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IEOR 130 Methods of Manufacturing Improvement

Solutions to Practice Examination Problems – Part II of Course Prof. Leachman

Fall, 2019 1. For a particular semiconductor product, the customer orders received to date are as follows, sorted by date due to ship to customer: Delivery date 1 2 3 4 5 6 7 Total orders due 120 90 120 125 180 60 0 The current finished goods inventory of the product is 80 units. The production plan for the product is as follows: Output date 1 2 3 4 5 6 7 Output quantity 120 150 100 100 100 100 100 (a) A new customer calls and asks for delivery of 10 units per period beginning in period 4 and ending in period 6. Can the company provide delivery as requested? No. See table under part (b) below. (b) Calculate the best delivery schedule the company can offer the customer. Date 1 2 3 4 5 6 7 Cum Supply 200 350 450 550 650 750 850 Cum orders 120 210 330 455 635 695 695 Delta 80 140 120 95 15 55 155 ATP 15 15 15 15 15 55 155 Cum RFQ 0 0 0 10 20 30 30 Cum Q 0 0 0 10 15 30 30 Uncum Q 0 0 0 10 5 15 0 Delta 15 15 15 5 0 25 125 New ATP 0 0 0 0 0 25 125 (c) Assuming the customer accepts the quote you calculated in part (b), calculate the new available-to-promise (ATP) quantities by period. Indicate whether you use cumulative ATP or incremental ATP. See above for new cumulative ATP.

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2. A semiconductor company sells a device type that has bin splits into high-speed and low-speed versions. These bin splits are 25% and 75%, respectively. Customers for the low speed version will accept shipment of the high-speed version in lieu of a low-speed version. Customers for the high-speed version will not accept version substitution. Currently the company has 100 units in inventory of the low-speed version, but no inventory of the high-speed version. The production plan for the product is as follows:

Output date 1 2 3 4 5 Output quantity before bin split 200 200 200 200 200

The total customer orders received are as follows, sorted by date due to ship to customer:

Delivery date 1 2 3 4 5 Total orders due for high-speed 40 40 60 40 40 Total orders due for low-speed 100 250 150 180 50

(a) Calculate the available-to-promise (ATP) quantities for both versions by period. Indicate

whether you use cumulative ATP or incremental ATP. Day 1 2 3 4 5 Cum Supply, high 50 100 150 200 250 Cum Supply, low 250 400 550 700 850 Cum Orders, high 40 80 140 180 220 Cum Orders, low 100 350 500 680 730 Delta, high 10 20 10 20 30 Cum ATP, high 10 10 10 20 30 Delta, low 150 50 50 20 120 Cum ATP, low 20 20 20 20 120 Cum ATP, low, if high ATP is absorbed: 30 30 30 40 150 (b) A new customer calls and asks for a delivery quote. The customer requests a delivery of 10

units of high-speed version and 20 units of low-speed version per day on day 2 and day 3. If these quantities cannot be met, the customer would like a quote of deliveries catching up to this schedule as quickly as possible. Calculate the best delivery quote that can be offered to the customer.

Day 1 2 3 4 5 Cum request, high 0 10 20 20 20 Cum quote, high 0 10 10 20 20 Uncum quote, high 0 10 0 10 0

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Revised high ATP 0 0 0 0 10 Cum ATP, low 20 20 20 20 120 Cum request, low 0 20 40 40 40 Cum quote, low 0 20 20 20 40 Uncum quote, low 0 20 0 0 20 (c) Suppose the fab experienced unexpected machine down when the new customer calls. The

fab management is forced to adjust the production plan so that the actual output will be as follows:

Output date 1 2 3 4 5 Output quantity before bin split 200 200 200 160 200

Can the company still ship as planned? How should it respond to the new customer in this case?

Day 1 2 3 4 5 Cum supply, high 50 100 150 190 240 Cum Orders, high 40 80 140 180 220 Delta, high 10 20 10 10 20 Cum ATP, high 10 10 10 10 20 Cum request, high 0 10 20 20 20 Cum quote, high 0 10 10 10 20 Uncum quote, high 0 10 0 0 10 Cum Supply, low 250 400 550 670 820 Cum Orders, low 100 350 500 680 730 Delta, low 150 50 50 -10 90 Cum ATP, low 0 0 0 0 90 Cum request, low 0 20 40 40 40 Cum quote, low 0 0 0 0 40 Uncum quote, low 0 0 0 0 40

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3. The process flow for making an advanced DRAM device requires the use of a 193nm scanner photolithography machine at the photo steps in layers 1, 5 and 9. The cycle times from fab start to the photo steps in layers 1, 5 and 9 are 0.2 weeks, 0.8 weeks and 1.8 weeks, respectively. The fab line yield is 100%. The UPH (units per hour) factors at the photo steps in layers 1, 5 and 9 are 50, 48 and 45, respectively. There are 3 193nm scanners in service. The CEE of the 193nm scanners is 72%. The fab operates 24 hours a day, 7 days per week. The 193nm scanner is believed to be the fab bottleneck equipment type. It is desired to develop a capacity constraint on weekly fab starts for the purposes of production planning. (a) Let xt denote the fab starts in week t. Express the 193nm scanner capacity constraint on xt, xt-1 and xt-2. You may assume the fab will operate every day of every week. You do not need to simplify your expression.

(b) In steady state, how many wafers may be started per week? In steady-state, sum up coefficients: 0.02 + 0.02083 + 0.02222 = 0.6305 0.6305 x <= 362.88 => x <= 5755.4 (c) Suppose 4,000 wafer starts were made last week, and 6,000 wafer starts were made the week before that. What is the maximum number of wafers that may be started this week without extending 193nm scanner cycle time?

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4. A small fabrication line operates two process flows, flow A and flow B. The important steps in each flow are as follows: Flow A Flow B Qualified CT up Qualified CT up Step UPH Machine_types to Step Step UPH Machine_types to Step 1 50 M1 0 1 50 M1 or M2 0 2 75 M1 or M2 1 2 75 M1 or M2 or M3 1 3 75 M1 or M2 2 3 75 M1 or M2 or M3 2 Note: “UPH” = units per hour, the speed of the machine when it performs that step. “CT up to step” is the average cycle time from fab start to arrival at the process step, expressed in weeks. Data on the photo machines is as follows: Machine_type Quantity CEE M1 2 0.65 M2 2 0.70 M3 3 0.75 The factory operates 168 hours per week and line yield losses are negligible. (a) Let xt denote the constant rate of starts of flow A in week t, and let yt denote the constant rate of starts of flow B in week t. Express capacity constraints on xt and yt in an arbitrary week t. You do not need to simplify numerical expressions.

(b) Suppose no starts of flow B are allowed. What is the maximum steady-state starts rate for flow A? In that case, which machine type(s) is (are) the bottleneck(s)? We have yt = 0 for all t and xt = x for all t for some x that we want to make as large as possible. Plugging in to the constraints above, the first constraint implies The second constraint implies The third constraint is not binding when x = 9,720. So 9,720 starts per week is the maximum for A. Both machine types M1 and M2 will be fully loaded. (c) Suppose no starts of flow A are allowed. What is the maximum steady-state starts rate for flow B? In that case, which machine type(s) is (are) the bottleneck(s)?

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Performing a similar analysis for B, the second constraint implies The third constraint implies The first constraint does not apply to B. So 17,820 starts per week is the maximum for B. All three machine types M1, M2 and M3 will be fully loaded. (d) The starts rates in the last two weeks for each flow were as follows: Week Flow A Flow B 0 9,000 12,000 -1 4,000 14,000 If in week 1 the starts for flow A are required to be zero, what is maximum amount of starts of flow B that should be allowed in week 1? Why is this number different from your answer to part (c)? Which machine type(s) will be the bottleneck(s) in week 1? Plugging in the WIP values to the constraints, we find the workload in week 1 from WIP in the second constraint is 173.33 hours vs. a capacity of 453.6 hours. This implies there is room for 14,013.33 starts of B in week 1. We also find the workload in week 1 from WIP in the third constraint is 520 hours vs. a capacity of 831.6 hours. This implies there is room for 15,580 starts of B in week 1. Considering the most limiting constraint, we can start 14,011.33 of B in week 1. Machine types M1 and M2 will be fully loaded.

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5. Consider a factory that has two products in production, A and B. Sales of product A are a steady 1,000 wafers per week and its remaining product life is two years. The average selling price per wafer is $5,000 and that price is declining 30% per year. Sales of product B are a steady 5,000 wafers per week and the remaining product life is one year. The average selling price per wafer is $2,000 and that selling price is declining 60% per year. (a) The cycle time to make product A is 30 days. The cycle time to make product B is 45 days. Compute the remaining lifetime revenue for each product. The lifetime revenue is given by

For product A, or , W = 1,000/7, P0 = 5,000, CT = 30 and H = 730. So R = $362,018,476. For product B, or , W = 5,000/7, P0 = 2,000, CT = 45 and H = 365. So R = $293,695,969. (b) There are 17 photo machines in the factory and they achieve 88% availability operating 168 hours per week. The mean and standard deviation of a down time event on photo machines at the factory are 9 and 6 hours, respectively. There are 17 photo steps to make product A and 20 photo steps to make product B. The process time to perform any photo step is 0.5 hours per 25-wafer lot and the standard cycle time per lot is 0.75 hours. Assume line yields for both products are 100%. Assuming utilization is balanced across all 17 photo machines, compute the utilization of availability of the photo machines. U/A = (0.5*17*1,000/25 + 0.5*20*5,000/25)/(0.88*17*168) = 0.9311 (c) Let m denote the number of photo machines qualified to perform a given photo step. Management is considering engineering work to qualify more photo machines at various photo steps to make Product A as follows: Step m # of Required Type steps engineering hours

in flow to add 1 more to m for all steps of that type

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Management also is considering engineering work to qualify more photo machines at various photo steps to make Product B as follows: Step m # of Required Type steps engineering hours

in flow to add 1 more to m for all steps of that type

1 2 8 160 2 3 12 80 Estimate the current total cycle time for each type of photo step on each product. You may assume ca2 =1, and c02 = 1 for all step types. The expected (average) queue time per step is

where

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For step type 1 on either product,

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The total entitlement cycle time for type 1 steps on product A is therefore ECT = 7*[QT + (1/A- 1)*PT + SCT] = (7)[8.88 + (1/0.88 – 1)*0.5 + 0.75] = 67.43 hours. The total entitlement cycle time for type 2 steps on product A is therefore ECT = 10*[QT + (1/A- 1)*PT + SCT] = (10)[5.79 + (1/0.88 – 1)*0.5 + 0.75] = 65.38 hours. The total entitlement cycle time for type 1 steps on product B is therefore ECT = 8*[QT + (1/A- 1)*PT + SCT] = (8)[8.88 + (1/0.88 – 1)*0.5 + 0.75] = 77.07 hours. The total entitlement cycle time for type 2 steps on product B is therefore ECT = 12*[QT + (1/A- 1)*PT + SCT] = (12)[5.79 + (1/0.88 – 1)*0.5 + 0.75] = 78.46 hours. (d) If all of the proposed work for Product A is undertaken, what additional revenue will the company obtain over the life of Product A? If all of the proposed work for Product B is undertaken, what additional revenue will the company obtain over the life of Product B? We re-compute QT with m increased by 1. For type 1 steps,

.

For type 2 steps,

The new cycle times are then as follows. The entitlement cycle time for type 1 steps on product A becomes ECT = 7*[QT + (1/A- 1)*PT + SCT] = (7)[5.79 + (1/0.88 – 1)*0.5 + 0.75] = 45.77 hours. The total entitlement cycle time for type 2 steps on product A becomes ECT = 10*[QT + (1/A- 1)*PT + SCT] = (10)[5.01 + (1/0.88 – 1)*0.5 + 0.75] = 50.07 hours. The total entitlement cycle time for type 1 steps on product B becomes ECT = 8*[QT + (1/A- 1)*PT + SCT] = (8)[5.79 + (1/0.88 – 1)*0.5 + 0.75] = 52.31 hours. The total entitlement cycle time for type 2 steps on product B becomes

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ECT = 12*[QT + (1/A- 1)*PT + SCT] = (12)[5.01 + (1/0.88 – 1)*0.5 + 0.75] = 60.08 hours. The reduction in cycle time for type 1 steps on product A is 67.43 – 45.77 = 21.67 hours. The reduction in cycle time for type 2 steps on product A is 65.38 - 50.07 = 15.31 hours. The total reduction in cycle time for product A is 36.98 hours. The reduction in cycle time for type 1 steps on product B is 77.07 – 52.31 = 24.76 hours. The reduction in cycle time for type 2 steps on product B is 78.46 – 60.08 = 18.37 hours. The total reduction in cycle time for product B is 43.14 hours. The revenue gain from a cycle time reduction DCT is given by

For product A, DR = $545,498. For product B, DR = $1,328,274. (e) Because of other engineering tasks and limited engineering staff, the company might not be able to undertake all of the photo qualification work above. Order the product-step-type projects listed above in the order of decreasing economic return per engineering hour expended. For type 1 steps on product A, DR/(eng hours) = $319,558/80 = $3,994. For type 2 steps on product A, DR/(eng hours) = $225,741/40 = $5,644. For type 1 steps on product B, DR/(eng hours) = $761,623/160 = $4,760. For type 2 steps on product B, DR/(eng hours) = $564,877/80 = $7,061. So the ranking is type 2 on B, type 2 on A, type 1 on B, type 1 on A.

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6. The management of a wafer fab with one process flow is considering the purchase of an additional process machine in order to reduce cycle time. At present, there is room in the fab for one more machine of either Type 1 or Type 2. Statistics about these machines are as follows: Type m MTTR cr2 PT (per A U/A SCT (per No. of steps

step) step) in process flow

1 3 5.0 10.0 1.0 0.8 0.83 1.2 6 2 2 2.5 7.5 0.5 0.8 0.80 1.5 20 (a) Estimate the cycle time per step for each machine type. You may assume fab line yield is 100%, ca2 =1, and c02 = 1. The formula in the notes for queue time is

QT =

And the formula for cycle time is CT = QT + (PT)(1/A -1) + SCT. From Excel calculations, for type 1, the queue time per step is 9.4 hours and the cycle time per step is 10.9 hours. For type 2, the queue time per step is 5.0 hours and the cycle time per step is 6.6 hours. (b) Considering the number of steps in the process flow performed by each machine type, estimate the total cycle time in the process flow contributed by each machine type. Considering all steps, the total cycle time for type 1 is 65.2 hours. Considering all steps, the total cycle time for type 2 is 132.0 hours. (c) Estimate the total cycle time reduction if one more machine of type 1 is installed, while keeping the fab starts rate constant. If one more machine is installed, m is increased by one and U/A changes to [m/(m+1)]*U/A. Re-computing the queuing formula, we find the total cycle time for type 1 drops to 18.3 hours. (d) Repeat (c) for one more machine of type 2. Re-computing the queuing formula, we find the total cycle time for type 2 drops to 45.0 hours. (e) The cost to install one more machine of type 1 is 1.0 million dollars. The cost to install one more machine of type 2 is 1.5 million dollars. Which installation would offer the greatest cycle time reduction per capital dollar expended?

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The CT ROI for type 1 is (46.9/1) = 46.9 hours per million dollars. The CT ROI for type 2 is (87.0/1.5) = 58.0. Hence type 2 has higher CT ROI. 7. The photolithography department in a semiconductor fabrication plant has 10 identical photo machines. Manufacturing lots move through a series of 100 steps in the plant, including five photolithography steps. At present, any of the ten machines can be used to perform any of the five photo steps. To improve the die yield, the photo process engineer proposes a “lot-to-lens dedication” policy, whereby each manufacturing lot would be required to visit the same photo machine for processing of photo steps 2, 3, 4 and 5 that the lot visited for photo step 1. It still would be the case that a lot could be processed by any of the ten machines at photo step 1. Suppose the photo machines average 90% availability, the factory works 168 hours per week, and the total hours of processing work that the ten photolithography machines perform is 1250 hours per week. (a) Assuming utilization of the ten machines is balanced, what is the utilization of availability for the photo machines? U/A = 125 / (0.9*168) = 82.67% This is the value of u to use in the queuing formula. (b) Assume all lots at all photo steps have the same process time. Assume the proposed change in policy would not have any effect on the lot inter-arrival times at each step nor any effect on equipment down times or process times. Estimate the RATIO of lot queue time at photo step 2 if lot-to-lens dedication is implemented to lot queue time at photo step 2 at present. Note that and all other terms in the expression for queue time would remain fixed. At present, m = 10, u = 0.8267, so Under lot-to-lens dedication, at step 2 we would have m =1. So

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Thus i.e., queue time at step 2 would increase by a factor of 16.685. (c) Estimate the percentage increase in total lot queue time for all five photo steps if lot-to-lens dedication is implemented. You may assume there is no line yield loss, i.e., each step has an equal production volume. Queue time at steps 2, 3, 4 and 5 would increase by a factor of 16.685. Queue time at step 1 would not increase. So total queue time would increase by a factor of [1 + 4*16.685]/5 = 13.548.

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8. The processing cycle for a diffusion furnace consists of three phases: load, run, and unload. During the “load” portion of the cycle, an operator transfers wafers from incoming lots into a “boat” accommodating 150 wafers. If the incoming lots include less than 150 wafers, the operator inserts dummy wafers to raise the total wafers in the boat up to 150. During the “run” portion of the cycle, the boat is mechanically inserted into the furnace, the wafers are cooked for a specified length of time, and then the boat is mechanically withdrawn from the furnace. During the “unload” portion of the cycle, the operator unloads the product wafers from the boat into lots to be sent to follow-on operations, and he/she unloads the dummy wafers for re-use in subsequent furnace runs as may be required. For a particular furnace, the “run” portion of the cycle takes exactly 6 hours every cycle. The theoretical times to perform the “load” and “unload” portions of the cycle are 0.5 hours each, but sometimes the operators take longer to complete these tasks. The average load time is estimated to be 0.6 hours (and the average unload time also is 0.6 hours). Last week this furnace completed 20 process cycles and experienced 4.5 hours of down time. The average batch size was 5.7 lots (i.e., 142.5 product wafers). (a) Estimate the utilization of total time, utilization of availability, and OEE of this furnace last week. Assume the factory is operated 24 hours per day, seven days per week. ThPT = 0.5 + 6 + 0.5 = 7 U = 7.2 * 20 / 168 = 85.7% U/A = 7.2 * 20 / (168 – 4.5) = 88.1% OEE = [(142.5/150) * 7 * 20]/168 = 79.2% (b) Identify the two reasons that rate efficiency was less than 100% for this furnace. Smaller-than-maximum batch sizes, and extra time operators require for load and unload. The equipment vendor offers a modification to the furnace whereby the furnace would be equipped with dual boats instead of a single boat. If equipped with dual boats, the operator could load boat B while the furnace was running boat A. After the run on boat A was completed, the furnace could immediately start the run on boat B. In parallel with the run on boat B, the operator could unload boat A. When loading and unloading are conducted in parallel with processing, the furnace is said to be “backloaded.” (c) If equipped with dual boats, what is the reduction in theoretical process time per cycle? From 7 hours to 6 hours. (d) Assuming the same number of process cycles were run with the same average batch size, estimate the OEE and utilization of availability last week if the furnace had been equipped with dual boats and all batches could be backloaded. OEE = [(142.5/150 * 6 * 20]/168 = 67.9% U/A = 6 * 20 /(168 – 4.5)] = 73.4%

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(e) Assuming the same number of process cycles were run with the same average batch size, estimate the reduction in cycle time last week if the furnace had been equipped with dual boats and all batches could be backloaded. Assume there are no alternative furnaces, i.e., this is the only one that can be used, and assume down time statistics and process time variability would be unchanged if dual boats are installed. Other data: c0 = 1, MTTR = 4.5, cr = 1.0, ca = 1, lot arrival rate = 0.679 lots per hour. CT = U/(1-U)[(ca2 + ce2)/2](PT/A) + (1/A – 1)PT + SCT + BT SCT = 7.2, A = 0.973, PTold = 7.2, Uold = 0.881, PTnew = 6, Unew = 0.734 BT = [(b-1)/2]*(1/lot_arrival_rate) = [(5.7-1)/2]*(1/0.679) = 3.461 ce2 = c02 + (1+SCVTTR2)A(1-A)(MTTR)/PT ce2old = 1 + (1+1)(0.973)(1-0.973)(4.5/7.2) = 1.033 ce2new = 1 + (1+1)(0.973(1-0.973)(4.5/6) = 1.039 CTold =[ (2.033)/2](54.63) + 7.4 + 3.461 = 66.387 CTnew = [(2.039)/2](17.01) + 7.37 + 3.461 = 28.173 DCT = 66.387 – 28.173 = 38.214 hours (f) Suppose the current revenue from one lot is $25,000 and is declining 25% per year. The current fab cycle time is 40 days. The remaining product lifetime is 3 years. Assuming last week’s processing rate is maintained, estimate the revenue gain from installation of dual boats in the furnace. DR = R0exp{-a(CT-DCT)}[(1-e-aH)/a] – R0exp{-aCT}[(1-e-aH)/a] = R0[exp{aDCT}-1] exp{-aCT}[(1-e-aH)/a] R0 = 25,000 * 20 * 5.7 / 7 = 407,143 H = 3(365) = 1,095 DCT = 38.2/24 = 1.592, CT = 40 a satisfies e-a365 = 0.75 or –a = (ln 0.75)/365 -> a = 7.882x10-4 [(1-e-aH)/a = 733.49, exp{-aCT} = 0.96896, exp{aDCT}-1 = 0.001256 DR = $407,143*0.001256*0.96896*733.49 = $363,328

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9. A wet bench consists of a series of tanks served by a robot arm. Two production lots (50 wafers total) form one batch that travels down the bench. The batch is dunked in each tank by the robot arm. One of the tanks contains sulfuric acid that strips an undesired film off the wafers. With repeated use, the acid bath contains more and more residue from previously stripped wafers, and there is increasing probability that the film on the wafers in the next batch may be inadequately stripped. An inspection step carried out after the wet bench step would detect this, in which case the batch must be re-worked. At some point the acid bath must be dumped and re-poured with fresh acid; this involves one hour of down time to the wet bench as well as the expenses for new sulfuric acid and disposing of the old acid. The process time in the sulfuric acid tank is 30 minutes per batch, whether for a first-time batch or a re-worked batch. Each batch is dunked for a lesser amount of time in the other tanks on the bench (pre-cleaning, rinsing and drying); the total time for a batch to traverse the bench is 90 minutes. The wet etch engineer estimates the probability that rework is required is a linear function of bath usage: P(n) = 0.05*n, where n is the number of first-time batches processed since the acid bath was re-poured and P(n) is the probability that the nth batch must be re-worked. You may assume that with probability one a batch that is reworked will be successfully stripped of the undesired film on the second pass through the tank, and that reworking does not cause the acid bath to deteriorate.

(a) Suppose our objective is maximum wet-bench capacity. What frequency of re-pour is best? (By “frequency”, we mean how many batches between re-pours of the sulfuric acid bath.)

We should maximize the output rate per cycle, where a cycle covers the period between re-pours. That is, we seek to maximize (number of batches per cycle, excluding rework) / (time required to complete cycle). Let G(n) be this function, where n is the number of batches in the cycle (excluding rework).

The time required to complete a cycle including n batches is

0.5n + 0.5 + 1 = 0.5n + 0.5 + 1 = 0.5n + 0.5(0.05) + 1.

Hence we seek to maximize

.

From Excel calculations, the output rate is maximized for n=9.

(b) Now suppose our objective is minimum cycle time. Assume the following data for the wet bench: m=1, ca = 1, ce = 1, the wet bench receives 250 lots per week (i.e., 125 batches per week, excluding rework), and the only down time is for re-pouring the acid bath. Now what frequency

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of re-pour is best? (Hint: You can calculate the availability and average rework rate as functions of the re-pour frequency. And be sure to include rework in utilization.)

where IAT is the interarrival time of lots, equal to 168/250 = 0.672 in this case. As a function of the number n of lots processed between repours, the average rework rate is

.

Excluding rework, U = (250)(0.5)(0.5) / (168) = 0.372. The availability of the bench depends on the length of the cycle between re-pours. This length is given by

T = (n – 1) * IATBatch + (1 + 0.05 * n) * PT + 1 where n is the number of lots (excluding reworks) between re-pours, IATBatch is the batch inter-arrival time (168/125 = 1.344 hours in this case), 0.05*n is the probability of reworking the nth batch, PT is the acid bath process time (0.5 hours in this case), and the re-pour time is one hour. The availability is then A = 1 – 1/T. From Excel calculations, cycle time is minimized when n = 6. (c) Now suppose our objective is maximum profit. What factors should be taken into account to decide the best frequency of re-pour? What other data would you request in order to make this determination? We can take into account the cost of acids and the revenue delta from cycle time. So we need the average revenue per lot, cost of acid, cost of re-pouring, in addition to previous data.

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10. A wafer fab operates a single process flow that has a photo step at the end of layers one through five. Layer six includes the remaining process steps after the last photo step. A detailed simulation of the fab has been run and stable steady-state statistics were collected as follows: Layer Avg. Wait time Std. Dev. Of Wait time Avg. No. of Lots in the layer (days) in the layer (days) in the layer 1 2.0 1.5 70 2 3.0 2.5 85 3 4.0 3.0 110 4 3.0 1.5 85 5 3.0 1.5 85 6 1.0 0.5 45 The average output rate in the steady-state statistics of the simulation is 20 lots per day. (a) What was the steady-state average total cycle time in the simulation? The total average WIP is 480. The average cycle time was therefore 480/20 = 24 days. (b) Suppose management sets the target cycle time for the process flow to be equal to the simulated average total cycle time. For an output rate of 20 lots per day, recommend target WIP levels in each layer. From the waiting time and the production rate we can determine the average buffer WIP in the simulation. Then comparing to the total WIP we can deduce the active WIP and hence the standard cycle time in each layer: Layer Waiting time Buffer WIP Total WIP Active WIP SCT 1 2 40 70 30 1.5 2 3 60 85 25 1.25 3 4 80 110 30 1.5 4 3 60 85 25 1.25 5 3 60 85 25 1.25 6 1 20 45 25 1.25 Total 8 The target cycle time is 24 days. So the total buffer time to be allocated is 24 – 8 = 16 days. The sum of standard deviations of the waiting times in layers 1 through 5 is 10. We assume no variance in SCT. So we can apply a k factor equal to 16/10 = 1.6 whereby TCTj = SCTj + k* sj Layer (j) SCTj sj TCTj TWj 1 1.5 1.5 3.9 78 2 1.25 2.5 5.25 105 3 1.5 3.0 6.3 126

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4 1.25 1.5 3.65 73 5 1.25 1.5 3.65 73 6 1.25 NA 1.25 25 (c) Suppose fab outs to date are on time, and suppose the target output rate of 20 lots per day will be maintained for many weeks. The fab operates two 12-hour shifts per day. Suppose it is the start of a shift and the current WIP levels in each layer are as follows: Layer WIP (no. of lots) 1 82 2 120 3 110 4 75 5 75 6 18 Calculate IPQs and Schedule Scores for each photo step. What is the priority order for scheduling the photo steps? Because there are two shifts per day, the target output rate per shift is 10. The IPQ = 10 + Target Downstream WIP – Actual Downstream WIP. Layer Target Actual Target D/S Actual D/S IPQ SS WIP WIP WIP WIP 1 78 82 402 398 14 -1.4 2 105 120 297 278 29 -2.9 3 126 110 171 168 13 -1.3 4 73 75 98 93 15 -1.5 5 73 75 25 18 17 -1.7 6 25 18 The priority of photo steps is 2, 5, 4, 1, 3. (d) In the etch area of the fab, the process time is 0.5 hours for all etch steps. There are two etch machines, M1 and M2. Each machine can perform any etch step. At the start of the shift, the etch area has the following data: Step IPQ (lots) Available WIP (lots) A 6 4 B 2 4 C 12 14 D 2 8 The following shift schedule has been prepared by the etch area supervisor:

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Machine M1: Run 4 lots of A, then run 2 lots of B, then run 6 lots of C Machine M2: Run 8 lots of C, then run 4 lots of D Can this schedule be improved upon? If so, suggest the best schedule you can, and briefly note the reasons why your schedule should be preferred over the schedule above. M1: Run 4 of A, then 4 of B, then 8 of D M2: Run 14 of C This schedule saves one changeover. Moreover, it schedules all the WIP, whereas the supervisor’s schedule left some WIP behind. Less changeovers means less risk of process instability. NOTE: A better version of this problem might be if the process times were 1 hour per lot. In that case, the following schedule could be proposed: Machine M2: Run 12 lots of C Machine M1: Run 4 lots of A, then run 2 lots of B, then run 6 lots of D Like the supervisor’s schedule, this schedule makes as much progress on IPQs as is possible. Like the supervisor’s schedule, both machines are fully loaded. But this schedule makes one less changeover, providing less risk of process instability.

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11. A fabrication process includes 4 visits to the photo bottleneck, one each at the end of layers 1, 2, 3 and 4. Recent statistics about the fabrication process are as follows:

Layer Avg. total WIP Avg. active WIP (no. of lots) (no. of lots) 1 80 70 2 125 100 3 135 110 4 70 60 5 25 20 The average production rate during the period of data collection was 10 lots per shift. No line yield losses were experienced.

(a) Recommend target cycle times by layer assuming line yields are 100% and the overall target cycle time is 50 shifts. Standard Cycle Time, SCT = Active WIP / prod_rate SCT = 7 + 10 + 11 + 6 + 2 = 36 Sum of ACTj - SCTj = 1+2.5+2.5+1.0=7.0 (last layer after last photo step not relevant) Target Buffer Time, TBT = 50 -36 = 14 Buffer factor k = (TCT - TBT) / 7 = 2 Target Cycle Time in Layer j, TCTj = SCTj + k * (ACTj - SCTj) j: 1 2 3 4 5 TCTj: 9 15 16 8 2 (b) Current WIP status is as follows. Data concerning the photo steps also is displayed.

Layer Actual WIP WIP at Photo Photo process Qualified (lots) (lots) time (hours/lot) Machines 1 90 20 0.4 A, B, C 2 140 20 0.5 B, C 3 158 5 0.6 A, B 4 87 18 0.4 C 5 15

Assume it is now the start of a shift and that fab-outs to date are on time. The target production rate continues to be 10 lots per shift. Calculate the ideal production quantity (IPQ) and the schedule score (SS) for each photo step for this shift.

IPQ=Target Downstream WIP + Prod_Rate – Actual WIP SSj = -IPQj / PR Layer Actual Target Actual Target IPQ SS

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WIP WIP Downstream Downstream WIP WIP

1 90 90 400 410 20 -2 2 140 150 260 260 10 -1 3 158 160 102 100 8 -0.8 4 87 80 15 20 15 -1.5 5 15 20 (c) Assume one shift lasts 8 hours. Using the data in part (b), determine an efficient shift schedule for the steppers. Assume only the WIP at photo may be scheduled and assume all 3 steppers will be available for all 8 hours. (An efficient schedule has the following properties: as much of the IPQs are completed as possible, steppers are utilized as much as possible, and changeovers of the steppers are minimized.)

Layer Actual WIP WIP at Photo Photo process Qualified (lots) (lots) time (hours/lot) Machines 1 90 20 0.4 A, B, C 2 140 20 0.5 B, C 3 158 5 0.6 A, B 4 87 18 0.4 C 5 15

Phase 1: Assign 20 lots of layer 1 to machine A (left: 0 lots layer 1, machine A full) Assign 15 lots of layer 4 to machine C (left: 3 lots layer 4, machine C 2 hrs) Assign 10 lots of layer 2 to machine B (left: 10 lots layer 2, machine B 3 hrs) Assign 5 lots of layer 3 to machine B (left: 0 lots layer 3, machine B full) Now all IPQs (excluding 3 lots unavailable on layer 3) are assigned Phase 2: Assign 3 lots of layer 4 to machine C (left: machine C 0.8 hrs) Assign 2 lots of layer 2 to machine C (left: 8 lots layer 2, machine C filled 0.2 hrs into next shift) Recap: Machine A: 20 lots layer 1 Machine B: 10 lots layer 2, 5 lots layer 3 Machine C: 18 lots layer 4, 2 lots layer 2 (0.2 hrs into next shift)

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12. A fabrication process includes three steps performed on the bottleneck equipment type. Layer 1 of the process ends at the first bottleneck step; layer 2 ends at the second bottleneck step; layer 3 ends at the third bottleneck step; and layer four includes all steps of the process after the third bottleneck step. The machines performing the bottleneck steps are inflexible whereby each bottleneck step must be performed by a different machine. Data on cycle times is as follows: Layer Actual Avg. Standard Theoretical i Cycle Time Deviation of Cycle Time (ACTi) ACTi (si) (SCTi) 1 8.4 2.2 4 2 10.9 2.8 5 3 13.5 3.0 7 4 7.2 1.6 4 Management of the factory operating the process has decided to set the target cycle time for the process to be 90% of the actual average cycle time (ACT). The total WIP level in the process in the most recent week was 400. (a) According to management’s target cycle time for the process, what is the target for the total WIP in the process? 360. (b) Compute efficient target WIP levels for each layer. Production rate = AW/ACT = 400/40 = 10. TCT = 0.9*40 = 36. SCT = 20, so we have 16 days of buffer time. s1 + s2 + s3 = 8, so we can have a buffer factor of k = 2. TCT and TW by layer: TCT1 = 4 + 2*2.2 = 8.4 TW1 = 84 TCT2 = 5 + 2*2.8 = 10.6 TW2 = 106 TCT3 = 7 + 2*3.0 = 13.0 TW3 = 130 TCT4 = 4 TW4 = 40 (c) Instead of your answer to (b), suppose management decides to set the target WIP level in each layer to be 90% of the actual WIP level in that layer. If management implements their target WIP levels, which of the three bottleneck steps will have the greatest idle time over the long run? And in this layer, will the amount of idle time be higher or lower than it would be using the target WIP levels in your answer in (b)? Explain. MTW1 = 75.6 (MTW1 – Active_WIP1)/s1 = 16.12 MTW2 = 98.1 (MTW2 – Active_WIP2)/s2 = 17.18 MTW3 = 121.5 (MTW3 – Active_WIP3)/s3 = 17.17 MTW4 = 64.8

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The first bottleneck step will have the most idle time over the long run, it has the least (relative) safety stock. 13. A simulation of a factory with a single process flow achieves steady-state. Various machines are used to perform the steps of the process flow; each machine performs one step on one production lot at a time. The steady-state statistics show that, on average, 63.5 out of 80 machines are busy processing lots. Summing across all steps in the process flow, the average total waiting time for each lot is 8.5 days, and the average total number of lots in queues is 80.5. (a) What is the production rate of the factory? Buffer WIP = 80.5, Total buffer time = 8.5 days, so from Little’s Law the production rate = 80.5/8.5 = 9.5 lots per day. (b) What is the level of active WIP in the factory? Active WIP = 63.5. (c) What is the standard cycle time of the process flow? Standard cycle time = (Active WIP) / (Production Rate) = 63.5 / 9.5 = 6.7 days (d) What is the actual cycle time of the process flow? Actual cycle time = Standard cycle time + Total waiting time = 6.7 + 8.5 = 15.2 days

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14. A wafer fab has a single process flow with five masking layers. The standard deviation of the cycle time between photo exposure operations has been computed as follows: Fab-in to Layer 1 exposure - 1.1 shifts (where one shift equals 8 hours) Layer 1 exposure to Layer 2 exposure - 1.9 shifts Layer 2 exposure to Layer 3 exposure - 2.5 shifts Layer 3 exposure to Layer 4 exposure - 2.5 shifts Layer 4 exposure to Layer 5 exposure - 2.0 shifts Layer 5 exposure to fab-out - 0.5 shifts The sum of standard cycle times for the process steps in each layer has been computed as follows: Layer 1 - 2 shifts Layer 2 - 4 shifts Layer 3 - 4 shifts Layer 4 - 4 shifts Layer 5 - 4 shifts Layer 6 - 2 shifts (a) The target cycle time for the entire fab process is 40 shifts. The target production rate is 10 lots per shift. Determine the target WIP level for each layer. TBT = 40 – (2 + 4 + 4 + 4 + 4 + 2) = 20 Ssi = (11 + 19 + 25 + 25 +20)/10 = 10 so we can have a k factor equal to 2. TCTi = SCTi + ksi TCT1 = 2 + 2*1.1 = 4.2 TCT2 = 4 + 2*1.9 = 7.8 TCT3 = 4 + 2*2.5 = 9 TCT4 = 4 + 2*2.5 = 9 TCT5 = 4 + 2*2 = 8 TCT6 = 2

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TWi = PR * TCTi TW1 = 42 TW2 = 78 TW3 = 90 TW4 = 90 TW5 = 80 TW6 = 20 (b) Currently, layer 5 has a WIP level of 110 lots. The photo stepper machines each can process two lots per hour when they are up; average availability of these machines is 90% and the average utilization (of total time) is 65%. If two steppers are qualified to perform the layer 5 exposure, estimate how long it will take to reduce the WIP in layer 5 to the target level. The excess WIP in layer 5 is 110 – 80 = 30 lots or 15 process hours. The idle time on the steppers is 90 – 65 = 25%. If all idle time is devoted to working off the excess WIP while maintaining the production rate, the time required to work off the excess WIP is 15/(0.25*2*24) = 1.25 days or 30 hours. (c) How many steppers would need to be qualified for layer 5 if the recovery time is to be within four hours, i.e., less than 0.5 shifts? A recovery time of 4 hours requires 24*15/(0.25*m*24) = 4 hours implies m = 15, i.e., 15 machines must be qualified for layer 5 to clear the excess WIP in half a shift. (d) Now suppose at the start of a shift the actual WIP levels (expressed in lots) are as follows: Layer Total WIP WIP On-Hand Qualified

in Layer at Photo step Steppers 1 38 10 A,B,C,D 2 82 16 C,D 3 90 18 A,B,C,D 4 94 20 D 5 80 15 B 6 18 - - Assuming fab-outs up until the start of the shift are exactly on time, determine the ideal production quantity (IPQ) and schedule score (SS) for each photo operation. (Each photo operation is the last operation in its layer.) Layer Actual Target Actual Target IPQ SS

WIP WIP Downstream Downstream WIP WIP

1 38 42 364 358 4 -0.4

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2 82 78 282 280 8 -0.8 3 90 90 192 190 8 -0.8 4 94 90 98 100 12 -1.2 5 80 80 18 20 12 -1.2 6 18 20 The priority ordering of photo steps is 4 and 5, 2 and 3, 1. (e) The fab has four total steppers (A,B,C,D). Each stepper is qualified to perform only certain photo operations, as specified in the table above. Assume that all four steppers will be available all 8 hours during the shift, and that the process time per lot is 0.5 hours for all photo operations. Suggest an efficient shift schedule for the four steppers. Assume that only WIP on-hand at photo may be scheduled for processing by the steppers. Phase 1: Try to complete IPQs Assign 12 lots of layer 4 to stepper D, takes 6 hours Assign 12 lots of layer 5 to stepper B, takes 6 hours Assign 8 lots of layer 2 to stepper C, takes 4 hours Assign 8 lots of layer 3 to stepper A, takes 4 hours Assign 4 lots of layer 1 to stepper A, takes 2 hours All IPQs completed. Phase 2: Assign more WIP, saving setups where possible Assign 4 more lots of layer 4 to stepper D, takes 2 hours (stepper D time used up) Assign 3 more lots of layer 5 to stepper B, takes 1.5 hours (layer 5 WIP used up) Assign 8 more lots of layer 2 to stepper C, takes 4 hours (stepper C time and layer 3 WIP used up) Assign 4 more lots of layer 3 to stepper A, takes 2 hours (stepper A time used up) Assign 1 lot of layer 1 to stepper B, takes 0.5 hours (stepper B time used up) Recap: 16 lots of layer 4 on stepper D 15 lots of layer 5, 1 lot of layer 1 on stepper B 16 lots of layer 2 on stepper C 12 lots of layer 3, 4 lots of layer 1 on stepper A 100% stepper utilization, one setup, all IPQs completed.