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# Operations & Manufacturing Strategies

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A project based on the book Factory Physics, written by Prof Hopp of the University of Michigan's Ross School of Business, which focuses on manufacturing and inventory based problems

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### Operations & Manufacturing Strategies

1. 1. OMS 605 1 Pro-Coat at HAL,Inc. A Case Study Write-up for the OMS 605 W-12 Term University of Michigan- Ross School of BusinessAniruddha Srinath Rehan Syed Sam Beck Yue Ma
2. 2. OMS 605 2 The layout Calculate for Practical Worst Case (PWC) Batch Size 60 Hours/Day 19.5 Process Std Dev or Load Process Conveyor Number Machine Time Time Trip Time of Setup Rate Time Name (min) (min) (min) Machines MTTF MTTR Avail Time (p/day) (min) Clean1 0.33 0 15 1 80 4 0.95 0 3377 36.5 Coat1 0.33 0 15 1 80 4 0.95 0 3377 36.5 Coat2 0.33 0 15 1 80 4 0.95 0 3377 36.5 Expose 103 10 - 5 300 4 0.99 10 3065 114.5 Develop 0.33 0 2.67 1 300 3 0.99 0 3510 22.7 Inspect 0.5 0.5 - 2 - - 1.00 0 4680 0.5 Bake 0.33 0 100 1 300 3 0.99 0 3510 121.0 MI 161 64 - 8 - - 1.00 0 3488 161.0 Touchup 9 0 - 1 - - 1.00 0 7800 9.0 538 Critical Process is Expose:-Critical Bottleneck = 2879/day Raw Process time= T0= 546 min= 546/(60*24)= 0.47 day Critical WIP= Critical bottleneck x Raw process time W 0= 2879 x 0.47= 1353 The Numbers- At a glance Critical WIP=W= 1500 (as stated in the class slide) Process Std Dev or Load Process Conveyor Number TH= w(w+W0-1)*rb= 1500/(1500+1343-1)*2879= 1520 Machine Time Time Trip Time of Setup However, Ken is only getting an average of around 1200 per day. Name (min) (min) (min) Machines MTTF MTTR Avail TimeClean1 0.33 0 15 1 80 4 0.95 0Coat1 0.33 0 15 1 80 4 0.95 0Coat2 0.33 0 15 1 80 4 0.95 0Expose 103 10 - 5 300 4 0.99 10Develop 0.33 0 2.67 1 300 3 0.99 0Inspect 0.5 0.5 - 2 - - 1.00 0Bake 0.33 0 100 1 300 3 0.99 0MI 161 64 - 8 - - 1.00 0Touchup 9 0 - 1 - - 1.00 0Note:- Present- TH= 304, WIP=1353-Consider Clean1, Coat1 & Coat2 together as part of a single equipment . For the equipment, consider the data of the conveyor- The exposing is done in batches (jobs, as per the problem) and the time is for the batch size.- The same analogy holds for touch up also. Aniruddha Srinath Rehan Syed Sam Beck Yue Ma
3. 3. OMS 605 3 Analysis of the 1200 Throughput situation VUT Calculator  Wallace J. Hopp and Mark L. Spearman, 2000 Load CCC Expose Load Develop Inspect Load Bake MI Touchup MEASURE: STATION: 1 2 3 4 5 6 7 8 9 10 Arrival Rate (parts/hr) ra 154.000 141.680 130.346 119.918 110.325 101.499 93.379 85.90837659 79.03570646 72.71284995 Arrival CV ca 2 0.000 0.000 0.468 0.632 0.357 0.358 0.473 0.460693456 0.7623802 0.739929982 Natural Process Time (hr) t0 0.006 0.750 0.029 0.006 0.045 0.008 0.006 1.667 0.045 0.003 Natural Process SCV c0 2 0.000 0.000 0.423 0.000 0.423 1.000 0.000 0.000 0.158 0.000 Number of Machines m 1 180 5 1 180.000 2 2 180.000 8 1 MTTF (hr) mf 80 26.66 300 300 300 200 200 300 200 200 MTTR (hr) mr 0 4 10 0 3 0 0 3 0 0 Availability A 1.000 0.870 0.968 1.000 0.990 1.000 1.000 0.99009901 1 1 Effective Process Time (failures only) te 0.006 0.863 0.030 0.006 0.045 0.008 0.006 1.683333333 0.044722222 0.0025 Eff Process SCV (failures only) c e2 0.000 1.210 22.245 0.000 1.745 1.000 0.000 0.035290658 0.158018595 0 Batch Size k 1 1 60 1 1 1 1 1 60 60 Setup Time (hr) ts 0.000 0.000 0.250 0.000 0.000 0.000 0.000 0 0 0 Setup Time SCV cs 2 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0 0 Arrival Rate of Batches ra /k 154.000 141.680 2.172 119.918 110.325 101.499 93.379 85.90837659 1.317261774 1.211880832 Eff Batch Process Time (failures+setups) te = kt0 /A+ts 0.006 0.863 2.024 0.006 0.045 0.008 0.006 1.683333333 2.683333333 0.15 Eff Batch Process Time Var (failures+setups) k*s0 2 /A2 + 2mr(1-A)kt0 /A+ss 2 0.000 0.900 1.167 0.000 0.004 0.000 0.000 0.1 0.018962963 0 Eff Process SCV (failures+setups) c e2 0.000 1.210 0.285 0.000 1.745 1.000 0.000 0.035290658 0.002633643 0 Utilization u 0.924 0.679 0.879 0.660 0.028 0.423 0.257 0.803402411 0.441831553 0.181782125 Departure SCV c d2 0.000 0.468 0.632 0.357 0.358 0.473 0.461 0.7623802 0.739929982 0.715479187 Yield y 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 Final Departure Rate ra *y 141.680 130.346 119.918 110.325 101.499 93.379 85.908 79.03570646 72.71284995 66.89582195 Final Departure SCV ycd2 +(1-y) 0.080 0.511 0.662 0.409 0.409 0.515 0.504 0.781389784 0.760735583 0.738240852 Utilization u 0.924 0.679 0.879 0.660 0.028 0.423 0.257 0.803402411 0.441831553 0.181782125 Throughput TH 141.680 130.346 119.918 110.325 101.499 93.379 85.908 79.03570646 72.71284995 66.89582195 Queue Time (hr) CTq 0.000 0.000 0.920 0.003 0.000 0.001 0.000 0.000228062 0.016261241 0.012329177 Cycle Time (hr) CTq+te 0.006 0.863 2.944 0.009 0.045 0.010 0.006 1.683561396 2.699594574 0.162329177 Cumulative Cycle Time (hr) S i (CTq(i)+te(i)) 0.006 0.869 3.813 3.821 3.866 3.876 3.882 5.565253056 8.26484763 8.427176808 WIP in Queue (jobs) ra CTq 0.000 0.001 119.930 0.404 0.000 0.143 0.011 0.019592463 1.285218674 0.896489625 WIP (jobs) ra CT 0.924 122.204 383.735 1.063 4.959 0.989 0.525 144.6320264 213.3643644 11.80341712 Cumulative WIP (jobs) S i (ra (i)CT(i)) 0.924 123.128 506.863 507.926 512.885 513.874 514.399 769.876 1010.453 1210.68 Throughput per day 2762.760 2541.739 2338.400 2151.328 1979.222 1820.884 1675.213 1541.196 1417.901 1304.469 ASSUMPTIONS1. Current arrival rate is 154 as that is the rate required to produce 3000 with no yield loss.2. Accordingly, our initial yield rate is 92% to get us closer to the daily throughput of 12003. The starving and blocking at the expose bottleneck brings down the 1304 projected throughput to the 1200 throughput given in the case.4. Cleaning, Coat 1, and Coat 2 are cominable into one conveyer5. Conveyers can be models as the loader and belt6. The parallel servers in the belt is equal to the capacity of the loader, which is 180 in this case. Aniruddha Srinath Rehan Syed Sam Beck Yue Ma
4. 4. OMS 605 4 T Options to work out TH=1304, CT=8.42 hrs, WIP=1211 TH=1353, CT=9.1 hrs, WIP=1500 1. Increase working hours 2. Cut MTTR of BN by half 6. Add operator to BN 3. Cut set-up time of BN 7. Have someone working BN processes during breaks 4. Maximise use of BN process & shift changes 5. Decrease Std variation of process time-Reducing variability decreases utilization.-However, that alone does not increase throughput.-We need lesser WIP in order to achieve greater throughput at thesame utilization level.-However, we also need to expand capacity significantly (like adding anotherexpose machine) in order to really reach 3000 units. `-This will require a huge cost investment as well as a large expansion in thebuilding.-We want to try the other options mentioned above. Aniruddha Srinath Rehan Syed Sam Beck Yue Ma
5. 5. OMS 605 51. Increase Working Hours to 21 hours- Get .5 hours by making members in each shift work 10 minutes overtime (during shift change) We can have workers from less utilized areas take a break when the line is running and ask them to work on the bottlenecks during breaksIncrease the cycle timeVUT Calculator  Wallace J. Hopp and Mark L. Spearman, 2000 Load CCC Expose Load Develop Inspect Load Bake MI TouchupMEASURE: STATION: 1 2 3 4 5 6 7 8 9 10Arrival Rate (parts/hr) ra 165.846 152.578 140.372 129.142 118.811 109.306 100.562 92.51671325 85.11537619 78.3061461Arrival CV ca 2 0.000 0.000 0.543 0.666 0.330 0.331 0.469 0.455976946 0.809392828 0.772712072Natural Process Time (hr) t0 0.006 0.750 0.029 0.006 0.045 0.008 0.006 1.667 0.045 0.003Natural Process SCV c0 2 0.000 0.000 0.423 0.000 0.423 1.000 0.000 0.000 0.158 0.000Number of Machines m 1 180 5 1 180.000 2 2 180.000 8 1MTTF (hr) mf 80 26.66 300 300 300 200 200 300 200 200MTTR (hr) mr 0 4 10 0 3 0 0 3 0 0Availability A 1.000 0.870 0.968 1.000 0.990 1.000 1.000 0.99009901 1 1Effective Process Time (failures only) te 0.006 0.863 0.030 0.006 0.045 0.008 0.006 1.683333333 0.044722222 0.0025Eff Process SCV (failures only) c e2 0.000 1.210 22.245 0.000 1.745 1.000 0.000 0.035290658 0.158018595 0Batch Size k 1 1 60 1 1 1 1 1 60 60Setup Time (hr) ts 0.000 0.000 0.250 0.000 0.000 0.000 0.000 0 0 0Setup Time SCV cs 2 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0 0Arrival Rate of Batches ra /k 165.846 152.578 2.340 129.142 118.811 109.306 100.562 92.51671325 1.418589603 1.305102435Eff Batch Process Time (failures+setups) te = kt0 /A+ts 0.006 0.863 2.024 0.006 0.045 0.008 0.006 1.683333333 2.683333333 0.15Eff Batch Process Time Var (failures+setups) k*s0 2 /A2 + 2mr(1-A)kt0 /A+ss 2 0.000 0.900 1.167 0.000 0.004 0.000 0.000 0.1 0.018962963 0Eff Process SCV (failures+setups) c e2 0.000 1.210 0.285 0.000 1.745 1.000 0.000 0.035290658 0.002633643 0Utilization u 0.995 0.731 0.947 0.710 0.030 0.455 0.277 0.865202596 0.475818596 0.195765365Departure SCV c d2 0.000 0.543 0.666 0.330 0.331 0.469 0.456 0.809392828 0.772712072 0.743098594Yield y 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920Final Departure Rate ra *y 152.578 140.372 129.142 118.811 109.306 100.562 92.517 85.11537619 78.3061461 72.04165441Final Departure SCV ycd2 +(1-y) 0.080 0.579 0.693 0.384 0.384 0.512 0.499 0.824641402 0.790895107 0.763650707Utilization u 0.995 0.731 0.947 0.710 0.030 0.455 0.277 0.865202596 0.475818596 0.195765365Throughput TH 152.578 140.372 129.142 118.811 109.306 100.562 92.517 85.11537619 78.3061461 72.04165441Queue Time (hr) CTq 0.000 0.000 2.763 0.004 0.000 0.002 0.000 0.001253058 0.023372309 0.014106915Cycle Time (hr) CTq+te 0.006 0.863 4.787 0.010 0.045 0.010 0.006 1.684586391 2.706705642 0.164106915Cumulative Cycle Time (hr) S i (CTq(i)+te(i)) 0.006 0.869 5.656 5.666 5.711 5.721 5.726 7.411039918 10.11774556 10.28185247WIP in Queue (jobs) ra CTq 0.000 0.006 387.913 0.580 0.000 0.178 0.014 0.11592882 1.98934286 1.104658142WIP (jobs) ra CT 0.995 131.609 672.011 1.290 5.340 1.089 0.567 155.8523961 230.382269 12.85058006Cumulative WIP (jobs) S i (ra (i)CT(i)) 0.995 132.604 804.615 805.905 811.245 812.334 812.901 968.7535676 1199.135837 1211.986417Throughput per day 2975.280 2737.258 2518.277 2316.815 2131.470 1960.952 1804.076 1659.750 1526.970 1404.812 Cumulative Change WIP 1211 0% Change TH 1404 8% Increase Aniruddha Srinath Rehan Syed Sam Beck Yue Ma
6. 6. OMS 605 62. Add workers to the expose process. In the previous step, we had shown the number of machines in expose as 5 as we were havingpeople from other processes pitch to complete the expose process. Increasing manpower here would create a more robust process. Add operators for expose- it is not increasing throughput but is improving my process VUT Calculator  Wallace J. Hopp and Mark L. Spearman, 2000 Load CCC Expose Load Develop Inspect Load Bake MI Touchup MEASURE: STATION: 1 2 3 4 5 6 7 8 9 10 Arrival Rate (parts/hr) ra 165.846 152.578 140.372 129.142 118.811 109.306 100.562 92.51671325 85.11537619 78.3061461 Arrival CV ca 2 0.000 0.000 0.543 0.666 0.330 0.331 0.469 0.455976946 0.809392828 0.772712072 Natural Process Time (hr) t0 0.006 0.750 0.029 0.006 0.045 0.008 0.006 1.667 0.045 0.003 Natural Process SCV c0 2 0.000 0.000 0.423 0.000 0.423 1.000 0.000 0.000 0.158 0.000 Number of Machines m 1 180 5 1 180.000 2 2 180.000 8 1 MTTF (hr) mf 80 26.66 300 300 300 200 200 300 200 200 MTTR (hr) mr 0 4 10 0 3 0 0 3 0 0 Availability A 1.000 0.870 0.968 1.000 0.990 1.000 1.000 0.99009901 1 1 Effective Process Time (failures only) te 0.006 0.863 0.030 0.006 0.045 0.008 0.006 1.683333333 0.044722222 0.0025 Eff Process SCV (failures only) c e2 0.000 1.210 22.245 0.000 1.745 1.000 0.000 0.035290658 0.158018595 0 Batch Size k 1 1 60 1 1 1 1 1 60 60 Setup Time (hr) ts 0.000 0.000 0.250 0.000 0.000 0.000 0.000 0 0 0 Setup Time SCV cs 2 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0 0 Arrival Rate of Batches ra /k 165.846 152.578 2.340 129.142 118.811 109.306 100.562 92.51671325 1.418589603 1.305102435 Eff Batch Process Time (failures+setups) te = kt0 /A+ts 0.006 0.863 2.024 0.006 0.045 0.008 0.006 1.683333333 2.683333333 0.15 Eff Batch Process Time Var (failures+setups) k*s0 2 /A2 + 2mr(1-A)kt0 /A+ss 2 0.000 0.900 1.167 0.000 0.004 0.000 0.000 0.1 0.018962963 0 2 Eff Process SCV (failures+setups) ce 0.000 1.210 0.285 0.000 1.745 1.000 0.000 0.035290658 0.002633643 0 Utilization u 0.995 0.731 0.947 0.710 0.030 0.455 0.277 0.865202596 0.475818596 0.195765365 Departure SCV c d2 0.000 0.543 0.666 0.330 0.331 0.469 0.456 0.809392828 0.772712072 0.743098594 Yield y 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 Final Departure Rate ra *y 152.578 140.372 129.142 118.811 109.306 100.562 92.517 85.11537619 78.3061461 72.04165441 Final Departure SCV ycd2 +(1-y) 0.080 0.579 0.693 0.384 0.384 0.512 0.499 0.824641402 0.790895107 0.763650707 Utilization u 0.995 0.731 0.947 0.710 0.030 0.455 0.277 0.865202596 0.475818596 0.195765365 Throughput TH 152.578 140.372 129.142 118.811 109.306 100.562 92.517 85.11537619 78.3061461 72.04165441 Queue Time (hr) CTq 0.000 0.000 2.763 0.004 0.000 0.002 0.000 0.001253058 0.023372309 0.014106915 Cycle Time (hr) CTq+te 0.006 0.863 4.787 0.010 0.045 0.010 0.006 1.684586391 2.706705642 0.164106915 Cumulative Cycle Time (hr) S i (CTq(i)+te(i)) 0.006 0.869 5.656 5.666 5.711 5.721 5.726 7.411039918 10.11774556 10.28185247 WIP in Queue (jobs) ra CTq 0.000 0.006 387.913 0.580 0.000 0.178 0.014 0.11592882 1.98934286 1.104658142 WIP (jobs) ra CT 0.995 131.609 672.011 1.290 5.340 1.089 0.567 155.8523961 230.382269 12.85058006 Cumulative WIP (jobs) S i (ra (i)CT(i)) 0.995 132.604 804.615 805.905 811.245 812.334 812.901 968.7535676 1199.135837 1211.986417 Throughput per day 2975.280 2737.258 2518.277 2316.815 2131.470 1960.952 1804.076 1659.750 1526.970 1404.812 Cumulative Change WIP 1211 0% Change TH 1404 8% Increase Aniruddha Srinath Rehan Syed Sam Beck Yue Ma
7. 7. OMS 605 73. Reduce set-up & repair time through out the line- This is decreasing my WIPReduce set up time & repair timeVUT Calculator  Wallace J. Hopp and Mark L. Spearman, 2000 Load CCC Expose Load Develop Inspect Load Bake MI TouchupMEASURE: STATION: 1 2 3 4 5 6 7 8 9 10Arrival Rate (parts/hr) ra 165.846 152.578 140.372 129.142 118.811 109.306 100.562 92.51671325 85.11537619 78.3061461Arrival CV ca 2 0.000 0.000 0.444 0.558 0.276 0.277 0.427 0.416611768 0.794256567 0.761002711Natural Process Time (hr) t0 0.006 0.750 0.029 0.006 0.045 0.008 0.006 1.667 0.045 0.003Natural Process SCV c0 2 0.000 0.000 0.423 0.000 0.423 1.000 0.000 0.000 0.158 0.000Number of Machines m 1 180 5 1 180.000 2 2 180.000 8 1MTTF (hr) mf 80 26.66 300 300 300 200 200 300 200 200MTTR (hr) mr 0 2 5 0 1.5 0 0 1.5 0 0Availability A 1.000 0.930 0.984 1.000 0.995 1.000 1.000 0.995024876 1 1Effective Process Time (failures only) te 0.006 0.806 0.029 0.006 0.045 0.008 0.006 1.675 0.044722222 0.0025Eff Process SCV (failures only) c e2 0.000 0.346 6.059 0.000 0.757 1.000 0.000 0.008910671 0.158018595 0Batch Size k 1 1 60 1 1 1 1 1 60 60Setup Time (hr) ts 0.000 0.000 0.125 0.000 0.000 0.000 0.000 0 0 0Setup Time SCV cs 2 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0 0Arrival Rate of Batches ra /k 165.846 152.578 2.340 129.142 118.811 109.306 100.562 92.51671325 1.418589603 1.305102435Eff Batch Process Time (failures+setups) te = kt0 /A+ts 0.006 0.806 1.870 0.006 0.045 0.008 0.006 1.675 2.683333333 0.15Eff Batch Process Time Var (failures+setups) k*s0 2 /A2 + 2mr(1-A)kt0 /A+ss 2 0.000 0.225 0.308 0.000 0.002 0.000 0.000 0.025 0.018962963 0Eff Process SCV (failures+setups) c e2 0.000 0.346 0.088 0.000 0.757 1.000 0.000 0.008910671 0.002633643 0Utilization u 0.995 0.683 0.875 0.710 0.030 0.455 0.277 0.860919415 0.475818596 0.195765365Departure SCV c d2 0.000 0.444 0.558 0.276 0.277 0.427 0.417 0.794256567 0.761002711 0.731837984Yield y 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920 0.920Final Departure Rate ra *y 152.578 140.372 129.142 118.811 109.306 100.562 92.517 85.11537619 78.3061461 72.04165441Final Departure SCV ycd2 +(1-y) 0.080 0.489 0.593 0.334 0.335 0.473 0.463 0.810716042 0.780122494 0.753290945Utilization u 0.995 0.683 0.875 0.710 0.030 0.455 0.277 0.860919415 0.475818596 0.195765365Throughput TH 152.578 140.372 129.142 118.811 109.306 100.562 92.517 85.11537619 78.3061461 72.04165441Queue Time (hr) CTq 0.000 0.000 0.574 0.004 0.000 0.002 0.000 0.000957157 0.022936646 0.013893145Cycle Time (hr) CTq+te 0.006 0.806 2.444 0.009 0.045 0.010 0.006 1.675957157 2.70626998 0.163893145Cumulative Cycle Time (hr) S i (CTq(i)+te(i)) 0.006 0.812 3.256 3.266 3.310 3.320 3.326 5.001815286 7.708085266 7.87197841WIP in Queue (jobs) ra CTq 0.000 0.000 80.547 0.485 0.000 0.171 0.013 0.088552977 1.952261294 1.087918606WIP (jobs) ra CT 0.995 123.019 343.082 1.196 5.314 1.082 0.566 155.0540477 230.3451874 12.83384052Cumulative WIP (jobs) S i (ra (i)CT(i)) 0.995 124.014 467.096 468.292 473.605 474.687 475.253 630.3067463 860.6519337 873.4857743Throughput per day 2975.280 2737.258 2518.277 2316.815 2131.470 1960.952 1804.076 1659.750 1526.970 1404.812 Cumulative Change WIP 874 28% Change TH 1404 8% Increase Aniruddha Srinath Rehan Syed Sam Beck Yue Ma
8. 8. OMS 605 84. Increase the yield to 94%. This increases my throughput but also requires me to further reduce my BN set-up time Increase yield & also reduce expose set up time VUT Calculator  Wallace J. Hopp and Mark L. Spearman, 2000 Load CCC Expose Load Develop Inspect Load Bake MI Touchup MEASURE: STATION: 1 2 3 4 5 6 7 8 9 10 Arrival Rate (parts/hr) ra 165.846 155.895 146.542 137.749 129.484 121.715 114.412 107.5475149 101.094664 95.02898412 Arrival CV ca 2 0.000 0.000 0.464 0.547 0.233 0.234 0.431 0.417161598 0.916188957 0.732581723 Natural Process Time (hr) t0 0.006 0.750 0.029 0.006 0.045 0.008 0.006 1.667 0.045 0.003 Natural Process SCV c0 2 0.000 0.000 0.423 0.000 0.423 1.000 0.000 0.000 0.158 0.000 Number of Machines m 1 180 5 1 180.000 2 2 182.000 6 1 MTTF (hr) mf 80 26.66 300 300 300 200 200 300 200 200 MTTR (hr) mr 0 2 2.5 0 1.5 0 0 1.5 0 0 Availability A 1.000 0.930 0.992 1.000 0.995 1.000 1.000 0.995024876 1 1 Effective Process Time (failures only) te 0.006 0.806 0.029 0.006 0.045 0.008 0.006 1.675 0.044722222 0.0025 Eff Process SCV (failures only) c e2 0.000 0.346 1.855 0.000 0.757 1.000 0.000 0.008910671 0.158018595 0 Batch Size k 1 1 60 1 1 1 1 1 60 60 Setup Time (hr) ts 0.000 0.000 0.125 0.000 0.000 0.000 0.000 0 0 0 Setup Time SCV cs 2 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0 0 0 Arrival Rate of Batches ra /k 165.846 155.895 2.442 137.749 129.484 121.715 114.412 107.5475149 1.684911066 1.583816402 Eff Batch Process Time (failures+setups) te = kt0 /A+ts 0.006 0.806 1.856 0.006 0.045 0.008 0.006 1.675 2.683333333 0.15 Eff Batch Process Time Var (failures+setups) k*s0 2 /A2 + 2mr(1-A)kt0 /A+ss 2 0.000 0.225 0.093 0.000 0.002 0.000 0.000 0.025 0.018962963 0 Eff Process SCV (failures+setups) c e2 0.000 0.346 0.027 0.000 0.757 1.000 0.000 0.008910671 0.002633643 0 Utilization u 0.995 0.698 0.907 0.758 0.032 0.507 0.315 0.989791689 0.753529671 0.23757246 Departure SCV c d2 0.000 0.464 0.547 0.233 0.234 0.431 0.417 0.916188957 0.732581723 0.691234317 Yield y 0.940 0.940 0.940 0.940 0.940 0.940 0.940 0.940 0.940 0.940 Final Departure Rate ra *y 155.895 146.542 137.749 129.484 121.715 114.412 107.548 101.094664 95.02898412 89.32724508 Final Departure SCV ycd2 +(1-y) 0.060 0.496 0.574 0.279 0.280 0.465 0.452 0.921217619 0.74862682 0.709760258 Utilization u 0.995 0.698 0.907 0.758 0.032 0.507 0.315 0.989791689 0.753529671 0.23757246 Throughput TH 155.895 146.542 137.749 129.484 121.715 114.412 107.548 101.094664 95.02898412 89.32724508 Queue Time (hr) CTq 0.000 0.000 0.766 0.005 0.000 0.002 0.000 0.159458224 0.383717576 0.017120438 Cycle Time (hr) CTq+te 0.006 0.806 2.622 0.010 0.045 0.010 0.006 1.834458224 3.067050909 0.167120438 Cumulative Cycle Time (hr) S i (CTq(i)+te(i)) 0.006 0.812 3.434 3.444 3.489 3.499 3.505 5.33933991 8.40639082 8.573511257 WIP in Queue (jobs) ra CTq 0.000 0.001 112.218 0.647 0.000 0.237 0.019 17.14933575 38.79179941 1.6269378 WIP (jobs) ra CT 0.995 125.693 384.195 1.405 5.791 1.252 0.648 197.2914231 310.062481 15.88128542 Cumulative WIP (jobs) S i (ra (i)CT(i)) 0.995 126.689 510.884 512.289 518.080 519.331 519.979 717.2703172 1027.332798 1043.214084 Throughput per day 3039.960 2857.562 2686.109 2524.942 2373.446 2231.039 2097.177 1971.346 1853.065 1741.881 Cumulative Change WIP 1043 14% Change TH 1741 34% Increase Aniruddha Srinath Rehan Syed Sam Beck Yue Ma
9. 9. OMS 605 9 In Conclusion, but in need of further action TH & WIP condition after implementation of proposals 1~4 Initial TH & WIP condition- After carrying out 4 changes in the HAL procoat line, we have been able to take our TH, WIP situation closer to the Best Case Line- However, we have concluded that we are still not able to meet Kens requirement of 3000 boards a day.- We have worked on the WIP front, reducing the WIP to the best extent possible.- If Ken and Mr Petitto still insist on the 3000 figure, they need to make significant investments in equipment (like the Expose equipment) and building, as well as increasing manpower, which is a business stand-off. Aniruddha Srinath Rehan Syed Sam Beck Yue Ma
10. 10. OMS 605 10THANK YOU FOR THE INITIATION INTO THE FACTORY PHYSICS CULT ! Aniruddha Srinath Rehan Syed Sam Beck Yue Ma