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A399 Pert CPM

The critical path is A-B-C-D-E-F with an expected project completion time of 55 days. Using the time estimates (te), the earliest and latest start and finish times are calculated for each activity along with any slack. Crashing plans are determined by finding the crash time and cost per day for each activity to see how many days can be reduced from the project by crashing certain activities. Crashing activities D and F would reduce the project completion time by 3 days at a cost of 2000 + 2000 + 2000 + 3000 = 9000.

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127 views3 pages

A399 Pert CPM

The critical path is A-B-C-D-E-F with an expected project completion time of 55 days. Using the time estimates (te), the earliest and latest start and finish times are calculated for each activity along with any slack. Crashing plans are determined by finding the crash time and cost per day for each activity to see how many days can be reduced from the project by crashing certain activities. Crashing activities D and F would reduce the project completion time by 3 days at a cost of 2000 + 2000 + 2000 + 3000 = 9000.

Uploaded by

Eyyy
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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hjb

EXERCISE ON PERT-CPM

1. The activities and time estimate for a particular project are as follows:

Immediate time
  in days    

ACTIVITY Predecessor To Tm Tp Te

A - 7 8 15 9

B A 10 11 12 11

C A,B 5 7 9 7

D B, C 11 14 17 14

E D 5 7 9 7

F E 5 7 11 7

A = (7+4*8+15)/6 = 9 D = (11+4*14+17) = 14

B = (10+4*11+12)/6 = 11 E = (5+4*7+9)/6 = 7

C = (5+4*7+9)/6 = 7 F = (5+4*7+11)/6 = 7.33 or 7

a. Construct the network diagram:


ES EF
B 9 20
11 9 20
LS LF

ES EF ES EF ES EF ES EF
A 0 9 D 27 41 E 41 48 F 48 55
START END
9 0 9 14 27 41 7 41 48 7 48 55
LS LF LS LF LS LF LS LF

ES EF
C 20 27
7 20 27
LS LF

What is the critical path and expected project completion time? ANS.

The critical path is A-B-C-D-E-F and the expected project completion time is 55 days.

b. Determine the expected projected completion time and critical path.


hjb

c. Using te, determine the earliest and latest activity times and slack times. Given ES=0 and
EF=LF.

Activity te LS ES LF EF Slack CP?

0 0 9 9 0
A 9 YES
(9-9) (ES = 0) (LS OF B) (9+0) (0-0)

9 9 20 20 0
B 11 YES
(20-11) (EF of A) (LS OF C) (11+9) (9-9)

20 20 27 27 0
C 7 YES
(27-7) (EF of B) (LS OF D) (7+20 (20-20)

27 27 41 41 0
D 14 YES
(41-14) (EF of C ) (LS OF E) (14+27) (27-27)

41 41 48 48 0
E 7 YES
(48-7) (EF of D) (LS OF F) (7+41) (41-41)

48 48 55 55 0
F 7 YES
(55-7) (EF of E) (EF = LF) (7 +48) (48-48)

d. Using te as the Normal time, compute determine an optimum crashing plan.

Activity Normal time crash time cost/day

A 9 6 12000

B 11 8 4000

C 7 4 7000

D 14 11 2000

E 7 5 7000

F 7 4 3000

ACTIVITY COST/DAY AVAIL.DAY

A 12,000 3
hjb

(9 – 6)
3
B 4,000
(11 – 8)
3
C 7,000
(7 – 4)
3
D 2,000
(14 – 11)
2
E 7,000
(7 – 5)
3
F 3,000
(7 – 4)

Length after crashing in days

Path ‘n=0 1 2 3 4 5

A - B -C -D - E - F 55 54 53 52 51 50

A-B-D-E-F 48 47 46 45 44 43

A - C - D - E -F 44 43 42 41 40 39

Activity Crashed D D D F F

Cost 2000 2000 2000 3000 3000

Deduct one day from the base time (n=0) every day in all three routes since Activity D
and Activity F which is to be crashed is among the activities in all three paths.

A - B -C -D - E – F A - C - D - E -F

9 + 11 + 7 + 14 + 7 + 7 = 55 9 + 7 + 14 + 7 + 7 = 44

A-B-D-E-F

9 + 11 + 14 + 7 + 7 = 48

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