IEOR 130 Methods of Manufacturing Improvement Fall, 2017, Prof. Leachman Solutions to Homework Assignment 8

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Sheryl Strickland
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1 IEOR 130 Methods of Manufacturing Improvement Fall, 2017, Prof. Leachman Solutions to Homework Assignment 8 1. Consider a factory operating at a steady production rate. Suppose the target WIP level is 200 lots. The average cycle time per lot from release to factory output is 50 days. (a) On average, how many lots leave the factory each day, and how many new ones are released into the factory? (Hint: Apply Little's Law.). WIP = (output rate)(cycle time) 200 = (output rate)(50) Output rate = 4 lots per day (b) Suppose photo is the bottleneck work center and it is visited 5 times by each lot as it moves through the process flow. The processing time at each photo step is 5 hours per lot. There are five identical machines in the photo work center averaging 90% availability. What is the utilization of total time and the utilization of available time in the bottleneck work center? Assume the factory operates 24 hours per day, seven days per week. U = (4 lots/day)(5 visits per lot)(5 hours per visit)/(5)(24 hours per day) = 83.33% U/A = /0.9 = 92.26% (c) The standard deviations of the cycle times between each photo exposure operation have been computed as follows: Fab-in to Layer days Layer 1 to Layer days Layer 2 to Layer days Layer 3 to Layer days Layer 4 to Layer days The sum of standard cycle times for the process steps in each layer are as follows: Layer 1-2 days Layer 3-4 days Layer 5-4 days Layer 2-4 days Layer 4-4 days Layer 6-2 days For a target cycle time of 50 days, determine the target WIP level in each layer. Sum of j = =

2 Standard Cycle Time, SCT = Sum of SCTj = 2 + 4(4) + 2 = 20 days Target Cycle Time, TCT = 50 days Buffer factor k = (TCT - SCT) / 10 = 3.0 Target Cycle Time in Layer j, TCTj = SCTj + k * j: TCT1 = * 1.1 = 5.3 TCT2 = * 1.9 = 9.7 TCT3 = * 2.5 = 11.5 TCT4 = * 2.5 = 11.5 TCT5 = * 2.0 = 10.0 TCT6 = 2 (no buffer in last layer since no more bottleneck steps) Target WIP, TW = (production rate)(target cycle time) = (4)(50) = 200 lots Target WIP in layer j, TWj = TCTj * (production rate) TW1 = 4 * 5.3 = 21.2, say 21 lots TW2 = 4 * 9.7 = 38.8, say 39 lots TW3 = 4 * 11.5 = 46 lots TW4 = 4 * 11.5 = 46 lots TW5 = 4 * 10.0 = 40 lots TW6 = 2 * 4 = 8 lots Note that we do not plan a buffer in Layer 6 since there are no visits to the bottleneck in this layer. (d) For a 50-day target cycle time, what is the multiplier of standard cycle time? What is the value of the multiplier in the various layers? Why is the multiplier different in different layers? SCT = 20 days, TCT = 50 days, so the multiplier is 2.5X. Multipliers by layer: Layer 1: 5.3/2 = 2.65 Layer 2: 9.7/4 = 2.43 Layer 3: 11.5/4 = 2.88 Layer 4: 11.5/4 = 2.88 Layer 5: 10/4 = 2.5 Layer 6: 2/2 = 1.0 The multipliers are different in the various layers because (1) not all layers have a bottleneck visit in them, and (2) the variability in the WIP level is different in the different layers. Some layers require larger buffers to ensure comparable productivity. 2

3 (e) What target cycle time would be equal to 2 times standard cycle time? A multiplier of 2X would mean a target cycle time of 40 days. 2. Consider a factory producing high-volume products. A particular type of processing machine performs several recipes with the following data: Actual Target Target Avg. Process On-hand WIP Recipe Downstream Downstream Fab Outs Time per and WIP arriving WIP WIP per shift Wafer (mins) this shift A B C D (a) Assuming the line yields are 100%, calculate the ideal production quantity (IPQ) for each recipe. (The IPQ is defined as the production quantity needed to bring the actual downstream WIP up to the target WIP level by the end of the shift, considering that one more shift's worth of fab outs must be added to the downstream WIP to replace the fab outs due in the current shift.) Calculate IPQs: IPQ(A) = = 110 IPQ(B) = = 25 IPQ(C) = = 230 IPQ(D) = = 30 (b) Calculate the schedule score for each recipe. (The schedule score is defined as the IPQ divided by the average fab out rate over the cycle time from recipe start to fab out.) According to the schedule score, what is the priority ordering of the recipes? Calculate schedule scores: SS(A) = / 50 = SS(B) = - 25 / 50 = SS(C) = /100 = SS(D) = - 30 /100 = The priority of recipes is C, A, B, D. (c) The fab line has 2 machines of the type used to process recipes A, B, C and D. Considering the available WIP and the IPQ's, how many lots of each recipe should be 3

4 allocated to each machine to be processed this shift? Assume all lots have 25 wafers, a shift lasts 8 hours, machine number one will be available all shift, and machine number two is due to receive a PM during the shift lasting 1.5 hours and will be otherwise available. Converting the Min {IPQ, Available WIP} for each recipe to machine hours of work, we have the following: Recipe A: 110 wafers rounds up to 125 wafers (5 lots), 125 * 2.5 = minutes Recipe B: 25 wafers (1 lot), 25 * 2.0 = 50 minutes of work Recipe C: 200 wafers (8 lots), 200 * 2.5 = 500 minutes of work Recipe D: 30 wafers rounds up to 50 wafers (2 lots), 50 * 2.0 = 100 minutes We have 480 minutes of time available on machine #1 and = 390 minutes on machine #2, for a total of 870 minutes available. We have minutes of work in the two top-priority recipes alone, another 50 minutes in the next priority recipe, and 100 minutes in the last-priority recipe. We can't finish everything this shift.. A sensible production schedule is as follows: Machine #1: Run 200 wafers (8 lots) of Recipe C, takes 500 minutes (20 minutes into the next shift) Machine #2: Run 125 wafers (5 lots) of Recipe A, takes minutes, PM takes 90 minutes, Run 25 wafers (1 lot) of Recipe B, takes 50 minutes, Run 50 wafers (2 lots) of Recipe D, takes 100 minutes (72.5 minutes into the next shift) All lots needed to complete IPQs except one will be started in the current shift (assuming machines are idle at start of shift). 3. 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 shifts (where one shift equals 8 hours) Layer 1 exposure to Layer 2 exposure shifts Layer 2 exposure to Layer 3 exposure shifts Layer 3 exposure to Layer 4 exposure shifts Layer 4 exposure to Layer 5 exposure shifts Layer 5 exposure to fab-out shifts The sum of standard cycle times for the process steps in each layer has been computed as follows: Layer 1-2 shifts Layer 3-4 shifts Layer 5-4 shifts Layer 2-4 shifts Layer 4-4 shifts Layer 6-2 shifts 4

5 (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. Let SCT(i), TCT(i), (i) denote the standard cycle time, target cycle time and standard deviation of cycle time in layer i. Now (i) = = 10.0 (Note that the standard deviation of cycle time for the string of operations after last photo visit is irrelevant.) Also, SCT(i) = 20, TCT = 40, and the production rate = 10. The total buffer time is 20 days, so TCT(i) = SCT(i) + 2* (i): TCT(1) = 2 + 2*1.1 = 4.2 TCT(2) = 4 + 2*1.9 = 7.8 TCT(3) = 4 + 2*2.5 = 9 TCT(4) = 4 + 2*2.5 = 9 TCT(5) = 4 + 2*2.0 = 8 TCT(6) = = 2 The target WIP in layer i, TW(i) = TCT(i)*Prodn_rate = TCT(i)*10. Hence TW(1) = 42 TW(2) = 78 TW(3) = 90 TW(4) = 90 TW(5) = 80 TW(6) = 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%. If two steppers are qualified to perform the layer 5 exposure, estimate how long will it take to reduce the WIP in layer 5 to the target level. Layer 5 currently has = 30 extra lots. Since the production time PT = 0.5 hrs, there is 15 production hours of excess WIP. The normal workload on the two steppers is (10 lots/shift)(0.5 hours/shift) = 5 production hours/shift. The stepper capacity is (0.9 availability)(16 stepper-hours per shift) = 14.4 stepper-hours per shift. The net capacity for WIP reduction is = 9.4 hours per shift. The recovery time is therefore 15 / 9.4 = 1.6 shifts. (c) How many steppers would need to be qualified for layer 5 if the recovery time is to be within two hours, i.e., less than 0.25 shifts? 5

6 To recover in two hours (0.25 shift), we would need a net capacity for WIP reduction of 60 hours per shift. This means a total capacity of 65 stepper-hours per shift, or 65 / (0.9*8) = 9.03 steppers, i.e., about 9 steppers qualified for layer 5. (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 A,B,C,D C,D A,B,C,D D B 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 (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. Layer SS IPQ On-Hand Layer Qualified WIP WIP Steppers A, B, C, D C, D A, B, C. D D B Schedule construction: 6

7 Phase 1: Try to complete IPQ's: Assign 14 lots from layer 4 to Stepper D (7 hours workload) Assign 12 lots from layer 5 to Stepper B (6 hours workload) Assign 10 lots from Layer 2 to Stepper C (5 hours workload) Assign 10 lots from Layer 3 to Stepper A (5 hours workload) Assign 6 lots from Layer 1 to Stepper A (3 hours workload) All IPQ's completed. Stepper time left: D has 1 hour, B has 2 hours, C has 3 hours, but A s time is used up WIP left: L5-3 lots (B), L4-6 lots (D), L3-8 lots (A, B, C, D), L2-6 lots (C, D), L1-4 lots (A, B, C, D) Phase 2: Assign remaining WIP to use up stepper capacity: First, consider same setup: Assign 3 more lots from Layer 5 to Stepper B (Layer 5 WIP gone) Assign 2 more lots from Layer 4 to Stepper D (Stepper D time gone) Assign 6 more lots from Layer 2 to Stepper C (Layer 2 WIP & Stepper C time gone) Next consider new setup: Assign 1 lot of Layer 3 (or Layer 1) to Stepper B (Stepper B time used up) Recap: Stepper A: 10 lots of Layer 3, 6 lots of Layer 1 Stepper B: 15 lots of Layer 5, 1 lot of Layer 3 (or Layer 1) Stepper C: 16 lots of Layer 2 Stepper D: 16 lots of Layer 4 4. 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

8 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 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* j Layer, j SCTj j TCTj TWj NA (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)

9 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 (d) In the etch area of the fab, the process time per lot is 1 hour 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 D 2 8 The following shift schedule has been prepared by the etch area supervisor: 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 reason why your schedule should be preferred over the schedule above. Machine M1: Run 4 lots of A, then run 2 lots of B, then run 6 lots of D Machine M2: Run 12 lots of C This schedule makes just as much progress towards completion of the IPQs as the supervisor s schedule. It does 4 lots of D beyond the IPQ for D whereas the supervisor s schedule does 2 lots of D and 2 lots of C beyond the IPQs for D and C, respectively. The reason it is preferred is because this schedule saves one changeover compared to the 9

10 supervisor s schedule. A reduced number of changeovers means less risk of process instability and equipment trouble and a higher expected productivity. 10

IEOR 130 Methods of Manufacturing Improvement Fall, 2018, Prof. Leachman Homework Assignment 8, Due Tuesday Nov. 13

IEOR 130 Methods of Manufacturing Improvement Fall, 2018, Prof. Leachman Homework Assignment 8, Due Tuesday Nov. 13 1. Consider a factory operating at a steady production rate. Suppose the target WIP level