MIT 2.810 Fall 2014 Solutions to Quiz 2 Name: _________________________________

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2.810 Manufacturing Processes and Systems

Quiz II (November 19, 2014)

Open Book, Open Notes, Computers with Internet Off

90 Minutes

Write your name on every page.

Clearly box your answers.

State any assumptions you have made.

Q1. Manufacturing Systems: Automobile Assembly Plant (20 points)

Q2. Product Mix on a Transfer Line (20 points)

Q3. Process Control (15 points)

Q4. TPS Cell (30 points)

Q5. Literature (15 points)

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TOTAL: 100 points

MIT 2.810 Fall 2014 Solutions to Quiz 2 Name: _________________________________

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Q1. Manufacturing Systems: Automobile Assembly Plant (20 points)

a. (8 points) An automobile assembly plant typically produces a car every minute. Using

information from “The Machine that Changed the World” reading, estimate the average

work-in-process (WIP) inventory in terms of vehicles in the assembly line for Japanese

assembly plants in Japan, and for All Europe assembly plants.

Answer:

From Chapter 4, Figure 4.7 of “The Machine that Changed the World”, we know that a

Japanese company’s Japan plant required 16.8 hours to assemble a vehicle, whereas the All

Europe one required 36.2 hours. We apply Little’s Law to determine the average inventory in

the system.

Japanese in Japan: L = λW. That is, L = 60*16.8 = 1008 vehicles.

All Europe: L = λW. That is, L = 60*36.2 = 2172 vehicles.

b. (12 points) Now consider the Japanese in Japan plant. We assume that there is an

assembly station for each vehicle and that the work statistics are identical at each station.

Furthermore, if everything worked perfectly the assembly line could actually produce one

car every 50 seconds. When an operator observes a problem on the line, he/she can pull

an Andon cord which stops the entire line. What fraction of time could an individual

station stop the line this way while still meeting the production rate of a car every

minute?

Answer:

We are given τ = 50 seconds, P = 1/60 seconds, and n = 1008. We assume the assembly

line has no buffers and that all processes have equal likelihood of stockouts. We therefore

use Buzzacott’s formula to determine the downtime.

Solving this, we get, MTTR/MTTF = 1.984 x 10-4

. Or MTTF/(MTTF + MTTR) =

99.98%. So the percentage of time that a station might be stopped this way is 0.02%.

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Q2. Product Mix on a Transfer Line (20 points)

Consider a two-machine line as shown below. The machines are perfectly reliable i.e., they do

not fail. There is no buffer between the two machines.

Figure 1: Two-machine zero-buffer line

The line is used to make two product types – product A and B which have an average demand of

4 and 5 parts per hour respectively. Each machine can process a part of either type in 5 minutes.

There is a setup change required for each machine when changing over from part A to B or vice

versa. The setup change times are given in Table 1 below.

Table 1: Setup change times

Setup Change Time required in minutes

Machine 1 Machine 2

From A to B 5 7.5

From B to A 5 7.5

a. (8 points) Assume that the line is constrained to make parts A and B in the ratio 4:5 every

hour. For the system shown in Figure 1, what is the maximum production rate of the line

with this constraint?

Answer:

Suppose we start with both machines set up to make type A parts. We first make four type A

parts. This requires 4 x 5 = 20 minutes. Then we do a setup change on both machines. This is

governed by the setup change on Machine 2 and it takes 7.5 minutes. Then both machines

work to make five Type B parts. This requires 5 x 5 = 25 minutes. We then do another setup

change to prepare to make type A parts which again requires 7.5 minutes. Thus, this cycle

requires 20 + 7.5 + 25 + 7.5 = 60 minutes. So we can make 9 parts an hour given the

constraint of making 4 Type A and 5 Type B parts every hour.

M2 M1

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b. (8 points) Now assume that the constraint is eased so that the line can make the parts in

the ratio 4:5 over a shift of 7.75 hours. What is the maximum production rate that you can

now achieve?

Answer:

To maximize the production rate, we limit ourselves to two changeovers. We first make 4N

type A parts, then do a changeover, then 5N type B parts, and then a changeover to part A.

We can add up the times for this sequence and add it up to 7.75 hours to determine what N

should be.

We need: 4N*5 + 7.5 + 5N*5 + 7.5 = 465 minutes. Therefore, we get, 45N = 450 or N = 10.

So we should make a batch size of 40 type A parts and then a batch size of 50 type B parts.

The production rate would be (40 + 50)/(465/60) = 11.61 parts an hour.

c. (4 points) What is the ultimate maximum production rate for the system making parts in

the ratio 4:5? And under what conditions would this occur?

Answer:

Again, we need to minimize the number of setups so we get the maximum production rate.

We look over an infinitely long time span and repeat the procedure of part b.

We need: 4N*5 + 7.5 + 5N*5 + 7.5 -> ∞. So the batch size tends to infinity for each part.

We can make 12 parts of each type in an hour. So the production rate approaches the

maximum possible rate of 12 parts an hour.

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Q3. Process Control (15 points)

Consider a process for which an X-bar control chart has been established.

a. (5 points) What is the probability that an observed data point would fall outside the

UCL?

Answer:

We assume that the UCL is at 3σ from the mean. From the standard normal table, we get

Φ(3) = 0.99865. So on each side, the area outside the 3σ limit is (1 - 0.99865) = 0.00135 or

0.135%. This is the probability that a point will fall outside the UCL.

b. (10 points) Assume that the process changes and there is a mean shift but the variance

stays the same. If the old control chart is being used and the mean shift is equal to plus

one standard deviation, what is the probability that an observed data point would fall

outside the UCL?

Answer:

This scenario is shown in the figure below. With the process now operating at 1σ from the

mean, the UCL is 2σ from the new mean. For a normally distributed process, the fraction of

parts outside the 2σ limit are (1 – Φ(2)) = (1 – 0.97725) = 0.02275 or 2.27%.

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Q4. TPS Cell (30 points)

Consider the TPS manufacturing cell shown below. There are 4 machines and 5 walking

segments. The manual time/machine times are indicated next to each machine in seconds. In the

present scenario, there is only one operator in the cell, and there are no de-couplers. The walking

time between the machines are shown next to the arrows.

a. (10 points) For the present scenario, determine the production rate, average inventory

and time spent by a part in the system.

Answer:

The time chart is as shown in the table.

Time in seconds

Operation Walking + Manual Machine

1 15 55

2 20 25

3 20 35

4 25 30

Finish 5

85 145

Thus, the production rate is constrained by the pace of the operator. It is λ = 1/85 part per

second. The average number of parts in the system is 5 (four parts in four machines and one

in the operator’s hands). The time spent in the system is, W = 5/(1/85) = 425 seconds.

1 2

4 3

5 s 5 s

5 s 5 s

5 s

10 s/ 55 s 15 s/ 25 s

15 s/ 35 s 20 s/ 30 s

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b. (20 points) The desired production rate is one part every 50 seconds. We need to add one

more worker to the cell. The cell could be split horizontally (operations 1 and 2 by one

worker, operations 3 and 4 by the other worker) or vertically (operations 1 and 4 by one

worker, and operations 2 and 3 by another worker). Which system would meet the

desired production rate? Are there any other changes required to the system?

Answer:

The two scenarios are analyzed below. In either case, we add a de-coupler between each pair

of machines. Also, machine 1 is the bottleneck since it requires 55 seconds. So we add

another machine there.

Horizontal Split

Time in seconds

Operation Walking + Manual Machine

1 15 55

2 20 25

12.5

3 17.5 35

4 25 30

Finish 15

105 145

Worker 1 47.5

Worker 2 57.5

Note that this system is still limited by worker 2 and we cannot meet a production rate of a

part every 50 seconds.

1 2

4 3

5 s 5 s

5 s

5 s

1

10 s/ 55 s

2.5 s

15 s/ 25 s 10 s/ 55 s

2.5 s

15 s/ 35 s 20 s/ 30 s

10 s

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Vertical Split

Time in seconds

Operation Walking + Manual Machine

1 15 55

2 17.5 25

3 20 35

7.5

4 30 30

Finish 5

95 145

Worker 1 50

Worker 2 45

This configuration would satisfy the production rate of one part every 50 seconds.

1 2

4 3

10 s/ 55 s

20 s/ 30 s 15 s/ 35 s

1

10 s/ 55 s 15 s/ 25 s

5 s

5 s

2.5 s 2.5 s

2.5 s

5 s 5 s 5 s

2.5 s

MIT 2.810 Fall 2014 Solutions to Quiz 2 Name: _________________________________

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Q5. Literature (15 points)

In your answers, state which reading and the page number of the information you are citing.

a. (5 points) What did Taiichi Ohno and others at Toyota think of the “Mass Producers”

management of their supply chains? Name at least two features utilized by the Mass

Producers that the Japanese thought irrelevant and/or unproductive.

Answer:

From Womack, chapter 3, page 58-59, we learn that Ohno thought that the make-vs-buy

decision was irrelevant and the bidding system seemed unsatisfactory. Ohno thought that

suppliers had little incentive to improve their processes and worked to the plan given to them

by the automaker. The suppliers were also competing amongst themselves which limited the

exchange of information.

b. (4 points) What argument did Henry Ford use to justify his plan to use more castings

and forgings instead of sheet metal stamping in the Model A?

Answer:

Hounshell, chapter 7, page 280-281: Henry Ford had reservations about using stampings in

his cars. He used the “thermal argument” saying that a cast part had to go through fewer

heating operations compared to production of sheet steel.

c. (3 points) How many people were employed at the River Rouge plant in the early 1930s?

(And just for fun, guess how many people work at the plant today. The plant now makes

the Ford F-150 truck and has a capacity of about 420,000 vehicles a year.)

Answer:

From Hounshell chapter 7 page 296 the River Rouge plant had 75,000 employees. On the

previous page, Hounshell says that the plant produced 1.4 million vehicles from January 1 to

September 1929. This indicates a productivity of 18.6 vehicles per employee. In the present

day, according to Wikipedia and Ford’s online sources, we know that the River Rouge

complex hires about 6,000 employees. This means a productivity of 70 vehicles per

employee.

d. (3 points) According to Womack et al, of the high volume automobile producers which

plant had the best quality performance in the world?

Answer:

From Womack’s chapter 4, page 87 we learn that Ford’s plant at Hermasillo, Mexico plant

had the best assembly quality.

MIT 2.810 Fall 2014 Solutions to Quiz 2 Name: _________________________________

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