CPMQ1. Information related to a project that involves the merger of two marketing firms (in days).ActivityImmediate Predecessor(s)Estimated Duration(days)

A-10

B-15

CA5

DB12

EC,D14

FB8

GD,F15

HE10

IE,G6

JF,I9

a. Draw the project network.b. Develop the project schedule (EST, EFT, LST, LFT).c. What are the critical activities?d. What is the project completion duration?

A1. Network Diagram

Project ScheduleActivityESTEFT

LSTLFT

A0101323

B015015

C10152325

D15271527

E27412842

F15231927

G27422742

H41514757

I42484248

J48574857

Critical ActivitiesA-C-E-HA-C-E-I-JB-D-E-HB-D-E-I-JB-D-G-I-JB-F-G-I-JProject Completion Time57 Days

Q2. A publisher has a contract with an author to publish a textbook. The activities associated with the production of the textbook are given below. The author is required to submit to the publisher a hard copy and a computer file of the manuscript. Develop the associated network for the project and calculate the project duration along with the critical paths.ActivityImmediate Predecessor(s)Estimated Duration(weeks)

A-3

B-2

C-4

D-3

EA,B2

FE4

GF2

HD1

IG,H2

JC,I4

A2. Critical ActivitiesA-E-F-G-I-J = 3+2+4+2+2+4 = 17 weeksB-E-F-G-I-J = 2+2+4+2+2+4 = 16 weeksC-J = 4+4 = 8 weeksD-H-I-J = 3+1+2+4 = 10 weeks

Since the length of the path A-E-F-G-I-J is the longest i.e. 17 weeks, hence it is the critical path.Critical PathA-E-F-G-I-J

Project Completion Time17 weeks

Q3. The following table gives the activities in a construction project and the time duration of each activity:ActivityImmediate Predecessor(s)Estimated Duration(days)

A-16

B-20

CA8

DA10

EB,C6

FD,E12

(i) Draw the activity network of the project. (ii) Find critical path. (iii) Find the total float and free-float for each activityA3. Critical ActivitiesA-C-E-F = 16+8+6+12 = 42 daysA-D-F = 16+10+12 = 38 daysB-E-F = 20+6+12 = 38 daysSince the length of the path A-C-E-F is the longest i.e. 42 days, hence it is the critical path.Critical PathA-C-E-F

Project Completion Time42 days

Total Float is the difference between the maximum time available to perform the activity and the activity durationFree Float is the portion of the total float within which an activity can be manipulated without affecting the floats of subsequent activities.

ActivityImmediate Predecessor(s)Estimated Duration(days)

Total FloatFree Float

A-1600

B-2044

CA800

DA1044

EB,C600

FD,E1200

PERTQ4. Consider the network information shown in question1 in the above section. The duration of some activities is not known with certainty. The estimates of these activities are shown below, assuming that the duration for the other activities remains unchanged.ActivityOptimisticMost Likely

Pessimistic

A81012

C357

D101214

G131517

H81012

a. What is the critical path?b. What is the projects expected completion time and variance?c. What is the probability that the project will be completed in 60 days or more?

A4. Activitytotm

tpte=(to+4tm+tp)/6Variance=((tp-to)/6)^2

A81012104/9

B150

C35754/9

D101214124/9

E140

F80

G131517154/9

H81012104/9

I60

J90

Critical ActivitiesA-C-E-H = 39 DaysA-C-E-I-J = 44 DaysB-D-E-H = 51 DaysB-D-E-I-J = 56 DaysB-D-G-I-J = 57 DaysB-F-G-I-J = 53 DaysSince the length of the path B-D-G-I-J is the longest i.e. 57 days, hence it is the critical path.

Critical Path B-D-G-I-JProject Completion Time57 DaysProbability that the project will be completed in 60 days or moreExpected time = 57 daysScheduled time= atleast60 daysTotal Variance = 4/9*5=2.22Z=(60-57)/2.2=1.363According to the z table, the z value 1.363 corresponds to a probability of 91.35%. So, there is a 91.35% probability that the project will be completed in 60 days or more.

Q5. A mother notes that when her son uses the telephone, he takes no less than 10 minutes for a call and sometimes as much as an hour. Twenty-minute calls are more frequent than calls of any other duration. If sons phone call were an activity in a PERT project:a) What would be the phone calls expected duration?b) What would be its variance?c) In scheduling the project, how much time would be allocated for the phone call?

A5. to = 10 minutes, tp = 60 minutes, tm = 20 minutesSo, te = (to+4tm+tp)/6 = (10+80+60)/6 = 25 minutesVariance = ((tp-to)/6)^2 = ((60-10)/6)^2 = 69.45In scheduling the project, expected duration of the phone call should be allocated i.e. 25 minutes.

Q6. The time estimate(in weeks) for the activities of a PERT network are given below: Activitytotm

tp

1-2117

1-3147

1-4228

2-5111

3-52514

4-6258

5-63615

(a) Draw the project network and identify all the paths through it. (b) Determine the expected project length. (c) Calculate the standard deviation and variance of the project length. (d) What is the probability that the project will be completed:1. At least 4 weeks earlier than expected time? 2. No more that 4 weeks later than expected time? (e) If the project due date is 19 weeks, what is the probability of not meeting the due date? (f) What should be the scheduled completion time for the probability of completion to be 90%?