# Essay on Operation Management

1571 Words7 Pages
| Cleaning Time | Waxing time | C1 | 9 | 6 | C2 | 6 | 2 | C3 | 5 | 7 | C4 | 3 | 5 | C5 | 8 | 4 | 41. (7 points) An automobile detailing shop has the following jobs waiting to be processed. All jobs (cars requiring detailing) must first be cleaned, then waxed. a. What processing sequence will minimize the makespan for these jobs? Job | Cleaning Time time | Cleaning Time end time | Waxing time time | Waxing time end time | Waxing time idle time | C4 | 3 | 3 | 5 | 8 | 3 | C3 | 5 | 8 | 7 | 15 | 0 | C1 | 9 | 17 | 6 | 23 | 2 | C5 | 8 | 25 | 4 | 29 | 2 | C2 | 6 | 31 | 2 | 33 | 2 | b. What is the minimum makespan for this set of five jobs? Makespan is 33 (POM SW output is attaced in excel) 42. (10 points)…show more content…
Sample Mean Range Sample | Mean | Range | 1 | 7.2 | 0.7 | 2 | 7.6 | 1 | 3 | 7.1 | 1.4 | 4 | 7.8 | 1.1 | 5 | 7.8 | 0.6 | 6 | 7.5 | 1.2 | _ a. Calculate the 3 X-chart and R-chart control limits. X bar = Average (Mean) = (7.2+7.6+7.1+7.8+7.8+7.5) / 6 = 7.50 Avge Range R bar = Avge (Range) = (0.7+1+1.4+1.1+0.6+1.2) / 6 = 1.00 Using Constant Table, we find the value of A2 for Subgroup of n=6 as A2 = 0.483 Now Upper control Limit UCL(X) = X + A2*R = 7.50+0.483*1 = 7.983 &amp; Lower control Limit LCL(X) = X - A2*R = 7.50 - 0.483*1 = 7.017 Now we calculate the Upper Control Limit for Range. Recall that when n &lt;7, LCL( R) = 0. Here n=6. So LCL(R) =0 &amp; UCL (R ) = D4*R Looking up Contant Table, we find for n=6, D4 = 2.282 So UCL (R ) = D4*R = 2.282*1 = 2.282 b. Calculate the mean (X) and range (R) for the following sample, which was taken from the same process at a later time. Item number: 1 2 3 4 5 Weight: 7.5 8.0 8.2 7.5 7.4 Mean (X) = Avge(Weights) = (7.5+8.0+8.2+7.5+7.4) / 5 = 7.72 Range (X) = Highest Value of Weight – Lowest value of weight = 8.2-7.4 = 0.8 Based on this sample and the control chart limits that you calculated in part (a), is the process in control? Why or why not? Based on X &amp; R UCL &amp; LCL found in (a ), we can say that Process is in Control. This is because, all Mean of samples