Question

Use the project description presented in the table below for the following questions: Activity Immediate Predecessor Optimistic Time (O.T.) Most likely Time (M.T.) Pessimistic Time (P.T.) A None 4 6 9 B A 3 5 7 C A 4 7 12 D B 4 8 10 E C 5 9 15 F D 4 4 5 G D, E 3 6 8 H F 5 6 10 I G 5 8 11 J H, I 5 5 5 Calculate the expected time for each activity (2 points) Calculate the variance for each activity (2 points) Using the information given, construct a network diagram (2 points) Use both the longest path method and ES/EF; LS/LF method on the diagram for the following questions and show your answers in the table below. What is the expected project completion time? (3 points) What are the critical activities? (3 points) What are the slack values for non-critical activities? (4 points) The Longest Path Method ES/EF/LS/LF Expected project finish time Critical activities A,B,D,E,G,I,J Non-critical activities (slack values) C,F,H Calculate the probability that the project will be completed in 40 weeks. (2 points) Calculate the probability

that the project will be completed in 43 weeks.

PLEASE DO NOT ROUND ANWSERS

Answer 1

Expected time, Te = (Pessimistic time + 4*Most likely time + Optimistic time)/ 6

Variance = ((Optimistic time - Pessimistic time)/ 6)²

Te and variance is calculated below.

Activity	Immediate Predecessor	OT	MT	PT	Expected time,Te	Variance, σ2
A	None	4	6	9	(4+4*6+9)/6=6.167	((9-4)/6)^2=0.694
B	A	3	5	7	5.000	0.444
C	A	4	7	12	7.333	1.778
D	B	4	8	10	7.667	1.000
E	C	5	9	15	9.333	2.778
F	D	4	4	5	4.167	0.028
G	D,E	3	6	8	5.833	0.694
H	F	5	6	10	6.500	0.694
I	G	5	8	11	8.000	1.000
J	H,I	5	5	5	5.000	0.000

Network diagram basis these details is drawn as below:

Longest path method:

To find the critical path, we will assess all possible paths and path with longest completion time will be critical path.

Possible paths are-

• A-B-D-F-H-J: 6.167+5+7.667+4.167+6.5+5= 34.501
• A-B-D-G-I-J: 37.667
• A-C-E-G-I-J: 41.667

Hence, path A-C-E-G-I-J with longest time of 41.667 is the critical path. Critical activities are A,C,E, G, I, J. Expected project completion time is same as critical path expected time of 41.667.

ES/ EF; LS/ LF method:

Earliest Start Time (ES) = the earliest time that an activity can begin

Earliest Finish Time (EF) = the earliest time that an activity can be completed

EF = ES + duration of activity        

Latest Start Time (LS) = the latest time that an activity can begin without lengthening the minimum project duration

Latest Finish Time (LF) = the earliest time than an activity can be completed without lengthening the minimum project duration

LF = LS + duration of activity   

Duration of the project = the difference between the maximum value of the "latest finish time" of projects and the minimum value of the "earliest start time" of projects

Project duration = max(LF) - min(ES)     

Slack = the amount of time that a project can be delayed without increasing the duration of the project

Slack = LF- LS or EF - ES

Critical path is the path with Total Slack = 0

Below is the updated table with the output:

Activity	Immediate Predecessor	Expected time,Te	Variance, σ2	Early Start (ES)	Early Finish (EF)	Late Start (LS)	Late Finish(LF)	Slack	Critical Path
A	None	6.167	0.694	0	6.167	0.000	6.167	0.000	Y
B	A	5.000	0.444	6.167	11.167	10.167	15.167	4.000
C	A	7.333	1.778	6.167	13.500	6.167	13.500	0.000	Y
D	B	7.667	1.000	11.167	18.834	15.167	22.834	4.000
E	C	9.333	2.778	13.500	22.834	13.500	22.834	0.000	Y
F	D	4.167	0.028	18.834	23.000	26.000	30.167	7.167
G	D,E	5.833	0.694	22.834	28.667	22.834	28.667	0.000	Y
H	F	6.500	0.694	23.000	29.500	30.167	36.667	7.167
I	G	8.000	1.000	28.667	36.667	28.667	36.667	0.000	Y
J	H,I	5.000	0.000	36.667	41.667	36.667	41.667	0.000	Y

Clearly, critical path with slack = 0 are A-C-E-G-I-J. Total expected time is 41.667.

Total variance of critical activities are 0.694+1.778+2.778+0.694+1+0= 6.944

Non critical activities with slack are- B(4), D(4), F (7.167), H (7.167).

Probability of project to be completed in 40 weeks- We will calculated z-statistics and use z table to find probability

Variance for critical path =σ² = 6.944. Hence, standard deviation =σ= 2.635

Mean expected time for critical path, µ = 41.667

X= 40

Calculating z-score,

z = (X-µ / σ) = (40 - 41.667)/ 2.635 = -0.63

(Looking at z-table given below, to find the probability. Under negative z value, look for row -0.6 and column 0.03 we will get probability value of 0.2643)

P(Z<-0.63) = 0.2643 or 26.43%

Hence, Probability of completing project within 40 weeks is 26.43%

Probability of completing project in 43 weeks, P(X<43),

z = (43 - 41.667)/ 2.635 =0.51

(looking at ztable, corresponding probability value is 0.6950)

P(Z<0.51) = 0.6950 or 69.50%

Hence, Probability of completing project in 43 weeks is 69.50%

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