Optimization II. © The McGraw-Hill Companies, Inc., 2004 Operations Management -- Prof. Juran2...

Optimization II

© The McGraw-Hill Companies, Inc., 2004

Operations Management -- Prof. Juran 2

OutlineOptimization Extensions • Multiperiod Models

– Operations Planning: Sailboats• Network Flow Models

– Transportation Model: Beer Distribution

– Assignment Model: Contract Bidding



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A B C D E F Gobjective: 6000 4000 154,800$ = profit

decision variables: (in 100s) skis snowboards18.6 10.8

constraints:Molding 3 2 77.4 <= 115.5Cutting 1 3 51 <= 51Van 2 1 48 <= 48Demand 0 1 10.8 <= 16Nonnegativity (snowboards) 1 0 18.6 >= 0Nonnegativity (skis) 0 1 10.8 >= 0



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A B C D E F GMicrosoft Excel 10.0 Answer ReportWorksheet: [Book1]downhillReport Created: 3/27/2006 6:28:16 AM

Target Cell (Max)Cell Name Original Value Final Value

$E$1 objective: 10,000$ 154,800$

Adjustable CellsCell Name Original Value Final Value

$C$4 skis 1 18.6$D$4 snowboards 1 10.8

ConstraintsCell Name Cell Value Formula Status Slack

$E$11 Nonnegativity (snowboards) 18.6 $E$11>=$G$11 Not Binding 18.6$E$12 Nonnegativity (skis) 10.8 $E$12>=$G$12 Not Binding 10.8$E$7 Molding 77.4 $E$7<=$G$7 Not Binding 38.1$E$8 Cutting 51 $E$8<=$G$8 Binding 0$E$9 Van 48 $E$9<=$G$9 Binding 0$E$10 Demand 10.8 $E$10<=$G$10 Not Binding 5.2



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A B C D E F G HMicrosoft Excel 10.0 Sensitivity ReportWorksheet: [s-downhill.xls]downhillReport Created: 3/27/2006 6:31:19 AM

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$C$4 skis 18.6 0 6000 2000 4666.67$D$4 snowboards 10.8 0 4000 14000 1000

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$E$11 Nonnegativity (snowboards) 18.6 0 0 18.6 1E+30$E$12 Nonnegativity (skis) 10.8 0 0 10.8 1E+30$E$7 Molding 77.4 0 115.5 1E+30 38.1$E$8 Cutting 51 400 51 13 27$E$9 Van 48 2800 48 27.2 26$E$10 Demand 10.8 0 16 1E+30 5.2

Most important number: Shadow PriceThe change in the objective function that would result from a one-unit increase in the right-hand side of a constraint



Sailboat Problem• Sailco must determine how many sailboats to produce

during each of the next four quarters. • At the beginning of the first quarter, Sailco has an

inventory of 10 sailboats.• Sailco must meet demand on time. The demand during

each of the next four quarters is as follows:

1st Qtr 2nd Qtr 3rd Qtr 4th Qtr

40 60 75 25



Sailboat Problem• Assume that sailboats made during a quarter can be

used to meet demand for that quarter. • During each quarter, Sailco can produce up to 50

sailboats with regular-time employees, at a labor cost of $400 per sailboat.

• By having employees work overtime during a quarter, Sailco can produce unlimited additional sailboats with overtime labor at a cost of $450 per sailboat.

• At the end of each quarter (after production has occurred and the current quarter’s demand has been satisfied), a holding cost of $20 per sailboat is incurred.

• Problem: Determine a production schedule to minimize the sum of production and inventory holding costs during the next four quarters.



Managerial FormulationDecision VariablesWe need to decide on production quantities, both regular and overtime, for four quarters (eight decisions).Note that on-hand inventory levels at the end of each quarter are also being decided, but those decisions will be implied by the production decisions.



Managerial Formulation

Objective FunctionWe’re trying to minimize the total labor cost of production, including both regular and overtime labor.



Managerial Formulation

ConstraintsThere is an upper limit on the number of boats built with regular labor in each quarter.No backorders are allowed. This is equivalent to saying that inventory at the end of each quarter must be at least zero.Production quantities must be non-negative.



Managerial FormulationNote that there is also an accounting constraint: Ending Inventory for each period is defined to be:

Beginning Inventory + Production – DemandThis is not a constraint in the usual Solver sense, but useful to link the quarters together in this multi-period model.



Mathematical Formulation Decision VariablesPij = Production of type i in period j.

Let i index labor type; 0 is regular and 1 is overtime.Let j index quarters; 1 through 4



Mathematical Formulation Objective Function

Minimize

1

0

4

1

4

1i j jjiji HIPCZ

Ci = Production Cost; $400 for regular, $450 for overtime

H = Holding Cost; $20 per boat per period

Define Dj to be demand in period j

Define Ij to be ending inventory for period j

ji

ijjj DPII

1

01



Mathematical Formulation

Constraints

jP0 50 For all j

jI 0 For all j

ijP 0 For all i, j



Solution Methodology 123

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A B C D E F G H I J K LProduction schedule Total cost -$5,700Month 1 2 3 4Regular time production 1 1 1 1 4 Month 1 2 3 4

<= <= <= <= Regular time unit cost 400 400 400 400Upper bound 50 50 50 50 Overtime unit cost 450 450 450 450

Unit holding cost 20 20 20 20Overtime production 1 1 1 1 4

Initial inventory 10Total Production 2 2 2 2 8

Regular time cost $1,600Demand 40 60 75 25 200 Overtime cost $1,800

Holding cost -$9,100Ending inventory -28 -86 -159 -182

>= >= >= >=0 0 0 0

=SUM(B3:E3)

=SUM(B7:E7)

=SUM(B9:E9)

=SUM(B11:E11)

=D13+E9-E11




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A B C D E F G H I J K L M N OProduction schedule Total cost -$5,700Month 1 2 3 4Regular time production 1 1 1 1 4 Month 1 2 3 4

<= <= <= <= Regular time unit cost 400 400 400 400Upper bound 50 50 50 50 Overtime unit cost 450 450 450 450




Holding cost -$9,100Ending inventory -28 -86 -159 -182

>= >= >= >=0 0 0 0

=SUM(I10:I12)

=SUMPRODUCT(I4:L4,B3:E3)=SUMPRODUCT(I5:L5,B7:E7)=SUMPRODUCT(I6:L6,B13:E13)



Solution Methodology




A B C D E F G H I J K LProduction schedule Total cost $77,350Month 1 2 3 4Regular time production 50 50 50 25 175 Month 1 2 3 4

<= <= <= <= Regular time unit cost 400 400 400 400Upper bound 50 50 50 50 200 Overtime unit cost 450 450 450 450




Holding cost $600Ending inventory 20 10 0 0

>= >= >= >=0 0 0 0



Solution Methodology It is optimal to have 15 boats produced on overtime in the third quarter. All other demand should be met on regular time. Total labor cost will be $76,750.



Sensitivity Analysis

Investigate changes in the holding cost, and determine if Sailco would ever find it optimal to eliminate all overtime. Make a graph showing optimal overtime costs as a function of the holding cost.




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A B CUnit holding cost Overtime cost Holding cost

0 $6,750 $05 $6,750 $150

10 $6,750 $30015 $6,750 $45020 $6,750 $60025 $6,750 $75030 $11,250 $30035 $11,250 $35040 $11,250 $40045 $11,250 $45050 $11,250 $50055 $15,750 $060 $15,750 $0




$0

$2,000

$4,000

$6,000

$8,000

$10,000

$12,000

$14,000

$16,000

$18,000

0 5 10 15 20 25 30 35 40 45 50 55 60

Unit Holding Cost

To

tal

Ov

ert

ime

Co

st

$0

$100

$200

$300

$400

$500

$600

$700

$800

To

tal

Ho

ldin

g C

os

t

Overtime cost

Holding cost



Sensitivity Analysis 161718192021222324252627282930

A B C D E F G H I J K L

Unit holding cost Total cost 1 2 3 4 Total 1 2 3 4 Total0 $76,750 50 50 50 25 175 15 0 0 0 155 $76,900 50 50 50 25 175 0 0 15 0 15

10 $77,050 50 50 50 25 175 0 0 15 0 1515 $77,200 50 50 50 25 175 0 0 15 0 1520 $77,350 50 50 50 25 175 0 0 15 0 1525 $77,500 50 50 50 25 175 0 0 15 0 1530 $77,550 40 50 50 25 165 0 0 25 0 2535 $77,600 40 50 50 25 165 0 0 25 0 2540 $77,650 40 50 50 25 165 0 0 25 0 2545 $77,700 40 50 50 25 165 0 0 25 0 2550 $77,750 40 50 50 25 165 0 0 25 0 2555 $77,750 30 50 50 25 155 0 10 25 0 3560 $77,750 30 50 50 25 155 0 10 25 0 35

Regular Overtime



Sensitivity Analysis Conclusions:It is never optimal to completely eliminate overtime. In general, as holding costs increase, Sailco will decide to reduce inventories and therefore produce more boats on overtime. Even if holding costs are reduced to zero, Sailco will need to produce at least 15 boats on overtime. Demand for the first three quarters exceeds the total capacity of regular time production.



Gribbin BrewingRegional brewer Andrew Gribbin distributes kegs of his famous beer through three warehouses in the greater News York City area, with current supplies as shown:

Warehouses Supply Hoboken 80

Bronx 145 Brooklyn 120



Bars Demand Der Ratkeller 80 McGoldrick's Pub 65 Night Train Bar & Grill 70 Stern Business School 85

On a Thursday morning, he has his usual weekly orders from his four loyal customers, as shown :



Tracy Chapman, Gribbin’s shipping manager, needs to determine the most cost-efficient plan to deliver beer to these four customers, knowing that the costs per keg are different for each possible combination of warehouse and customer:

Ratkeller McGoldrick's Night Train Stern Hoboken $4.64 $5.13 $6.54 $8.67

Bronx $3.52 $4.16 $6.90 $7.91 Brooklyn $9.95 $6.82 $3.88 $6.85



a) What is the optimal shipping plan?b) How much will it cost to fill these four orders?c) Where does Gribbin have surplus inventory?d) If Gribbin could have one additional keg at one of the

three warehouses, what would be the most beneficial location, in terms of reduced shipping costs?

e) Gribbin has an offer from Lu Leng Felicia, who would like to sublet some of Gribbin’s Brooklyn warehouse space for her tattoo parlor. She only needs 240 square feet, which is equivalent to the area required to store 40 kegs of beer, and has offered Gribbin $0.25 per week per square foot. Is this a good deal for Gribbin? What should Gribbin’s response be to Lu Leng?



Managerial Problem Formulation

Decision VariablesNumbers of kegs shipped from each of three warehouses to each of four customers (12 decisions).

ObjectiveMinimize total cost.

ConstraintsEach warehouse has limited supply.Each customer has a minimum demand.Kegs can’t be divided; numbers shipped must be integers.



Mathematical Formulation

Decision VariablesDefine Xij = Number of kegs shipped from warehouse i to customer j.

Define Cij = Cost per keg to ship from warehouse i to customer j.

i = warehouses 1-3, j = customers 1-4



Mathematical Formulation Objective

Minimize Z =

ConstraintsDefine Si = Number of kegs available at warehouse i.

Define Dj = Number of kegs ordered by customer j.

Do we need a constraint to ensure that all of the Xij are integers?

3

1

4

1i jijijCX

4

1jiij SX

3

1ijij DX



1

23

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A B C D E F G H ITotal Cost 74.97

Shipping Plan Ratkeller McGoldrick's Night Train Stern

Hoboken 1 1 1 1 4 <= 80Bronx 1 1 1 1 4 <= 145

Brooklyn 1 1 1 1 4 <= 120

3 3 3 3= = = =80 65 70 85

CostsHoboken 4.64$ 5.13$ 6.54$ 8.67$

Bronx 3.52$ 4.16$ 6.90$ 7.91$ Brooklyn 9.95$ 6.82$ 3.88$ 6.85$

=SUM(B4:E4)

=SUM(E4:E6)

=SUMPRODUCT(B4:E6,B12:E14)



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A B C D E F G HTotal Cost 1469.55

Shipping Plan Ratkeller McGoldrick's Night Train SternHoboken 0 0 0 35 35 <= 80

Bronx 80 65 0 0 145 <= 145Brooklyn 0 0 70 50 120 <= 120

80 65 70 85= = = =80 65 70 85

CostsHoboken 4.64$ 5.13$ 6.54$ 8.67$

Bronx 3.52$ 4.16$ 6.90$ 7.91$ Brooklyn 9.95$ 6.82$ 3.88$ 6.85$



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A B C D E F G HAdjustable Cells

Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease

$B$8 Hoboken to Der Ratkeller 0 0.360 4.640 1E+30 0.360$C$8 Hoboken to McGoldrick's Pub 0 0.210 5.130 1E+30 0.210$D$8 Hoboken to Night Train Bar & Grill 0 0.840 6.540 1E+30 0.840$E$8 Hoboken to Stern Business School 35 0.000 8.670 0.210 0.760$B$9 Bronx to Der Ratkeller 80 0.000 3.520 0.360 1E+30$C$9 Bronx to McGoldrick's Pub 65 0.000 4.160 0.210 1E+30$D$9 Bronx to Night Train Bar & Grill 0 1.960 6.900 1E+30 1.960$E$9 Bronx to Stern Business School 0 0.000 7.910 0.760 0.210$B$10 Brooklyn to Der Ratkeller 0 7.490 9.950 1E+30 7.490$C$10 Brooklyn to McGoldrick's Pub 0 3.720 6.820 1E+30 3.720$D$10 Brooklyn to Night Train Bar & Grill 70 0.000 3.880 0.840 1E+30$E$10 Brooklyn to Stern Business School 50 0.000 6.850 1.82 0.840

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$B$11 Der Ratkeller Demand 80 4.280 80 0 35$C$11 McGoldrick's Pub Demand 65 4.920 65 0 35$D$11 Night Train Bar & Grill Demand 70 5.700 70 45 35$E$11 Stern Business School Demand 85 8.670 85 45 35$F$8 Hoboken Supply 35 0.000 80 1E+30 45$F$9 Bronx Supply 145 -0.760 145 35 0$F$10 Brooklyn Supply 120 -1.820 120 35 45



Where does Gribbin have surplus inventory?

The only supply constraint that is not binding is the Hoboken constraint. It would appear that Gribbin has 45 extra kegs in Hoboken.



If Gribbin could have one additional keg at one of the three warehouses, what would be the most beneficial location, in terms of reduced shipping costs?



According to the sensitivity report,

• One more keg in Hoboken is worthless.

• One more keg in the Bronx would have reduced overall costs by $0.76.

• One more keg in Brooklyn would have reduced overall costs by $1.82.



Gribbin has an offer from Lu Leng Felicia, who would like to sublet some of Gribbin’s Brooklyn warehouse space for her tattoo parlor. She only needs 240 square feet, which is equivalent to the area required to store 40 kegs of beer, and has offered Gribbin $0.25 per week per square foot.

Is this a good deal for Gribbin?

What should Gribbin’s response be to Lu Leng?



Assuming that the current situation will continue into the foreseeable future, it would appear that Gribbin could reduce his inventory in Hoboken without losing any money (i.e. the shadow price is zero).

However, we need to check the sensitivity report to make sure that the proposed decrease of 40 kegs is within the allowable decrease.

This means that he could make a profit by renting space in the Hoboken warehouse to Lu Leng for $0.01 per square foot.



Lu Leng wants space in Brooklyn, but Gribbin would need to charge her more than $1.82 for every six square feet (about $0.303 per square foot), or else he will lose money on the deal.

Note that the sensitivity report indicates an allowable decrease in Brooklyn that is enough to accommodate Lu Leng.



As for the Bronx warehouse, note that the allowable decrease is zero. This means that we would need to re-run the model to find out the total cost of renting Bronx space to Lu Leng.

A possible response from Gribbin to Lu Leng:

“I can rent you space in Brooklyn, but it will cost you $0.35 per square foot. How do you feel about Hoboken?”



Contract Bidding ExampleA company is taking bids on four construction jobs. Three contractors have placed bids on the jobs. Their bids (in thousands of dollars) are given in the table below. (A dash indicates that the contractor did not bid on the given job.) Contractor 1 can do only one job, but contractors 2 and 3 can each do up to two jobs.

Job 1 Job 2 Job 3 Job 4 Contractor 1 50 46 42 40 Contractor 2 51 48 44 — Contractor 3 — 47 45 45



Formulation

Decision VariablesWhich contractor gets which job(s).ObjectiveMinimize the total cost of the four jobs.ConstraintsContractor 1 can do no more than one job.Contractors 2 and 3 can do no more than two jobs each.Contractor 2 can’t do job 4.Contractor 3 can’t do job 1.Every job needs one contractor.



Formulation Decision VariablesDefine Xij to be a binary variable representing the assignment of contractor i to job j. If contractor i ends up doing job j, then Xij = 1. If contractor i does not end up with job j, then Xij = 0.

Define Cij to be the cost; i.e. the amount bid by contractor i for job j.

ObjectiveMinimize Z =

3

1

4

1i jijijCX



Formulation Constraints for all j.

for i = 1.

for i = 2, 3.

13

1

i

ijX

14

1

j

ijX

24

1

j

ijX

01,34,2 XX



Solution Methodology1234567

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A B C D E F G HJob 1 Job 2 Job 3 Job 4

Contractor 1 50 46 42 40Contractor 2 51 48 44 10000Contractor 3 10000 47 45 45

Assignment of contractors to jobsJob 1 Job 2 Job 3 Job 4 Total Max

Contractor 1 0 0 0 0 0 <= 1Contractor 2 0 0 0 0 0 <= 2Contractor 3 0 0 0 0 0 <= 2

Total 0 0 0 0= = = =

Required 1 1 1 1

Total cost ($1000s) 0=SUMPRODUCT(B2:E4,B8:E10)

=SUM(E8:E10)

=SUM(B8:E8)




Notice the very large values in cells B4 and E3. These specific values (10,000) aren’t important; the main thing is to assign these particular contractor-job combinations costs so large that they will never be in any optimal solution.



Optimal Solution

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A B C D E F G HJob 1 Job 2 Job 3 Job 4

Contractor 1 50 46 42 40Contractor 2 51 48 44 10000Contractor 3 10000 47 45 45

Assignment of contractors to jobsJob 1 Job 2 Job 3 Job 4 Total Max

Contractor 1 0 0 0 1 1 <= 1Contractor 2 1 0 1 0 2 <= 2Contractor 3 0 1 0 0 1 <= 2

Total 1 1 1 1= = = =

Required 1 1 1 1

Total cost ($1000s) 182



Conclusions

The optimal solution is to award Job 4 to Contractor 1, Jobs 1 and 3 to Contractor 2, and Job 2 to Contractor 3. The total cost is $182,000.




1. What is the “cost” of restricting Contractor 1 to only one job?

2. How much more can Contractor 1 bid for Job 4 and still get the job?



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A B C D E F G HMicrosoft Excel 9.0 Sensitivity ReportWorksheet: [03b-02-bids.xls]Sheet1Report Created: 12/30/01 5:11:36 PM

Adjustable Cells

Final Reduced Objective Allowable AllowableCell Name Value Cost Coefficient Increase Decrease

$B$8 Contractor 1 Job 1 0 1 50 1E+30 1$C$8 Contractor 1 Job 2 0 1 46 1E+30 1$D$8 Contractor 1 Job 3 0 0 42 1 3$E$8 Contractor 1 Job 4 1 0 40 3 1E+30$B$9 Contractor 2 Job 1 1 0 51 1 1E+30$C$9 Contractor 2 Job 2 0 1 48 1E+30 1$D$9 Contractor 2 Job 3 1 0 44 1 1$E$9 Contractor 2 Job 4 0 9958 10000 1E+30 9958$B$10 Contractor 3 Job 1 0 9949 10000 1E+30 9949$C$10 Contractor 3 Job 2 1 0 47 1 1E+30$D$10 Contractor 3 Job 3 0 1 45 1E+30 1$E$10 Contractor 3 Job 4 0 3 45 1E+30 3

Constraints

Final Shadow Constraint Allowable AllowableCell Name Value Price R.H. Side Increase Decrease

$B$11 Total Job 1 1 51 1 0 1$C$11 Total Job 2 1 47 1 1 1$D$11 Total Job 3 1 44 1 0 1$E$11 Total Job 4 1 42 1 0 1$F$8 Contractor 1 Total 1 -2 1 1 0$F$9 Contractor 2 Total 2 0 2 1E+30 0$F$10 Contractor 3 Total 1 0 2 1E+30 1



Conclusions

The sensitivity report indicates a shadow price of –2 (cell E29).

(Allowing Contractor 1 to perform one additional job would reduce the total cost by 2,000.)

The allowable increase in the bid for Job 4 by Contractor 1 is 3. (He could have bid any amount up to $43,000 and still have won that job.)



Con. 3Con. 2Con. 1

Network Representation

Job 3Job 2 Job 4Job 1



Optimal Solution

Con. 1

Job 3

Con. 2 Con. 3

Job 2 Job 4Job 1

4047

44

51



Summary

• Optimization Extensions • Multiperiod Models

– Operations Planning: Sailboats• Network Flow Models

– Transportation Model: Beer Distribution

– Assignment Model: Contract Bidding

Optimization II. © The McGraw-Hill Companies, Inc., 2004 Operations Management -- Prof. Juran2...

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