Question

The following table represents a plan for a project:

	PREDECESSOR JOB(S)	TIMES (DAYS)
JOB NO.		a	m	b
1	–	2	4	6
2	1	3	3	15
3	1	3	4	11
4	1	1	2	9
5	2	2	6	7
6	3	3	7	8
7	4	4	6	8
8	5,6	2	3	4
9	8	1	4	7
10	7	3	4	5
11	9,10	5	6	7

  

b. Indicate the critical path.

	1-2-5-8-9-11
	1-4-7-10-11
	1-2-5-8
	1-3-6-8-9-11

c. What is the expected completion time for the project? (Round your answer to 2 decimal places.)

Expected completion time            days

d. You can accomplish any one of the following at an additional cost of $2,500 and if you will save $1,700 for each day that the earliest completion time is reduced, which action, if any, would you choose?

1. Reduce job 2 by three days.

	No
	Yes

2. Reduce job 6 by two days.

	Yes
	No

3. Reduce job 10 by two days.

	Yes
	No

e. What is the probability that the project will take more than 29 days to complete? (Round your answer to 2 decimal places.)

Probability           

Answer 1

from the given project

JOB NO.	predecessor job	shortest time a	Most likely time m	longest time b	Expected time a+4m+b /6	variance in time ((b-a) /6 )2
1	-	2	4	6	4	0.11
2	1	3	3	15	5	4
3	1	3	4	11	5	1.78
4	1	1	2	9	3	1.78
5	2	2	6	7	5.5	0.69
6	3	3	7	8	6.5	0.69
7	4	4	6	8	6	0.44
8	5,6	2	3	4	3	0.11
9	8	1	4	7	4	1
10	7	3	4	5	4	0.11
11	9,10	5	6	7	6	0.11

b) the critical path from 1--2--5---8----9------11 the duration =23,5 days

the critical path from 1---4----7----10-----11 the duration = 19 days

the critical path from 1----2---5----8 the duration = 13.5 days

the critical path from 1---3-----6-----8-----9----11 the duration = 24.5 days

c) Expected completion time of the project = 28.5 days

d) reduce job 2 by 2 days

e) expected completion time = M =28.5

dtandard deviation in completion time =1.9493

standard normal variate Z= (29 - 28.5) / 1.9493 = 0.2563

probability that the project will take more than 29 days P(D> 29) = P( Z> 0.2565 ) = 0.5+ P(0<Z<0.2565)

= 0.5 + 0.01 =0.51

51% chance that the project duration may be extend 29 days