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. (Z-Table is needed)

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 Mean, Standard Deviation, z-value to 2 decimal places and Probability to 4 decimal places.)

Path	Mean	Std. Dev.
a-d-e-h	24.33	1.35
a-f-g	15.50	1.25
b-i-j-k	14.83	1.01
c-m-n-o	26.17	1.66

  
Probability ($1,000)       =     0.4099

Probability ($500)      =      0.6742

I require assistance with the bolded portion. It says it's incorrect.

Answer 1

Step 1: For each activity, find the expected time Tx = (a + 4m+b)/6, and Variance oʻ = [(b-a)/6)* Step 2: Find the Earliest start and finish and Latest start and finish times and Slack for each activity. Earliest Start is calculated by forward path method as latest of the predecessors Earliest Finish Latest Start is calculated using reverse path method as earliest of the successors Latest Start. Earliest Finish = Earliest Start + Activity Duration Latest Start=Latest Finish - Activity Duration Slack = Latest Finish - Earliest Finish, or Latest Start - Earliest Start Activity Schedule is given below Activity Predecesso b ce ESS EF LF Slack Critical Activity D Expected (Ts) 4.00 8.00 9.17 3.17 4.50 7.00 2.17 3.33 4.00 5.33 8.17 1.00 7.50 9.50 0.00 variance (o) 0.44 0.44 0.69 0.25 0.69 0.44 0.03 0.44 0.11 0.44 1.36 0.00 0.69 0.69 0.00 0.00 4.00 12.00 21.17 4.00 8.50 0.00 2.17 5.50 9.50 0.00 8.17 9.17 16.67 26.17 1.83 4.00 5.83 12.00 13.83 21.17 23.00 24.33 14.67 8.50 19.17 15.50 11.332 .17 13.50 15.50 16.83 9 .50 20.83 14.83 0.00 8.17 8.17 9.17 9.17 1 16.67 16.67 26.17 26.17 26.17 5.83 1.83 13.83 1.83 23.00 1.83 26.17 | 1.83 19.17 10.67 26.17 10.67 13.50 11.33 16.83 11.33 20.83 | 11.33 26.17 1 1.33 8.17 0.00 9.17 0.00 16.67 0.00 26.17 0.00 26.17 0 .00 C Critical Critical Critical M Crities End H,G,K,O Critical Step 3: Find the critical path. Critical path is that path in the network diagram, which has maximum length (duration) as per expected time () The expected duration of critical path is the sum of expected duration of each activity on critical path. Similarly, variance of critical path is the sum of individual variance of each activity on critical path. Critical path Project paths a-d-e-h a-t-e b-i-j-k c-m-n-o Length Variance 24.33 0.25 15.50 1.14 14.831.03 26.172 .75 S.D. 0.50 1.07 1.01 1.66 Critical path Expected duration, variance and standard deviation of Critical path Expected Completion Time (Tx) = 26.17 weeks Expected variance (v) = 2.75 weeks Standard Deviation =Vv=O 1.66 weeks 225 Step 4: Probability that the project will be completed in specified time 26 weeks Target Time (T5) = Calculate z score as below z score = TS-TX = = -0.1005 (26- 26.1667) 1.658 See the probablity from the standard normal table corresponding to z score (or using excel formula NORMSDIST) Probability that the project will take AT MOST 26 weeks to complete ENORMSDIST(-0.1005) 46.00% 27 weeks Target Time (T5) = Calculate z score as below z score = TS-TX = (27-26.1667) 1.658 = 0.5025 See the probablity from the standard normal table corresponding to z score (or using excel formula NORMSDIST) Probability that the project will take AT MOST 27 weeks to complete ENORMSDIST(0.5025) 69.23%

Probability of finishing within 26 weeks of start and earning $ 1000 bonus = 46%

Probability of finishing within 27 weeks of start and earning $ 500 bonus = 69.23%

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