Question

PART I: PERT

Consider the following table:

ACTIVITY      PREC. ACT.              TIME ESTIMATES     (days)        

to                       tml                 tp                       te

    A                      ---                           2                      3                      4                     

    B                      A                            2                      4                      6

    C                      A                            4                      6                      8

    D                      A                            4                      5                      6

    E                       B                            6                      7                      8

    F                       D                            1                      3                      5

    G                 E, C, F                        3                      5                      7

1.   Construct a PERT network. (20 points)

[NOTE: This is the most crucial part of the exercise. Make sure you have the correct network before proceeding to the next sections.]

2.   By using ES, EF, LF, LS, find the slack of each activity and clearly state the critical path. (10 points)

3.   Find the completion time ( ) and the standard deviation (σ) for the project— round off the standard deviation to the nearest hundredth: two decimal places. (20 points)

[NOTE: Sigma is only computed on the critical path.]

4.   What is the completion time at 68% and 95% confidence levels? (20 points)

5.   What is the likelihood of this project being completed in more than 20 days? (15 points)

6.   Between 16 and 18 days? (15 points)

Answer 1

Formula:

Expected Activity Duration using Beta Distribution / Formula = (O + (4 X M) + P)/6

Variance =Standard Deviation^2

Standard Deviation=(Pessimistic Time -Optimistic Time)/6

1	B	C	D	E	F	G	H	I	J
2	Activity	Predecessor	Optimistic time - O	Most Likely Time - M	Pessimistic Time - P	Expected Activity Duration using Beta Distribution / Formula = (O + (4 X M) + P)/6	Variance	Standard Deviation	Activity on Critical Path - Yes or No
3	A	---	2	3	4	3	0.11	0.33	Yes
4	B	A	2	4	6	4	0.44	0.67	Yes
5	C	A	4	6	8	6	0.44	0.67	No
6	D	A	4	5	6	5	0.11	0.33	No
7	E	B	6	7	8	7	0.11	0.33	Yes
8	F	D	1	3	5	3	0.44	0.67	No
9	G	E,C,F	3	5	7	5	0.44	0.67	Yes
10			Total Variance of the Critical Path				1.11
11			Standard Deviation of the Critical Path				1.05

Formula:

1	B	C	D	E	F	G	H	I	J
2	Activity	Predecessor	Optimistic time - O	Most Likely Time - M	Pessimistic Time - P	Expected Activity Duration using Beta Distribution / Formula = (O + (4 X M) + P)/6	Variance	Standard Deviation	Activity on Critical Path - Yes or No
3	A	---	2	3	4	=(D3+(4*E3)+F3)/6	=I3^2	=(F3-D3)/6	Yes
4	B	A	2	4	6	=(D4+(4*E4)+F4)/6	=I4^2	=(F4-D4)/6	Yes
5	C	A	4	6	8	=(D5+(4*E5)+F5)/6	=I5^2	=(F5-D5)/6	No
6	D	A	4	5	6	=(D6+(4*E6)+F6)/6	=I6^2	=(F6-D6)/6	No
7	E	B	6	7	8	=(D7+(4*E7)+F7)/6	=I7^2	=(F7-D7)/6	Yes
8	F	D	1	3	5	=(D8+(4*E8)+F8)/6	=I8^2	=(F8-D8)/6	No
9	G	E,C,F	3	5	7	=(D9+(4*E9)+F9)/6	=I9^2	=(F9-D9)/6	Yes
10			Total Variance of the Critical Path				=SUM(H3+H4+H7+H9)
11			Standard Deviation of the Critical Path				=SQRT(H10)

1)

Network diagram:

7 E 14 3 B 7 0 7 0 4 0 7 0 14 7 3 14 G 19 9 с 6 0 5 0 con w 5 0 14 14 A 3 19 0 3 0 3 0 3 D 8 8 F 11 3 5 3 3 3 11 3 14 6 11

2)

Legends are as stated below to find ES, EF, LS, LF and Total float of each activity

Activity Id Early Finish-Early Start + Duration Duration Early Start=Highest Early finish time of all predecessor activities Slack or total float=Late start-Early Start Late Start=Late finish- Duration Slack or total float=Late finish-Early finish Resource Late finish=Lowest Late start time of all followers activities

here, critical path is A-B-E-G as activities on this path have zero total float

3)

Project completion time=19 days

Standard Deviation =1.05 as shown in the excel sheet with formula

4)

Completion time at 95% confidence level

=(19-(1.96*1.05/Square root of 4)) to (19+(1.96*1.05/Square root of 4))

= (19-(1.96*1.05/2))to (19+(1.96*1.05/2))

= (19-1.03)to (19+1.03)

=17.97 to 20.03

Completion time at 68% confidence level

=(19-(0.468*1.05/Square root of 4)) to (19+(0.468*1.05/Square root of 4))

=(19-(0.468*1.05/2))to (19+(0.468*1.05/2))

= 18.754 to 19.245

5)

Project will take at least 19 days time to complete, so μ= 19 Days, Standard Deviation = 1.05

Given, X = 20 Days

Thus, Value of Z score = (X- μ)/ Standard Deviation = (20-19)/1.05 = 0.95 (Rounded up to 2 Decimal Numbers)

Thus, using Z table, The probability that the project will complete within 20 days = 82.894%

The probability that the project will take more than 20 days=(100-82.894)%=17.106%

6)

Project will take at least 19 days time to complete, so μ= 19 Days, Standard Deviation = 1.05

Given, X = 20 Days

Thus, Value of Z score = (X- μ)/ Standard Deviation = (20-19)/1.05 = 0.95 (Rounded up to 2 Decimal Numbers)

Thus, using Z table, The probability that the project will complete within 20 days = 82.894%

The probability that the project will take more than 20 days=(100-82.894)%=17.106%

Project will take at least 19 days time to complete, so μ= 19 Days, Standard Deviation = 1.05

Given, X = 16 Days

Thus, Value of Z score = (X- μ)/ Standard Deviation = (16-19)/1.05 = -2.86 (Rounded up to 2 Decimal Numbers)

Thus, using Z table, The probability that the project will complete within 16 days = 0.00212=.212%

Given, X = 18 Days

Thus, Value of Z score = (X- μ)/ Standard Deviation = (18-19)/1.05 = -0.95 (Rounded up to 2 Decimal Numbers)

Thus, using Z table, the probability that the project will complete within 18 days=.17106=17.106%

Thus, using Z table, the probability that the project will complete Between 16 and 18 days=(17.106-0.212)=16.89%