HW-03-31-10s-Chs19-21

HW-03-31-10s-Chs19-21 - IE 131 Solutions for Problems due...

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IE 131 Solutions for Problems due Mar 31 (Chs 19, 20, and 21) 19.1 In a mechanical assembly operation, the first work unit required 7.83 min to complete and the learning rate for mechanical assembly is 84% in the Crawford model. Using this learning curve model, determine the unit times to produce (a) the second unit, (b) the 10 th unit, and (c) the 100 th unit. (d) What are the total cumulative time and the average cumulative time for the 100 units? Solution : (a) m = ln(0.84)/ln 2 = -0.25154 T 2 = 7.83(2) -0.25154 = 7.84(0.84) = 6.58 min (b) T 10 = 7.83(10) -0.25154 = 7.84(0.5604) = 4.39 min (c) T 100 = 7.83(100) -0.25154 = 7.84(0.3140) = 2.46 min (d) E ( TT 100 ) = 7.83{(100.5 1-0.25154 – 0.5 1-0.25154 )/(1 – 0.25154) } E ( TT 100 ) = 7.83{(31.517 – 0.595)/0.74846} = 7.83(41.314) = 323.49 min = 5.39 hr 100 T = 323.49/100 = 3.235 min 19.2 In a certain manual operation, the first work unit required 7.83 min to complete and the learning rate is known to be 84% in the Wright model. Using this learning curve model, determine the average cumulative times to produce (a) the second unit, (b) the 10 th unit, and (c) the 100 th unit. (d) Find the total cumulative time for the 100 units and the unit time for the 100th unit. Solution : (a) m = ln(0.84)/ln 2 = -0.25154 2 T = 7.83(2) -0.25154 = 7.84(0.84) = 6.58 min (b) 10 T = 7.83(10) -0.25154 = 7.84(0.5604) = 4.39 min (c) 100 T = 7.83(100) -0.25154 = 7.84(0.3140) = 2.46 min (d) TT 100 = 7.83(100) 1-0.25154 = 7.83(31.40) = 245.86 min = 4.098 hr TT 99 = 7.83(99) 1-0.25154 = 7.83(31.164) = 244.02 min = 4.067 hr T 100 = 245.86 – 244.02 = 1.84 min 19.5 A worker produces 13 parts during the first day on a new job, and the first part takes 46 min (unit time). On the second day, the worker produces 20 parts, and the 20th part on the second day takes 18 min. Using the Crawford learning curve model, determine (a) the learning rate and (b) the total cumulative time to produce all 33 parts. Solution : (a) Given T 1 = 46 min and T 13+20 = T 33 = 18 min T 33 = T 1 (33) m 18 = 46(33) m 0.3913 = 33 m m ln 33 = ln 0.3913 3.4965 m = -0.9383 m = -0.2683 LR = 2 -0.2683 = 0.830 = 83% (b) E ( TT 33 ) = 46{(33.5 1-0.2683 – 0.5 1-0.2683 )/(1-0.2683)} E ( TT 33 ) = 46{(13.056- 0.602)/0.7317} = 783 min = 13.05 hr 19.16 An automobile-customizing firm makes alterations of new cars to satisfy individual requirements of its clients. Jobs include customizing of cars for police and fire departments, limousine fleets, etc. For a certain order for 4 customized cars, it took 100 hours to remake the first
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HW-03-31-10s-Chs19-21 - IE 131 Solutions for Problems due...

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