Problems

Q. 2003 / 6(b) : Activities of a project are described below. Calculate the following

1. Total Time for completing the project2. Critical Path3. ES, EF, LS and LF times for all activities4. Total Float and Free Float available for non-critical activities

ACTIVITY ESTIMATED DURATION

IMMEDIATE PREDECESSORS

A 3 -B 1 AC 2 A, BD 7 -E 8 A, DF 3 BG 1 E, FH 2 D

Answer

ACT DUR PRED SUC. ES EF LS LF TF FFA 3 - B,C,E 0 3 4 7 4 0B 1 A C,F 3 4 11 12 8 0C 2 A, B - 4 6 14 16 10 0D 7 - E,H 0 7 0 7 0 0E 8 A, D G 7 15 7 15 0 0F 3 B G 4 7 12 15 8 8G 1 E, F - 15 16 15 16 0 0H 2 D - 7 9 14 16 7 0

ES = Zero for activities without predecessors = EF of predecessors= Maximum/highest EF of predecessors

EF = ES + Duration LF = Maximum / Highest of EF for activities without successors

= LS of successors = Minimum / Least of LF for more than one successorsLS = LF - Duration TF = LF-EF or LS-ES, where TF is zero those activities are critical activities

= LF-ES-DurationFF = ES of successor – EF of activity ( if there are more than one successor then

take the minimum of ES)

Total time taken for the project is 16 days and the critical path is D-E-G as the total floats for these activities are zero.

Q 2003/ 7(b)

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The table below gives activities, their normal estimated durations and their relationships for a small project. It also indicates the minimum possible durations if crashed and extra cost incurred for such crashing. The management desires the project to be completed in a shorter duration than normally expected, but the extra funds available for this purpose is limited to Rs.500,000/- only. What would be the shortest duration of the project after crashing under the budget limitations ?

ACTIVITY IMMEDIATE PREDECESSOR

DURATION COST (in Rs.’000)NORMAL CRASH NORMAL CRASH

A - 5 3 400 600B - 5 1 300 500C A 10 5 400 700D B 7 2 400 600E A 6 2 300 500F C,D 11 5 600 930G C,D 6 4 300 600H E,F 5 1 200 400I G 4 1 200 500

Answer :

ACT PRED.

SUCC

DUR COST COST SLOP

E

ES EF LS LF TF FFN C N C

A - C,E 5 3 400 600 100 0 5 0 5 0 0B - D 5 1 300 500 50 0 5 3 8 3 0C A F,G 10 5 400 700 60 5 15 5 15 0 0D B F,G 7 2 400 600 40 5 12 8 15 3 3E A H 6 2 300 500 50 5 11 20 26 15 15F C,D H 11 5 600 930 55 15 26 15 26 0 0G C,D I 6 4 300 600 150 15 21 21 27 6 0H E,F - 5 1 200 400 50 26 31 26 31 0 0I G - 4 1 200 500 100 21 25 27 31 6 0

3100 5330

SR.NO. CRASH ACTIVITY

BY DAYS

P1 P2 P3 P4 P5 ACT.COST

CRASHCOST

0 - - 16 31* 25 28 22 31001 F & H & C 3 13 22 22 22 22 495

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Crash Cost – Normal CostCost Slope = ------------------------------(Cost / Duration) Normal Dur. – Crash Dur.

A C

E

B D

F

G

H

I

TECH. SEQ. ND CDP1 : AE H 16 6P2 : ACFH 31 14P3 : ACGI 25 13P4 : BDFH 28 9P5 : BDGI 22 8

The extra fund available to crash the project is limited upto Rs.500,000. Considering the technological sequence the Normal Duration (ND) of Path P2 : ACFH is 31 days. Hence, we take activities in this path for crashing first. In the path the cost slope of activity A is 100, C is 60, F is 55 and H is 50. Considering 3 days crashing of each activity F, H & C the crashing cost comes to 3 x (50+55+60) = Rs.495,000 which is within the limitation of Rs.500,000.

Hence the project can be crashed by 9 days to finish in 22 days.

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Q. 2003 / 8(b)

The progress observed at the end of the seventh day from the beginning on a 12 day duration project is a s given in the table. The actual cost incurred till date ie reported to be Rs.3100.00

Draw a Gantt Chart for the project and find the project performance on the basis of cost and schedule performance indices. (assume the activity costs are incurred uniformly over its duration)

Activity ImmediatePredecessor

Estimated duration in

days

Budgeted Cost of activity

% of Completion at end of 7

daysA - 3 600 100B - 1 200 100C A 4 800 75D B 4 700 100E B 5 500 95F D 2 200 80G E 3 500 50H C 4 400 0I F 2 600 0j G 3 300 0

Answer

Act. Pred. Dur. Bud. Cost

% of Completion at end of 7 days

BCWP % Scheduled Completio

n

BCWS

A - 3 600 100 600 100 600B - 1 200 100 200 100 200C A 4 800 75 600 100 800D B 4 700 100 700 100 700E B 5 500 95 475 100 500F D 2 200 80 160 100 200G E 3 500 50 250 33.33 167.7H C 4 400 0 0 0 0I F 2 600 0 0 0 0j G 3 300 0 0 0 0

4800 2985 3167.7

BCWS (Budgeted Cost of Work Scheduled) = % Scheduled completion X Total Budgeted Cost = 3167.7

BCWP (Budgeted Cost of Work Performed )= % Actual Completed X Total Budgeted Cost = 2985

ACWP ( Actual Cost of Work Performed)= % Actual Completed X Actual Cost incurred till date of reporting= 3100

Gantt Chart

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Act. 0 1 2 3 4 5 6 7 8 9 10 11 12

ABCDEFGHIJ

Date of reporting

Cost Performance Indices (CPI) = BCWP/ACWP = 2985 / 3100 = 0.96Schedule Performance Indices (SPI) = BCWP/BCWS = 2985 / 3167.7 = 0.94Cost of Completion = Budgeted Cost / CPI = 4800 / 0.96 = 5000Time of Completion = Scheduled Duration/SPI = 12/0.94 = 12.76 i.e. 13 days

Q. 2003 / 2(b)

Following data is available about actual sales for the past 14 years.

YR 1 2 3 4 5 6 7 8 9 10 11 12 13 14Sales 2.3 2.2 2.0 2.25 2.60 3.0 3.5 4.1 3.8 4.0 4.3 4.2 4.8 5.2

Find the “Forecast” for the yr 15 using “Two Years” as well as “three years” moving averages. Which of the two forecasts is more reliable on the basis of Mean Squared Error (MSE) criterion ?

Q.2003 / 3(b)

Q. 2003 / 4 (b)

Q.2002 / 4 (b) : DECISION TREE ANALYSIS

Motor City Auto.Co. must decided whether or not to introduce a new car, which features a radically new pollution control system. They must also decide whether or

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not first to see if results of test marketing a limited production show promise or not. The test marketing of limited production will cost Rs.4.0 crores.

The marketing department has estimated that : If the new car achieve high acceptance by the public, company profits will

increase by Rs.25 crores. Low acceptance will reduce company profits by Rs.15 crores. Not introducing the car will not affect the profit.

Probabilities for different outcomes through alternate actions are : If test marketing is done, the possibility of high acceptance is 0.40 The assumed probability for a favourable result from test marketing is 0.50 The conditional probability for high acceptance after a favourable result is

0.64 If the car is introduced inspite of unfavourable test marketing results ,

probability for low acceptance is 0.84

Construct DECISION TREE and determine the optimal course of action.

Answer :

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D1

Not-Introduce

Introduce

HA (0.40)

LA (0.60)D4

C 4

Not-Introduce

Introduce

Not-Introduce

Introduce

HA (0.64)

LA (0.36)

HA (0.16)

LA (0.84)

Favourable (0.5)

Un-favourable (0.5)

C1

D2

C 2

D3

C 3

Test

NoTest

Evaluation of Chance points :

C2 = (0.64 x 25) + (0.36 x –15) = +10.6, hence D2 = +10.6C3 = (0.16 x 25) + (0.84 x –15) = - 8.6, hence D3 = 0C4 = (0.40 x 25) + (0.60 x –15) = + 1.0, hence D4 = + 1.0

C1 = (0.50 x D2) + (0.50 x D3) = (0.50x10.6) +(0.50x 0) = +5.3

Hence D1 : If decision for test marketing then = Profit – Test marketing cost = 5.3-4.0= Rs. 1.3 crore

If no test marketing the profit is Rs.1.0 crores

Hence the decision is to go for test marketing and then introduce the new car.

Q 2002 / 6 (b)

A new pharmaceutical product is to be released on a deadline for which 44 days are left. Activities involved in product launch with their interdependencies and probabilistic times for completion are given in the table. Draw a project network and find the probability of completing the project in time.

Activity Predecessor Time DelaysOptimistic Most Likely Pessimistic

A - 6 10 14B A 1 2 3C A 16 20 30D B 3 5 7E D 2 3 4F C 7 10 13G D 1 2 3H G 1 3 5I C,G 2 2 2J I 2 3 4K H 1 1 1L J, K 1 2 3

Extract from “Z” value table with some cumulative probabilities for normal distribution curve is given.

Answer : Project estimation Maximum Time or Pessimistic Time (tp)Minimum Time or Optimistic Time (to)Most Likely Time (tm)

Average Time (ta ) = (to + 4 tm + tp ) / 6 ------- Classical Formula

Act. Pred. Succ. Duration ES EF LS LF TF Var Var2

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(to) (tm) (tp) (ta )A - B,C 6 10 14 10 0 10 10 10 0 1.33 1.77B A D 1 2 3 2 10 12 26 28 16C A F,I 16 20 30 21 10 31 10 31 0 2.33 5.43D B E,G 3 5 7 5 12 17 28 33 16E D - 2 3 4 3 17 20 38 41 21F C - 7 10 13 10 31 41 31 41 0 1.0 1.0G D H,I 1 2 3 2 17 19 33 35 16H G K 1 3 5 3 19 22 35 38 16I C,G J 2 2 2 2 31 33 36 38 5J I - 2 3 4 3 33 36 38 41 5K H L 1 1 1 1 22 23 38 39 16L J, K - 1 2 3 2 36 38 39 41 3

Calculate average time considering the Classical formula and this is to be considered for calculating ES,EF,LS & LF.

As highest EF is 41, the LF of all activities with no successors i.e last activities E,F,J & L will be 41. Then the LS can be calculated.

ACF are critical activities as there is no float available. As the distribution would be a normal distribution we have to calculate the variance of critical activities only.

Variance of critical activities = (tp - to ) / 6

Is the standard deviation for the normal distribution curve. I.e. ơ = 2.86

= 41 +1ơ = 43.86 -1ơ =38.14+2ơ = 46.72 -2ơ = 35.28+3ơ = 49.58 -3ơ = 32.42

There is very remote chance that the project will go beyond +3ơ = 49.58 i.e. 50 days i.e. 99.9 % chance . The earliest time period during which the project can be completed is 32 days i.e. -3ơ = 32.42 i.e. 0.001 % chance. To find out the possibility that the project can be completed in 44 days :

Z = (Desired Duration- Minimum Duration)/ ơ = (44-41)/2.86 = 1.049Z value of 1.049 from the normal distribution table is 0.8531 and hence the possibility of completing the project within 44 days is 85.31%.

Q. 2002 / 7(b)

The table below defines activities within a small project :Activity Predecessor Duration (days) Cost (Rs.’000)

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________ _____ (var)2 = 8.196 = 2.86

-3ơ -2ơ -1ơ +1ơ +2ơ +3ơ

Normal Crash Normal CrashA - 6 4 24 34B - 4 3 12 22C A 5 3 20 28D A 7 4 29 47E B 6 5 26 34F B 8 5 34 52G C,E 10 6 27 47H D,F 9 7 34 48

The project attracts a penalty of Rs.10,00 per day for the project completion beyond 18 days. Draw the project network by PDM(AON) method and find out the number of days delay intended beyond the stipulated date to minimize the total project cost.

Answer :

ACT PRED.

SUCC

DUR COST COST SLOP

E

ES EF LS LF TF FFN C N C

A - C,D 5 4 24 34 5 0 6 0 6 0 0B - E,F 4 3 12 22 10 0 4 1 5 1 0C A G 5 3 20 28 4 6 11 7 12 1 0D A H 7 4 29 47 6 6 13 6 13 0 0E B G 6 5 26 34 8 4 10 6 12 2 1F B H 8 5 34 52 6 4 12 5 13 1 1G C, E - 10 6 27 47 5 11 21 12 22 1 0H D, F - 9 7 34 48 7 13 22 13 22 0 0

206

Precedence Diagram ( Activity on Node)

Penalty is Rs.10,00.00 per day beyond 18 days


BY DAYS

P1 P2 P3 P4 ACT.COST

PENALTYCOST

TOTAL COST

0 - - 21 22* 20 21 206 40 246

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A

B

C

D

E

F

G

H


TECH. SEQ. ND CDP1 : A-C-G 21 13P2 : A-D-H 22 15P3 : B-E-G 20 14P4 : B-F-H 21 15

1 A 1 20 21* 20 21* 206+5 30 2412 H 1 20* 20* 20* 20* 211+7 20 2383 G & H 1 19 19 19 19 218+1

210 240

1st crashing : P2 (A-D-H) as this sequence takes maximum no. of days. Activity A is to be crashed as this activity has the minimum slope amongst the three activities involved in this sequence.

2nd Crashing : P2 & P4 as this sequence has the maximum no. of days after the first crashing. Activity H is to be crashed as this activity has the minimum cost slope.

3rd crashing : All paths as all have equal no. of days. This is to test whether the total cost increases or decreases after the previous crashing. Activity G & H are to be crashed as the cost slopes are minimum in these cases.

As the total cost goes up in 3rd crashing, it is advisable to adopt Sr.No.2. ie. Pay penalty for 2 days and complete the project in 20 days instead of 18 days.

Q.2002 / 8 (b)

The following information is available at the end of day 40 of a new plant erection project. Determine if the project is under control based on Earned Value evaluation system, and if not, what is the likely extent of cost and time over-runs at completion.

Activity Predecessor Duration Total BudgetIn Rs.’000

Act. Cost till date

Act. % of completion

A - 10 300 250 100B A 8 400 450 100C A 12 350 380 100D C 0 0 0 0E B,D 18 405 400 70F E 16 450 - 0

Answer :

Act. Pred. Dur. Total BudgetIn Rs.’000

Act. Cost till date

Act. % of completion

EV=BCWP

% Sched. Completion

BCWS

A - 10 300 250 100 300 100 300B A 8 400 450 100 400 100 400C A 12 350 380 100 350 100 350D C 0 0 0 0 0 0 0E B,D 18 405 400 70 283.5 100 405F E 16 450 - 0 0 0 0

1905 1905 1333.5 1455

ACWP ( Actual Cost of Work Performed) = % Actual Completed X Actual Cost incurred till date of reporting= 1905 ………… given in the problem

We have to calculate BCWP i.e. Earned value of the project and BCWS.

BCWP (Budgeted Cost of Work Performed )= % Actual Completed X Total Budgeted Cost = 1333.5

BCWS (Budgeted Cost of Work Scheduled)

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= % Scheduled completion X Total Budgeted Cost

0 10 10 18 22 40 40 56

10 22 22 22

From the above diagram it can be seen that at the end of 40 th day all activities upto should be completed as per schedule. Hence, percentage of scheduled completion will be 100 % for all activities upto E, although the actual % of completion of E is only 70%.

Cost Performance Indices (CPI) = BCWP/ACWP = 1335.5/1480 = 0.90Schedule Performance Indices (SPI) = BCWP/BCWS = 1335.5/1455 = 0.92Cost of Completion = Budgeted Cost / CPI = 1905 / 0.90 = 2116.7Time of Completion = Scheduled Duration/SPI = 56/0.92 = 60.9 i.e. 61 days

Hence the cost over-run would be Rs.211,700.00 and the duration for completion would be 61 days instead of 56 days.

Q. 2001 / 6 (b)

Following table lists the various details of project activitiesActivity Predecessor DURATION in WEEKS

(to) (tm) (tp)A - 1 1 7B - 1 4 7C - 2 2 8

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A 10

B 8

C 12

E 18

D 0

F 16

D A 1 1 1E B 2 5 14F C 2 5 8G D, E 3 6 15

1. Draw the project network using the AOA convention and identify all paths through it.

2. Find expected average duration and variance for each activity.3. What is the expected duration of the project with 50% chance of completion4. What would be your level of confidence to accept the project deadline of 20

weeks

Note : A normal distribution curve covers 50.0%, 84.1% and 97.7% area to its left for 0, +1 and +2 SD from the mean respectively.

Answer :

(i) Project Network Diagram using AOA convention

(ii) Average Time (ta ) = (to + 4 tm + tp ) / 6 & Variance = (tp - to) / 6

Act. Pred.

DURATION in WEEKS ES EF LS LF TF Var Var2

(to) (tm) (tp) (ta )A - 1 1 7 2 0 2 7 9 7 1B - 1 4 7 4 0 4 0 4 0 1 1C - 2 2 8 3 0 3 9 12 9 1D A 1 1 1 1 2 3 9 10 7 0E B 2 5 14 6 4 10 4 10 0 2 4F C 2 5 8 5 3 8 12 17 9 1G D, E 3 6 15 7 10 17 10 17 0 2 4

(iii) As the distribution curve is normal curve, variance of only critical activities are considered for calculating the standard deviation.

As per the distribution curve the expected duration of the project with 50% chance of completion is 17 days.

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G(7)

F(5)

E(6)

D(1)

C(3)

B(4)

A(2)

1

2

3

4

5

6

Critical Path : B-E-G = 17 weeks

Other Paths are :A-D-G : 10 days & C-F = 8 weeks

________ __

Standard Deviation : ơ = (var)2 = 9 = 3.0

99.9 %

97.3 %

50 %

84.5 %

-3ơ -2ơ -1ơ +1ơ +2ơ +3ơ

= 17 +1ơ = 20+2ơ = 23

(iv) Project Deadline of 20 weeks means +1ơ as the distribution curve is a normal curve and the standard deviation is 3. At +1ơ i.e. 20 days the distribution curve covers 84.1% area to its left indicating that the level of confidence would be 84.1%.

Q. 2001/7(b)

Project Network using AOA Convention


BY DAYS

P1 P2 P3 Crash Cost/day

Cumm. Crash Cost

0 - - 20* 16 171 E 4 16 16 17* 15000 600002 C 1 16 16 16 30000 900003 F 1 15 15 15 40000 1300004

Q. 2001 / 8 (b)

Gantt Chart with monthly and cumulative cash flows

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F(2)

dummy(0)

E(10)

D(5)

C(15)

B(8)

A(9)

1

2

3

4

5 6

Critical Path : P1 : B-E-F = 20 Days

Normal Duration : 20 daysCrash Duration : 12 daysOptimum duration cannot be found out as overheads/day is not given

Other Paths are : P2-ADF & P3-CF

The cost of a project rises gradually, then rises steeply and towards the completion of the project slows down again, thus resembling the alphabet S.

Q.2000/5(b) DECISION TREE ANALYSIS Investible Surplus = Rs. 100 crores

Investing this amount in existing business will given assured return of 8%.Alternatively , the amount can be invested in diversification, which if:

1. successful will give estimated return of 17% and2. unsuccessful will give estimated return of 2%.

In a decision tree analysis, Decision points are mutually exclusive and are represented by a small box , whereas chance points are represented as

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Diversification

Existing business

2%unsuccessful

successful

D1

C1

8%

17%

Let ‘a’ be the chance of success, then the chance of failure is (1-a)

Evaluation of Chance C-1 = (a x 17%) + (1-a) x 2%

Minimum of amount of risk to diversify would be that the evaluation of C1 equals assured return of Investible surplus in old business i.e. 8%

Hence, 17a+2(1-a) = 8 which implies a= 0.4

Hence probability of success would be 0.4 and failure would be 0.6, in case the Investible surplus is utilized in diversification.

Q. 2000 / 7 (b) : Project Network on AOA Convention

Calculation of Total Floats and Free Floats

ACT. PRED. SUCC. DUR ES EF LS LF TF FFA - B,C 3 0 3 0 3 0 0B A D 5 3 8 7 12 4 0C A E,F 7 3 10 3 10 0 0D B G 4 8 12 12 16 4 4

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Critical Path : A- C- E –G –H - LTotal Time taken for completion of Project is 33 days

L(4)

K(2)

H(8)

G(5)

F(4)

E(6)

D(4)

C(7)

B(5)

A(3)1 2

3

4

5 6 7

8

9

E C G 6 10 16 10 16 0 0F C H 4 10 14 17 21 7 7G D , E H,K 5 16 21 16 21 0 0H G, F L 8 21 29 21 29 0 0K G L 2 21 23 27 29 6 6L H, K - 4 29 33 29 33 0 0

Project Network on AON Convention

0 3 3 8 8 12 16 21 21 23

0 3 7 12 12 16 16 21 27 29

3 10 10 16 29 33

3 10 10 16 29 33

10 14 21 29

17 21 21 29

Q.2000/8 (b)

BCWS (Budgeted Cost of Work Scheduled) = % Scheduled completion X Total Budgeted Cost

BCWP (Budgeted Cost of Work Performed )= % Actual Completed X Total Budgeted Cost

ACWP ( Actual Cost of Work Performed)= % Actual Completed X Actual Cost incurred till date of reporting

Activity

% Completion Cost in Rs. Lakhs BCWS BCWP Cost Variance

Scheduled

Actual Budgeted

Actual

(A) (B) (C ) (D) (AXC) (BXD)1 100 100 10.0 12.0 10 10 02 100 100 12.0 12.5 12 12 03 70 60 18.0 12.0 12.6 10.8 -1.84 55 50 25.0 13.0 13.75 12.5 -1.255 30 25 20.0 6.0 6 5 -16 10 0 15.0 0 1.5 0 -1.57 0 0 10.0 0 0 0 08 0 0 8.5 0 0 0 09 0 0 6.5 0 0 0 010 0 0 5.0 0 0 0 0

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F 4

A 3

B5

C 7

D 4

E 6

G5

H 8

K 2

L 4

DUR

TF

FF

ES EF

LS LF

A

TOTAL 130.0 55.5 55.85 50.3 -5.55

1. BCWS = Rs.55.85 lakhs2. BCWP = Rs.50.3 lakhs3. Cost Variance = BCWP-ACWP = 50.3 – 55.5 = -5.20 4. CPI = BCWP/ACWP = 50.3/ 55.5 = 0.915. SPI = BCWP/BCWS = 50.3/ 55.85 = 0.906. Cost of Completion = Budgeted Cost / CPI = 130/0.91 = Rs.142.86 lakhs7. Time of Completion = Sch. Duration/SPI = 35/0.90 = 38.89 i.e. 39 weeks.

Hence the project is likely to be completed in 39 weeks instead of the scheduled 35 weeks and the project cost is likely to escalated upto Rs.142.86 lakhs.

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Q. 1999/6(b)

The longest path is the critical path i.e. (1-2) – (2-3) – (3-5) – (5-8) = 36 days

Average Time (ta ) = (to + 4 tm + tp ) / 6 & Variance = (tp - to) / 6

Act DURATION in WEEKS Var Var2

(to) (tm) (tp) (ta )1-2 3 6 15 7 2 41-6 2 5 14 6 2 42-3 6 12 30 14 4 162-4 2 5 8 5 1 13-5 5 11 17 11 2 44-5 3 6 15 7 2 46-7 3 9 27 11 4 165-8 1 4 7 4 1 17-8 4 19 28 18 4 16

A standard deviation of +1.0 to the mean value under Normal Distribution Curve covers 84.1% area to its left i.e. +1ơ = 41, where = 36 & ơ = 5

Hence the expected time for completion of project with 84.1 % probability is 41 days.

Q.1999/ 7 (b)

ACT. PRED DUR SUCC ES EF LS LF TF RES.

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(7)

(6)

(14) (11)

(5)

(11)

(7)

(4)

(18)

As the distribution curve is normal curve, variance of only critical activities are considered for calculating the standard deviation. ________ __

Standard Deviation : ơ = (var)2 = 25 = 5.0

99.9 %

97.3 %

50 %

84.1 %

-3ơ -2ơ -1ơ +1ơ +2ơ +3ơ

1 2 3

4

5

6 7

8

A - 8 C 0 8 2 10 2 4B - 7 D,E,F 0 7 0 7 0 8C A 6 G 8 14 10 16 2 5D B 8 G 7 15 8 16 1 4E B 4 H 7 11 11 15 4 8F B 8 I 7 15 7 15 0 6G C,D 5 J 15 20 16 21 1 5H E 6 J 11 17 15 21 4 4I F 6 J 15 21 15 21 0 5J G,H,I 10 - 21 31 21 31 0 6

Gantt Chart

Resource Graph

Q.1999/ 8 (b)

Total Project Budget Rs.250.00 CroresScheduled Completion 80 weeksDate of Reporting End of 50 weeks

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If the resources are limited to 20 workers then the days of over allocation of resources are on the 8th

day , 9th day , 10th day and 11th day.

BCWP Rs.170.00 CroresACWP Rs.180.00 CroresBCWS Rs. 187.00 Crores

Cost Variance = BCWP-ACWP = 170 – 180 = - 10 CroresCPI = BCWP/ACWP = 170/180 = 0.94SPI = BCWP/BCWS = 170/187 = 0.91Cost of Completion = Budgeted Cost / CPI = 250/0.94 = Rs.265.96 CroresTime of Completion = Sch. Duration/SPI = 80/0.91 = 87.91 i.e. 88 weeks.

Till the date of reporting the actual cost of work performed has exceeded the Budgeted by Rs.10 Crores. The cost of completion would exceed the budgeted cost of completion by Rs.15.96 crores and the time of completion would exceed the scheduled completion period of 80 weeks by 8 weeks.

Notes :

Network Relationships : Start to Start (SS)Finish to Start (FS)Finish to Finish (FF)Start to Finish (SF)

SS : Start of the successor depends upon the start of the predecessorFS : Start of the successor depends upon the Finish of the predecessorFF : Finish of the successor depends upon the Finish of the predecessorSF : Finish of the successor depends upon the start of the predecessor

In SS relationship, there can be a lag i.e. SS+Lag meaning the successor starts 3 days after the start of the predecessor activity. It can also be represented as % i.e SS+20% meaning the successor starts after 20% of the predecessor activity is completed.

0 10 3 8

In FS Relationship, there can be lag or a lead,

0 10 LAG 13 18 0 10 LEAD 7 12

Q. 1998 / 6(b)

ACTIVITY DURATION PREDECESSOR RELATIONSHIP LEAD/LAGA 5 - - -B 3 A FS -2C 4 A FS +3D 7 C SS -

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A 10

B5

SS + 3

A 10

B5

FS + 3A

10

B5

FS - 3

E 5 C FF +4F 3 B FS -

D FS -1G 2 E FS -H 6 F FS -

G FS -2I 3 G FS -J 4 H SS +4K 2 I FS -

J FF +1

0 5 3 6 14 12 17 23 21 25

0 5 11 14 14 17 17 23 21 25

8 12 8 15

8 12 8 15

11 16 16 18 18 21 24 26

12 17 17 19 21 24 24 26

Critical Path : A C D F H J K

Q.1998/7(b)ACTIVITY NORMAL DUR. CRASH DUR. NORMAL COST CRASH COST

1-2 5 2 600 9002-4 6 3 700 10001-3 4 2 100 2003-4 7 4 400 8004-7 9 5 600 9203-5 12 3 1600 19604-6 10 6 1500 1800

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FS-2 FS FS SS+4

FS+3

SS

FF+4

FS-1

FF+1

FS FS

FS-2

FS

A

5

C

4

B

3

D

7

F

3

H

6

J

4

E

5

G

2

I

3

K

2

6-7 7 4 400 4907-9 6 4 300 4205-9 12 7 400 850

Draw network diagram and find Critical Path. In case the project duration is required to be crashed by two days, which activities will get crashed.

ACT. PRED SUCC ND CD ES EF LS LF TF NC CC slope

1-2 - 2-4 5 2 0 5 0 5 0 600 900 1002-4 1-2 4-7,4-6 6 3 5 11 5 11 0 700 1000 1001-3 - 3-4,3-5 4 2 0 4 0 4 0 100 200 503-4 1-3 4-7,4-6 7 4 4 11 4 11 0 400 800 1334-7 2-4,3-4 7-9 9 5 11 20 19 28 8 600 920 803-5 1-3 5-9 12 3 4 16 10 22 6 1600 1960 404-6 2-4,3-4 6-7 10 6 11 21 11 21 0 1500 1800 756-7 4-6 7-9 7 4 21 28 21 28 0 400 490 307-9 4-7,6-7 - 6 4 28 34 28 34 0 300 420 605-9 3-5 - 12 7 16 28 22 34 6 400 850 90

There are two critical paths in the problemCP 1 : (1-2-4-6-7-9) and CP 2 : (1-3-4-6-7-9) and Project Duration is 34 daysOther Paths are P3 – (1-2-4-7-9) = 26 days and P4 (1-3-5-9) = 28 days

In case the project is to be crashed by two days, then it is advisable to crash that activity in the critical path which has the minimum cost slope. Ie. Activity 6-7 has a cost slope of 30 and by crashing this activity by two days the increase in cost would be 60.

Q.1998/8(b)

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(6)

(12)

(7)(10)

(12)

(9)

(7)(4)

1 2

3

4

5

6

7

9

(5) (6)


The peak requirement of resources are on the Day 7 where the resource required is 12.

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