OPERATION & PRODUCTION IN INDUSTRIAL MANAGEMENT

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OPERATION & PRODUCTION IN INDUSTRIAL MANAGEMENT (BPT 1113)
SEMESTER II SESSION 2012/2013
EXERCISE #2
1. If Caroline has times of 8.4, 8.6, 8.3, 8.5, 8.7 and 8.5 and a performance rating of 110%, what is
the normal time for this operation? Is she faster or slower than normal?
2. Coco Jelly is famous for its boxed candies which are sold primarily to businesses. One operator
had the following observed times for gift wrapping in minutes: 2.2, 2.6, 2.3, 2.5 and 2.4. The
operator has a performance rating of 105% and an allowance factor of 10%. What is the standard
time for gigt wrapping?
3. Peter, a loan processor at Swiss Bank, has been timed performing four work elements, with the
results shown in the following table. The allowances for tasks such as this are personal, 7%;
fatigue, 10% and delay, 3%.
Task
Performance
Observations (minutes)
Element
Rating (%)
1
2
3
4
5
1
110
0.5
0.4
0.6
0.4
0.4
2
95
0.6
0.8
0.7
0.6
0.7
3
90
0.6
0.4
0.7
0.5
0.5
4
85
1.5
1.8
2.0
1.7
1.5
a) What is the normal time?
b) What is the standard time?
4. A time study of a factory worker has revealed an average observed time of 3.2 minutes with a
standard deviation of 1.28 minutes. These figures were based on a sample of 45 observations. Is
this sample adequate in size for the firm to be 99% confident that the standard time is within 5% of
the true vale? If not, what should the proper number of observations?
5. A total of 300 observations of Bob, an assembly line worker were made over a 40-hour work week.
The sample also showed that Bob was busy working (assembling the parts) during 250
observations.
a) Find the percentage of time Bob was working.
b) If you want a confidence level of 95% and if 3% is an acceptable error, what size should the
sample be?
ANSWER
1. NT = Avg  PR = 8.5  1.10 = 9.35 seconds; worker is faster than normal
2. Avg =
2.2  2.6  2.3  2.5  2.4
 2.4 minutes
5
 NT = 2.4  1.05 = 2.52
2.52
 ST  1  0.10  2.8 minutes
3.
Observations (minutes)
(actual time)
Task
Element
Performance
Rating
1
2
3
4
5
110%
95%
90%
85%
0.5
0.6
0.6
1.5
0.4
0.8
0.4
1.8
0.6
0.7
0.7
2.0
0.4
0.6
0.5
1.7
0.4
0.7
0.5
1.5
1
2
3
4
(a)
Solutions
Actual
Time Average
NormalTime
0.46
0.68
0.54
1.70
Total
0.506
0.646
0.486
1.445
3.083
Normal time 3.083 minutes
Normal time
1  Total allowance
3.083

 3.85 minutes
1.0  0.20
(b) Standard time 
4.
10.27
2
2
2
 (2.58)(1.28) 
 zs 
 3.30 
n  

 0.16   426
 hx 
 (0.05)(3.20) 
where z  2.58, s  1.28, h  0.05, x  3.20
 Sample
5.
(a)
size 45 is not adequate. They need 381 more observations.
250
 .833  83.3%
300
(b) n 
Z 2 p(1  p)
h2
 (at 95% confidence level and 3% acceptable error)
n
(1.96)2 (0.167)(0.833)
(0.03)2
(3.84)(0.167)(0.833)

 593.7  594
0.0009
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