MIT 2.810 Fall 2015 Homework 10 Solutionsweb.mit.edu/2.810/www/files/homeworks/2015_homeworks/sol/hw10... · MIT 2.810 Fall 2015 Homework 10 Solutions 1 ... M1 M2 M3 M4 M5 M6 MTTF

MIT 2.810 Fall 2015 Homework 10 Solutions

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MIT 2.810 Manufacturing Processes and Systems Homework 10 Solutions

Manufacturing Systems, Transfer Lines & TPS

November 9, 2015 Problem 1. Consider a transfer line as shown in figure 1 below.

Figure 1: Transfer line layout

The machine parameters are as given in the table below.

M1 M2 M3 M4 M5 M6 MTTF (hours) 30 40 20 20 10 50 MTTR (hours) 1 1 1 1 1 1 Processing time (hours) 1 1 1 1 1 1

The buffer has infinite storage capacity. Assume operation-dependent failures.

(a) Line evaluation Estimate the production capacity of this system. Assume that the profit per part is $10 and that you sell every part you make. Maintenance costs (not included in the profit calculation) are $1 per hour. Calculate the net profit for this system. Answer: The first three machines and the last three machines behave like zero-buffer three-machine lines. We apply Buzzacott’s formula to calculate the production rate for each three-machine line:

𝑃 =1𝜏 ×

11+ !""#

!""#!!

𝑃! =11×

1

1+ !!"

+ !!"

+ !!"

=1

1.1083 = 0.902

𝑃! =11×

1

1+ !!"

+ !!"

+ !!"

=1

1.1083 = 0.855

We can now treat these lines as two machines with an infinite buffer between them. Then, the production rate is the production rate of the bottleneck = P2 = 0.855. Profit = 0.855 (parts/hr)*10($/part) - 1($/hour) = 7.55 ($/hour).


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(b) Line improvement: Part I

Identify the bottleneck machine for this line. Suppose that you have a budget of $40,000 to improve the line. Each $10,000 increases the MTTF of a machine by 10 hours. How will you allocate this budget so that you can achieve a production rate of 0.91?

Answer: We improve the system by identifying the bottleneck and improving it by 10 units until it no longer is the bottleneck. One iteration can be as follows:

1. Increase MTTF of machine M5 from 10 to 20. It is now tied with M4 as the bottleneck. The production rate of the second three-machine line (M4-M5-M6) is 0.8928.

2. Increase MTTF of machine M4 from 20 to 30. Now M5 is the bottleneck. The production rate of the second three-machine line (M4-M5-M6) is 0.9063.

3. Increase MTTF of machine M5 from 20 to 30. The production rate of the second three-machine line (M4-M5-M6) is 0.9202.

Now, this three-machine line is capable of meeting the required demand of 0.91 parts per hour. However, the M1-M2-M3 system has a production rate of 0.9022, which is less than the demand. We continue iterating as follows:

4. Improve the MTTF of machine M3 from 20 to 30. It is now tied with M1 as the bottleneck. The production rate of the first three-machine line (M1-M2-M3) is 0.916.

Thus, this system now meets the required demand rate. The cost of these improvements is: $10,000 * 4 = $40,000. Thus, we are within our budget.

(c) Line improvement: Part II Each 0.005 units decrement in the failure rate of a machine increases the maintenance costs by 3%. Recall that the failure rate, p = 1/MTTF. For example, if you reduce the p-value for machine M2 from (1/40 = 0.025) by 0.005 to 0.02, the maintenance costs would become 1.03 ($/hour). If you could improve only one machine, which machine would you apply this improvement to? Set up the equation for the profit of such a system, in terms of the number of decrements, n. What are the upper and lower limits on the value of n? Set up an optimization problem to find the value of n which maximizes profit. Extra credit: Plot the profit function, and find (manually or using software) the optimum value of n, the resulting failure probability rate for the chosen machine, and the maximum profit.


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Answer: M4-M5-M6 is the bottleneck three-machine line. We could pick any machine to apply the improvement to since the denominator simply consists of a sum of p-values. We focus our attention on machine M5. It has a failure rate, p5 of 0.1. Let n be the number of 0.005 decrements we make to failure rate p5. We can write the production rate and profit of the system as a function of n as follows:

P(𝑛) =1𝜏 ×

1

1+ !.!"!

+ !.!!!×!.!!"!

+ !.!"!

=1

1.17− 0.005𝑛

Then the profit per hour (not including maintenance costs) is given by:

Pr(𝑛) =10

1.17− 0.005𝑛 Note that p5 is 0.1. Thus, n can be at most 20 so that p5 is non-negative, and n cannot be negative. Thus, n lies between 0 and 20. The maintenance cost increases by 3% for every n decrement in p5. That is, the maintenance cost can be written as 1.03n. Then, the net profit function becomes:

𝑁𝑒𝑡 Pr (𝑛) =10

1.17− 0.005𝑛 − 1.03!

Extra Credit: The profit function is plotted in figure 2 below. Solving this problem gives n = 10.16 for which, p5 becomes 0.04917. The production rate becomes 0.8935 and the net profit is $7.58/hour.

The function is concave in the region of interest for n in [0, 20]. Thus, we can have two values of n which give the same net profit. A higher value of n means that the failure rate is less. In that case, we would expect less inventory to accumulate in the buffer.


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Figure 2: The net profit as a function of n

(d) Inventory costs As the market matures, we are looking to cut our production costs. Inventory costs, which were so far not considered in our profit calculations, are now considered important. Inventory cost per part per hour is $0.01. You have estimated the production rates and average inventory in the system for various sizes of buffers. These are given in the table below.

Buffer size Production rate Average inventory 1 0.7866 0.611 5 0.8006 3.077 10 0.8127 6.218 20 0.8274 12.74 50 0.8449 34.345 100 0.8522 76 200 0.8545 170 400 0.8547 370

1. Considering the inventory costs, find the buffer size (out of the options in the table)

which maximizes your profit. 2. What answer would you get for the optimum buffer size using Gershwin’s

Approximation?


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Answer:

1. Buffer size (A)

Prod rate (B)

Avg. inventory (C)

Profit not considering inventory costs (D = B*10 - $1/hr)

Inventory costs based on $0.01/part/hr (E = 0.01*C)

Net Profit ($/hr) (F = D – E)

1 0.7866 0.611 6.866 0.00611 6.85989 5 0.8006 3.077 7.006 0.03077 6.97523 10 0.8127 6.218 7.127 0.06218 7.06482 20 0.8274 12.74 7.274 0.1274 7.1466 50 0.8449 34.345 7.449 0.34346 7.10554 100 0.8522 76 7.522 0.76 6.762 200 0.8545 170 7.545 1.7 5.845 400 0.8547 370 7.547 3.7 3.847

Based on the inventory costs, we find that a buffer of size 20 gives us the maximum net profit.

Figure 3: Inventory costs and net profit

Note: Here, we have the first three-machine line producing parts at a faster rate than the second three-machine line. Recall from the “Time, Rate, and Variability” lecture that for an M/M/1 queue with λ > µ, the inventory in the system with an infinite buffer tends to infinity over time. In this part of the problem, we are considering finite buffers. Here, we witness the phenomenon called blocking. When the buffer gets full, the upstream three-machine line gets “blocked” and can only make parts at the rate at which the downstream three-machine line can process parts from the buffer. Thus, we expect the upstream three-machine line to be idle quite often. Because

0

1

2

3

4

5

6

7

8

400 200 100 50 20 10 5 1 Buffer Size

Profit not incl. inventory costs

Inventory Costs

Net Profit


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of the assumption of operation-dependent failures, the upstream three-machine line fails less frequently than it would without blocking.

2. Let us now use Gershwin’s Approximation to find the optimum buffer size. As per the approximation,

𝑁∗ = 2 𝑡𝑜 6 ∗𝑀𝑇𝑇𝑅 ∗ 𝜇 Here, MTTR1 = 1 and MTTR2 = 1. Thus, the mean MTTR is 1. Also, µ for each three-machine line is 1. Thus, we get, N* to be between 2 to 6.


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Problem 2. Production Flow Issues (a) Estimate the production rate, inventory, and time in the system for the system shown in

figure 2, made up of eight identical process steps each which is capable of producing 100 parts a day when operating. The two buffers are of infinite capacity. The MTTF is 10 days and the MTTR is 1 day for each machine.

Figure 2: Transfer line layout with infinite buffers

(b) Please estimate the number of parts in buffer b1 at any time t. State assumptions. (c) Please estimate the number of parts in buffer b2 at any time t. State assumptions.

Answer:


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Problem 3. Toyota Production System (TPS) Cell Consider the TPS manufacturing cell shown below. There are 4 machines and 5 walking segments each 5 seconds. The manual time/machine times are indicated next to each machine in seconds.

Figure 3: TPS cell showing walking times and manual/machine times (in seconds)

(a) Please estimate λ, L and W for the system. State any assumptions.

(b) In order to obtain this performance, machine 1 (a lathe) is running at the maximum material removal rate. At this rate, the cutting insert wears out every 11 minutes of cutting time and require 1 minute to replace. Due to preventative maintenance, the machines never unexpectedly break down. Due to the much longer tool replacement rates for the other machines, they all are replaced during scheduled maintenance. With this new information, re-estimate the production rate of the system. State any assumptions.


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(c) You suggest running the lathe at a slower speed. From experimental results, you find that a

different set of cutting parameters increases the cutting time from 55 to 65 seconds, but the inserts can now be replaced during the scheduled maintenance. What is the production rate of the system at these new conditions? State any assumptions.

The walking time + manual time don’t change from the calculation in part (a), so the average production rate is still 1 part/85 seconds = 0.71 parts/min. The cutting time for machine #1 increases from 55 seconds to 65 seconds, still remaining below the average system production rate of 85 seconds per part. Since the insert doesn’t need to be replaced outside scheduled maintenance (when all machines are down), we don’t need to account for that replacement time. Then the system production rate remains at 0.71 parts/min.


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Problem 4. Output of Photovoltaic (PV) System Estimate the power output for a PV system that is 20% efficient (sunlight to electricity). During sunny periods, the solar radiation is 1000 W/m2, for light clouds—500 W/m2, and heavy clouds—100 W/m2. Meteorological data shows that over the long term the time ratio for the three conditions during daytime hours is 1 hour/30 minutes/10 minutes, respectively. What is the average electric power output of the system during the daytime? What would the result be if we included daytime (12 hrs) and nighttime (12 hrs)? State any assumptions. Answer: Output of 20% efficient PV system during: Sunny periods (60 min): 1000 W/m2 * 0.20 = 200 W/m2 Light clouds (30 min): 500 W/m2 * 0.20 = 100 W/m2 Dark clouds (10 min): 100 W/m2 * 0.20 = 20 W/m2 Weighted daytime average (100 min): (60/100)* 200 W/m2 + (30/100)* 100 W/m2 + (10/100)* 20 W/m2 = 120 + 30 + 2 = 152 W/m2. The day and night average (assuming 12 hours duration for each) would be: 152/2 = 76 W/m2.

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MIT 2.810 Fall 2015 Homework 10 Solutionsweb.mit.edu/2.810/www/files/homeworks/2015_homeworks/sol/hw10... · MIT 2.810 Fall 2015 Homework 10 Solutions 1 ... M1 M2 M3 M4 M5 M6 MTTF