questions must be answered by an inventory-control decision rule?

Any inventory control model or rule must establish (1) when items should be ordered, and (2) how many shouldbe ordered.

6. Discuss the assumptions that are inherent in production setup cost, ordering cost, and carrying cost. How validare they?

Investigation of ordering and production setup cost will likely show that a single, unique cost does not exist foreach product, nor is it linearly related to the number of order (as implied in the equations or inventory models).In the purchasing department, for example, an employee is paid either a salary or an hourly rate for a normalwork week. The cost for that employee is sometimes divided among the number of items or orders for which hehas responsibility, resulting in an averaged or allocated cost for each order he places. However, when weconsider an inventory ordering cost based on the number or orders per year (as is done in most inventorymodels), reducing the number of orders the individual places does not necessarily decrease the net cost to thefirm since his weekly pay remains the same. What happens is really an increase in the ordering cost for each ofthe remaining items within his responsibility.

Nonlinearity of costs also occurs in production setups. Consider the time for making a setup in preparation for aproduction run. Setup time is roughly based on an expected frequency of making this particular product run.However, as the frequency increases, familiarity with the setup allows some shaving of the setup time.Moreover, if the setup is repeated often, an investment in specialized equipment or the construction of jigs maybecome warranted, reducing the setup time even more.

The terms carrying or holding costs for maintaining goods in inventory include a multitude of cost elements. Todetermine the nature and amounts of these costs can be a challenging feat. Fortunately, total inventory costcurves tend to be dish shaped and can, therefore, tolerate some error. The holding costs associated with


11-3

insurance, obsolescence, and personnel who are handling materials are extremely difficult to ascertain on anitem-by-item basis, yet each requires realistic analysis. Warehouse storage costs of an item, for example, may bebased on a ratio of its required square footage and the entire available warehouse space, but this may not be anaccurate representation since it is an allocation of cost rather than true cost. Take the warehouse that is too large,or is used to stock products in an off season or depressed period. Allocation based on a share of total warehousecost will result in a high cost for storage, when, in fact, excess storage space should create pressure for higher—not lower—order quantities.

In the simple inventory model, holding costs are based on the average inventory on hand. “Average” inventorypresumes that, as stock is depleted, other product lines will be moved in to occupy the space. It may be thatcosts should be based on maximum inventory, especially if these is an excess of space, or if the needs of an itemare so specialized that no other products can use the space (for example, due to environmental requirements).Each remaining cost may be similarly challenged. Breakage, pilferage, deterioration, and insurance costs are notconstant but, rather, vary with inventory size. As the value of inventory increases, insurance rates are lower,more refined handling procedures can be installed to reduce breakage, some environmental control andmaintenance can be used to reduce deterioration, and better security procedures can reduce theft.

These challenges to determining true costs are not intended to discourage the use of inventory models. Theintent, rather, is to prevent the use of any model without clear knowledge of its requirements and assumptions.Indeed, each application must consider the operating conditions and needs of the firm. An appropriate modelcan then be developed in a fashion similar to those covered in this chapter.

7. “The nice thing about inventory models is that you can pull one off the shelf and apply it so long as your costestimates are accurate.” Comment.

Unfortunately, there is no model or set of models universally applicable to all inventory situations. As stated inthe chapter several times, each situation is different and requires a model to suit those conditions. Studentsfrequently try to memorize specific models rather than the process of building any inventory model. See also theanswers to question 8 below.


11-4

8. Which type of inventory system would you use in the following situations?

a. Supplying your kitchen with fresh food.

b. Obtaining a daily newspaper.

c. Buying gas for your car.

To which of these items do you impute the highest stockout cost?

(a) Supplying kitchen with food—both a periodic model and order quantity. Generally, a household will shoponce weekly for the majority of items (periodic), then pick up items such as bread and milk as the supplyruns low (fixed quantity with reorder point).

(b) Obtaining a daily newspaper—a daily newspaper is obviously a periodic model. One does not usually waituntil he has finished one daily paper before buying the next day’s paper.

(c) Buying gas for your car—generally, this is a hybrid type model wherein a reorder point is signaled whenthe gas indicator is low, then the tank is filled. Many people, however, have a fixed quantity purchasewhen the reorder point is reached, such as “put in 10 gallons or $10.00 worth.” Still others (drawing uponour own experience) use a periodic ordering system on their wife’s car, such as taking it out and filling itevery Sunday after church (or in Chase’s case, after the football game).

The highest stockout cost for most well-fed, well-read individuals would be running out of gas in your car. Thecost could range from practically zero if one runs out in front of a gas station—to being late for an appointmentor causing an accident on the highway.

9. What is the purpose of classifying items into groups, as the ABC classification does?

Using a classification scheme such as this one allows a greater portion of time to be spent in controlling specificgroups or classes or items. For the ABC grouping, greater control is afforded those items which comprise thegreatest dollar volume in usage. The result of this classification is a reduction in the overall inventory size and,therefore, decreased costs for the same level of satisfying inventory demands.


11-5

Problems

1. Cu = $10 - $4 = $6Co = $4 - $1.50 = $2.50

7059.650.2

6

uo

u

CC

CP , NORMSINV(.7059)=0.541446

Should purchase 250 + .541446 (34) = 268.4 or 268 boxes of lettuce.

2. Cu = $125Co = $250

333.125250

125

uo

u

CC

CP , NORMSINV(.333)=-0.43164

Should purchase 25 + (-.43164)(15) = 18.5254. Super Discount should overbook 19 passengers on the flight.

3.

Cu = 25 – 15 = 10, Co = 15-10 = 5, Cu/(Co+Cu) = 10/(5+10) = .6666

Increase Q as long a Pr[demand < Q] < .6666

Pr[demand < 1]=0, Q: Is this (strictly) less than 0.66..? A: Yes.

Pr[demand < 2]=0.2, Q: Is this (strictly) less than 0.66..? A: Yes.




Pr[demand < 6]=0.9, Q: Is this (strictly) less than 0.66..? A: No.

Therefore the optimal order quantity = 5.

4.

Cu = 2.15 – .2 = 1.95, Co = .2, Cu/(Co+Cu) = 1.95/(.2+1.95) = .90697

From the standard normal table, Z-value is 1.325.

Combined demand has mean 2000*4=8000, and standard deviation = sqrt(4*5002) = 1000.

Using the above, the optimal production quantity is 8000+1.325*1000 = 9325.


11-6

5. Cu = 17.99 – 6.75 = 11.24, Co = 6.75 – 0.99 = 5.76

a. If the company decided to bake 15 apple pies each day, what would be her expected

profit?

If demand is 5 (probability 0.1), we will sell 5 apple pies and have 10 leftover.

If demand is 10 (probability 0.2), we will sell 10 apple pies and have 5 leftover.

If demand is at least 15 (probability 0.7), we will sell all the 15 apple pies.

The expected profit is

0.1(5*11.24 – 10*5.76) + 0.2(10*11.24 – 5*5.76) + 0.7(15*11.24) = $134.60

b. Based on the demand distribution above, how many apple pies should the company

bake each day to maximize her expected profit?

Cu = 17.99 – 6.75 = 11.24, Co = 6.75 – 0.99 = 5.76

Increase Q so long as P(demand < Q) < 11.24/(11.24+5.76) = .663

The optimal quantity of apple pies is 20.

6.

a. What is the key probability (service rate)?

Co = 8 – 4 = $4

Cu = 20 – 8 = $12

The key probability (service level) is: 12/ (12+4) = 0.75

b. How many T-shirts should she produce for the upcoming event?

Demand Probability CumulativeProbability

300 0.05 0.05400 0.10 0.15500 0.40 0.55600 0.30 0.85700 0.10 0.95800 0.05 1.00

She should produce 600 T-shirts.


11-7

7. Co = $0.25, Cu = $0.75

a) How many copies of San Pedro Times should you buy each morning?

Service level = ($0.75)/($0.75 + 0.25) = 0.75. Z-value for 75% is 0.67.Q = 250 + 0.67 * 50 = 283.5

b) Based on part (a), what is the probability that you will run out of stock?

25%

8.

100

25)1000(22

H

DSQopt

= 22.36 22

9. Service level P = .95, D = 5000, d = 5000/365, T = 14 days, L = 10 days, = 5 per day, and I =150.

IzLTdq LT )(

22 )5)(1014()( LTLT = 24.495

From Standard normal distribution, z = 1.64

150)495.24(64.1)1014(365

5000q = 218.94 219


11-8

10. Service level P = .98, d = 150, T = 4 weeks, L = 3 weeks, = 30 per week, and I =500 pounds.

IzLTdq LT )(

22 )30)(34()( LTLT = 79.4


q 150(4+3) + 2.05(79.4) – 500 = 712.77 713 pounds

11.

a.)10(33.

250)25750(22

H

DSQopt = 1975.23 1975


LzLdR = 515(1) + (1.64)25 = 556

b. Holding cost = 10)33(.2

1975

2H

Q= $3,258.75

Ordering cost = )250(1975

25750S

Q

D= $3,259.49

c. Holding cost = 10)33(.2

2000

2H

Q= $3,300.00

Ordering cost = )250(2000

25750S

Q

D= $3,218.75

Total annual cost with discount is $6,518.75 – 50(25750/2000) = $5,875.00, without discount it is$6,518.24. Therefore, the savings would be $643.24 for the year.


11-9

12.

IzLTdq LT )(

22 )1)(230()( LTLT = 5.657

98% S.L. From Standard normal distribution, z = 2.05

35)657.5(05.2)230(5 q = 136.60 137chips

The most he would ever order would be when on-hand was zero.

)657.5(05.2)230(5 q = 171.60 172 chips

13. a.)10(20.

150)10000(22

H

DSQopt = 1224.74 1225 units

ssLdR = (10000/52)(4) + 55 = 824.23 824 units

b. IssLTdq )( = (5000/52)(3+1) + 5 – I = 390 – I

14.

a.2

10)1000(22

H

DSQopt = 100 units


11-10

15.

10.

20.31)15600(22

H

DSQopt = 3120 units

22 )90(4 LL = 180 units


LzLdR = 300(4) + (2.05)180 = 1200.0 + 369 = 1569

If safety stock is reduced by 50 percent, then ss = 185 units.

Lzss ,180

185

L

ssz

= 1.03, so the service probability is 84.8%

16. Service level = .98, d = 100 per day, T = 10 days, L = 6 days, = 25 per day, and I =50.

IzLTdq LT )(

22 )25)(610()( LTLT = 100


50)100(05.2)610(100 q = 1755 units

17.

a.5

10)2000(22

H

DSQopt = 89.44 89

b. Ordering cost = )10(89

2000S

Q

D= $224.72

c. Holding cost = )5(2

89

2H

Q= $222.50


11-11

18.65.

100)13000(22

H

DSQopt = 2,000 units

22 )40(4 LL = 80 units


LzLdR = 250(4) + (2.05)80 = 1000 + 164 = 1164

If safety stock is reduced by 100 units, then ss = 64 units.

Lzss ,80

64

L

ssz

= .80

From Standard normal distribution, z = .80, service probability is 79%

19. a.Item number Annual usage Class

18 61000 A4 50000 A

13 42000 A10 15000 B11 13000 B2 12000 B8 11000 B

16 10200 B14 9900 B5 9600 C

17 4000 C19 3500 C20 2900 C3 2200 C7 2000 C1 1500 C

15 1200 C9 800 C6 750 C

12 600 C

b. If item 15 is critical to operations, it may be desirable to reclassify it from C to A to ensure morefrequent reviews.


11-12

20.

a.)3(20.

10)5000(22

H

DSQopt = 408.25 408 bottles

b. 22 )30(3 LL = 52 units


LzLdR = 100(3) + (1.64)52 = 300.00 + 85.28 = 385.28 385 bottles

21.

a.4

5)2400(22

H

DSQopt = 77.46 77 sets

b. 22 )4(7 LL = 10.583 sets


LzLdR = (2400/365)(7) + (2.05)10.583 = 46.03 + 21.70 =67.73 68 sets

Order 77 sets when the on-hand inventory level reaches 68 sets.

22. Service level = .98, d = 60 units per day, T = 10 days, L = 2 days, = 10 units per day, and I =100units.

IzLTdq LT )(

22 )10)(210()( LTLT = 34.64


100)64.34(05.2)210(60 q = 691 units


11-13

23. Service level = .99, d = 2000 capsules per day, T = 14 days, L = 5 days, = 800 capsules per day, and I=25000 units.

IzLTdq LT )(

22 )800)(514()( LTLT = 3487 capsules


25000)12.3487(3263.2)514(2000 q = 21,112 capsules

Original problem 24 removed…

24. a.Demand(dozen)

Probabilityof demand

Probability ofselling nth unit

Expected numbersold Sold (rev.)

Unsold(rev.)

Totalrevenue Cost Profit

1800 0.05 1.00 1800 $1242.00 $0.00 $1242 $882 $360

2000 0.10 0.95 1990 1373.10 2.90 1376 980 396

2200 0.20 0.85 2160 1490.40 11.60 1502 1078 424

2400 0.30 0.65 2290 1580.10 31.90 1612 1176 436

2600 0.20 0.35 2360 1628.40 69.60 1698 1274 424

2800 0.10 0.15 2390 1649.10 118.90 1768 1372 396

3000 0.05 0.05 2400 1656.00 174.00 1830 1470 360

b. The optimal number to make would be 2,400 dozen. This yields an expected profit of $436.

c. Cu = $0.69 - $0.49 = $0.20Co = $0.49 - $0.29 = $0.20

P =20.20.

20.

uo

u

CC

C= .50

Demand (dozen) Probability of demand Cumulative Probability (P)1800 0.05 .052000 0.10 0.152200 0.20 0.352400 0.30 0.652600 0.20 0.852800 0.10 0.953000 0.05 1.00

Produce 2,400 dozen cookies.


11-14

25.

)30(25.

50)3500(22

H

DSQopt = 216.02 216 mufflers

22 )6(2 LL = 8.49 mufflers


LzLdR = (3500/300)(2) + (1.28)8.49 = 23.33 + 10.87 = 34.20 34 sets

Order 216 sets when the on-hand inventory level reaches 34 sets.

26.a. The obvious choice is ABC analysis.

b.

Item number Annual usage Class

q 90000 Ak 80000 Af 68000 At 32000 Bn 30000 Be 24000 Bg 17000 Bc 14000 Br 12000 Ba 7000 B or Cs 3000 Cj 2300 Cd 2000 Co 1900 Ci 1700 C

m 1100 Cb 1000 Ch 900 Cp 800 Cl 400 C

27. Service level = .98, d = 5000/365 boxes per day, T = 14 days, L = 3 days, = 10 boxes per day, and I =60 boxes.

IzLTdq LT )(

22 )10)(314()( LTLT = 41.23 boxes


11-15


60)23.41(05.2)314)(365/5000( q = 257.40 257 boxes

28.

)500(20.

100)500(22

H

DSQopt = 31.62 32 refrigerators

L 10 refrigerators


LzLdR = (500/365)(7) + (1.88)10 = 9.59 + 18.8 = 28.39 28 refrigerators

Order 32 refrigerators when the on-hand inventory level reaches 28 refrigerators.

29.

)35(20.

20)1000(22

H

DSQopt = 75.59 76 tires

22 )3(4 LL = 6 tires


LzLdR = (1000/365)(4) + (2.05)6 = 10.96 + 12.3 = 23 tires

Order 76 tires when the on-hand inventory level reaches 23 tires.

30. Service level = .99, d = 600 hamburgers per day, T = 1 day, L = 1 day, = 100 hamburgers per day,and I = 800 hamburgers.

IzLTdq LT )(

22 )100)(11()( LTLT = 141.42 hamburgers


800)42.141(326.2)11(600 q = 728.94 729 hamburgers


11-16

31.Item number Average monthly

demandPrice per unit Monthly usage Class

5 4000 21 84000 A3 2000 12 24000 A or B4 1100 20 22000 B7 3000 2 6000 B9 500 10 5000 B1 700 6 4200 B or C8 2500 1 2500 C

10 1000 2 2000 C6 100 10 1000 C2 200 4 800 C

32.

a.50.

10)365)(20(22

H

DSQopt = 540.37 540 cans

LdR = 20(14) = 280 cans

b. 22 )15.6(14 LL = 23.01 cans

99.5% S.L. From Standard normal distribution, z = 2.57

LzLdR = 20(14) + (2.57) 23.01 = 280.00 + 59.14 = 339 cans

Order 540 cans when the on-hand inventory level reaches 339 cans.

33. Service level = .98, d = 20 gallons per week, T = 1 week, L = 1 week, = 5 gallon perweek, and I = 25 gallons.

IzLTdq LT )(

22 )5)(11()( LTLT = 7.07 gallons


25)07.7(05.2)11(20 q = 29.49 29 gallons


11-17

34.Quantity range Cost (C) EOQ Feasible

Less than 100 pounds $20 per pound 219 pounds No100 to 999 pounds $19 per pound 225 pounds Yes

1,000 or more pounds $18 per pound 231 pounds No

Note: EOQ =iC

DS2

Therefore, calculate total cost at Q=225, C=$19; and at Q=1000, C=$18

19)25(.2

22540

225

3000)19(3000

219,225 iCQ

SQ

DDCTC CQ = $58,068

18)25(.2

100040

1000

3000)18(3000

218,1000 iCQ

SQ

DDCTC CQ = $56,370

The best order size is 1,000 units at a cost of $18 per pound.

35.Quantity range Cost (C) EOQ Feasible

Less than 2500 pounds $0.82 per pound 4277 pounds No2500 to 4999 pounds $0.81 per pound 4303 pounds Yes5,000 or more pounds $0.80 per pound 4330 pounds No

Note: EOQ =iC

DS2

Therefore, calculate total cost at Q=4303, C=$0.81, and at Q=5000, C=$0.80

14.41197$

)81.0)(20(.2

430330

4303

50000)81.0(50000

281.0,4303

iCQ

SQ

DDCTC CQ


11-18

00.40700$

)80.0)(20(.2

5000)30(

5000

50000)80.0(50000

280.0,5000

iCQ

SQ

DDCTC CQ

The best order size is 5,000 units at a cost of $0.80 per pound.

36.

a. What is optimal order quantity of cotton?

Total demand (D) = 3*12 months * 3000 units * 0.5 pound = 54000 pounds

Total purchasing cost per unit (C) = $2.5 + $0.20 = 2.70

Set up cost (S) = $100

Holding cost per unit = 2.70 * 20% = $0.54

pounds447254.0

)100)(000,54)(2(Q

b. How frequently should the company order cotton?

Number of orders per year = Total demand / EOQ = 54000 / 4472 = 12.08

That is the SYM orders 12.08 times a year.

The company orders every 12 months / 12.08 = 0.99 months or about once a month.

c. Assuming that the first order is needed on April 1st, when should SYM place the order?

The delivery lead-time is 2 weeks. Thus, the company needs to order the cotton two weeks in advance,which is about March 15th.

d. How many orders will SYM place during the next year?

See answer to question b.

e. What is the resulting annual holding cost?

Annual holding cost = H*Q/2 = 0.54 * 4472 / 2 = $1207 per year.

f. What is the resulting annual ordering cost?

Annual ordering cost = Number of orders * ordering cost = 12.08 * 100 = $1208

g. If the annual interest cost is only 5% how will it affect the annual number of orders, the optimal batchsize and the average inventory? (You are not expected to provide a numerical answer to this question. Justdescribe the direction of the change, explain your answer)

If the holding cost is lower the batch size is larger, thus, the average inventory is larger. The numberof orders would be smaller.


11-19

37.

a. What is the cost per shipment and annual holding cost per book?

It is important to notice that the set up cost is $50 (This is the cost that is independent ofthe quantity shipped). The variable cost is $2 + $10 per unit. The annual holding cost is12*20% = $2.40 / book.

b. What is the optimal shipment size?

The annual demand is 250*50 = 12,500 (assuming 50 weeks per year).

pounds7224.2

)50)(500,12)(2(Q

c. What is the average throughput time?

To calculate the throughput time, we first calculate the number of orders:

months694.03.17

12Orders BetweenTime

3.17722

500,12OrdersofNumber

The average time a unit stays in the system is half of that, i.e., 0.347 months.

38. D = 10, T = 15, L = 2, 90% S.L. z = 1.28, 74.246)215( 2 LT

a. 67.201)74.24(28.1)17(10)( LTZLTd

b. If the service probability requirement is 95%, the optimal target level (your answer in part a) will(select one): INCREASE

39.

a. To manage inventory, the company is using: Continuous Review System

b. pounds200,15.0

)10)(000,36)(2(Q

c. 90% S.L. z = 1.28, 54.33)15(5 22 LL

543)54.33)(28.1()5)(100( LzLdR


11-20

ANALYTICS EXERCISE: Inventory Management at Big10Sweaters.com

1. You are curious as to how much Rhonda and Steve made in their business last year. You do nothave all the data, but you know that most of their expenses relate to buying the sweaters and havingthem monogrammed. You know they paid themselves $50,000 each and you know the rent, utilities,insurance, and a benefit package for the business was about $20,000. About how much do you thinkthey made “before taxes” last year? If they must make their payment to the venture capital firm, andthen pay 50% in taxes, what was their increase in cash last year?

Last Year’s Pre-Tax Profit

Unit UnitSales Sale Price Cost Revenue Cost Margin

Ohio 2,300 $120 73.88 $276,000 $169,924 $106,076Michigan 1,468 $120 73.88 $176,160 $108,456 $67,704Purdue 890 $120 73.88 $106,800 $65,753 $41,047

eBay 342 $50 60.88 $17,100 $20,821 ($3,721)Totals 5,000 $576,060 $364,954 $211,106

Overhead: $120,000Net Profit: $91,106

If they pay 25% to the venture capital firm, this is $22,776.50, their profit before taxes is $68,329.50.They then pay $34,164.75 in taxes leaving them with an increase in cash of about $34,165. Majorpoint here is to show how relevant these decisions are to the success of the firm.

2. What was your reasoning behind using the aggregate demand forecast when determining the sizeof your order rather than the individual school forecasts? Should you rethink this or is there a soundbasis for doing it this way?

Here we argue that the aggregate forecast should be more accurate than the individual person’sforecast. You can easily calculate the coefficient of variation (CV) in the individual forecasts andcompare that to the aggregate forecast to prove this (the CV for the individual forecasts are between10% and 14%, and the CV for the aggregate forecast is less than 6%). A big assumption here is thatthe forecasts at each school are independent and that they are not biased. If this is true the errors willtend to cancel each other out. If there is major bias in the forecasts (for example, they are all high orlow), then we have a problem and it might be better to use the individual forecasts. In our analysishere, we assume the forecasts are independent and not biased but we also calculate the orders byindividual school.

3. How many sweaters should you order next year? Break down your order by individual school.Document your calculations in your spreadsheet. Calculate this based on the aggregate forecast andalso the forecast by individual school.

Here the single period model is applicable. The cost of underestimating demand is the lost profit. Inthis case a sweaters would be sold for $120 and it would cost $73.88 (supplier plus subcontractorcost), so the return is $46.12 per sweater. The cost of overestimating demand is the differencebetween the supplier cost of $60.88 and the eBay price of $50 which is $10.88.

The critical probability then is Cu/(Co+Cu) = 46.12/(10.88+46.12) = .809123


11-21

Now, using the aggregate demand forecast, which has a mean of 7,400 and standard deviation of 430units, you should order NORMINV(.809123,7400,420) = 7,776 units. Based on the forecast dataOhio State gets 33.78% or 2,627 units, Michigan get 23.87% or 1,856 units, Purdue get 13.51% or1,051 units, Michigan State gets 21.85% or 1,699 units, and Indiana gets 6.98% or 543 units.

If we base this on the individual forecasts we would order the following:Ohio State = NORMINV(.809123,2500,300) = 2,762Michigan = NORMINV(.809123,1767,252) = 1,987Purdue = NORMINV(.809123,1000,100) = 1,087Michigan State = NORMINV(.809123,1617,126) = 1,727Indiana = NORMINV(.809123,517,76) = 583

Total order size would be 8,146. There is a difference of 8,146 - 7,767 = 379 sweaters.

4. What do you think they could make this year? They are paying you $40,000 and you expect yourbenefit package addition would be about $1,000 per year. Assume that they order based on theaggregate forecast.

We base this on the expected average sales from our forecast and an aggregate order size of 7,767sweaters. Assuming sales are as forecast, the safety stock would be sold on eBay. We also needto adjust overhead to account for the $41,000 increase due to the new employee.

This Year’s Expected Pre-Tax Profit

Unit UnitSales Sale Price Cost Revenue Cost Margin

Ohio 2,500 $120 73.88 $300,000 $184,700 $115,300Michigan 1,767 $120 73.88 $212,040 $130,546 $81,494Purdue 1000 $120 73.88 $120,000 $73,880 $46,120

Michigan State 1617 $120 73.88 $194,040 $119,464 $74,576Indiana 517 $120 73.88 $62,040 $38,196 $23,844

eBay 366 $50 60.88 $18,300 $22,282 ($3,982)Totals 7,767 $906,420 $569,068 $337,352

Overhead: $161,000Net Profit: $176,352

Using the same logic as before, the venture capital people get $44,088 leaving us with $132,264to pay taxes on. Taxes would be $66,132. This creates an increase in our cash of about $66,132.


11-22

5. How should the business be developed in the future? Be specific and consider changes related toyour supplier, the monogramming subcontractor, target customers, and products.

This is open ended, so you may get many different ideas. Here you can begin the discussion bytalking about the core competencies of this firm. Actually, this firm does not have much thatcould not be quickly duplicated. They have their website, a marketing channel through the gameprograms, and the unique design of their monogram. So in developing the business they shouldthink about ways they could make better use of these capabilities and assets. Here are somethoughts:

Supplier – Here it would be good to try to reduce cost, reduce the minimum order quantity, andreduce the lead time associated with the order. Any of these would be desirable. If it werepossible to reduce the minimum order quantity and the lead time, then instead of a single order,multiple orders could be place during the season. One order could cover the initial half of theseason and a second for the rest, for example. This should allow for more accurate forecasts andless product sold through eBay. They might consider using a domestic (US) supplier or possiblyeven consider subcontracting the making of the sweaters to locals. A quick web search showsthat automated machines at fairly low cost are now available that might be used.

Monogramming subcontractor – Might consider doing this in-house. They have a pretty gooddeal right now, though, since the subcontractor is providing space for inventory and shipping theproduct to the customer.

Target customers – They could expand this to the rest of the Big Ten teams. Other sports,particularly international venues, such as soccer could be developed.

Products – Many similar products that would be personalized could be developed such assweatshirts, jackets, blankets, and blazers. These would possibly use the same or similarsuppliers, and have the same requirements related to monogramming. Getting into totallydifferent kinds of products, such as coolers, might be another idea. It’s probably important to tryto exploit the idea of high end products that are attractive as gifts.

Chapter 12 - Lean Supply Chains

12-1

CHAPTER 12LEAN SUPPLY CHAINS

Review and Discussion Questions1. Is it possible to achieve zero inventories? Why or why not?

In reality, zero inventories are a challenging, if not impossible, goal for most organizations. Theconcept is theoretical because the ideal production unit is one. Nothing is made until the customerexpresses an unmet need for the product. In reality, inventories will always exist due to the timingbetween the expressed need and the actual delivery of the completed unit(s). Nevertheless, this goalaids in understanding of the lean concepts, and remains a reference point to continually remember inthe on-going improvement process.

2. Stopping waste is a vital part of lean. Identify some sources of waste in your home or dorm anddiscuss how they may be eliminated.

Waste can include work in process, raw materials, and finished goods that are not being directlyworked on or being shipped to the customer. Any processes or procedures not needed to complete theproduct or deliver the service are wastes. Material sitting in stores and queues are also sources ofwaste as is excess or inefficiencies. Through applications of lean principles of streamlining flows andonly performing work as it is needed, these wastes can be reduced and possibly eliminated.

Answers will vary to this question, but some obvious choices can be found in the refrigerator, withbills waiting to be paid, and, of course, laundry. For example, if laundry were done in small lots on aregular basis, fewer clothes would be needed. Of course, many people might not consider this animprovement.

3. Why must lean have a stable schedule?

Because any changes in the final product schedule are magnified backward along the line, a stableschedule is necessary. This schedule must be frozen at some point. Also, because suppliers andvendors are delivering in small batches just as materials are needed, they need accurate informationabout the build schedule so they can plan their corresponding deliveries.

4. Will lean work in service environments? Why or why not?

Lean is already achieving successes in a number of service environments. As services identify theircomponents that resemble an assembly line and are repetitive in nature, the concepts will work. Also,the philosophy of reducing waste and streamlining flows to eliminate waste can work in any setting.

5. Discuss ways to use lean to improve one of the following: a pizza restaurant, a hospital, or an autodealership.

Any number of answers would be correct. For example, in a pizza restaurant, streamlining the pizzapreparation and baking operations would speed the product to the customer. Fast and efficientcustomer ordering and payment would allow the system to process more customers. Possibly lettingcustomers refill their own drinks or serve themselves would speed processing. In a hospital orautomobile dealership, procedures can be streamlined and altered to serve the customer.


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6. Which objections might a marketing manager have against uniform plant loading?

Uniform plant loading might upset a marketing manager who is planning a special promotionalcampaign for a specific product. If production did not make enough of the units during thepromotional period, backorders or lost sales might result. Also because some products have differentlife cycles and sales patterns, this smoothing might hinder the marketing activities.

7. What are the implications for cost accounting of lean production?

Cost accounting can benefit lean analysis, but outdated measures tied to labor rates and productivityno longer apply. Overhead is the key variable to measure under lean. Labor is only a small part ofthe entire production dollar. Also labor and machinery may be idle under lean because goods areonly produced as needed. Labor and machinery variances may not reflect the lean philosophy.

8. What are the roles of suppliers and customers in a lean system?

Lean involves customers and suppliers as an integral part of the process. Customers provide productenhancement, modification, and usage data. Suppliers work with the manufacturing organization tocoordinate delivery and raw material or other input production. Both groups may sit on lean teamsand participate in improvement activities, as all groups will benefit from changes.

9. Explain how cards are used in a Kanban system.

Cards in a kanban system represent a visual work order. As material is moved from the line to thecustomer, the last operator in the process goes to the next workstation up the line and pulls a bin ofwork for further processing. This employee removes a card from the bin and leaves it at the previousstation. This card represents a work order for this station to make or process more products. Thissequence continues in a backward fashion through the line and back to the suppliers.

10. In which ways, if any, are the following systems analogous to Kanban: returning empty bottles to thesupermarket and picking up filled one; running a hot dog stand at lunchtime; withdrawing moneyfrom a checking account; raking leaves into bags?

All the systems represent work orders when the empty containers are returned. The empty bottles atthe supermarket will be picked up by the soda bottler and represent a need to clean and refill thebottles and return them to customers. A hot dog stand at lunchtime has hungry customers as workorders to process. The customers in line represent needs for the hot dogs. Withdrawing money froma checking account serves as a receipt and also a tickler to the individual to deposit more money inthe account at the next pay period. Raking leaves into bags is also a kanban. Once a bag is filled, theindividual pulls an empty bag from the box and continues to fill bags until the yard is free of leaves orno empty bags remain.


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11. Why is lean so hard to implement in practice?

A key implementation difficulty is the lack of emphasis on lean on an on-going basis. The leanimprovements are slow, take time, and are never ending. Initial enthusiasm may wane over time.Other problems in implementation include poor supplier quality, a lack of employee commitment, andproblems in reducing machine set up times.

12. Explain the relationship between quality and productivity under the lean philosophy.

Under JIT, quality and productivity are key and equal partners. As quality improves, so doesproductivity, as only good units are assembled. No work is wasted on preparing inferior qualityitems. Both are necessary in the lean philosophy.

13. How would you show a pull system in VSM Symbols between the blanking and CNC stages of thebolt manufacturing solved problems? Exhibits 12.11, 12.12 & 12.13.

Problems

1.

D = 10 gauges per hourL = 2 hoursS = .20C = 5 gaugesK = DL(1+S)/CK = 10(2)(1+0.20) / 5 = 4.8 5 Kanban card sets

2.

D = 4 transmissions per hourL = 1 hourS = .50C = 4 transmissionsK = DL(1+S)/CK = 4(1)(1+0.50) / 4 = 1.50 2 Kanban card sets

3.

D = 2,400 bottles/2 hours = 1200/60 minutes = 20 per minuteL = 40 minutesS = .10C = 120 bottlesK = DL(1+S)/CK = 2(40)(1+0.10) / 120 = 7.33 8 Kanban cards


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4.

D = 16 catalytic converters per hourL = 2 hoursS =.125C = 10 catalytic convertersK = DL(1+S)/CK = 16(2)(1+0.125) / 10 = 3.6 4 Kanban cards

5.

D = 32 catalytic converters per hourL = 1 hourS =.125C = 8 catalytic convertersK = DL(1+S)/CK = 32(1)(1+0.125) / 8 = 4.5 5 Kanban cards

CASE: Quality Parts Company – Teaching Note

QUESTIONS:

1. Which of the changes being considered by the manager of Quality Parts Company gocounter to the JIT philosophy?

Almost all of the recommended changes run counter JIT principles: Using MRP to “keep theskids filled” implies the use of inventory as a motivator to push production. Adding externalinspectors is counter the JIT practice of in-process inspection. Setting up a rework line onlyinstitutionalizes the acceptance of rework. Labor and machine utilization are not objectives ofJIT. The focus should be more on flexibility and reducing the waste of overproduction. Theinstallation of high rise shelving indicates an acceptance of wasteful inventory.

2. Make recommendations for JIT improvements in such areas as scheduling, layout, Kanban,task groupings, and inventory. Use quantitative data as much as possible; state necessaryassumptions.

Answers will vary. The students might be encouraged to use the Lotfi and Pegels software todevelop layouts. Machines and operations might be located in U-shaped layouts according to theassembly line balance.

3. Sketch the operation of a pull system for quality for Quality Parts Company’s currentsystem.

Answers will vary. The U-shaped layout is a useful tool. Machining cells might also be utilized.


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4. Outline a plan for the introduction of JIT at Quality Parts Company.

The plan will depend on the specific recommendations. Likely steps include acceptance ofrecommendations, development of an implementation schedule, training, team development,waste reduction, retooling, reallocation of workspace, and implementation of workflows. Topdown direction in the change should be emphasized. Shigeo Shingo estimates that mostcompanies will need five years to implement JIT.

Case: Value Stream Mapping – Answers to Questions

1. Eliminating the queue of work dramatically quickens the time it takes a part to flowthrough the system. What are the disadvantages of removing those queues?

The big consideration is whether or not these machines should operate independently or not. Thebuffers allow the machines to be scheduled, at least to some extent, independently of one another.Totally eliminating the buffers and moving to “one-piece flow” is like setting up an assembly linein this area.

2. How do you think the machine operators would react to the change?

No doubt they would not like the change since they probably enjoyed the independence that theybefore.

3. What would you do to ensure that the operators were kept busy?

This is a major issue since the cycle time for the first machine is only .5 minutes and the secondmachine on 1.2 minutes. Probably you could just assigned one worker to both machines.

CASE: Pro Fishing Boats – Teaching Note

1. Create a value stream map of this supply chain. What other information is needed?


12-6

Here is a simplified map. Currently, we do not know about the components using by Manufacturing Inc.All we know is that there are 12 weeks’ worth of one component and 4 weeks’ worth of the other.There is about a 27.3 weeks cumulative lead time from arrival of the raw materials to Manufacturing Inc.to the Pro Fishing Warehouse. Most of this is transportation time and waiting delay. If we add thecomponent shipped from the US to China this cumulative lead time may be double this time.The lead time analysis may not be very accurate. It might be good to actually track some components andproduct through the network to see what the actual time is and how much variability there is in the flowtime. This could reveal much about how the network actually operates.

2. Where is there risk for supply chain disruptions or stoppages to the flow of materials?

Of course, there are many all along this supply chain. The supply of parts coming from the US could bedisrupted. The plant in China could experience a problem. The complex port operations in Shanghai andLA could be delayed. In a sense, it is surprising that they can get away with only 6 weeks of inventory inthe Pro Fishing warehouse.

3. Where do opportunities reside in improving supply chain operations and how has VSM helpedto reveal these?

There may be some significant opportunities to reduce cost in this supply chain. Pro Fishing couldanalyze the total cost of the network, not just the cost of parts quoted by Manufacturing, Inc. They shouldalso look at the efficiency (cost) of the 2nd tier suppliers in the network.

Chapter 13 - Global Sourcing and Procurement

13-1

CHAPTER 13GLOBAL SOURCING AND PROCUREMENT

Review and Discussion Questions1. What recent changes have caused supply chain management to gain importance?

Changes include:

a. Competitive pressures from foreign firms.

b. Elevation of product quality to a very high level of importance.

c. International marketing and international purchasing.

d. Trends towards choosing sole-source suppliers and long term relationships.

e. Product varieties and ranges are rapidly changing, and speed of delivery to market isessential.

f. Product life cycles have shortened necessitating knowledge and control of inventoriesin the various pipelines.

g. Adoption of JIT production has changed supplier relationships and has also increasedthe focus on reducing inventories.

h. Trends in the legal system hold manufacturers liable for product failures, even thoughcauses of failure may lie outside of the production system itself.

i. Use of EDI in purchasing.

j. The growth of supplier development.

2. With so much productive capacity and room for expansion in the United States, why would acompany based in the United States choose to purchase items from foreign firm? Discuss thepros and cons.

The use of foreign firms can provide a U.S. firm more alternatives in selecting a supplier.The pros are: more choices, potentially reduced costs in the areas of materials, transportation,production, and distribution, and potentially moving closer to a foreign market. The cons are:the distance is generally increased; communications problems are increased due to distance,culture, and technology; and there may be problems with customs, government regulations,political stability, etc.

3. Describe the differences between functional and innovative products.

Functional products are staples that people buy in a wide range of retail outlets. Typically,they do not change much over time, have low profit margins, stable predictable demand andlong life cycles. Innovative products, on the other hand, give customers additional reasons to


13-2

buy. Fashionable clothes and personal computers are examples of innovative products.Innovative products have short life cycles, high profit margins, and volatile demand.

4. What are characteristics of efficient, responsive, risk-hedging and agile supply chains? Can asupply chain be both efficient and responsive? Risk-hedging and agile? Why, or why not?

Efficient supply chains are designed to minimize cost that requires high utilization,minimizing inventory, and selecting vendors based primarily on cost and quality, anddesigning products that are produced at minimum cost. Market-responsive supply chains aredesigned to minimize lead time to respond to unpredictable demand, thus minimizingstockout costs and obsolete inventory costs. Risk sharing supply chains are those that shareresources so that risks in the supply chain can be shared. Agile are those supply chains thatare flexible while still sharing risks of shortages across the supply chain. Generally, thesesupply chains carry excess capacity and higher buffer stocks. Vendor in responsive supplychains would be selected for speed, flexibility, and quality. It is possible to be both efficientand responsive, and both Risk-hedging and Agile, but Exhibit 13.4 helps illustrate whysupply chains are generally not both.

5. As a supplier, which factors would you consider about a buyer (your potential customer) tobe important in setting up a long-term relationship?

The financial stability and credit worthiness of the company is of primary importance. Thereputation of the company vis-à-vis their supplier is also very important. For example, is thisa company that is fair with its suppliers and honors its payables in a timely fashion? Is thetechnological match between supplier and customer sufficient? Will delivery schedules andquantities be stable, facilitating smooth operations?

6. Describe how outsourcing works. Why would a firm want to outsource?

Outsourcing is the act of moving some of a firm's internal activities and decisionresponsibilities to outside providers. The terms of the agreement are established in a contract.Outsourcing goes beyond the more common purchasing and consulting contracts because notonly are the activities transferred, but also resources that make the activities occur aretransferred. Reasons for outsourcing are listed in Exhibit 13.6. Some of the major categoriesfrom this exhibit include organizational, improvement, financial, revenue, cost, and employeedriven reasons.


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Problems

1.Year: 0 1 2 3

Demand 200,000 300,000 500,000Cost of Capital 0.15

Purchase OptionPurchase Cost PerUnit 0.1 $20,000.00 $30,000.00 $50,000.00

Shipping/Unit 0.01 $2,000.00 $3,000.00 $5,000.00Inventorycharge/Unit 0.005 $1,000.00 $1,500.00 $2,500.00

Monthly charge 20 $240.00 $240.00 $240.00Total Purchase Cost $23,240.00 $34,740.00 $57,740.00

Make OptionDirect Material 0.05 $10,000.00 $15,000.00 $25,000.00Direct Labor 0.03 $6,000.00 $9,000.00 $15,000.0050% Surcharge 0.015 $3,000.00 $4,500.00 $7,500.00Indirect Labor 0.011 $2,200.00 $3,300.00 $5,500.0050% Surcharge 0.0055 $1,100.00 $1,650.00 $2,750.00Overhead 100% DL 0.03 $6,000.00 $9,000.00 $15,000.00Total Variable Manufacture Cost $28,300.00 $42,450.00 $70,750.00Investment Engineer $30,000.00Equipment $10,000.00

Cost Comparison AnalysisMake Cost – Buy Cost $40,000.00 $5,060.00 $7,710.00 $13,010.00Discount factor 1 0.86957 0.75614 0.65752NPV (Make – Buy) $40,000.00 $4,400.00 $5,829.87 $8,554.29Total NPV(Make - Buy) $58,784.15

Alternative: Option NPV CalculationsBuy $143,226.27 $40,000.00 $24,608.70 $32,098.30 $46,519.27Make $84,442.11 $20,208.70 $26,268.43 $37,964.99Difference $58,784.15

Continuing to make in-house would cost us over $58,000 more in current dollars thanbuying from the supplier. We should accept the bid.


13-4

2.Requirement (annual forecast) 12,000.00 unitsWeight 22 pounds per engineOrder processing cost $125.00 per orderInventory carry cost 20% of average inventory

Lot Size (order quantity) 1,000 Units - given in the case

Supplier 1 2Unit Price $510 $505Annual Purchase Cost $6,120,000 $6,060,000One-Time Tooling Cost $22,000 $20,000Orders per year 12 12Order Processing Cost $1,500 $1,500Inventory carry cost $51,000 $50,500

Distance 125 100 miles

Weight per load 22,000Transportation (Less-than-truckload)$1.20 per 2,000 lbs. per mile $19,800 $15,840

Total Cost $6,214,300 $6,147,840 $66,460 differenceWe would prefer supplier #2.

Required lot size for truckload 1818 Units (40,000 lbs. max. load/22 lbs. per engine)

Supplier 1 2Unit Price $500 $505Annual Purchase Cost $6,000,000 $6,060,000One-Time Tooling Cost $22,000 $20,000Orders per year 6.6 6.6Annual Order Processing Cost $825 $825Annual Inventory carry cost $90,900 $91,809

Distance 125 100 miles

Weight per load 40,000Transportation (truckload)$0.80 per 2,000 lbs. per mile $13,200 $10,560

Total Cost $6,126,925 $6,183,194 $56,269 differenceYes, it would make sense to order in truckload lots as we can reduce total costs. While carrying costs increase,purchase and transportation costs decrease by a greater amount. Note that if ordering in truckload lots,supplier #1 becomes the lowest choice option.

In future years the cost would be reduced by the one-time tooling cost included here.


13-5

3.

6.148350

52*1000*00.1

ValueInventoryAggregateAverage

SoldGoodsofCostTurnoverInventory

The problem tells us that we sell 4,000 QUARTER pound burgers a week, therefore we sell 1,000pounds a week, and each pound of hamburger costs $1.00. The problem also tells us that onaverage, the store has 350 pounds of inventory on hand. By dividing the Cost of Goods Sold byAverage Aggregate Inventory Value, We can figure the Inventory Turns. This means that theirinventory turns 148.6 times a year.

350.52*52*1000*00.1

35052*

SoldGoodsofCost

ValueInventoryAggregateAverageSupplyofWeeks

On average the restaurant has about a third of a week’s supply on hand.

4.Q1 Q2 Q3 Q4

Sales

United States 300 350 405 375

Canada 75 60 75 70

Europe 30 33 20 15

COGS (Total) 280 295 340 350

Inventory

Raw Materials 50 40 55 60

WIP and FG 100 105 120 150

DC Inventory

United States 25 27 23 30

Canada 10 11 15 16

Europe 5 4 5 5

Total Inventory 190 187 218 261

Inventory Turnover 1.5 1.6 1.6 1.3

Using the end-of-quarter inventory numbers as a substitute for the average inventory level, wehave the following quarterly and annual inventory turn values. Average inventory for the annualfigure is based on the average of the 4 quarterly inventory numbers.

Q1 Q2 Q3 Q4 Annual280/190 = 1.474 295/187 = 1.578 340/218 = 1.560 350/261 = 1.341 1265/214 = 5.911


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If you were given the assignment to increase inventory turnover, what would you focus on?Why?

To increase the inventory turns, a firm needs to reduce the amount of inventory or increase salesor both. To increase turns, the item most readily within our control is the amount of inventorythat the firm has on hand. The raw materials, WIP, and FG inventories are the most obvioustargets for reduction.

The company reported that it used 500M worth of raw material during the year. Onaverage, how many weeks supply of the raw material are on hand at the factory?

797.852*1265214

52*SoldGoodsofCost

ValueInventoryAggregateAverageSupplyofWeeks

The 500M does not come into play in this problem.Analytics Exercise: Global Sourcing Decisions – Grainger

1. Evaluate the current China/Taiwan logistics costs. Assume a current total volume of 190,000CBM and the 89% is shipped direct from the supplier plants in containers. Use the data fromthe case and assume that the supplier loaded containers are 85% full. Assume thatconsolidation centers are run at each of the four port locations. The consolidation centersonly use 40’ containers and fill them to 96% capacity. Assume that it costs $480 to ship a 20’container and $600 to ship a 40’ container. What is the total cost to get the containers to theUnited States? Do not include United States port costs in this part of the analysis.

Basic Data

Total Current Volume (CBM) 190,000

Direct Ship Percentage 0.89

Direct Ship Volume (CBM) 169,100

Consolidation Center Volume 20,900

Shipping Cost Calculations

Direct Ship by Container Type 20' 40'

Volume (%) 21% 79%

Volume (CBM) 35511 133589

Container Capacity Used 85% 85%

Consolidation Center by Container Type

Volume (%) 100%

Volume (CBM) 20900

Container Capacity Used 96%

Container Capacity (CBM) 34 67


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Containers Shipped 1,229 2,671

Shipping Cost per Container $ 480.00 $ 600.00

Shipping Costs by Container Size $ 589,920 $ 1,602,600

Total Shipping Cost $ 2,192,520

Consolidation Center Operating Cost Calculations

Number of Centers 4

Annual Fixed Cost per Center $ 75,000

Total Annual Fixed Cost $ 300,000

Variable Cost per CBM $ 4.90

Total Annual Variable Cost $ 102,410

Total Annual Consolidation Center Costs $ 402,410

Total China/Taiwan Logistics Cost $ 2,594,930

2. Evaluate an alternative that involves consolidating all 20’ volume and using only a singleconsolidation center in Shanghai/Ningbo. Assume that all the existing 20’ volume and theexisting consolidation center volume is sent to this single consolidation center by suppliers.This new consolidation center volume would be packed into 40’ containers filled to 96% andshipped to the United States. The existing 40’ volume would still be shipped direct from thesuppliers at 85% capacity utilization.

Basic Data

Total Current Volume (CBM) 190000

Direct Ship Percentage 0.7031

Direct Ship Volume (CBM) 133589

Consolidation Center Volume 56411

Shipping Cost Calculations

Direct Ship by Container Type 20' 40'

Volume (%) 0% 100%

Volume (CBM) 0 133589

Container Capacity Used 85% 85%

Consolidation Center by Container Type

Volume (%) 100%

Volume (CBM) 56411

Container Capacity Used 96%


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Container Capacity (CBM) 34 67

Containers Shipped 0 3223

Shipping Cost per Container $ 480.00 $ 600.00

Shipping Costs by Container Size $ - $ 1,933,800

Total Shipping Cost $ 1,933,800

Consolidation Center Operating Cost Calculations

Number of Centers 1

Annual Fixed Cost per Center $ 75,000

Total Annual Fixed Cost $ 75,000

Variable Cost per CBM $ 1.40

Total Annual Variable Cost $ 78,975

Total Annual Consolidation Center Costs $ 153,975

Total China/Taiwan Logistics Cost $ 2,087,775

Assuming the new consolidation center has the same fixed cost as before (questionable given theincrease in volume), the new approach saves $507,155 per year.

3. What should be done based on your analytics analysis? What have you not considered thatmay make your analysis invalid or that may strategically limit success? What do you thinkGrainger management should do?

Consolidating the 20’ volume and using only a single Consolidation Center looks veryattractive from this analysis. However, there are other issues to be considered.

- For one, we have not considered the increased cost to the suppliers that currently packtheir own 20’ containers. These suppliers will need to bear the cost of shipping theirgoods to the Shanghai/Ningbo consolidation center. This cost will probably be pushedback to Grainger in the long run.

- There will also be some added cost for the suppliers that currently ship to consolidationcenters directly. These will all need to use the Shanghai/Ningbo now, which might notbe as close as their current consolidation center.

- The cost calculations also assume that the Shanghai/Ningbo center can handle theincreased workload and the fixed cost will remain the same. Neither of theseassumptions is guaranteed (or even likely).

We may want to seriously consider using two consolidation centers with the other being inYantian/Hong Kong. It may be attractive to have consolidation centers in bothShanghai/Ningbo and Yantian/Hong Kong since these are the most heavily used ports.Assumptions regarding the consolidation center fixed costs would need to be tested as well.

Chapter 14 - Location, Logistics, and Distribution

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CHAPTER 14LOCATION, LOGISTICS, AND DISTRIBUTION

Review and Discussion Questions1. What motivations typically cause firms to initiate a facilities location or relocation project?

There are a variety of reasons, both positive and negative. Firms may wish to move closer tomarkets or sources of supply. Cost reduction is another major reason. A company might need aneducated work force, and therefore move near a major university. On the negative side, firmsrelocate to avoid costly regulation or unionization. Students should be able to cite a large numberof reasons.

2. List five major reasons why a new electronic components manufacturing firm should move intoyour city or town?

Answers will vary depending upon the location. Possibilities include quality of education, taxadvantages, proximity of supply, proximity of markets, or favorable quality of life.

3. How do facility location decisions differ for service facilities and manufacturing plants?

In many ways, the decisions are similar. However, since the customer is often involved in theproduction of the service, proximity to the customer is of greater importance. Services oftenutilize multiple sites to remain close to the customer. Market needs impact services locationdecisions. Alternatively, resource considerations have much more impact on manufacturing.

In addition, the cost of establishing a service facility is relatively low when compared tomanufacturing. Manufacturing decisions are often based on cost minimization while servicedecisions are based on profit maximization.

4. What are the pro and cons of relocating a small or midsized manufacturing firm (that makesmature products) from the United States to China?

According to the product life cycle concept, mature products require more of a cost orientationdue to price competition and low product differentiation. In such industries, many firms areevaluating the cost/benefit trade-offs associated with moving operations to China. Labor costsare significantly lower in China. Land is relatively cheap and there is less regulation. The lowcost of producing goods in China is the primary reason for the boom in Chinese manufacturing.In many cases, this reduced cost more than offsets the problems that companies can face whenmoving operations to China. These issues include a less-educated workforce, quality problems,higher logistics costs, longer transportation times, and an extended supply chain to manage.

5. If you could locate your new software development company anywhere in the world, which placewould you choose, and why?

Answers will vary depending upon preferences. However, student should provide sound businessreasons for their responses.


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Problems

1.

d1x = 150 d1y = 75 V1 = 8,000

d2x = 100 d2y = 300 V2 = 6,200

d3x = 275 d3y = 380 V3 = 7,000

7.176200,21

)000,7)(275()200,6)(100()000,8)(150(

i

iixx V

VdC

5.241200,21

)000,7)(380()200,6)(300()000,8)(75(

i

iiyy V

VdC

2.

d1x = 300 d1y = 320 V1 = 4,000

d2x = 375 d2y = 470 V2 = 6,000

d3x = 470 d3y = 180 V3 = 3,000

8.373000,13

)000,3)(470()000,6)(375()000,4)(300(

i

iixx V

VdC

9.356000,13

)000,3)(180()000,6)(470()000,4)(320(

i

iiyy V

VdC


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3.

a.

From/To New York Fort Worth San Diego Minneapolis SupplyBoulder 7 11 8 13 100,000Macon 20 17 12 10 100,000Gary 8 18 13 16 150,000Requirements 50,000 70,000 60,000 80,000

b.

Total profit = $4,240,000. Note that an alternate optimal solution exists that changes theshipping so that Fort Worth is supplied entirely by Gary (70,000 units, no units from Macon);San Diego receives 50,000 units from Macon and 10,000 units from Gary. New York andMinneapolis are supplied as shown above.

Production should be:Boulder 10,000 unitsMacon 100,000 unitsGary 150,000 units

From/To New York Fort Worth San Diego MinneapolisFactorySupply

Boulder 7 11 8 13 100,000Macon 20 17 12 10 100,000Gary 8 18 13 16 150,000Requirements 50,000 70,000 60,000 80,000 350,000

CandidateSolution

TotalShipped

Boulder - - - 10,000 10,000Macon 50,000 50,000 - - 100,000Gary - 20,000 60,000 70,000 150,000Total Supplied 50,000 70,000 60,000 80,000 260,000

ProfitBoulder - - - 130,000Macon 1,000,000 850,000 - -Gary - 360,000 780,000 1,120,000Total Profit Total Cost 4,240,000$


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4.

a. This is the optimal solution, with a total cost of $720.

From/To D E F G Supply

A 9 8 6 5 50B 9 8 8 0 40C 5 3 3 10 75Requirements 50 60 25 30 165

CandidateSolutionA 25 0 25 0 50B 10 0 0 30 40C 15 60 0 0 75Total Shipped 50 60 25 30 165

CostA 225 0 150 0B 90 0 0 0C 75 180 0 0Total Costs $720

b. Place a sufficiently small cost into the A to D cell to force a shipment from A to D. Thesolver can provide the value that will cause shipments to go from A to D, and in this case,any cost less than $10 will cause cars to be sent that way.

5. a. The key here is properly calculating the total cost per 1,000 pounds for each source anddestination pair. Some students will convert the given capacity to the equivalent numberper 1,000 lbs. Here it is left as given and the spreadsheet does the conversion.

MaximumCapacity Prod. Costs

(1,000 lbs.)Plant (100,000 lbs.)

Philadelphia 7.5 $325.00

Atlanta 9 $275.00

St. Louis 12 $305.00

Salt Lake City 10.3 $250.00


14-5

Transport Costs per 1,000 lbs.

From/To NYC Birmingham TerreHaute Dallas Spokane San Diego

Philadelphia $45 $52 $56 $62 $78 $85

Atlanta $55 $42 $58 $59 $80 $82

St. Louis $57 $60 $50 $54 $65 $70

Salt Lake City $72 $71 $67 $57 $52 $60

Total Demand(x 1,000 lbs.) 525 415 925 600 325 400

Combined Costs per 1,000 lbs.


Philadelphia $370 $377 $381 $387 $403 $410

Atlanta $330 $317 $333 $334 $355 $357

St. Louis $362 $365 $355 $359 $370 $375

Salt Lake City $322 $321 $317 $307 $302 $310

Solution (x 1,000 lbs.)

From/To NYC Birmingham TerreHaute Dallas Spokane San Diego Supplied

(x 100,000)

Philadelphia 60 0 0 0 0 0 0.6Atlanta 465 415 0 20 0 0 9St. Louis 0 0 925 275 0 0 12

Salt Lake City 0 0 0 305 325 400 10.3Received 525 415 925 600 325 400


14-6

Total Costs


Philadelphia $22,200 $0 $0 $0 $0 $0

Atlanta $153,450 $131,555 $0 $6,680 $0 $0

St. Louis $0 $0 $328,375 $98,725 $0 $0

Salt Lake City $0 $0 $0 $93,635 $98,150 $124,000

TOTAL: $1,056,770

b. We are using almost none of the supply from Philadelphia. If the Philadelphia capacitycannot be used to satisfy demand elsewhere, then we should consider closing Philadelphiaand adding slightly to capacity at another plant to pick up the slack.

Analytics Exercise: Distribution Center Location

1. Relative to the United States distribution network, calculate the cost associated with running theexisting system. Assume that 40% of the volume arrives in Seattle and 60% in Los Angeles andthe port processing fee for federal inspection at both locations is $5.00 per CBM. Assume thateverything is transferred to the Kansas City distribution center by rail where it is unloaded andquality checked. Assume that all volume is then transferred by truck to the 9 existing warehousesin the United States.

K.C Distribution Center OnlyPort to DC Costs

Volume Port Distance K.C. Unload Total

Pct CBM Processing to K.C. Rail Costs and Q.C. Costs

Basic Data 190,000 $5.00 $0.0018 $3.00

Seattle Port 40% 76,000 $ 380,000 1,870 $ 255,816 $ 228,000 $ 863,816

L.A. Port 60% 114,000 $ 570,000 1,620 $ 332,424 $ 342,000 $ 1,244,424

$ 2,108,240

DC to Warehouse CostsTruck

Miles FreightWarehouse Demand from KC Costs


14-7

$0.022

Kansas City 20900 0 $ - 11.0%

Cleveland 17100 800 $ 300,960 9.0%

New Jersey 24700 1200 $ 652,080 13.0%

Jacksonville 15200 1150 $ 384,560 8.0%

Chicago 22800 520 $ 260,832 12.0%

Greenville 15200 940 $ 314,336 8.0%

Memphis 17100 510 $ 191,862 9.0%

Dallas 22800 500 $ 250,800 12.0%

Los Angeles 34200 1620 $ 1,218,888 18.0%

Total: 190000 $ 3,574,318

Total Cost of Current System: $ 5,682,558

2. Consider the idea of upgrading the Los Angeles warehouse to include a distribution centercapable of processing all the volume coming into the United States. Assume that containerscoming into Seattle would be inspected by federal officials (this needs to be done at all portlocations) and then immediately shipped by rail in their original containers to Los Angeles. Allvolume would be unloaded and quality checked in Los Angeles (the quality check cost %5.00 perCBM when done in Los Angeles). 18% of the volume would then be kept in Los Angeles fordistribution through that warehouse and the rest transshipped by rail to the Kansas Citywarehouse. The cost to transship to Kansas City would be $0.0018 per CBM. The material sent toKansas City would not need to go through the “unload and quality check process,” and would bestored directly in the Kansas City distribution center. Assume that the remaining volume wouldbe transferred by truck to the 8 remaining warehouses in the United States at a cost of $0.0220per CBM.

K.C and L.A. Distribution CentersPort to L.A. DC Costs

Volume Port Distance L.A. Unload Total

Pct CBM Processing to L.A.. Rail Costs and Q.C. Costs

Basic Data 190,000 $5.00 $0.0018 $5.00

Seattle Port 40% 76,000 $ 380,000 1,140 $ 155,952 $ 380,000 $ 915,952

L.A. Port 60% 114,000 $ 570,000 0 $ - $ 570,000 $ 1,140,000

$ 2,055,952

L.A. DC Operating Costs: $ 350,000

Total L.A. DC Costs: $ 2,405,952Transshipment Costs to K.C.

Volume Distance

Pct CBM L.A. to K.C. Rail Cost

82% 155,800 1,620 $454,313


14-8

K.C. DC to Warehouse CostsTruck

Miles FreightWarehouse Demand from KC Costs

$0.022

Kansas City 20900 0 $ -

Cleveland 17100 800 $ 300,960

New Jersey 24700 1200 $ 652,080

Jacksonville 15200 1150 $ 384,560

Chicago 22800 520 $ 260,832

Greenville 15200 940 $ 314,336

Memphis 17100 510 $ 191,862

Dallas 22800 500 $ 250,800

Los Angeles 0 1620 $ -

Total: 155800 $ 2,355,430

Total Cost of New System: $ 5,215,695

Annual Savings of New System: $ 466,863

3. What should be done based on your analytics analysis of the United State distribution system?Should the new Los Angeles distribution center be added? Is there any obvious change thatGrainger might do to make this option more attractive?

Based on the economic analysis this looks attractive as the yearly $466,863 savings would defraythe $1,500,000 investment to upgrade Los Angeles in 3.21 years (ignoring the time value ofmoney).

The obvious thing to do would be to ship everything to Los Angeles that is sourced inChina/Taiwan. This would remove the cost of shipping from Seattle to Los Angeles which is$155,952. The total new savings is $622,815 and the new payback period is 2.41.

4. Is this strategically something that Grainger should do? What have they not considered that maybe important?

Rationalizing the network seems to make sense. One thing not considered in this analysis is thecosts of the Seattle facility that would not be incurred under the new system. If the facility couldbe sold, or the lease terminated, the payback period would be less than calculated above. Apossibly negative result of the new system would be the risk of utilizing a single inbound port.Any issues with the L.A. port would directly affect Grainger. Another thing is that is difficult topredict is how economic forces may change in the next 2 to 3 years. Shipping cost may begreatly impacted by increases in oil prices. Further, labor costs in China and Taiwan mayincrease significantly thus making using these areas less attractive for sourcing product. Sincethe returns have such long payback periods, it might be reasonable to just leave this part of thesystem as it is.

Appendix A - Linear Programming Using the Excel Solver

A-1

APPENDIX ALINEAR PROGRAMMING USING THE EXCEL SOLVER

1. Excel Solution

X Y TotalDecision 6 0Profit $3 $1 $18

ResourcesX Y Used Capacity

A 12 14 72 <= 85B 3 2 18 <= 18C 1 0 <= 4

Graphical solution--the problem requested the Excel solution, but the following graphical solutionis provided for classroom use if desired.


A-2

2.

A B TotalDecision 15 10Cost $2 $4 $70

ResourcesA B Used Capacity

A 4 6 120 >= 120B 2 6 90 >= 72C 1 10 >= 10



A-3

3. a. Maximize Z = 20X1 + 6X2 + 8X3

s.t. 8X1 + 2X2 + 3X3 < 800

4X1 + 3X2 < 480

2X1 + X3 < 320

X3 < 80

X1 , X2 , X3 > 0

b. Excel solution

X1 X2 X3 TotalDecision 45 100 80Profit $20 $6 $8 $2,140

ResourcesX1 X2 X3 Used Capacity

Milling 8 2 3 800 <= 800Lathes 4 3 480 <= 480Grinders 2 1 170 <= 320Sales 1 80 <= 80

c. Solution is X1 = 45 S1 = 0 Z = $2140

X2 = 100 S2 = 0

X3 = 80 S3 = 150

S4 = 0

d. S1 = 0 implies milling machines at capacity

S2 = 0 implies lathes at capacity

S3 = 150 implies grinders not at capacity, with 150 hours available

S4 = 0 implies that X3 is at maximum sales capacity

e. The shadow price for the milling machine department is $2.25 per hour. Since it only cost$1.50 per hour to work overtime in this department, it is worthwhile to do so. Theallowable increase in overtime is 400; however, only 200 hours are available. Therefore, itis recommended that 200 hours of overtime in the milling machine department be used.


A-4

4. a.

Let A = pounds of food A

B = pounds of food B

Minimize : z = .75A + .15B

s.t. 600A + 900B < 3,600 Maximum calories

600A + 900B > 1,800 Minimum calories

200A + 700B < 1,400 Maximum starch

400A + 100B > 400 Minimum protein

A < 2 Maximum amount of A


A-5

5. Add constraint 100B < 150, and change objective function to z = 1.75A + 2.50B

6. a.

Let A = gallons of fuel A to mix

B = gallons of fuel B to mix

Minimize z = 1.20A + 0.90B

s.t. A + B > 3,000 fuel demand

A + B < 4,000 Maximum storage

A < 2,000 Maximum fuel A available

B < 4,000 Maximum fuel B available

10A - 5B > 0 Blend 80 octane minimum*

A, B > 0

*Note, blend constraint can be stated as (90A + 75B)/(A + B) > 80


A-6

b.

A B TotalDecision 1000 2000 3000Cost $1.20 $.90 $3,000

ResourcesA B Used Capacity

Min demand 1 1 3000 >= 3000Max Storage 1 1 3000 <= 4000Max Fuel A 1 1000 <= 2000Max Fuel B 1 2000 <= 4000Blend 10 -5 0 > 0



A-7

7. Let: F = dollars spent on food

S = dollars spent on shelter

E = dollars spent on entertainment

Maximize Z = 2F + 3S + 5E

s.t. F + S + E < 1500 Total Budget

F + S < 1000 Maximum on Food and Shelter

S < 700 Maximum on Shelter alone

E < 300 Maximum on Entertainment

F S E TotalDecision 300 700 300Profit 2 3 5 4,200

ResourcesX1 X2 X3 Used Capacity

total budget 1 1 1 1300 <= 1500$ on food and shelter 1 1 1000 <= 1000$ on shelter 1 700 <= 700$ on entertainment 1 300 <= 300

8. Produce 50 barrels of Expansion Draft and 50 barrels of Burning River. The total revenue willbe $1400.

Expansion Draft Burning River TotalDecision 50 50Sales $20 $8 $1400

ResourcesX1 X2 Used Capacity

Corn 8 2 500 <= 500Rice 0 6 300 <= 300Hops 4 3 350 <= 400


A-8

9. Run zone for 6 hours and man for 4 hours at a cost of $384.

Zone Man Total

Decision 6 4Cost $48 $24 $384

ProcessesX1 X2 Produced Demand

BCP1 3 1 22 >= 20BCP2 1 1 10 >= 10BCP3 1 0 6 >= 6

10. She should plant 700 acres in corn and 100 acres in soybeans.

Corn Soybeans Wheat TotalDecision 700 100 0 800Profit per acre $2,000 $2,500 $3,000 $1,650,000

ResourcesCorn Soybeans Wheat Used Capacity

Labor (workers) 0.1 0.3 0.2 100 <= 100Fertilizer (tons) 0.2 0.1 0.4 150 <= 150Acres Planted 1 1 1 800 <= 900

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