The University of Melbourne
Melbourne Business School
Operations (PT)
Assoc. Prof. Kannan Sethuraman
Mock Final Exam
Answer Key
100 points
Exam Duration: 3.5 hours
(This time can be used for reading and writing).
Instructions
1. This is an open book, open notes test.
2. You may use your calculators but mobile phones are strictly not permitted for use during the exam.
Laptop computers are permitted. Computers should not be used for browsing through the
internet or communicating with one another. You can use the laptops to access the notes and
readings distributed and for performing simple computations in lieu of a calculator. However, you
must show all your work in the answers to gain credit.
3. The exam will consist of two parts: Part A and Part B. Part A will be quantitative in nature where
you will need to solve problems and the answers can be either handwritten or typed but if you
plan to type it, you will be expected to type out all formulas appropriately. However, part B will
require you to write an extended response to a case question or other qualitative explorations and
you would be expected to fully type your answers to this part. You will be required to submit two
separate pdf documents (one for Part A and another for Part B) via Canvas Turnitin at the
conclusion of the exam.
4. Responses should be presented in a concise, logical, and legible fashion. Developmental work
leading to each particular result must also be shown in a complete, logical, legible manner for credit
to be given. If your handwriting or presentation can’t be followed easily, credit will be lost. As a
general rule, you must show supporting work in words, in numbers, or both depending on
the type of question. You must show your supporting work without which credit will not be
given. If you think the information provided in the question is inadequate, please make reasonable
assumptions for answering the question(s). You need to clearly state the assumptions with a brief
justification.
5. The performance in this test will account for 40% of the total marks for this subject.
1
1
Part A
(Part A has 5 questions and you can either type up your responses or write them
in plain paper. If written by hand, you will need to scan the sheets in order and
upload them as a pdf. Please ensure each page has your student ID written on it
and number the pages as well).
2
Question 1
Process Capacity at Metal Box Inc.
Time allocated: 45 minutes
(23 points)
Metal Box Inc. manufactures metal window boxes in 5 basic colors in a small plant in Reservoir,
a northern suburb in Melbourne. The manufacturing process consists of 3 basic operations:
stamping, painting, and assembly, as shown below.
Each window box is made up of three pieces: a base (one unit of part A) and two sides (two units
of part Bs).
Stamping Operation: The firm currently owns a single stamping machine. The required parts
for a window box are fabricated at the stamping machine which requires a setup time of 120
minutes whenever it switches between the two part types. Once the machine is setup, the run
time per unit for part A is 1 minute while the run time per unit for part B is 30 seconds. Currently,
the stamping machine rotates its production between a batch of 360 units for part A and a batch
of 720 units for part B. Completed parts move from the stamping machine onto the painting
station only after the entire batch (comprising of 360 units of part A and 720 units of part B)
finishes fabrication at the stamping machine.
Painting Operation: At the painting station, parts are painted by a robot in 1 of 5 colors. The
robot takes 30 seconds to paint a single unit of part A and 10 seconds to paint a single unit of
part B. The robot can easily switch between painting the two parts (i.e., A and B) without
incurring any setup time. However, a switch in color does require 20 minutes for setup. Once a
piece finishes painting, it must wait 120 minutes to dry before moving to assembly. Assume that
there is no space limitation at the buffer between paint and assembly. The painting robot is
currently programmed to change color every time it finishes 360 component sets (i.e., 360 of
part A and 720 of part B) and all parts move to assembly as a batch.
Assembly Operation: At assembly, parts of the same color are assembled manually to form the
finished product. Each unit of the finished product requires a base unit (part A) and two sides
(two part Bs), as well as a number of small purchased components for its assembly. Each product
requires 27 minutes of labor time to assemble. Currently, a single worker completes all the
required tasks for a given assembly of a finished product. There are 12 assembly workers and
they work on a given batch in parallel.
3
The factory runs one 8 hour shift per day, five days per week. There are currently 15 workers:
12 assembly workers, 2 operators for the stamping machine, and 1 operator for the robot. There
is sufficient demand to sell every box the system can make.
1a.
Determine the capacity at each of the three operations: stamping, painting and assembly.
Which operation represents the bottleneck for the current process? Calculate the
maximum daily production rate for the entire system in terms of the # of units of finished
product produced per day.
(6 points)
1b.
Suppose we begin the day with no work in process. What will be the flow time for the
first batch of completed units? What other factors might affect the flow time for the
subsequent batches as the day progresses?
(4 points)
1c.
If you have the option to change the batch size, would you consider increasing the batch
size at the stamping operation? Explain qualitatively why you may consider increasing the
batch size for the stamping operation. What impact would doubling the batch sizes at the
stamping operation have on the total daily capacity? Show calculations to justify your
answer. Explain qualitatively what other, possibly negative, effects this doubling may result
in.
(5 points)
1d.
Management is considering investing in a setup reduction program for the stamping
machine. Suppose the setup time could be reduced to 20 minutes, with some additional
investment cost. What would happen to the flow rate if the batch sizes remained the
same as in part (a) (i.e., 360 for part A and 720 for part B)?
(4 points)
1e.
Evaluate the proposal in 1d carefully and if altering the batch size option exists, what
would do you do now and why? Explain through calculations the optimal batch size should
be under this scenario.
(4 points)
4
Answer Key
(i) In order to identify the bottleneck, we must calculate the capacity for each of the three
process steps. When setup times are involved, it is useful to perform these calculations in terms
of a production batch, rather than individual items.
A production batch consists of 360 units of part A and 720 units of part B. This will enable you
to produce 360 units of finished products.
Stamping:
Time to complete a production batch
= 120 minutes/batch setup for part A + 360 pieces/batch x (1 minutes/unit of A)
+ 120 minute/batch setup for part B + 720 pieces/batch x (0.5 minute/unit of B)
= 960 minutes
= 2 days
Therefore, the daily capacity of the stamping process = 360 / 2 days = 180 units/day.
Painting:
Time to complete a production batch at the paint operation
= 20 minute/batch setup + 360 pieces x (0.5 minute/unit of A) + 720 pieces ( 1/6
minute/unit of B)
= 320 minutes = 0.667 days.
Therefore, the paint capacity = 360 product units/0.667 days = 540 product units/day
Assembly:
Since assembly is labor paced, we must compare the labor content needed to make one finished
unit against the total number of labour hours available per day.
Labour content per day = 27 minute /product
Total labour available per day = 12 workers x 60 minutes/hr x 8 hours/day
Daily capacity = 5760 minutes/ (27 minute/finished unit)
5
= 213 units of finished product/day
Hence the bottleneck is at the stamping operation as it has the least capacity of the three
operations. Therefore, the total system can’t produce more than 180 finished units/day in steady
state.
It is sometimes suggested that we could increase the capacity of stamping, and therefore the
entire process, if we reversed the sequence of A and B in successive runs, e.g., in first batch
produce in order A, B, in the second batch B, A, then A, B, etc. This is an interesting idea and
saves one setup per batch, but one might argue that really we’ve just doubled the effective batch
size.
We don’t need to consider the 2 hours dry time for paint unless there is limited space to hold
parts while they dry. Since the problem doesn’t give us a space constraint, it is reasonable to
assume that drying space is not a binding constraint. In any event, even if parts had to reside in
the paint station while they dried, this would only increase paint time per batch from 320 minutes
to 440 minutes, and paint would still have the lowest time per batch of the three steps.
(ii)
The total manufacturing lead time for a batch equals the total of the times for stamping,
painting, drying and assembly:
= 960 minutes + 320 minutes + 120 minutes + 810 minutes = 2210 minutes.
The 810 minute assembly time assumes that all 12 assembly workers work on a given batch in
parallel, each assembling 30 product units. If a single worker was required to assemble a complete
batch then this time would grow from 810 minutes to 27 x 360 = 9720 minutes.
If we could setup paint in the last 20 minutes of the stamping run, then the manufacturing lead
time would be reduced by 20 minutes.
As the day progresses, the manufacturing lead time for other batches would depend upon the
colours that are handled in any given day as well as other unexpected delays in any of the
machines. Similarly, the lunch breaks and other breaks for the workers will also play a role. Also,
machine breakdowns, employee absenteeism and other disruptions due to unexpected events
will have an impact on the actual manufacturing lead for a batch.
(iii)
If the batch size is doubled at the stamping process, we would expect capacity at stamping
stage to increase. Since stamping is also the global bottleneck, we would expect the
capacity of the entire system to increase. We can recalculate the capacity of stamping
operation with the new batch size to determine the actual effect of this change.
The revised time to complete a batch at stamping = 120 minutes/batch setup for part A + 720
pieces/batch x (1 minutes/unit of A) + 120 minute/batch setup for part B + 1440 pieces/batch x
(0.5 minute/unit of B)
6
= 1680 minutes = 3.5 days
Therefore, daily capacity = 720 / 3.5 days = 205.7 units/day.
The stamping still continues to remain as the global bottleneck for the system.
While increasing the batch size does increase the daily capacity of the bottleneck station, it may
have other negative effects on the process. Obvious effects include increased WIP and therefore
increasing the flow time and decreasing the flexibility to respond to changes in demand.
(iv)
If the setup time is reduced to 20 minutes at stamping, we would expect capacity at
stamping to increase. We can calculate the change precisely as we did in part (iii).
The revised time to complete a batch at stamping = 20 minutes/batch setup for part A + 360
pieces/batch x (1 minutes/unit of A) + 20 minute/batch setup for part B + 720 pieces/batch x
(0.5 minute/unit of B)
= 760 minutes = 1.583 days.
Therefore, daily capacity of the stamping operation = 360/1.583 days = 227.4 units/day.
Note that the capacity of stamping is now greater than the capacity of assembly stage. Assembly
becomes the new bottleneck and the capacity for the entire process = 213 units/day.
v) If the batch sizes were allowed to change, it would now be beneficial to reduce the batch sizes
at stamping. However, we would need to be careful not to reduce the batch size to the point
where stamping becomes a bottleneck again. Batch size reduction is attractive because it allows
increased responsiveness and reduced WIP. It gives you better chance to make supply meet
demand better without unnecessary inventory piling up between the two steps.
If you equate the capacity of assembly to the capacity of the stamping, you can find the optimal
batch size.
12/27 = B/(40+2B)
480 + 24 B = 27 B Which implies B = optimal batch size = 480/3 = 160.
At 160, both stamping and assembly will have exactly the same capacity.
7
Question 2: Western Pennsylvania Milk Company
(10 points)
Estimated time: 20 minutes
The Western Pennsylvania Milk Company is producing milk at a fixed rate of 5000
gallons/hour The company’s clients request 100,000 gallons of milk over the course of one
day. This demand is spread out uniformly from 8 a.m. to 6 p.m. If there is no milk available,
clients will wait until enough milk is produced to satisfy their requests.
The company starts producing at 8 a.m. with 25,000 gallons in finished goods inventory. At
the end of the day, after all demand has been fulfilled, the plant keeps on producing until the
finished goods inventory has been restored to 25,000 gallons.
When answering the following questions, treat trucks/milk as a continuous flow process.
Begin by drawing a graph indicating how much milk is in inventory and how much milk is
“back-ordered” over the course of the day.
2a) Draw an inventory buildup diagram to show how the milk inventory changes during the
course of the day. Mark on your diagram, the inventory buildup rate as well as how the
inventory is depleted at various points in time. Also show on the diagram, the time of the
day when the clients start waiting for their requests to be filled as well when they stop waiting.
(5 marks)
2b) Assume that the milk is picked up in trucks that hold 1,250 gallons each. What is the
maximum number of trucks that are waiting?
(2 marks)
2c) Assume that the plant is charged $50 per hour per waiting truck. What are the total
waiting time charges per day?
(3 marks)
8
25000 gallons
25000 gallons
5000 gallons/hour
8 am
1 pm
6 pm
11 pm
4 am
5000 gallons/hour
25000 gallons backlog
We start the day with 25,000 gallons of milk in inventory. From 8am onwards, we produce
5,000 gallons, yet we ship 10,000 gallons. Thus inventory is depleted at a rate of 5000 gallons
per hour, which leaves us without milk after 5 hours (at 1pm). From then onwards, clients
will have to wait. This situation gets worse and worse and by 6pm (last client arrives), we are
short 25,000 gallons.
2a.
The clients will have to start waiting from 1pm.
Clients will stop waiting when we have worked off our 25,000 gallon backlog that
we are facing at 6pm. Since we are doing this at a rate of 5,000 gallon per hour,
clients will stop waiting at 11pm (after 5 more hours).
2b.
At 6pm, we have a backlog of 25,000 gallons, which is equivalent to 20 trucks.
2c
The waiting time is the area in the triangle
- width: beginning of waiting (1pm) to end of waiting (11pm)=10 hours
- height: maximum number of trucks waiting: 20 (see part c above)
Hence, we can compute the area in the triangle as:
0.5*10hours*20trucks=100 truck* hours
The cost for this waiting is 50$/truck* hour * 100 truck hours=5000$
9
Question 3: Home and Garden
Estimated time: 25 minutes
(12 marks)
The Home and Garden (HG) chain of superstores imports decorative planters from Italy. Weekly
demand for planters average 1500 units with a standard deviation of 800. Each planter costs $10.
HG incurs a holding cost of 25% per year to carry inventory. Each order shipped from Italy incurs
a fixed transportation and delivery cost of $10,000 irrespective of the number of planters
ordered. Consider 52 weeks in the year.
3a) Determine the optimal order quantity of planters for HG.
(2 marks)
3b) If the delivery lead time from Italy is 4 weeks and HG wants to ensure that within 100 order
cycles, stock-out does not occur in more than 20 cycles, determine the safety stock HG should
carry?
(3 marks)
3c) Recently, customers of HG have started complaining about the frequent stock-out of
planters. In response, the manager at HG is considering increasing the service level but does not
know how much the increase may cost in extra inventory. Determine how the safety inventory
requirement will change when he alters the service level to 85%, 90%, 95% and 99%? Draw
inferences plotting the relationship between service level and safety inventory requirements.
(3 marks)
3d) Fastship is a new shipping company that promises to reduce the delivery lead time for
planters from 4 weeks to 1 week using a faster ship and expedited customs clearance. Using
Fastship will add $0.2 to the cost of each planter. Assume the service level desired to be the same
as the one you used in part (b) of this question. Should HG go with Fastship? Why or why not?
Quantify the impact of the change.
(4 marks)
10
Answer key for question 3
Home and Garden
Parameters given:
o Weekly demand for planters d is normally distributed
▪ with mean 𝜇𝑑 = 1500 and standard deviation 𝜎𝑑 =800.
o
o
o
o
Average annual demand for planters = R
Shipping cost/order
=K
Cost of a planter
Holding cost
=H
= 1500 x 52 = 78,000 units
= $10,000 /order
=c
= $10/planter
= 25% of the cost of the planter
= 0.25 x $10 = $2.50 /unit/year
a) Optimal order quantity can be determined using the EOQ formula:
Q* =
2RK / H =
2  78,00010,000/ 2.5 = 24,979 units
b) Determination of safety stock
•
•
•
•
•
Desired service level = 80%
Weekly demand distribution is given.
Lead time = LTt = 4 weeks.
Standard deviation of demand during lead time
=𝜎𝐿𝑇𝐷 = 𝜎𝑑 × √𝐿𝑇𝑡 = 4  800 = 1600
Z corresponding to 80% service level = NORMSINV(0.8) = 0.8416
Safety stock needed for this level of service
= NORMSINV(0.8) x 𝜎𝐿𝑇𝐷 = 0.8416 x 1600 = 1347 units.
c) Taking different service level requirement, we can recompute safety stock needed.
Service level
80%
85%
90%
Safety stock
1346.5
1658.3
2050.5
95%
2631.8
99%
3722.2
11
You can see that as the desired service level increases, there is a disproportionate increase
in the safety stock needed. There is diminishing value to the investment as you aspire for
100% service level.
d) In this, the cost of the planters increases to $10.2 and hence the EOQ will change and all
the associated costs will change.
Old supplier
New supplier
Annual demand
Cost/unit
Cost of goods purchased
Cost of ordering
Cost of average inventory
Cost of safety stock
78,000
$10
$780,000
$31,224.99
$31224.99
$3366.49
78,000
$10.20
$795,600.00
$31,535.69
$31,535.69
$1,716.91
Total costs
$845,816
$860,388.30
Please see the attached spreadsheet for calculations.
Since the costs have increased in the second option, even though the lead time comes down and
the safety stock requirements come down, the costs of the planters increase substantially and
hence this option is not preferred.
12
Question 4. Lindt Chocolates
Estimated time: 25 minutes
(15 points)
Government regulations mandate that Swiss chocolate bars sold in packages of ½
kilogram cannot weigh less than 500 grams, the specified weight on the package. If regulations
are violated, the company is fined. The chocolate machine at Lindt Chocolate Company fills
packages with a standard deviation of 5 grams regardless of the mean setting. To be sure that
government regulations are met with high probability, the machine operator decides to set the
mean at 515 grams.
a) While underweight chocolate bars might prompt FDA (Food and Drug Administration)
action, overweight bars certainly cost Lindt Chocolate Company in terms of higher
material costs. Therefore, quality control manager Ms. Elizabeth Taylor wants to monitor
the chocolate filling machine to ensure that its mean does not deviate from the 515 grams
to which it is set. She plans to weigh 25 randomly selected bars at regular time intervals
and plot the average weight on a chart. Following industry practice of setting 3 sigma
limits, what control limits should the operator use for the X-bar chart?
(3 points)
b) If the process is found to be in control, approximately what fraction of chocolate bars will
weigh less than 500 grams (this is the fraction that would violate FDA regulations)?
(3 points)
c) Based on the control chart limits obtained from part (a), Liz plots sample averages every
day to track whether the process is in control or not. On Monday, August 1, 2011, Ms.
Taylor found the average weight of a sample of 25 bars to be 506 grams. The company’s
legal staff is pleased that this performance is better than the FDA requirement of 500
grams. What action if any should Ms. Taylor take? Would you agree with the statement
by the legal staff? Justify your answer using appropriate calculations.
(4 Points)
d) Clearly, producing an excess average chocolate weight of 15 grams just in order to
prevent regulation fines is costly in terms of chocolate “given away for free”. Lindt
Chocolate Company management wants to reduce the average excess weight to 3 grams
while staying in line with regulation “practically always” which means 99.87% of the time.
In what sense will this require improved process technology? Give an explanation in words
as well as a specific numeric answer.
(5 points)
13
Answer Key to Lindt Chocolates
LTL = lower tolerance limit 500 gms.
In other words, the government regulations stipulates that the chocolate bars sold in packages
of ½ kg cannot weigh less than 500 gms, the specified weight on the package.
a) The mean of the process is set to be 515 gms. The standard deviation of the process is given
as 5 gms.
m x = procesmean = 515gms
s x = processstandard deviation =5gms
Ms. Elizabeth wants to monitor the chocolate filling machine to ensure that its mean does not
deviate from the intended value of 515 gms. She plans to weigh 25 randomly selected bars at
regular time intervals and plat the average weight of samples on a chart to study whether it
is in control.
n = sample size = 25
With n = 25, and  = 5 grams, we can now determine the ideal control limits on subgroup
averages of 25 samples as:
3 x
3´ 5
= 518 grams
25
n
3
3´ 5
Lower Control Limit = LCL =  x − x = 515 = 512 grams
25
n
Upper Control Limit = UCL =  x +
= 515 +
b) For the given process mean and standard deviation, we can look at what fraction will weigh
less than 500 gms.
Z LTL =
LTL - m x
sx
=
500 - 515
= -3
5
From above calculations, we can see that the FDA requirement of 500 gms is 3 standard
deviation left of the process mean. We can use NORMSDIST(-3) to find what area of the
normal curve will be to the left of LTL or 500 gms. This equals 0.135%. In other words,
0.135% of the bars produced will violate the FDA requirement of 500 gms. If you make 1000
bars, roughly 1 bar will weigh less than the FDA requirement of 500 gms.
c) Note that we are plotting sample averages and they will have a lower standard deviation
than the process mean.
14
If σx is the population mean then the sample averages would have a standard deviation = σx/√n
where n represents the sample size. Hence it would not be right to compare the sample
averages with the specification limits as the sample average being above FDA requirement of
500 gms does not guarantee the samples to be above the value of 500 gms.
In this case, Ms. Taylor has found the average weight of a sample of 25 bars to be 506 gms. If
we look at the control limits that we determined in part b of this question, they were 518
and 512 gms. The value 506 gms is clearly outside the control limits which indicates that the
mean of the process is showing some anomalous behavior and it requires some investigation.
We need to investigate and determine if there are any assignable causes for this deviation and
need to eliminate them and bring the process back to control.
d) The firm is concerned about giving too much chocolate for free. Particularly, in the way they
are currently operating, it is clear that their mean value is set at 515 gms and it means they
are giving on an average 15 gms more than what the intended weight was.
If they wish to reduce the average excess weight to 3 gms, it means that they wish to redesign
the process to have a average weight of 500+3 = 503 gms.
Now the lower tolerance limit or the FDA requirement still remains at 500 gms. If they wish
99.87% of the products produced to be in line with regulation, we have to lower the standard
deviation so that the FDA limit of 500 gms is 3 standard deviation away from the new mean
of 503 gms.
This will require a substantial lowering of the variability in their process. Currently, they have
a standard deviation of 3 gms and it will need to be lowered to 1 gm so that they can achieve
this new capability they are desiring.
15
Problem 5: Car rental company
Estimated time: 20 minutes
10 points
The airport branch of a car rental company maintains a fleet of 50 SUVs. The interarrival time
between requests for an SUV is 2.4 hours, on average, with a standard deviation of 2.4 hours.
There is no indication of a systematic arrival pattern over the course of a day. Assume that, if all
SUVs are rented, customers are willing to wait until there is an SUV available. An SUV is rented,
on average, for 3 days, with a standard deviation of 1 day.
a) What is the average number of SUVs parked in the company’s lot?
(2 points)
b) Through a marketing survey, the company has discovered that if it reduces its daily rental
price of $80 by $25, the average demand would increase to 12 rental requests per day
and the average rental duration will become 4 days. Is this price decrease warranted?
Provide an analysis to justify your recommendation.
(3 points)
c) What is the average time a customer has to wait to rent an SUV? Please use the initial
parameters rather than the information in part b)
(2 points)
d) How would the waiting time change if the company decides to limit all SUV rentals to
exactly 4 days? Assume that if such a restriction is imposed, the average interarrival time
will increase to 3 hours, with the standard deviation changing to 3 hours.
(3 points)
16
Answer Key for Car rental company
a) This is a multi-server problem and hence we will be using the generalised
Kingman equation.
• Here we approach the problem by treating the rental cars as servers.
Hence m = 50.
• Arrival Process
o Hence, a = average inter-arrival time
= 2.4 hours/customer
o It is also given that the standard deviation of the interarrival
time = 2.4 hours and hence the co-efficient of variation for
arrival process 𝐶𝑉𝑎 = 1
• Service Process
o Average rental period is given as 3 days or 72 hours
o Hence, service time = p = 72 hours/customer
o It is also given that the standard deviation of service time is
given to be = 1 day or 24 hours
1
o Hence, 𝐶𝑉𝑝 = = 0.333
3
• So, the utilization of the cars = u =
𝑝
𝑚𝑎
=
72
50×2.4
= 60%
• In other words, on an average, 60% of the cars are in use and hence
there will be 20 cars available for rent at any time.
17
Kingman Equation template calculations are shown below:
Inputs
Interarrival Time (a)
Mean
Std. Deviation
2.4000
2.4000
No. of Servers (m)
Processing Time (p)
Mean
Std. Dev.
72.0000
24.0000
50
Outputs
Coefficients of Variation
CVa = coefficient of variation of inter arrivals
CVp = coefficient of variation of processing time
1.0000
0.3333
Arrival Rate (jobs/unit time)
Processing Rate (jobs served per unit time)
Processing rate factor
Utilization ρ
0.4167
0.6944
1.4400
0.6000
Utilization factor
Variability factor
Expected wait time = T q
Using Little's Law
Flow rate=Min (arrival rate,processing rate) =R
Inventory in queue =Iq = (1/a)*Tq
Total Flow time =T=Tq+p
Inventory in service =Ip=m xρ
Inventory in system = I=Iq+Ip
0.0239
0.5556
0.0192 hours
0.4167
0.0080
72.0192
30.0000
30.0080
b) We assume that the interarrival times for the new situation continues to
follow an exponential distribution.
If the average demand is increased to 12 rentals per day, then the
interarrival time = 2 hours/customer
Standard deviation of the interarrival time will also equal 2 hours as
opposed to 2.4 hours earlier.
If the average rental duration increases to 4 days, then p = 96 hours.
If the standard deviation of rental duration continues to be 1 day as before,
𝑝
96
then the utilization of the cars = u =
=
= 96%
𝑚𝑎
18
50×2
This means that 48 cars are rented on average as opposed to 30 cars earlier. The
initial average revenue per day = $80 x 30 = $2400 and with the proposed changes,
the average revenue per day = $55 x 48 = $2640. Therefore, the company is better
off by introducing the daily rental price of $55.
The revised Kingman Equation template would like the table below:
Inputs
Interarrival Time (a)
Mean
Std. Deviation
2.0000
2.0000
No. of Servers (m)
Processing Time (p)
Mean
Std. Dev.
96.0000
24.0000
50
Outputs
Coefficients of Variation
CVa = coefficient of variation of inter arrivals
CVp = coefficient of variation of processing time
1.0000
0.2500
Arrival Rate (jobs/unit time)
Processing Rate (jobs served per unit time)
Processing rate factor
Utilization ρ
0.5000
0.5208
1.9200
0.9600
Utilization factor
Variability factor
Expected wait time = T q
Using Little's Law
Flow rate=Min (arrival rate,processing rate) =R
Inventory in queue =Iq = (1/a)*Tq
Total Flow time =T=Tq+p
Inventory in service =Ip=m xρ
Inventory in system = I=Iq+Ip
17.2432
0.5313
17.5880
19
0.5000
8.7940
113.5880
48.0000
56.7940
c) Using the wait time formula and looking at the table given for the initial
set of values, we can see that the average wait time = 0.192 hours or 1.15
minutes.
d) The new service time = 4 days = 96 hours and the standard deviation of
service time = 0 as they expect all customers to rent for exactly 4 days.
With this restriction, the interarrival time = 3 hours and the standard
deviation of interarrival time = 3 hours.
You can see that the expected wait time has come down to 0.046 hours.
Using these values, the revised Kingman equation template values would look
as shown below:
Inputs
Interarrival Time (a)
Mean
Std. Deviation
3.0000
3.0000
No. of Servers (m)
Processing Time (p)
Mean
Std. Dev.
96.0000
0.0000
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Outputs
Coefficients of Variation
CVa = coefficient of variation of inter arrivals
CVp = coefficient of variation of processing time
1.0000
0.0000
Arrival Rate (jobs/unit time)
Processing Rate (jobs served per unit time)
Processing rate factor
Utilization ρ
0.3333
0.5208
1.9200
0.6400
Utilization factor
Variability factor
Expected wait time = T q
Using Little's Law
Flow rate=Min (arrival rate,processing rate) =R
Inventory in queue =Iq = (1/a)*Tq
Total Flow time =T=Tq+p
Inventory in service =Ip=m xρ
Inventory in system = I=Iq+Ip
0.0479
0.5000
0.0460
20
0.3333
0.0153
96.0460
32.0000
32.0153
Part B
(Part B has 2 questions and you must type up your responses and submit the
response as a pdf. Please ensure each page has your student ID written on it and
number the pages as well).
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Question 6: Toyota Group and the Aisin Fire
Estimated Time: 60 minutes
20 points
To the Rescue: Toyota's Fast Rebound After Fire at Supplier Shows Why It Is Tough --- Its Affiliates, Going
All Out, Built an Unfamiliar Part Within a Matter of Days --- Like an Amish Barn-Raising
By Valerie Reitman, Staff Reporter of The Wall Street Journal, 2049 words, 8 May 1997
KARIYA, Japan -- No one knows what sparked the fire that roared through Aisin Seiki Co.'s
Factory No. 1 here before dawn on Saturday, Feb. 1, leveling the huge auto-parts plant. But one
thing is clear: The crisis-control efforts that followed it dramatically illustrate one reason Toyota
Motor Corp. is among the world's most admired and feared manufacturers.
The fire incinerated the main source of a crucial brake valve that Toyota buys from Aisin and
uses in most of its cars. Most Toyota plants kept only a four-hour supply of the $5 valve; without
it, Toyota had to shut down its 20 auto plants in Japan, which build 14,000 cars a day. Some
experts thought Toyota couldn't recover for weeks.
But five days after the fire, its car factories started up again. Only recently have Toyota and its
suppliers revealed how they did it. The secret lay in Toyota's close-knit family of parts suppliers.
In the corporate equivalent of an Amish barn-raising, suppliers and local companies rushed to the
rescue. Within hours, they had begun taking blueprints for the valve, improvising tooling systems
and setting up makeshift production lines.
By the following Thursday, the 36 suppliers, aided by more than 150 other subcontractors, had
nearly 50 separate lines producing small batches of the brake valve. In one case, a sewing-machine
maker that had never made car parts spent about 500 man-hours refitting a milling machine to
make just 40 valves a day.
"Toyota's quick recovery," says Yoshio Yunokawa, general manager of Toyoda Machine Works
Ltd., a Toyota-group maker of machine tools and steering systems, "is attributable to the power
of the group, which handled it without thinking about money or business contracts."
That is a common approach in Japan's keiretsu, the almost tribal groups of companies that supply
and support behemoths such as Toyota. Japanese auto makers' blood pacts with their suppliers
largely explain how they can slash their costs to the bone and stay globally competitive.
And the fealty the parts makers paid to Toyota during its crisis helps indicate why Japan's auto
companies return the loyalty -- often to the detriment of U.S. and other foreign parts makers
seeking market share here. Toyota and Aisin didn't bother to approach any foreign companies
during the crisis, a Toyota spokesman says, because "there were no foreign suppliers in a position
to help us."
Aisin (pronounced "eye-sheen") is an archetypal supplier in Toyota's group. Founded during
World War II to make aircraft engines, it is based in Kariya, an industrial warren occupied by
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many other big Toyota suppliers. Toyota holds 23% of Aisin's stock. Aisin's president is Kanshiro
Toyoda, scion of the Toyoda family that founded the auto maker.
Long a supplier of engine and brake parts, 80% of which it sells to Toyota, Aisin has won almost
all of Toyota's contracts for brake-fluid-proportioning valves -- "P-valves" in industry parlance.
The fist-sized valves control pressure on rear brakes and help prevent skidding. For most parts,
Toyota has at least two suppliers. But over the years, it turned to Aisin to produce all but 1% of
its P-valves because of Aisin's high quality and low cost. The supplier shipped parts to Toyota
plants under a just-in-time inventory system: several times a day, just enough valves for a few
hours of production.
Depending on a single source and holding essentially no inventory is a calculated risk, concedes
Kiyoshi Kinoshita, Toyota's general manager of production control. But it also is what keeps
Toyota's production lean; Aisin gets major economies of scale that it passes on to Toyota in
lower prices.
Toyota acknowledges that it didn't figure in the risk of fire. Aisin executives speculate that sparks
from a broken drill bit may have ignited wooden platforms. Just after 4 that morning, flames swept
through an air duct and ignited the roof. Graveyard-shift workers escaped as 36 fire engines
arrived, mostly from nearby Toyota-group companies. Even as the fire burned, Aisin officials
organized a committee to assess the damage, notify customers and labor unions and, following
Japanese custom, visit neighbors to apologize. A subcommittee ordered 320 cellular phones, 230
extra phone lines and several dozen sleeping bags for executives who were expected to live at
headquarters in the coming days.
At 8 a.m., Aisin asked Toyota to help. Kosuke Ikebuchi, a Toyota senior managing director, was
tracked down at a golf-course clubhouse; he left his wife there and rushed to Toyota
headquarters to help set up a "war room" to direct the damage-control operation. Toyota quickly
sent more than 400 engineers to Aisin. In reacting to such a crisis, Mr. Kinoshita says, "we're like
the U.S. military."
When the last embers died just before 9 a.m., the damage at Factory No. 1 began to grow clear
-- along with an apparent Achilles' heel in Toyota's lean corporate physique. Most of the factory's
506 highly specialized machines, which make other brake parts as well as P-valves, were charred
and useless. Toyota estimated that more than two weeks would be needed just to restore a few
milling machines to partial production, and six months to order new machines. That was too
long: Auto plants were on overdrive to meet strong domestic demand and serve the brisk-selling
U.S. market.
Moreover, a Toyota shutdown would damage local economies. Firms supplying the 20,000 parts
in the average Toyota, along with hundreds of businesses such as utilities and trucking companies
would be hurt without Toyota orders. Each day Toyota is down, a state agency calculates, cuts
Japan's annual industrial output by 0.1 percentage point.
Prospects for a quick comeback seemed to dim as the Saturday wore on. Toyota production
officials were dismayed to learn they needed 200 P-valve variations. And chances that anyone else
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would quickly take up production looked distant: The valves have many complex tapered orifices
that require highly customized jigs and drills. The production department told Toyota President
Hiroshi Okuda, who was in the Middle East, that they would close most plants from Monday until
at least Thursday -- the longest suspension in company history.
On Saturday afternoon, Toyota and Aisin summoned officials from some of the major parts
suppliers to a second war room, at Aisin headquarters. It quickly became a hectic scene, with
officials shouting out for copies of the blueprints of different P-valves while Toyota executives
divvied up valve-making assignments, recalls Osamu Natsume, sales division head at Toyoda
Machine Works. "It was chaos," he says.
Then the parts makers had to assemble the tools, dies, drills and other fixtures for machining
systems that would normally take several months to perfect. "We had to work, no matter how
hard," says Tetsuro Yamakage, production-engineering manager at Toyoda Machine. He
immediately raced to find the 30 kinds of cutters, knives and special drills needed to make the
valves. But there still weren't enough suppliers. So, Toyota purchasing officials called more parts
makers to a Sunday afternoon meeting. These officials, like those that had met on Saturday, were
like family -- people who work closely with Toyota from the start of a car's design. "It was crucial
because we knew each other, we knew the face of the people," Mr. Ikebuchi says.
One familiar face was Masakazu Ishikawa, a former Toyota manager whose division had designed
Toyota P-valves and who now is executive vice president of Somic Ishikawa Inc., a supplier of
brake parts and suspension ball joints. Mr. Ishikawa called Somic's top production engineers from
his cellular phone on the two-hour ride home from Aisin and asked them to meet at the office
at 8 p.m. Sunday. They stayed there until after midnight to plot strategy: They would farm out
some of their current factory work to free up machines to make the Toyota parts.
At 6 a.m. Monday, Somic's four designers began an eight-stage design process. "We'd literally
never done anything like this before," says Isoo Suzuki, production-engineering director. Staying
up 40 hours, Mr. Suzuki and his engineers designed jigs. Then they called in some chits from
Somic's chain of suppliers. For example, Somic got a machine-tool maker, Meiko Machinery Co.,
to turn down other orders and put 30 workers on round-the-clock shifts to make the jigs it
needed.
Somic drafted technical and administrative staffers to help man the machines. On Feb. 6, right on
schedule, it delivered its first P-valves to Toyota. So many suppliers were rushing to please Toyota
that they sparked an unofficial race. Taiho Kogyo Co. would have been first, says Nobuo Fukuma,
a vice president of the bearing maker, if it hadn't had to search nationwide for special machine
tools. Taiho was forced to alter imported tools, he complains, because "the toolmakers didn't
give us priority" and shipped to bigger parts makers instead. Although Taiho's first batch of 500
P-valves was ready on Thursday, less than a day behind two bigger Toyota affiliates, Mr. Fukuma
was despondent. "We had wanted to be first," he says.
Others were delighted just to have been called on. Brother Industries Ltd., which usually makes
things like sewing and fax machines, got calls Sunday night from Aisin and Toyota. A sense of
panic at Toyota had inspired it to go far afield, and Toyota knew Brother had a small machine-
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tools unit. Brother cobbled together a P-valve production line by adapting a computerized milling
machine that usually makes sewing-machine and typewriter parts. Each valve took Brother 10
minutes to make and five minutes to inspect -- hardly a cost-effective use of its resources. The
firm helped out, says Yoshihiko Tsuzuki, a general manager, "because it was an emergency."
Early in the week after the fire, even Toyota's Mr. Ikebuchi had doubts about the goal of resuming
production in all plants by Friday. But the supplier group came through. Trucks bearing the first
1,000 usable P-valves rolled in late Wednesday. On Thursday, 3,000 more arrived, and on Friday,
5,000. Slowly, Toyota's assembly lines started up again.
All told, Toyota lost production of 72,000 vehicles. But with overtime and extra shifts, Toyota
officials say, it has already nearly recouped the lost output. The fire and its aftermath have left
Toyota executives convinced that they have the right balance of efficiency and risk. "Many people
say you might need to scatter production to different suppliers and plants, but then you have to
think of the costs" of setting up expensive milling machines at each site, Mr. Ikebuchi says. "We
re-learned that our system works."
In fact, the fire may have made the system even more efficient. Nisshin Kogyo Co., which was
making the other 1% of Toyota's P-valves, says that during the crisis it raised production efficiency
30% by speeding up production. Mr. Kinoshita says the fire spurred Toyota to begin an effort to
trim the number of its parts variations, a project that should eventually cut costs. And sole-source
suppliers are moving quickly to build fail-safe mechanisms. Somic, which makes all of Toyota's
steering linkages, is revamping its system so it can easily shift to another site if disaster strikes.
Suppliers never asked Toyota or Aisin what they would be paid for rushing out the valves, says
Somic's Mr. Ishikawa. "We trusted them." Indeed, as the first valves arrived at Toyota factories,
Aisin told the suppliers it would pay for everything, from drills and overtime pay to lost revenue
and depreciation. And Toyota promised the suppliers a bonus totaling about $100 million "as a
token of our appreciation," says Mr. Okuda, its president. He adds that the auto maker will
certainly remember the companies that pitched in during its crisis.
Based on the above article, please answer the following questions:
(Note that all answers will be rewarded for brevity and precision)
6a) Does a single-sourcing strategy make sense for such a key component? Discuss the rationale
behind Toyota’s approach of following a single sourcing strategy for such a critical component
and debate whether this is consistent with their lean approach? Discuss the pros and cons of such
an approach?
(8 marks)
6c) Finally, what are the underlying mechanisms that exist in Toyota’s supply chain that help
the firm quickly recover from the sudden supply disruption? Discuss the key takeaways from
this episode.
(12 marks)
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I am not providing a detailed answer key for this case. I have provided a few ppts and a detailed
case study on this topic which gives you a great synthesis on this episode.
We will discuss the case during the review session.
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7. Real Life Examples of Delayed Differentiation
15 minutes
10 points
Delayed differentiation or Postponement is a popular concept in operations management
where a firm starts by making a generic product and differentiates it at a later stage into specific
end-products. This approach is extensively used in industries faced with high demand uncertainty
and helps firms in those industries to minimize market mediation costs (costs resulting due to
mismatch between supply and demand).
A commonly cited example is that of Benetton who knitted their sweaters in grey (or in undyed
form), and then dyed into different colors only when the season/customer color
preference/demand is known.
Provide two original examples drawn from your experience or from other sources (one in a
manufacturing and another in a service setting) as to how delayed differentiation approach
can be deployed. Ensure that you have explained how such an approach helps the firm to minimize
market mediation costs. No credit will be given for mentioning examples discussed in
class or discussed in articles from the reading pack.
7a) Example from the manufacturing sector:
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(5 points)
7b) Example from the service sector:
(5 points)
I am not providing answers for this question as the examples should be based on your experience.
We have had some amazing responses on this topic through the Canvas discussion forum.
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