IEOR 130 Methods of Manufacturing Improvement Solutions to HW #5 Fall 2018, Prof. Leachman

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1 IEOR 130 Methods of Manufacturing Improvement Solutions to HW #5 Fall 2018, Prof. Leachman 1. Data in a factory has been collected on the performance of five types of machines, as displayed in the following table. The machines are all used in the same process flow. Shown in the table for each machine type are the theoretical production time per lot (TPT) in hours, the number of lots (NL) processed, the actual production time (PT) in hours and the down time (DT) in hours. Total time for each machine type is 24 hours. Machine TPT NL PT DT Type A B C D (a) Calculate the availability, the utilization of total time and the utilization of availability for each machine type. Calculate the OEE of each machine type. Assume quality efficiency is 100% for all machine types. M/C A: Avail = 0.896, Util = 0.875, Util of avail = / = 0.977, OEE = 20/24 = M/C B: Avail = 0.938, Util = 0.896, Util of avail = / = 0.955, OEE = 21/24 = M/C C: Avail = 0.979, Util = 0.833, Util of avail = / = 0.851, OEE = 38*0.5 / 24 = M/C D: Avail = 0.875, Util = 0.833, Util of avail = / = 0.952, OEE = 36*0.5 / 24 = 0.75 (b) Suppose it is possible to reduce down time by 0.5 hours per day for any of the four machine types. However, it is not possible to reduce down time for more than one of the machine types, i.e., you must pick one machine type and reduce down time of that machine type. For which one do you think it would be most beneficial to fab performance? Why? How will fab performance improve? Machine A is the bottleneck; reducing its down time will increase fab throughput. Reducing down time at the other machines primarily serves to reduce cycle time. If other machines are upstream from machine A in the process flow, reducing their down time might help to reduce idle time of Machine A, but this is a weaker effect.! 1

2 (c) Suppose it is possible to increase rate efficiency of any of the four machine types. For which one do you think it would be most beneficial to fab performance? Why? How will fab performance improve? Machine A is the bottleneck; increasing its rate efficiency will increase fab throughput. Increasing rate efficiency at the other machines primarily serves to reduce cycle time. If other machines are upstream from machine A in the process flow, improving their rate efficiency might have a minor positive impact on reducing idle time of Machine A, but this is a very weak effect. 2. A wafer fabrication plant produces 800 wafers per day of a memory device. The line yield is 100%. The steppers are the bottleneck equipment. At present, the OEE of the steppers averages 0.667, the down time averages 10% and the idle time averages 5%. The quality efficiency for the steppers is 100%. Considering all photo steps to make the memory device, the total theoretical process time per wafer is 0.5 hours. (a) How many steppers are operating in the plant? OEE = 800*(0.5) / N*24! N = (800)(0.5) / (.667)(24) = 25 (b) What is the rate efficiency of the steppers? OEE = U*RE*QE = U*RE = (0.85)*RE! RE = OEE / U = (.667)/(.85) =.7847 (c) What rate efficiency would be sufficient to reduce the number of operating steppers by one without reducing the output? (.85)*RE*24*24 = 800*(0.5)! RE = (800)(0.5) / (.85)(24)(24) = Two alternative process machines are under consideration. Statistics about machine performance and maintenance are as follows: Machine Rework Scrap Avg. PM Avg. Machine Avg. Process time Rate Rate Hours/Week Failures (Hours) (Hours/lot-pass) MTBF MTTR A B Assume the fab operates 168 hours per week. The machines process one lot at a time. When rework is required, the lot must be processed a second time. The average process time for the second pass is the same as for the first pass. The chance of a third pass is negligible.! 2

3 (a) What is the average availability of each machine type? AvailA = 1 16/168 8/208 = = AvailB = 1 18/168 6/106 = = (b) The two machines have the same price. From a productivity point of view, which one would you recommend? Explain. Throughput = Avail * [ 1 / (Expected process time)] * (1 Scrap rate) ThroughputA = * ( ) * { 1 / [(.55)(.99) + (1.1)(.01)]} = ThroughputB = * ( ) * { 1 / [(.5)(.98) + (1.0)(.02)]} = 1.629! B is a better choice. 4. An I Line stepper is the bottleneck of a small fabrication facility, and management desires to maximize its output rate. When equipped with a brand new bulb, the lamp intensity of the stepper is approximately 1,000 mw/cm2. The intensity declines after every wafer exposure until the bulb is replaced. The photo engineer has estimated the lamp intensity function to be LI(n) = 1,000(0.9994) n, where n refers to the nth wafer exposed since the bulb was replaced and LI(n) is the lamp intensity realized for the nth wafer. The total machine down time to replace the bulb and re-qualify the stepper is four hours. Other stepper data: wafer exchange time XT = 13 seconds, initial alignment time AT = 27.5 seconds, and move and align time MT = 0.5 seconds. There is only one exposure step performed by the stepper, with NE = 60 and EE = 1,000 mj. No blading is required. Consider three possible frequencies for changing the bulb: once every 100 wafers, once every 1,000 wafers, or once every 5,000 wafers. Which frequency would maximize n n i 1 a a = a stepper output rate? Explain. (Useful fact from algebra:! i= 1 1 a.) We calculate the average total down time (for bulb change) plus processing time (in seconds) per wafer for each alternative (n = 100, 1000, 5000): p(n)= (DT+ TPTW(n) )/n where : TPTW (n) = total processing time for n wafers. The total processing time for the n wafers is! 3

4 TPTW(n)= n (XT + AT + (NE-1)MT) + NE * EE * Σn (1/ LI(n)) = n ( *0.5) + 60* 1000 * Σn (1/ 1000(0.9994) n ) = n ( *0.5) + 60* Σn (1/ (0.9994) n ) TPTW(100)= TPTW(1000)= TPTW(5000)= p(100)= p(1000)= p(5000)= ! we should choose the smallest p(n), i.e., 1000 wafers is the frequency providing the highest throughput rate. 5. In a 24-hour period, a particular photo exposure machine had 2.5 hours of down time. Its rate efficiency was 85 percent. It completed processing of 35 lots, but 5 of these were the second run on a lot previously processed (i.e., 5 lots were reworked). The average process time per machine cycle was 0.5 hours. (One lot is processed per machine cycle, and a rework lot takes the same amount of time as a first-pass lot.) (a) Estimate the availability, the utilization of availability and the OEE of the machine for that 24-hour period. Let APT denote the average process time per lot. A = (24 2.5)/24 = , U = (35)*(0.5)/24 = U/A = / = ThPT = APT * RE = (0.5)(0.85) = OEE = (0.425)(30)/(24) = (b) An engineering change to the machine is under consideration. The change will reduce the rework rate to about 6 percent, but it will increase the average process time by 2 percent. (Theoretical process time will not be changed.) Assuming the 24-hour period reflected average conditions, estimate what the OEE score will be if the engineering change had been implemented. Assume the number of fab starts will be increased just enough so that the utilization is held constant. APT! (1.02)(0.5) = 0.51 # of lots processed (including rework) = (24)(0.7292)/(0.51) = 34.3 # of first-pass lots = (34.3) / (1.06) = OEE = (32.37)(0.425)/24 = ! 4

5 (c) Now suppose it is not known how low the rework rate will be after the engineering change, but it is known that average process time would rise by 2 percent. Find the upper limit on the rework rate after the change such that the engineering change will not decrease OEE from its value in part (a). Once again, assume the fab starts would be adjusted so that the utilization is held constant. [34.3/(1+r)] * (0.425)/24 = = 1 + r implies r! 14.3% (was 5/30 = 16.7%) 6. A semiconductor factory producing 0.8 micron products utilizes 5X G Line stepper machines to perform photolithography. The machine works as follows. One or two cassettes of 25 wafers each may be placed in the stepper by the operator. The operator must enter a recipe into the machine and place the proper reticle (i.e., the photomask) in the machine before starting the machine. The machine accommodates up to six reticles in an internal magazine. A new cassette, recipe and/or reticle can be tendered to the machine while it is running on another cassette/reticle/recipe combination. Mechanical apparatus inside the machine move wafers one at a time out of the input cassette and onto an X-Y stage where the exposures will take place. The machine performs the lithography exposures on the first wafer, then the completed wafer is moved to an empty cassette while the second wafer is moved onto the X-Y stage. The fixed time to move a completed wafer off the stage and replace it with the next wafer to be processed is called the wafer exchange time (XT). After a wafer is moved onto the X-Y stage, the machine performs an aligning procedure to properly orient the wafer level and straight with the camera lens. The fixed time to perform this initial alignment is called the alignment time (AT). The 5X stepper performs multiple exposures, one at a time. (The name "stepper" comes from the fact that it performs multiple "steps" across the wafer surface.) Each exposure encompasses a small number of die. The number of exposures (NE) required to fully process a particular wafer type is a given factor which depends on the size of the die and the exposure field size of the machine. The process recipe calls for a certain exposure energy (EE) to be achieved, measured in milli-joules (mj). The time required to perform this exposure (ET) equals EE divided by the lamp intensity (LI) of the machine, where LI is measured in milli-watts per square centimeter (mw/sq cm). LI depends on the condition of the exposure bulb and the quality of the optical path inside the machine. After a single exposure is completed, the X-Y stage moves the wafer over for the exposure of the next field. The fixed time for the mechanical apparatus to move the wafer! 5

6 and re-align the camera is called the move time (MT). There is no MT before exposing the first die; all initial motion of the wafer on the X-Y stage is included in AT. For the die types produced in this factory, there are no "test keys" printed on the wafers. The entire wafer surface is printed with product dice. After exposure of all wafers in a cassette are completed, the cassette can be removed from the machine and the exposed wafers are then sent through a development process. (Similar to the way the machine can hold two cassettes of inbound wafers, there is room in the machine for two outbound cassettes. A full outbound cassette can be removed from the machine while it continues to process wafers into the other cassette.) After developing, the wafers are inspected under a microscope. If for some reason the exposure was mis-aligned with earlier mask layers, the current layer can be stripped off in an acid bath, and the wafers sent back through the stepper for a second try. This repetition is known as rework. Let RW denote the fraction of total wafer operations performed on the 5X stepper that are rework. The operational procedures followed in the factory sometimes result in the machine being completely flushed out of wafers and falling idle. For quality control purposes, the machine might be deliberately stopped after processing one wafer and held idle while waiting for the results of the after-develop inspection. When the machine is to be used again after becoming idle, the machine is inactive while the operator puts in a new reticle, puts in a new cassette, enters the appropriate recipe, and starts the machine. It then takes the machine some time to bring the first wafer from the inbound cassette to the X-Y stage in order to start the basic processing cycle described above. The sum of all these delay times is called a setup time (ST). Let WPS denote the average number of wafers processed between setups. The industrial engineer for the fab has collected the following data about a particular 5X stepper in the factory: Parameter Theoretical Average LI 770 mw/sq cm 720 mw/sq cm MT 0.46 sec 0.47 sec AT 27.0 sec 29.0 sec XT 12.0 sec 12.0 sec ST 150 sec (a) Provide a formula for the theoretical processing time (seconds per wafer) as a function of the number of exposures per wafer (NE), the exposure energy (EE), and the numerical parameters above.! 6

7 TPW = XT + AT + NE(EE/LI) + (NE-1)MT, where XT = 12.0, AT = 27.0, LI = 770, MT = 0.46 (b) Provide a formula for the average processing time (seconds per wafer). Assume the average rework rate RW and the average number of wafers per setup WPS parameters are given. Assume rework does not result in extra stepper setups. TPW = [ XT + AT + NE(EE/LI) + (NE-1)MT ] (1+RW) + ST/WPS, where XT = 12.0, AT = 29.0, LI = 720, MT = 0.47, ST = 150 (c) The Photo Process Engineering Manager would like some help prioritizing the efforts of her staff. She would like to know: Which would reduce average processing time more, a 5% reduction in exposure time, or a 5% reduction in the rework rate? Assume the average number of exposures NE is 60, the current average exposure time is 0.3 seconds, and the current average rework rate is 2.6%. Answer her question and explain. Let ET = EE/LI. We compare TPW1 and TPW2, where TPW1 = [ XT + AT + NE(0.95)(ET) + (NE-1)MT ] (1 + RW) + ST/WPS, and TPW2 = [ XT + AT + NE(ET) + (NE-1)MT ] ( RW) + ST/WPS Equivalently, we can compare (1 + RW)(NE)(0.95)(ET) + RW [ XT + AT + (NE-1)MT ] with ( RW)(NE)(ET) (RW) [ XT + AT + (NE-1)MT ], or we can compare NE(0.95)(ET) + RW [ XT + AT + (NE-1)MT ] with (NE)(ET) RW [ XT + AT + (NE-1)MT ]. Plugging in numerical values, we compare 60(0.95)(0.3) + (0.026) [ ] = 20.5 with 60(0.3) + (0.95)(0.026) [ ] = Thus reducing the exposure time 5% is better. (d) Suppose the average processing time for the stepper is reduced. Would this translate into a higher OEE score for the stepper? Explain. The OEE will rise only if the stepper is the fab bottleneck. Otherwise,! 7

8 the reduction in process time is just turned into idle time. The reduction also will reduce cycle time. 7. At present, the process specifications for the operation of a metal etcher require the operator to perform a test run of the machine on a single blank wafer before processing each production lot. The blank wafer must be inspected for particles using a measurement machine that counts particles on the wafer surface. During this time, the metal etcher is held idle. If the particle count is low enough, the operator may proceed to process the production lot through the metal etcher. If the particle count is too high, the metal etcher machine is declared out-of-control and a cleaning of the machine must be performed before any production lots can be processed. The relevant parameters are as follows: Parameter Average Theoretical Time to process blank wafer 10.2 minutes 9.1 minutes Time to inspect blank wafer 8.5 minutes 6.2 minutes Time to process production lot 30.5 minutes 28.0 minutes Cleaning time per 24 hours 2.0 hours Other down time per 24 hours 1.2 hours Production lots processed per day 25.0 (a) What is the current OEE of the metal etcher? Assume that, theoretically, test runs are not necessary. OEE = 25(28.0) / 24(60) =.4861 (b) Estimate the current availability of the metal etcher. Suppose we define utilization (of total time) for the metal etcher to include processing time plus the enforced idleness during inspection of test runs. Estimate the current utilization (of total time) of the metal etcher. Availability = 1 - (3.2)/24 = = Utilization of total time = 25( ) / 24(60) = (c) Optimistically, what is the potential increase in OEE if test runs are eliminated but no other improvements are made? That is, what is the ratio of potential OEE to current OEE? (To estimate this increase, assume there will be no change in idle time, cleaning time, other down time or in rate of quality after the test runs are eliminated. Moreover, assume the metal etcher is the fab bottleneck by a very wide margin.) Assume same utilization: (n)(30.5) / 24(60) = 0.854! n = (0.854)(24(60) / 30.5 = 40.3! OEE = 40.3(28.0) / 24(60) =.7836 up by a ratio of or up by 30 percentage points! 8

9 (d) Pessimistically, there might be a major drop in die yield if the test runs are not performed. By what factor would the die yield need to drop to offset the gains in wafer throughput? Do you think such a drop is likely? (DY1)(25) = (DY2)(40.3) 25/40.3 = DY2 / DY = DY2/DY1 A 38% drop in yield seems very unlikely. (e) How would the trade-off change if the metal etcher was not the fab bottleneck? Would we be more inclined or less inclined to eliminate test runs? Explain very briefly. If metal etch is not the bottleneck, then 40.3 lots may be impossible, so the wafer throughput gain would be less. The die yield drop would still be the same, so the trade-off becomes less favorable.! 9