Answer Happy

factory physics
Consider two different machines A and B that could be used at a station. Machine A has a mean effective process time te of 1.0 hour and an SCV ² of 0.25. Machine B has a mean effective process time of 0.85 hour and an SCV of 4. In the tables below TH is the throughput that is given, te, is the mean effective process time, c² is the effective SCV of the process times, u is the utilization given by (TH) (te), CT is the expected time in queue given by (c²+²) 2 CTq CT is simply the sum of CT and te. - U (1-u) te for the single machine From HW#2, we have computed the following: Part (a) Machine TH-Throughput (jobs/hr) te(hour) c²² m u-utilization (4 decimal place) CT (hr) (4 decimal place) CT (hr) (4 decimal place) A 0.92 1 0.25 1 0.9200 7.1875 8.1875 B 0.92 0.85 4 1 0.7820 7.6227 8.4727

Now, with only one machine at each station, let the arrival rate be 0.95 job per hour with c 1. Recompute the average time spent at the stations for both machine A and machine B. Compare with Part(a). See below. Part (c) Machine TH-Throughput (jobs/hr) te (hour) c² m u-utilization (4 decimal place) CT (hr) (4 decimal place) CT (hr) (4 decimal place) A 0.95 1 0.25 1 a) Compute utilization for Machine A and Machine B (5 points: 2.5 point each) b) Compute CT for Machine A and Machine B (10 points: 5 point each) c) Compute CT for Machine A and Machine B (10 points: 5 point each) B 0.95 0.85 4 1 = d) Compare to Part (a), what is the percentage increased in Cycle Time at Machine A? (5 points) [Note: A modest increase in TH, yields a large increase in CT.]