Problem 4 7. Three recent college graduates have formed a partnership and have o
ID: 332420 • Letter: P
Question
Problem 4 7. Three recent college graduates have formed a partnership and have opened an advertising firm. Their first project consists of activities listed in the following table a. Draw the precedence diagranm b. What is the probability that the project can be completed in 24 days or less? In 21 days or less? **Prior to completing part b., you must create an ES/LS table and determine the average time for completion. You will not receive full credit unless you submit these parts. TIME IN DAYS Immediate Activity Predecessor Optimistic Most Likely Pessimistic 12 15 End E, G,I d. The partners have decided that shortening the project by two days would be beneficial, as long as it doesn't cost more than about $20,000. They have estimated the daily crashing costs for each activity in thousands, as shown in the following table. Which activities should be crashed and what further analysis would they probably want to do? Activity First Crash Second Crash $10 $8 10 10Explanation / Answer
PLEASE FIND BELOW ANSWER TO PROBLEM 4 , 7:
Please find below table which will help to calculate average time for completion of project :
Activity
Optimistic, a
Most likely, b
Pessimistic, c
Expected time, te
Variance, ( b - a)^2/36
A
5
6
7
6.00
0.11
B
8
8
11
8.50
0.25
C
6
8
11
8.17
0.69
D
9
12
15
12.00
1.00
E
5
6
9
6.33
0.44
F
5
6
7
6.00
0.11
G
2
3
7
3.50
0.69
H
4
4
5
4.17
0.03
I
5
7
8
6.83
0.25
Following may be noted :
Expected time . te = ( Optimistic time + 4 x Most likely time + Pessimistic time ) /6
Variance = ( Pessimistic time – Optimistic time)^2/36
The precedence diagram of all the tasks as follows :
A
B
D
C
H
F
E
I
G
END
The various possible paths and their expected durations calculated on basis of summing up expected durations of relevant individual activities as follows :
A-C-E = 6 + 8.17 + 6.33 = 20.5 days
B-H-I = 8.5 + 4.17 + 6.83 = 19.5 days
D-F-G = 12 + 6 + 3.5 = 21.5 days
Since D-F-G has the longest duration , it forms the critical path. Average duration of the project is same as duration of the critical path .
Therefore , average time for completion of the project =m = 21.5 days
Variance of the critical path
= Sum of variances of D , F and G
= 1 + 0.11 + 0.69
= 1.80
Therefore , standard deviation of the critical path = Sd = Square root ( Variance of critical path ) = Square root ( 1.80 ) = 1.34
Let z value for probability that project will complete in 21 days or less = Z1
Z value for probability that project will complete in 24 days or less = Z2
Therefore ,
M + Z1 x sd = 21
Or, 21.5 + 1.34.Z1 = 21
Or, Z1 =( 21 – 21.5)/1.34
Or, Z1 = - 0.373 ( 0.37 rounded to nearest whole number )
Probability for Z1 = - 0.37 as derived from standard normal distribution table = 0.35569
Similarly,
M + Z2 x Sd = 24
Or, 21.5 + 1.34.Z2 = 24
Or, 1.34.Z2 = 2.5
Or, Z2 = 2.5/1.34
Or, Z2 = 1.8657 ( 1.87 rounded to 2 decimal places )
Probability for Z2 = 1.87 as derived from standard normal distribution table = 0.96926
PROBABILITY THAT A PROJECT CAN BE COMPLETED IN 24 DAYS OR LESS = 0.96926
PROBABILITY THAT A PROJECT CAN BE COMPLETEDIN 21 DAYS OR LSS = 0.35569
Activity
Optimistic, a
Most likely, b
Pessimistic, c
Expected time, te
Variance, ( b - a)^2/36
A
5
6
7
6.00
0.11
B
8
8
11
8.50
0.25
C
6
8
11
8.17
0.69
D
9
12
15
12.00
1.00
E
5
6
9
6.33
0.44
F
5
6
7
6.00
0.11
G
2
3
7
3.50
0.69
H
4
4
5
4.17
0.03
I
5
7
8
6.83
0.25
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