IEOR 130 Methods of Manufacturing Improvement Fall, 2016, Prof. Leachman Solutions to Homework Assignment 10.

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1 IEOR 130 Methods of Manfactring Improvement Fall, 016, Prof. Leachman Soltions to Homework Assignment Consider a fab prodcing a NAND flash device. Prodction volme is 50,000 wafer starts per week. Acconting for line yield and die yield, otpt is 600 die per wafer start. The cycle time is crrently 50 days. Average selling price is crrently $10.00 per die bt is declining 50% per year. The remaining economic life of the device is years. (a) Compte the vale of a 1-day redction in cycle time in terms of the lifetime revene gains. The revene gain from a one-day redction in cycle time is expressed as The vale of satisfies R H CT 1 e e WYP e e 365 or = The vale of one day of cycle time is then or R ,000 e e ( )(50) R ( )(714.86)( )(0.9094)( ) or R $9,58, e ( )(730) (b) In addition to the revene gains, what costs do yo expect cold be redced if cycle time is redced? Assme there is no redction in process time and there is no change in the eqipment set. Cycle time redction reslts in a one-time redction in working capital reqirements (de to the redction in WIP). Cycle time redction may also increase the speed of yield improvement, reslting in higher average die yield and hence lower die cost. (c) Compte the vale of a 10-day redction in cycle time in terms of the lifetime revene gains. The revene gain from a 10-day redction in cycle time is expressed as 1

2 R The vale of 10 days of cycle time is then or R H 10 CT 1 e e WYP e. 1 0 (10) ,000 e e ( )(50) R ( )(714.86)( )(0.9094)( ) or R $303,449, e ( )(730) (d) Is the vale of cycle time redction approximately linear in the nmber of days of redction? Comparing (a) and (c), we see that the vale of cycle time redction grows slightly faster than linear in the nmber of days of redction.. A particlar machine type performs three steps in a fabrication process flow. There are 5 machines of this type and the average process time per lot is 1 hor for each step. Assme all 5 machines are qalified to perform each of the three steps. The fab starts 00 lots per week. The fab operates 4 hors per day, 7 days per week. Assme line yield losses are negligible. Other statistics: Availability A = 0.85 MTTR = 5.0 c = 1 0 c r = 4 ca = 1 SCT = 1. hors per step (a) What is the tilization of availability for this machine type? We are given n = 5, A = 0.85, PT = 1. If the fab starts 00 lots per week, then the machine type mst process 3(00) = 600 lot-steps per week (three steps per lot). The tilization of availability is therefore (1)(600) / (5)(168)(0.85) = (b) Compte the total entitlement cycle time for this machine type (i.e., the sm of entitlement cycle times for each step). What percent of cycle time is standard cycle time? What percent is waiting time? The expected (average) qee time per step is

3 ca ce ( m1) 1 PT ) A where MTTR ce c0 (1 cr ) A(1 A). PT We compte Then ce 5 1 (1 4)(0.85)(1 0.85) (51) ( ) 1 (.59375)( )(1.1765) per step. The total entitlement cycle time is therefore ECT = 3*[ + (1/A 1)*PT + SCT] = (3)[ (1/0.85 1)*1 + 1.] = hors. The total waiting time for all three steps is WT = 3*[ + (1/A 1)*PT] = 3*[ (1/0.85 1)*1] = hors Abot 31% of entitlement cycle time is standard cycle time and abot 69% of entitlement cycle time is waiting time. (c) How mch wold cycle time be redced if the fab installed a sixth machine? If we add a sixth machine, m becomes 6, and the tilization of availability drops to (1)(600) / (6)(168)(0.85) = The expected qee time per step becomes (61) ( ) 1 (.59375)(0.094)(1.1765) per step. The entitlement cycle time becomes (3)[ (1/0.85 1)*1 + 1.] = hors, i.e., a drop of 5.55 hors. (d) Assme there are still 5 machines. The engineer finds that the yield is not as good on one of the machines as on the other for when performing the third process step. He 3

4 proposes to disqalify it from the third step. How mch wold cycle time increase if it is disqalified? Assme the workload still cold be distribted across the machines so as to balance tilization at Then for the step with only for machines qalified, the expected qee time is (41) ( ) 1 (.59375)( )(1.1765) The entitlement cycle time becomes ()( (1/0.85 1)*1 + 1.) + ( (1/0.85 1)*1 + 1.) = = hors, i.e., an increase of abot 0.79 hors. 3. Consider the sitation in part (c) of problem. Sppose it is decided to install a sixth machine. Frther, sppose the engineer proposes that two machines be dedicated to each process step. (a) Assme the process times for all three steps are statistically the same. Compare the tilization of availability for the engineer s proposal vs. the case where any of 6 machines can perform any of the three process steps. Becase the process times are the same for all three steps, dedicating pairs of machines to steps wold leave the tilization of availability the same, , for all three sbgrops. (b) Compare the entitlement cycle time for the case of 6 total machines, dedicated to each step, to that for the case where any of 5 machines can perform any of the three process steps. Note that the cycle time is higher for the first case, even thogh it has 6 machines. Why? For the case where any machine can do any step, the average qee time per step was compted in (c) to be hors. This reslted in an entitlement cycle time of hors for all three steps. For the case where only two machines can do each step, we have (1) ( ) 1 (.59375)(0.9955)(1.1765) per step. This makes for an total entitlement cycle time of (3)[ (1/0.85 1)*1 + 1.] = hors. This compares to hors for the original case, an increase of abot hors. Cycle time goes p becase there are only machines to choose from instead of 5. Even thogh the drop in tilization helps cycle time, the redced nmber of qalified machines per step hrts cycle time more. 4. In the same fab, consider another work station consisting of ten identical diffsion frnaces. Only one fabrication step is performed by the work station on 600 lots per 4

5 week. The batch size is flexible between a minimm of 3 lots and a maximm of 5 lots. The process time is 8 hors and the standard cycle time is 9.5 hors. Other data: Availability A = 0.96 MTTR =.0 c = 1 0 c r = 16 ca = 1 As a fnction of average batch size, determine the entitlement cycle time. (Yo can examine cycle time for increments of 0.5 lots in the average batch size, from 3 p to 5.) What average batch size minimizes cycle time? The overall cycle time is estimated as where ECT = SCT + BT + (1/A 1)*PT + BT b 1 (600 /168) and ca ce ( m1) 1 ) PT A. In this case, tilization of availability depends on the average batch size b, i.e., Ths (600 / b)(8).976. (10)(0.96)(168) b CT b 1 ca 9.5 (1/ ) *8 (600 /168) ce ( m1) 1 PT ) A where is compted from b as above. Trying b = 3.0, 3.5, 4.0, 4.5, 5.0, we find the following: CT(3.0) = CT(3.5) =

6 CT(4.0) = CT(4.5) = CT(5.0) = In this case, rnning fll frnace loads minimizes cycle time. 5. The photolithography department in a semicondctor fabrication plant has 10 identical photo machines. Manfactring lots pass throgh a series of 100 steps in the plant, inclding five photolithography steps. At present, any of the ten machines can be sed to perform any of the five photo steps. To improve the process yield, the photo process engineer proposes a lot-to-lens dedication policy, whereby each manfactring lot wold be reqired to visit the same photo machine for processing of photo steps, 3, 4 and 5 that the lot visited for photo step 1. It still wold be the case that a lot cold be processed by any of the ten machines at photo step 1. Sppose the photo machines average 90% availability, the factory works 168 hors per week, and the average total hors of processing work for the ten photolithography machines to perform is 150 hors per week. a. Assming tilization of the ten machines is balanced (i.e., each machine is allocated 15 hors of processing work per week), what is the tilization of availability for the photo machines? Utilization of Availability = 15/(0.9*168) = 0.87 b. Assme all lots at all photo steps have the same process time. Assme the proposed change in policy wold not have any effect on the lot inter-arrival times at each step nor any effect on eqipment down times or process times. Estimate the ratio of lot qee time at photo step if lot-to-lens dedication is implemented to lot qee time at photo step at present. The qeing formla for average qee time is as follows: c a c e ( m1) 1 PT ) A where ca is the coefficient of variability in lot inter-arrival time, ce is the coefficient of variability in service time, is the tilization of availability, m is the nmber of machines allowed to perform service on the lot, PT is the average process time, and A is the machine availability. To answer the qestion, we don t need to know the vales of all of these parameters. Only the middle term of is changing, so it sffices to compte the ratio of the middle terms. Crrently, m = 10. Under the proposal, m = 1. So 6

7 proposed crrent (11) (1 0.87) (101) (1 0.87) c. By what percentage will the total photo qee time change if lot-to-lens dedication is implemented? Qee time at steps, 3, 4 and 5 wold increase by a factor of Qee time at step 1 wold not increase. So total qee time wold increase by a factor of [1 + 4*16.67]/5 =