Question

The following table represents a plan for a project:

	PREDECESSOR JOB(S)	TIMES (DAYS)
JOB NO.		a	m	b
1	–	2	4	9
2	1	2	2	11
3	1	2	4	6
4	1	3	7	8
5	2	2	3	4
6	3	1	2	6
7	4	3	5	10
8	5,6	1	3	5
9	8	1	2	12
10	7	3	4	5
11	9,10	1	5	6

  

b. Indicate the critical path.

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

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 $1,500 and if you will save $1,000 for each day that the earliest completion time is reduced, which action, if any, would you choose?

1. Reduce job 5 by two days.

	Yes
	No

2. Reduce job 6 by two days.

	No
	Yes

3. Reduce job 7 by four 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

Ans:

a) Expected Time= (a+4m+b)/6

Job Calculation Expected Time 1 (2+4*4+6)/6 2 3 4 (3+3*4+15y/6 5 (3+4*4+1176 5 (1+2*4+9)/6 3 (2+4*6+7)/6 5.5 (3+4*7+8)/6 6.5 (4+4*6+8)/6 5 6 7 00 (2+4*3+4)763 9 (1+4*4+71/6 4 10 (3+4*4+5)/6 4 11 (5+4*6+7)/6 6

b)

1−2−5−8−9−11 = 4+4.5+5.5+3+4+6 = 27

1−3−6−8−9−11 = 4+5+6.5+3+4+6 = 28.5

1−4−7−10−11 = 4+3+6+4+6 = 23

c) Critical Path: 1-3-6-8-9-11

The expected duration is 28.5.

d)

1. Reduce job 2 by three days.

No, because activity 2 doesn't lie in the critical path.

2. Reduce job 6 by two days.

No, because crashing 6 by two days will crash the critical more than specific needed crashing.

3. Reduce job 10 by two days.

No, because required crashing is of 1.5 days, that is, if above this the critical path is crashed than different path will become a critical path including the further crashing cost will increase the overall project cost.

e)

Job Calculation Expected Time 1 (6-2,622infoy 2 0.435 2 (15-36(info) 24 3 (11-376 info)^2 1.77 4 (9-16 info) 2 1.77 (7-296.info/"2 0.6944 6 (8-3)6info) 2 0.6944 7 (8-462 info) 2 0.444 8 (4-2)6info) 2 0.111 9 (7-16 info 21 10 (5-36 info) 2 0.111 11 (7-56info/"2 0.111 11.1408

SD=√11.1408=3.337

Z = (Given Time−Critical Path Time)/ Sd = (29−28.5) / 3.337=0.1498

P value = 0.4404 = 44.04%