04 - PERT
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Transcript of 04 - PERT
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P R T
I.K. Gunarta
Department of Industrial EngineeringITS
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4
The technique is based on theassumption that an activitys duration
follows a probability distributioninstead of being a single value.
The probabilistic information about the
activities is translated into probabilisticinformation about the project.
Project Evaluation and Review Technique (PERT)
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PERT
Three time estimates are required to
compute the parameters of an activitys
duration distribution:
pessimistic time (tp ) - the time the activity wouldtake if things did not go well
most likely time (tm ) - the consensus best estimateof the activitys duration
optimistic time (to ) - the time the activity wouldtake if things did go well
. . . more
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From these three time estimates aboutan activity, two probability distribution
parameters are calculated: the mean(te ) and the variance (Vt ).
te = ( to + 4tm + tp ) / 6
Vt = [ ( tp - to ) / 6 ]2
PERT
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Steps in PERT Analysis
Draw the network.
Analyze the paths through the network
and find the critical path. The length of the critical path is the
mean of the project duration
probability distribution which isassumed to be normal.
. . . more
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Steps in PERT Analysis
The standard deviation of the projectduration probability distribution is
computed by adding the variances ofthe critical activities (all of the activitiesthat make up the critical path) andtaking the square root of that sum
Probability computations can now bemade using the normal distributiontable.
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Contoh
Aktivitas A M B Te [(b a)/6]2
A 17 29 47 30 25
B 6 12 24 13 9C 16 19 28 20 4
D 13 16 19 16 1
E 2 5 14 6 4
F 2 5 8 5 1
Berapa probabilitas proyek selesai dalam waktu kurang dari 67 hari?
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Network
0 A
B
C
D
E
F
30
30
30
30
13
43
50
20
43
16
59
50
6
56
59
5
64
6459
59
59
43
53
4330
5333
300
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Possible Project Duration
Ts = 67Te = 64
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Solusi
V Path = VA + VB + VD + VF
V Path = 25 + 9 + 1 + 1
path = 6
Z = (67 64)/
= + 0.5
P = 0.69
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Immed. Optimistic Most Likely Pessimistic
Activity Predec. Time (Hr.) Time (Hr.) Time (Hr.)
A -- 4 6 8
B -- 1 4.5 5C A 3 3 3D A 4 5 6E A 0.5 1 1.5F B,C 3 4 5G B,C 1 1.5 5
H E,F 5 6 7I E,F 2 5 8J D,H 2.5 2.75 4.5K G,I 3 5 7
PERT Example
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PERT Example
ACTIVITY PREDECESSOR OPTIMISTIC
TIME
MOST
LIKELY
TIME
PESSIMISTIC
TIME
A - 4 6 8
B - 1 4,5 5
C A 3 3 3
D A 4 5 6
E A 0,5 1 1,5
F B, C 3 4 5
G B, C 1 1,5 5
H E, F 5 6 7
I E, F 2 5 8
J D, H 2,5 2,75 4,5
K G, I 3 5 7
Probability the project will be completed within 24 days?
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PERT Network
A
D
C
B
F
E
G
I
H
K
J
PERT Example
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Activity Expected Time and Variance
Activity Expected Time Variance
A 6 4/9
B 4 4/9C 3 0D 5 1/9E 1 1/36
F 4 1/9G 2 4/9H 6 1/9I 5 1J 3 1/9K 5 4/9
PERT Example
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Earliest/Latest Times
Activity ES EF LS LF Slack A 0 6 0 6 0 *critical
B 0 4 5 9 5
C 6 9 6 9 0 *D 6 11 15 20 9E 6 7 12 13 6F 9 13 9 13 0 *G 9 11 16 18 7
H 13 19 14 20 1I 13 18 13 18 0 *J 19 22 20 23 1K 18 23 18 23 0 *
PERT Example
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Probability the project will be completedwithin 24 hours
Vpath = VA + VC + VF + VI + VK= 4/9 + 0 + 1/9 + 1 + 4/9
= 2
path = 1.414
z = (24 - 23)/ = (24-23)/1.414 = .71
PERT Example
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Probability the project will be completedwithin 24 hours
From the Standard Normal Distributiontable:
P(z < .71) = .5 + .2612 = .7612
23 24
.5000
.2612
PERT Example