Question

Use a PERT network to establish the expected duration for the project described in the table...

Use a PERT network to establish the expected duration for the project described in the table below:    (SHOW the PERT NETWORK)

Activity

Predecessors

Duration

a        m       b

Exp.

Duration

Var.

Crash

Week 1

Crash

Week 2

A

-

3

6

9

700

100

B

A

6

7

14

600

200

C

A

4

6

14

400

        600

D

B, C

4

6

8

200

                   

        600

E

B, C

6

9

12

400

300

F

C

4

6

14

200

900

G

D, E

3

5

7

400

400

H

E

3

6

9

600

100

I

G

8

8

8

600

J

H

4

10

16

700

300

a.   List the activities on the CRITICAL PATH:………………………  

b. What is the project expected duration?...........

c.   In your table list ALL expected durations and ALL variances.

d.   List is the slack for:   J ……..            I ………            F……….           C ………

e.   List the latest starting time for:    G ……..               F ………                E ……….

f.   The company indicated that they need to have a 90% probability of completing in 39 weeks.      

    What will become of the expected duration to make sure that we meet the 90% probability to

     complete within 39 weeks?......................

   

g.   Because of the new expected duration, (see f), how many weeks are to be crashed?.............

h.   List specifically which activities you would crash, and for how many weeks:           

Activity

# Weeks

Crashing Cost

         

0 0
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Answer #1

8 6 B 5 G - 8 10 Start А. С E Н. J End 6 7 9 6 F 7 a) Critical path is the path with longest completion time. We will assess

We will use below approach for Slack and Late finish as well as validating critical path and project duration.

Earliest Start Time (ES) = the earliest time that an activity can begin

Earliest Finish Time (EF) = the earliest time that an activity can be completed

EF = ES + duration of activity        

Latest Start Time (LS) = the latest time that an activity can begin without lengthening the minimum project duration

Latest Finish Time (LF) = the earliest time than an activity can be completed without lengthening the minimum project duration

LF = LS + duration of activity   

Duration of the project = the difference between the maximum value of the "latest finish time" of projects and the minimum value of the "earliest start time" of projects

Project duration = max(LF) - min(ES)     

Slack = the amount of time that a project can be delayed without increasing the duration of the project

Slack = LF- LS or EF - ES

Critical path is the path with Total Slack = 0

Calculations are below:

Expected time = (Optimistic + 4*Most Likely + Pessimistic)/ 6
Variance = ((Pessimistic - Optimistic)/ 6)2
Activity Predecessors Duration Exp. Duration Variance Early Start (ES) Early Finish (EF) Late Start (LS) Late Finish (LF) Slack Critical Path
a m b
A - 3 6 9 6.0 1.00 0 6.0 0.0 6.0 0.0 Y
B A 6 7 14 8.0 1.78 6.0 14.0 6.0 14.0 0.0 Y
C A 4 6 14 7.0 2.78 6.0 13.0 7.0 14.0 1.0
D B,C 4 6 8 6.0 0.44 14.0 20.0 18.0 24.0 4.0
E B,C 6 9 12 9.0 1.00 14.0 23.0 14.0 23.0 0.0 Y
F C 4 6 14 7.0 2.78 13.0 20.0 29.0 39 19.0
G D,E 3 5 7 5.0 0.44 23.0 28.0 24.0 29.0 1.0
H E 3 6 9 6.0 1.00 23.0 29.0 23.0 29.0 0.0 Y
I G 8 8 8 8.0 0.00 28.0 36.0 29.0 39.0 3.0
J H 4 10 16 10.0 4.00 29.0 39.0 29.0 39.0 0.0 Y
a. Activities on Critical path is A-B-E-H-J
b. Project expected duration is 39 weeks
c. Table with expected duration and all variances ->> Updated above
d. Slack: J=0, I=3, F=19, C=1
e. Latest starting time for G=24, F=29, E=14

Variance of critical path = 1 + 1.78 +1 + 1+ 4 = 8.78

Standard deviation of critical path = √8.78 = 2.96

f)

For probability we will use z statistics,

Z= (X- Expected time)/ Standard deviation

For 90% probability, looking at z table as given below we will get z value of 1.29

Z= (X- Expected time)/ Standard deviation

1.29 = (39-Expected time)/ 2.96

Expected time = 35.18

Target expected duration to male sure that we meet 90% probability to complete within 39 weeks is 35.18 weeks. (~ 35 week rounded)

g.

To meet new expected duration, 4 weeks duration (approaximately) need to be crashed

(Earlier expected duration was 39 and now required is ~35 weeks)

h.

Now to crash project duration by 4 weeks, we will crash activities on critical path A-B-E-H-J starting with activities with lowest crashing cost.

- Activity E has lowest crashing cost for Week 1 (400), so we crash E by 1 week. New project duration is 39-1=38 weeks. Crashing cost is $400. Critical path is same A-B-E-H-J.

- Next we will crash activity E by second week (300).New project duration is 38-1=37 weeks. Crashing cost is $400+300=700. Critical path is same A-B-E-H-J.

- Next we will crash activity H by 1 Week (600). New project duration is 37-1=36 weeks. Crashing cost is $700+600=$1300. Critical path is same A-B-E-H-J.

- Next we will crash activity H by second Week (100). New project duration is 36-1=35 weeks. Crashing cost is $1300+100=$1400. Critical path is same A-B-E-H-J.

Activity # Weeks Crashing Cost
E 1 400
E 1 300
H 1 600
H 1 100
1400

So total $1400 of crashing cost is needed to reduce project duration to 35 weeks.


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Table of Standard Normal Probabilities for Negative Z-scores Table of Standard Normal Probabilities for Positive Z-scores v -

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