Question

1. An IT project requires 6 activities. The table below displays the expected time and budgeted cost for each activity.

Activity	Expected time (weeks)	Budgeted Cost	Immediate Predecessors
A	3	$3,000	None
B	4	$6,000	None
C	2	$7,000	A
D	5	$9,000	B
E	3	$3,000	C
F	6	$8,000	B,D

Develop a budget for the 15 weeks of the project using ES times.

2. The following represents a project with four activities. All times are in weeks.

Activity	Immediate Predecessor	Optimistic Time	Most Likely Time	Pessimistic Time
A	-	2	8	14
B	-	8	8	8
C	A	6	9	18
D	B	5	11	17

The critical path is BàD. The expected completion time is 19 days.   Assume the normal distribution is appropriate to use to determine the probability of finishing by a particular time. What is the probability that the project is finished in 22 weeks or more? (Round to two decimals.)

Answer 1

Problem 1:

Activity	A	B	C	D	E	F
Activity time	3	4	2	5	3	6
Predecessor	-	-	A	B	C	B, D
Activity Cost	3000	6000	7000	9000	3000	8000
Activity Cost per day	1000	1500	3500	1800	1000	1333.333
Earliest Start Time	0	0	3	4	5	9
Period	Cost schedule according to ES time	Total in Period	Cumulative from start
1	1000	1500					2500	2500
2	1000	1500					2500	5000
3	1000	1500					2500	7500
4		1500	3500				5000	12500
5			3500	1800			5300	17800
6				1800	1000		2800	20600
7				1800	1000		2800	23400
8				1800	1000		2800	26200
9				1800			1800	28000
10						1333.33	1333.33	29333.33
11						1333.33	1333.33	30666.67
12						1333.33	1333.33	32000
13						1333.33	1333.33	33333.34
14						1333.33	1333.33	34666.67
15						1333.33	1333.33	36000

Problem 2:

Critical Path: B – D

Critical Path Duration = 19 days

Variance of activity = σ2 = [(p-o)/6]2

Time Variance of B = ((8 -8) / 6)2 = 0

Time Variance of D = ((17 - 5) / 6)2 = 4

Variance of critical path = sum of variance of the critical activities = 0 + 4 = 4

Standard Deviation of Critical Path = σ = √variance = √4 = 2

Expected completion time of project = T = 19 weeks

Desired completion duration of project = X = 22 weeks

Considering the project completion time follows normal distribution,

The z-score for X = 22 is calculated as follows:

z = (X – T)/σ or

z = (22 – 19)/2 = 1.5

From Normal distribution table:

P(z <= 1.5) = [=NORMSDIST(1.5) = 0.9332

Thus, P(X <= 22) = P(z <= 1.5) = 0.9332

There is 93.32% chance that the critical path will finish in 22 weeks.