1. Lead time - The time from when a customer places an order to when it is delivered. Reducing lead time is important in make-to-order. 2. Work in progress (WIP) - The amount of unfinished orders at each stage of production. Lower WIP allows for more flexibility. 3. On-time delivery - The percentage of orders delivered by the promised date. Higher on-time rates increase customer satisfaction. 4. Process bottlenecks - Any steps that slow the overall flow of orders through production. Eliminating bottlenecks can speed up lead times. 5.

1. Operations Management
Examples

2. Agenda
• Takira Motors: Creating Assembly and
Process Chart
• Grouping Activities
• Bakery Example

3. Predecessor Components
Sequencing the operations

5. Assembly
chart

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

7. Assignment-2 Q2
Cycle Time and Deciding on stages
0.1 min. 1.0 min. A Simple Precedence
a Diagram
b
c d e
0.7 min. 0.5 min. 0.2 min.
Arrange tasks shown in the above Figure into three workstations.
Use a cycle time of 1.0 minute
Assign tasks in order of the most number of followers

8. Raw Pack Finished
Bread Making WIP
Material Cycle Time: Goods
Cycle Time:
3/4Hr /100
1 Hr/100 loaves
loaves
packaging operation will be idle for quarter-hour periods (15 mins)
Bread Making
Cycle Time: The packaging operation is
1 Hr/100 loaves now the bottleneck
Raw WIP Pack Finished
Material Cycle Time: Goods
3/4Hr /100
loaves
Bread Making
Cycle Time: If we operated the packaging operation
1 Hr/100 loaves 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).

9. 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?

10. • 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)

11. 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.

12. 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

13. During Period End of Period
IN Out To Serve Tables Wait Wait Time
11:30-11:45 15 0
11:45-12:00 35 0
12:00-12:15 30 15
12:15-12:30 15 20
12:30-12:45 10 20
12:45-1:00 5 20
1:00-1:15 0
1:15-1:30 0
Total Served
Formula Data
The first 15 tables empty after 30 mins.
After that the restaurant can serve 20 tables every 15 minuets.

14. During Period End of Period
IN Out To Serve Tables Wait
11:30-11:45 15 0 15 15 0
11:45-12:00 35 0 50 40 10
12:00-12:15 30 15 65 40 25
12:15-12:30 15 20 60 40 20
12:30-12:45 10 20
12:45-1:00 5 20
1:00-1:15 0
1:15-1:30 0
Total Served
Formula Data
. 12:00, we will have to keep 10 parties waiting: 15+35-40
At
Between 12:00 – 12:15, we have another 30 parties coming in and 15
leaving. So we will have 15+35+30-40 -15 = 25 waiting
12:15-12:30 we have 15 parties coming in and 20 leaving, so we will have
15+35+30+15-40-15-20 = 20 waiting

15. During Period End of Period
IN Out To Serve Tables Wait
11:30-11:45 15 0 15 15 0
11:45-12:00 35 0 50 40 10
12:00-12:15 30 15 65 40 25
12:15-12:30 15 20 60 40 20
12:30-12:45 10 20 50 40 10
12:45-1:00 5 20 35 35 0
1:00-1:15 0 20 15 15 0
1:15-1:30 0 15 0 0 0
Total Served 110
20 per 15 Table+IN- IF((To serve<40),To To serve-
Formula Data mins OUT serve,40) Table
.

16. During Period End of Period
IN Out To Serve Tables Wait Wait Time
11:30-11:45 15 0 15 15 0 0
11:45-12:00 35 0 50 40 10 7.5
12:00-12:15 30 15 65 40 25 18.75
12:15-12:30 15 20 60 40 20 15
12:30-12:45 10 20 50 40 10 7.5
12:45-1:00 5 20 35 35 0 0
1:00-1:15 0 20 15 15 0 0
1:15-1:30 0 15 0 0 0 0
Total Served 110
20 per 15 Table+IN- IF((To serve<40),To To serve-
Formula Data mins OUT serve,40) Table Wait*15/20
Expected wait time = customers waiting x 0.75 minute
Double 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?

17. 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

18. No of
customers Av Time
8-9 AM 2,000 45
9-10 AM 4000 30
10-11 AM 6000 30
11-12 noon 5000 30
12-1 PM 4000 30
1-2 PM 3500 30
2-3 PM 3000 45
3-4 PM 3000 45
4-5 PM 3000 45
5-6 PM 4000 45
6-7 PM 3000 45
7-8 PM 1500 45
How do we estimate number of passengers
carried by a bus per trip? Per hour?

19. No of Passenger Min No of Max No of
customers Av Time Hours Buses Buses
8-9 AM 2,000 45 1500 18.75 30
9-10 AM 4000 30 2000 25 40
10-11 AM 6000 30 3000 37.5 60
11-12 noon 5000 30 2500 31.25 50
12-1 PM 4000 30 2000 25 40
1-2 PM 3500 30 1750 21.875 35
2-3 PM 3000 45 2250 28.125 45
3-4 PM 3000 45 2250 28.125 45
4-5 PM 3000 45 2250 28.125 45
5-6 PM 4000 45 3000 37.5 60
6-7 PM 3000 45 2250 28.125 45
7-8 PM 1500 45 1125 14.0625 22.5

20. Process Throughput Time Reduction
• Perform activities in parallel
• Change the sequence of activities
• Reduce interruptions

21. 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.

22. Process Performance Metrics

23. 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.)

24. 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.

26. 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.

27. MAC: Make to Stock Customer
Orders
RAW Finished
Material Cook Assemble Goods
Deliver
Burger King:
Hybrid Assemble
RAW
Material WIP Customer Sta
Cook Orders Deliver
nd
ard
Wendy’s: Make to Customer Finished
Order Orders Goods
Assemble
RAW
Material
Cook Assemble Deliver
Chili

30. Make to Order
Process Analysis Issues and Metrics?
• ????
• Assignment