Question

Pik-Fast, a chain of convenience stores, is planning to open a new store.  The activities required, their immediate predecessors, and the optimistic, most likely, and pessimistic estimates of their times (in days) are provided below.

Activity	Description	Immediate Predecessors	Optimistic Time (days)	Most Likely Time (days)	Pessimistic Time (days)	TE	ES	EF	LF	LS	Slack	Standard Dev	Variance
A	Select manager	--	10	12	20	13					0	1.67
B	Select site	--	8	15	22	15					0	2.33
C	Hire workers	A	17	21	31	22					0	2.33
D	Basic training	C	3	4	5	4					0	0.33
E	Secure permits	B	12	18	30	19					0	3.00
F	Construct building	E	26	30	40	31					0	2.33
G	Purchase equipment	--	30	45	60	45					0	5.00
H	Install equipment	F,G	4	6	8	6					0	0.67
I	Advertise opening	B	15	55	89	54					0	12.33
J	Final training	D,H	2	3	4	3					0	0.33
K	Stock store	H,I	4	6	8	6					0	0.67
L	Open store	J,K	1	1	1	1					0	0.00

b) Identify the critical path and projected project completion time using the expected completion times.

c)  Calculate the expected slack times for any two paths not including the critical path. For the path slack time, compute this compared to the critical path and that also considers the slack time for each activity on that path.

d) Find the probability that the critical path will be completed within 80 working days.

e)  Illustrate or explain why the probability that the critical path will be finished in 80 days or less is not necessarily the probability that the project will be finished in 80 days or less.  Assume that times for each activity will not change from what is provided below and no new activities are added nor are the predecessor relationships changed.

Answer 1

For each activity, calculate the expected time u-(a 4m b)/6 For each activity, Variance o -(b-a)/36. Standard deviation of each activity is square root of Variance + + Find the Early Start (ES), Early Finish (EF), Late Start (LS), Late Finish (LF) times and Slack for each activity. Early Start is calculated by forward pass method as latest of the predecessors Early Finish Late Finish is calculated using reverse pass method as earliest of the successors Late Start. Early Finish Early Start + Activity Duration Late Start= Late Finish - Activity Duration Slack Late Finish - Early Finish, or Late Start Early Start Activity Predecessor Optimistic (a) Pessimistic Variance Most Slack TE ES LS EF LF likely (m) (o) (b) 2.78 A 10 12 20 13 0 35 13 48 35 B 8 0 15 22 15 5.44 0 15 15 5.44 C A 17 21 31 22 13 48 35 70 35 D C 3 4 5 4 0.11 35 70 39 74 35 B E 12 30 19 9.00 15 15 34 34 18 E 26 30 40 31 5.44 34 34 65 65 0 G 30 45 60 45 25.00 0 20 45 65 20 4 6 8 6 0.44 Н F,G 65 65 71 71 0 B 15 55 89 54 152.11 15 17 69 71 2 D,H J 2 3 4 3 0.11 71 74 74 77 3 K Н 4 8 6 0.44 71 71 77 77 0 L JK 1 1 1 1 0.00 77 77 78 78 ONm

b)

Critical path is the longest path. It can be identified in the following table. All the activities having Slack time of 0 constitute the critical path.

So, critical path is: B-E-F-H-K-L

Expected completion time = LF time of the last activity of critical path = 78 days

c)

The expected slack times are shown in the right-most last column of the above table

d)

Expected time of critical path = 78 days

Variance of time of critical path = 5.44+9+5.44+0.44+0.44+0 = 20.78

Standard deviation of time of critical path = sqrt(20.78) = 4.56

Expected time of z value = (80-78)/4.56 = 0.4388

F(z) = NORMSDIST(0.4388) = 0.6696

Probability of completion of critical path within 80 working days = 0.6696 ~ 67 %

e) Probability of that the critical path will be finished within 80 days is 0.6696, whereas probability that all the paths will be finished within 80 days is 0.5796. Both are different.

Step 1: List all possible paths through the network alongwith their respective expected times and variances The expected time of each path is the sum of expected time of each activity on that path Similarly, variance of each path is the sum of individual variance of each activity on that path Path Expected time Std dev Variance A-C-D-J-L B-E-F-H-J-L B-E-F-H-K-L G-H-J-L G-H-K-L 2.91 43 8.44 75 20.44 4.52 78 20.78 4.56 55 25.56 5.06 5.09 58 25.89 Critical path is the longest path (maximum expected time). Critical path are: Project Completion Time (H) 78 days Step 2: Finding probability of completing the project in given time Target time (x) days For target time, x = 80 days, caculate value of z (x-H)/o and P(z) NORMSDIST(z) for each path 80 X Expected time z=(x-H)/o Path P(z) Var SD A-C-D-J-L B-E-F-H-J-L B-E-F-H-K-L G-H-J-L G-H-K-L 43 8.44 2.91 12.73 1 4.52 1.11 75 20.44 0.8656 0.6696 78 20.78 4.56 0.44 4.95 55 25.56 5.06 1 58 25.89 5.09 4.32 1 Probability of completing the project within 80 days 1*0.8656*0.6696*1 = 0.5796