Question

A project manager has compiled a list of major activities that will be required to install a computer information system in her firm. The list includes estimated completion times for activities and precedence relationships. (Use Table B1 and Table B2.)

(table B1)

(table b2)

Immediate Predecessor	Estimated Times (weeks)
A	—	2-4-6
D	A	6-8-10
E	D	7-9-12
H	E	2-3-5
F	A	3-4-8
G	F	5-7-9
B	—	2-2-3
I	B	2-3-6
J	I	3-4-5
K	J	4-5-8
C	—	5-8-12
M	C	1-1-1
N	M	6-7-11
O	N	8-9-13
End	H, G, K, O

If the project is finished within 26 weeks of its start, the project manager will receive a bonus of $1,000; and if the project is finished within 27 weeks of its start, the bonus will be $500. Find the probability of each bonus. (Round path means and path standard deviations to 2 decimal places. Round probabilities to 4 decimal places. Do not round any other intermediate calculations.)

Path	Mean	Std. Dev.
a-d-e-h
a-f-g
b-i-j-k
c-m-n-o

Probability ($1,000)

Probability ($500)

Answer 1

Path A-D-E-H Expected completion time 4.00 +8.00 9.17 3.17- 24.34 Variance 16/36 16/36 25/36 9/36 66/36 Standard deviation 66 1.354 36 26 24.34 1.354 Using Appendix B Table B : Probability of completion weeks 0.8907 26 Path A-F-G: Expected completion time 4.00 4.50 7.00 15.50 Variance 16/36 25/36 16/36 57/36 Standard deviation1.258 36 26 15.50_8 35 1.258-8.35 Probability of completion 26 weeks = 1.0000 (z value> +3.00-treat probability of completion as being = 1.0000)

Path B-I-J-K: Expected completion time-2.17 + 3.33 4.00+ 5.33 14.83 Variance 1/36 16/36 4/3616/36 37/36 Standard deviation1.014 37 36 26 14.83 1.014 11.02 Probability of completion < 26 weeks = 1.0000 (z value > +3.00-treat probability of completion as being 1.0000) Path C-M-N-O Expected completion time = 8.17 + 1.00 + 7.50 + 9.50 = 26.17 Variance 49/36 0/36 25/36 25/36 99/36 Standard deviation991.658 36 26-26.17 1.658 -0.10 Using Appendix B Table B: Probability of completion s 26 weeks = 0.4602 Conclusion: The probability of receiving a bonus of $1,000 = Probability (T < 26 weeks) = 0.8907 x 1.0000 x 1.0000 x 0.4602 0.4099

The project manager will receive a bonus of $500 if the project is inished withın 27 weeks P (T £27): Path A-D-E-H 27-24.34 1.354 = 1.96 Using Appendix B Table B: Probability of completion weeks-0.9750 27 Path A-F-G: 27-15.50 1.2589.14 Probability of completion < 27 weeks = 1.0000 (z value > +3 .00-treat probability of completion as being = 1.0000) Path B-I-J-K: 27 14.83 1.01412.00 Probability of completion < 27 weeks = 1.0000 (z value > +3.00-treat probability of completion as being-1.0000) Path C-M-N-O: 27-26.17 1.6580.50

Using Appendix B Table B : Probability of completion weeks 0.6915 27 Conclusion: The probability of receiving a bonus of S500- Probability (T < 27 weeks) = 0.9750 x 1.0000 x 1.0000 x 0.6915 = 0.6742