Operations Management

Examples

Agenda

• Takira Motors: Creating Assembly and Process Chart

• Grouping Activities• Bakery Example

Predecessor ComponentsSequencing the operations

Assembly chart

Cycle time is the time taken by the longest activity (here it is 7 min). Throughput time is 46 min.

Assignment-2 Q2 Cycle Time and Deciding on stages

A Simple Precedence Diagrama b

c d e

0.1 min.

0.7 min.

1.0 min.

0.5 min. 0.2 min.

Arrange tasks shown in the above Figure into three workstations.Use a cycle time of 1.0 minuteAssign tasks in order of the most number of followers

Raw Material Bread Making

Cycle Time:1 Hr/100 loaves

WIPFinished Goods

PackCycle Time:3/4Hr /100

loaves

Raw Material

Bread MakingCycle Time:

1 Hr/100 loaves

Bread MakingCycle Time:

1 Hr/100 loaves

WIP PackCycle Time:3/4Hr /100

loaves

Finished Goods

packaging operation will be idle for quarter-hour periods (15 mins)

The packaging operation is now the bottleneck

If we operated the packaging operation for three eight-hour shifts, and bread making for two shifts each day, then the daily capacity of each would be identical at 3,200 loaves a day (800 loafs x 4 shifts).

Packaging 3 shifts Bread making 2 shifts: work-in-process inventory

• If both bread-making operations start at the same time, at the end of the first hour, then the first 100 loaves move into packaging while the second 100 loaves wait—work-in-process inventory.

• The waiting time for each 100-loaf batch increases until the baking is done at the end of the second shift.

• What is the time that the bread is sitting in work-in-process?

• Average wip during first two shifts, inventory builds from 0 to 1,200 loaves (1,600 x .75). (Or 800 loaves x 2 shifts x .75 pkg.) = 1200/2

• Average wip during third shift is again 1200/2• The overall average WIP over the 24-hour period is

simply 600 loaves of bread.• Packaging process in batches.

1) 0.75 hour per 100 loaves 2) Throughput rate of 133.3 loaves/hour (100/0.75)• Little’s Law calculates the average time that loaves

are in work-in-process is 4.5 hours (600 loaves/133.33 loaves/hour)

Our Restaurant• Assume that we have designed our buffet so customers take an

average of 30 minutes to get their food and eat.• Assume the restaurant has 40 tables. Each table can

accommodate four people.• The Cycle Time for the restaurant, when operating at capacity, is

0.75 minute (30 minutes/table ÷ 40 tables).• The restaurant could handle 80 customer parties per hour (60

minutes ÷ 0.75 minute/party) is the capacity or customer parties per hour.

• Assume that our customers eat in groups (or customer parties) of two or three to a table.

• How many customers can the restaurant serve if the average customer party is 2.5?

• During lunch time, Customers arrive as per the schedule given.

Solution• If the average customer party is 2.5 individuals, then

the average seat utilization is 62.5 percent (2.5 seats/party ÷ 4 seats per table) when the restaurant is operating at capacity.

• The Cycle Time for the restaurant, when operating at capacity, is 0.75 minute (30 minutes/table ÷ 40 tables).

• The restaurant could handle 80 customer parties per hour (60 minutes ÷ 0.75 minute/party) is the capacity or customer parties per hour.

• 20 tables empty during each 15-minute interval

During Period End of PeriodIN Out To Serve Tables Wait Wait Time

11:30-11:45 15 011:45-12:00 35 012:00-12:15 30 1512:15-12:30 15 2012:30-12:45 10 2012:45-1:00 5 201:00-1:15 01:15-1:30 0Total ServedFormula Data

The first 15 tables empty after 30 mins. After that the restaurant can serve 20 tables every 15 minuets.

During Period End of PeriodIN Out To Serve Tables Wait

11:30-11:45 15 0 15 15 011:45-12:00 35 0 50 40 1012:00-12:15 30 15 65 40 2512:15-12:30 15 20 60 40 2012:30-12:45 10 2012:45-1:00 5 201:00-1:15 01:15-1:30 0Total ServedFormula Data

.At 12:00, we will have to keep 10 parties waiting: 15+35-40Between 12:00 – 12:15, we have another 30 parties coming in and 15 leaving. So we will have 15+35+30-40 -15 = 25 waiting12:15-12:30 we have 15 parties coming in and 20 leaving, so we will have 15+35+30+15-40-15-20 = 20 waiting

During Period End of PeriodIN Out To Serve Tables Wait

11:30-11:45 15 0 15 15 011:45-12:00 35 0 50 40 1012:00-12:15 30 15 65 40 2512:15-12:30 15 20 60 40 2012:30-12:45 10 20 50 40 1012:45-1:00 5 20 35 35 01:00-1:15 0 20 15 15 01:15-1:30 0 15 0 0 0Total Served 110

Formula Data20 per 15 mins

Table+IN-OUT

IF((To serve<40),To serve,40)

To serve- Table

.

During Period End of PeriodIN Out To Serve Tables Wait Wait Time

11:30-11:45 15 0 15 15 0 011:45-12:00 35 0 50 40 10 7.512:00-12:15 30 15 65 40 25 18.7512:15-12:30 15 20 60 40 20 1512:30-12:45 10 20 50 40 10 7.512:45-1:00 5 20 35 35 0 01:00-1:15 0 20 15 15 0 01:15-1:30 0 15 0 0 0 0Total Served 110

Formula Data20 per 15 mins

Table+IN-OUT

IF((To serve<40),To serve,40)

To serve- Table Wait*15/20

Expected wait time = customers waiting x 0.75 minuteDouble up parties at the tables, thus getting a higher seat utilization. Might be the easiest solution to the problem. If 25 out of the 40 tables were doubled up, our problem would be solved… (40+20)*2.5 > 40*4.. So?

Transit Bus Operation • Logistics refers to the movement of things

such as materials, people, or finished goods.• A single bus takes exactly two hours to

traverse the route during peak traffic. • Wait time for the customer: Maximum is 2

Minimum is 0, Average would be one hour.• Bus capacity: seating 50 Standing 30

No of customers Av Time

8-9 AM 2,000 459-10 AM 4000 3010-11 AM 6000 3011-12 noon 5000 3012-1 PM 4000 301-2 PM 3500 302-3 PM 3000 453-4 PM 3000 454-5 PM 3000 455-6 PM 4000 456-7 PM 3000 457-8 PM 1500 45

How do we estimate number of passengers carried by a bus per trip? Per hour?

No of customers Av Time

Passenger Hours

Min No of Buses

Max No of Buses

8-9 AM 2,000 45 1500 18.75 309-10 AM 4000 30 2000 25 4010-11 AM 6000 30 3000 37.5 6011-12 noon 5000 30 2500 31.25 5012-1 PM 4000 30 2000 25 401-2 PM 3500 30 1750 21.875 352-3 PM 3000 45 2250 28.125 453-4 PM 3000 45 2250 28.125 454-5 PM 3000 45 2250 28.125 455-6 PM 4000 45 3000 37.5 606-7 PM 3000 45 2250 28.125 457-8 PM 1500 45 1125 14.0625 22.5

Process Throughput Time Reduction

• Perform activities in parallel

• Change the sequence of activities

• Reduce interruptions

Productivity and Efficiency

• Productivity is output to input. It may be total factor productivity or partial factor productivity. – The ratio is generally taken in monitory terms.

• Efficiency is the ratio of actual output of a process relative to some standard or best level of output.

Process Performance Metrics

Little’s Law

• Little’s Law—states a mathematical relationship between throughput rate, throughput time, and the amount of work-in-process inventory.

• Little’s Law estimates the time that an item will spend in work-in-process inventory, which can be useful for calculating the total throughput time for a process.

• Little’s Law = Throughput time = Work-in-process Throughput rate

• (Throughput rate is the output rate that the process is expected to produce over a period of time.)

Example• if the assembly line has six stations with one unit of work-

in-process at each station, and the throughput rate is 2 units per minute (60 seconds/30 seconds per unit),

• then the throughput time is three minutes (6 units/2 units per minute).

• This formula holds for any process that is operating at a steady rate.

• Steady rate—means that work is entering and exiting the system at the same rate over the time period being analyzed.

• For example, our assembly line has 120 units entering and 120 units exiting the process each hour. Not 150 units entering the system each hour but only 120 units exiting.

The McDonaldization of Society (1993)George Ritzer

• Efficiency – the optimal method for accomplishing a task. In this context, Ritzer has a very specific meaning of "efficiency". Here, the optimal method equates to the fastest method to get from point A to point B.

• Calculability – objective should be quantifiable (e.g., sales) rather than subjective (e.g., taste). McDonaldization developed the notion that quantity equals quality, and that a large amount of product delivered to the customer in a short amount of time is the same as a high quality product.

• Predictability – standardized and uniform services. “Their tasks are highly repetitive, highly routine, and predictable”.

• Control – standardized and uniform employees, replacement of human by non-human technologies.

RAW Material Cook Assemble

CustomerOrders

Finished Goods

Deliver

RAW Material Cook

Assemble

CustomerOrders

Assemble

Deliver

Finished Goods

WIP Standard

RAW Material

CustomerOrders

Cook

Chili

Assemble Deliver

Wendy’s: Make to Order

Burger King: Hybrid

MAC: Make to Stock

Make to OrderProcess Analysis Issues and Metrics?

• ????• Assignment