on this question I missed on the (AFG) path for the mean and standard deviation.

Question

on this question I missed on the (AFG) path for the mean and standard deviation. can you show me the correct solution?

Problem 17-12

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.

Activity 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.)

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 a-d-e-h a- b-i-j-k C-m-n-O Mean Std. Dev Probability ($1,000) Probability ($500)

Explanation / Answer

c-m-n-o is the critical path.

Probability ($1000) = Prob (Completion <= 26) = NORM.DIST(26,26.17,1.66,TRUE) = 0.4592
Probability ($1000) = Prob (Completion <= 27) = NORM.DIST(27,26.17,1.66,TRUE) = 0.6915

Activity Predecessor Successor Optimistic Most likely Pessimistic Expected SD variance ES EF LS LF Slack (a) (m) (b) te=(a+4m+b)/6 (b - a)/6 =SD^2 =LF - EF A - D,F 2 4 6 4.00 0.67 0.44 0 4.00 1.83 5.83 1.83 D A E 6 8 10 8.00 0.67 0.44 4.00 12.00 5.83 13.83 1.83 E D H 7 9 12 9.17 0.83 0.69 12.00 21.17 13.83 23.00 1.83 H E - 2 3 5 3.17 0.50 0.25 21.17 24.33 23.00 26.17 1.83 F A G 3 4 8 4.50 0.83 0.69 4.00 8.50 14.67 19.17 10.67 G F - 5 7 9 7.00 0.67 0.44 8.50 15.50 19.17 26.17 10.67 B - I 2 2 3 2.17 0.17 0.03 0 2.17 11.33 13.50 11.33 I B J 2 3 6 3.33 0.67 0.44 2.17 5.50 13.50 16.83 11.33 J I K 3 4 5 4.00 0.33 0.11 5.50 9.50 16.83 20.83 11.33 K J - 4 5 8 5.33 0.67 0.44 9.50 14.83 20.83 26.17 11.33 C - M 5 8 12 8.17 1.17 1.36 0 8.17 0.00 8.17 0.00 M C N 1 1 1 1.00 0.00 0.00 8.17 9.17 8.17 9.17 0.00 N M O 6 7 11 7.50 0.83 0.69 9.17 16.67 9.17 16.67 0.00 O N - 8 9 13 9.50 0.83 0.69 16.67 26.17 16.67 26.17 0.00

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