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
10.25154
– 0.5
10.25154
)/(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)
10.25154
= 7.83(31.40) = 245.86 min = 4.098 hr
TT
99
= 7.83(99)
10.25154
= 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
10.2683
– 0.5
10.2683
)/(10.2683)}
E
(
TT
33
) = 46{(13.056 0.602)/0.7317} = 783 min = 13.05 hr
19.16 An automobilecustomizing 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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 Spring '08
 Groover
 sigma level, Appendix 21A, DUPM, Ndu

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