Programme Evaluation and
Review Technique
CA Final Advance Management Accounting, Paper 5, Chapter 14
Dr. Kavita Sharma
Discussion plan
Introduction to critical path analysis
Basics about techniques of PERT/CPM
Steps of critical path analysis
Errors in logical sequencing
Construction of Network
Network Analysis
• Identification of Critical Path
• Calculation of Floats
Resource Analysis
• Crashing, Resource Levelling, Resource Allocation
PERT
Resource Analysis
Step-1
• Project Planning
• Network construction
Step -II
• Project scheduling
• Identification of critical path
Step-III
• Project Controlling
• Resource requirement
Adding of resources
Adding resource to select activities
Increase in project cost
Decision on adding resources
Cost – duration relationship
Types of cost
Direct and Indirect cost
Project cost






Direct cost- cost associated with individual activity
Indirect costs – include administration or supervision
cost
Both direct and indirect costs are inversely related
Indirect costs increases with increase in project
duration, direct cost increases with the decrease in
the activity time
Time-cost trade off is one of the problem related to
resource analysis.
Resource leveling and resource allocation are the
other two problems
Time-cost trade-off
Conception is: cutting of duration of
some of the activities
Requirement of additional
resources
Crashing
Crashing
Crash time
trade off
• Added resources result in added cost
• Identify the activities that cost least to crash
Crash time
• Minimum possible activity duration time
• Further attempt for time reduction only increases cost
Crash cost
• Cost corresponding to crash time
• Minimum direct cost required to achieve the crash
performance on an activity
Normal cost
• Absolute minimum of direct cost required to perform the
activity
Normal Time
• Shortest activity duration to perform the activity under
minimum direct cost constraint.
Time – cost curve
A
Direct cost
Crash cost
Crash time cost
point
Actual cost curve
Normal cost
B
Crash
time
Activity Time
Normal
time
Normal time cost
point
Slope of cost curve

= incremental cost
𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐 −𝑁𝑁𝑁𝑁𝑁𝑁 𝑐𝑐𝑐𝑐
𝑁𝑁𝑁𝑁𝑁𝑁 𝑡𝑡𝑡𝑡=𝐶𝐶𝐶𝐶𝐶 𝑡𝑡𝑡𝑡

=

Rate of increase in the cost of performing the activity
per unit decrease in time
Crash the activity with least incremental cost

Time cost optimization alogritham
Step-I
• Schedule the project with all its activities at their normal duration
• Identify critical path
Step -II
• Calculate cost slope for activities that can be crashed by adding
more resources
• Rank the activities in ascending order of cost slope
Step-III
• As per ranking crash the activities on critical path
• The activity with minimum slope/incremental cost is crashed first
Step -IV
• As the critical path duration is reduced, critical path is identified
again
• Other paths may become critical and there may be multiple paths
Step -V
• Continuing with crashing till further crashing is either not possible
or does not result in the reduction of project duration
Step-VI
• Compute total project cost by adding corresponding fixed cost to
direct cost
Points of special reference



Crashing of only critical path activities
Change in critical path of the project after crashing
If multiple critical path, reduce activity on each of the
critical path


Choose activities from each path
Choose least cost combination of activities
Exercise-I
Activity
Normal
Crash
Time (days)
Cost (Rs)
Time (days)
Cost (Rs)
1-2
6
600
4
1,000
1-3
4
600
2
2,000
2-4
5
500
3
1,500
2-5
3
450
1
650
3-4
6
900
4
2,000
4-6
8
800
4
3,000
5-6
4
400
2
1,000
6-7
3
450
2
800
1. Per day fixed cost is Rs 100.
2. Crash the activities and determine the optimum project
completion time and cost
Solution-normal time duration
3(1)
2
1
5
4
8(4)
3(2)
6
3
Critical path : 1-2-4-6-7
Project duration :22 days
Associated total project cost = Direct cost + Indirect Cost
= 4,700 + (100× 22)
= 6,900
7
Project duration with crash time
3(1)
2
1
5
4
8(4)
3(2)
6
3
Path
Normal Length
Crash length
1-2-5-6-7
16
9
1-2-4-6-7
22 *
13*
1-3-4-6-7
21
12
7
Revised critical path-IInd/IIIrd Crashing
3(1)
2
1
5
4
8(4)
3(2)
6
7
3
*Crash 6-7 : project duration=20 days
Project Cost = DC+IC
= 4900+350 +100×20 =7250
**Crash 4-6 by 4 days: project duration 16
days
5,250+550×4 +100×16 = 9050
Crashing
alternatives
Incremental cost
1-2,1-3
200+700=900
1-2,3-4
200+550= 750
1-3,2-4
700+500=1200
2-4,3-4
500+550=1050
4-6
550**
6-7
350*
Incremental Cost –Ist crashing
Activity
Normal
Incremental cost
Crash
𝑐𝑐𝑐𝑐𝑐 𝑐𝑐𝑐𝑐 − 𝑁𝑁𝑁𝑁𝑁𝑁 𝑐𝑐𝑐𝑐
𝑁𝑁𝑁𝑁𝑁𝑁 𝑡𝑡𝑡𝑡 − 𝐶𝐶𝐶𝐶𝐶 𝑡𝑡𝑡𝑡
Time (days)
Cost (Rs)
Time
(days)
Cost (Rs)
1-2
6
600
4
1,000
1-3
4
600
2
2,000
700
2-4
5
500
3
1,500
500
2-5
3
450
1
650
100
3-4
6
900
4
2,000
550
4-6
8
800
4
3,000
550
5-6
4
400
2
1,000
300
6-7
3
450
2
800
350
200*
Crash activity 1-2 by one day.
Project cost corresponding to 21 days = DC+IC
= 4,700+200×1+100×21= 7,000
Crashing –IV/V
3(1)
2
1
5
4
8(4)
3(2)
6
7
3
*Crash 1-2,3-4 by 1 day each: project
duration=15 days
Project cost =DC+IC
= 7450+750+100×15= 9700
Project cost:
**Crash 2-4,3-4 by 1 day each: project
duration 14 days
8200 + 1050 + 100×14 = 10,650
Crashing
alternatives
Incremental cost
1-2,1-3
200+700=900
1-2,3-4
200+550= 750*
1-3,2-4
700+500=1200
2-4,3-4
500+550=1050**
4-6
600
6-7
350
Crashing-VI
3(1)
2
1
5
4
8(4)
3(2)
6
7
3
*Crash 1-3, 2-4 by 1 day each: project
duration=13 days
Project cost:
10,450 + 100×13 = 11,750
Crashing
alternatives
Incremental cost
1-2,1-3
200+700=900
1-2,3-4
200+550= 750
1-3,2-4
700+500=1200*
2-4,3-4
500+550=1050
4-6
600
6-7
350
Time cost relationship
Crashing
Minimum duration of
13 days with the
associated cost of Rs
11750
3(1)
2
1
Project
duration
(days)
Direct
cost
Indirect
cost
Total cost
22
4700
2200
6900
Ist (1-2)
21
4900
2100
7000
IInd (6-7)
20
5250
2000
7250
IIIrd (4-6)
16
7450
1600
9050
IV (1-2), (34)
15
8200
1500
9700
V (2-4)(3-4)
14
9250
1400
10650
VI (1-3) (24)
13
10450
1300
11750
5
4
3
Determination of time cost relationship
8(4)
3(2)
6
7
Exercise -II
Activity
Normal
Time (days)
Crash
Crashing
Time (days)
Cost (Rs per
day)
1-2
9
4
20
1-3
8
2
25
1-4
15
3
30
2-4
5
1
10
3-4
10
4
15
4-5
2
4
40
Given: overhead cost Rs 60 per day
Critical path with normal time duration
2
15(10)
2(1)
4
1
5
3
Path
Normal Length
Crash length
1-2-4-5
16
10
1-4-5
17
11
1-3-4-5
20*critical path
12*critical path
Time cost relationship
Project
duration
Crashing
Crashing
cost
Cumulative
crashing cost
Overhead cost
(@Rs 60 per day)
Total cost
20
-
-
-
1200
1200
19
3-4
15
15
1140
1155
18
3-4
15
30
1080
1110
17
3-4
15
45
1020
1065
16
1-4, 3-4
45
90
960
1050
15
4-5
40
130
900
1030*
14
1-3, 1-4, 2-4
65
195
840
1035
13
1-3, 1-4, 2-4
65
260
780
1040
12
1-3, 1-4, 1-2
75
335
720
1055
Optimum crashing up to 15 days
Time-cost function for the project
1400
1200
1000
800
cum cost
indirect cost
total cost
600
400
200
0
0
5
10
15
20
25
Resource Levelling and Allocation
Resource Levelling
Performance of activities require resource allocation
Resources are limited in supply
Maximum utilization
Adjust starting times
Levelling ensures constancy of resource usage
Purpose – to reduce peak resource requirement and smooth out period – to –
period assignments within a constraint on the project duration
Resource utilization is smoothed out without shortening the project time
Resource Allocation
Allocation of available resources to determine shortest
project duration
Explore adequacy of resources
Resource overlaps
Activities may require same or varying amount of resources
Use floats as criterion for allocation of resources
Resource allocation is based on certain assumptions
Programme Evaluation and Review
Technique -PERT
Probability considerations
Conditions of uncertainity
Optimistic time = 𝑡𝑜 ( shortest possible time)
Pessimistic time = 𝑡𝑝 (longest possible time)
Most likely time = 𝑡𝑚 (estimate of normal time)
Estimated time is represented as probability
distribution
Properties of probability distribution of
activity times
𝑡𝑜 and 𝑡𝑝 specify the range
𝑡𝑚 may not coincide with the mid point (𝑡𝑜 + 𝑡𝑝 )/2
𝑡𝑚 may occur left or to the right of the mid point
Duration of each activity time may follow Beta (β) distribution
with uni-model point occurring at 𝑡𝑚 and its end points 𝑡𝑜 and
𝑡𝑝
Probability of meeting the scheduled time
 Expected value of activity duration is
𝑡𝑒 =
1
3
[2 𝑡𝑚 + (𝑡𝑜 + 𝑡𝑝 )/2]
𝜎=
1
6
(𝑡𝑝 - 𝑡𝑜 )
1
6
= [𝑡𝑜 + 4 𝑡𝑚 + 𝑡𝑜 ]
𝜎2
1
[
6
2
=
(𝑡𝑝 − 𝑡𝑜 )]
For a given project, if the critical activities are 1,2,3….k, we have
𝑇𝑒 = 𝑡𝑒𝑒 + 𝑡𝑒𝑒 + 𝑡𝑒𝑒 + ⋯ 𝑡𝑒𝑒 , and
𝑉𝑇 = 𝜎𝑇 = 𝜎 2 = 𝜎 21 + 𝜎 2 2 + 𝜎 2 3 + ⋯ 𝜎 2 𝑘
Estimation of time for project completion
𝑇𝑒
= Expected project completion time
 𝑉𝑇 = Variance
 Invoke Central Limit Theorem – state that the sum of
several independent activity duration will tend to be
normally distributed
 Get probability distribution of times for completing
the project, which is normally distributed with
μ= 𝑇𝑒 and σ = 𝜎𝑇

Prob 𝑍 <
𝑇𝑇 −𝑇𝑇
𝜎𝜎
Some specific points




Case of multiple critical paths – select the one with
largest variance
Assumed the completion of non-critical activities
Calculate separate probabilities in case of paths
with lesser difference between their project
completion estimates
Possibility of change in critical path
Case Example -III
Activity
Predec
essor
Time Estimation (weeks)
To
Tm
Tp
Te
Earliest
Latest
𝜎2
ES
EF
LS
LF
Total
Float
s
A
-
2
3
4
3
1/9
0
3
0
3
0
B
-
8
8
8
8
0
0
8
1
9
1
C
A
7
9
11
9
4/9
3
12
3
12
0
D
B
6
6
6
6
0
8
14
9
15
1
E
C
9
10
11
10
1/9
12
22
18
28
6
F
C
10
14
18
14
16/9
12
26
12
26
0
G
C,D
11
11
11
11
0
14
25
15
26
1
H
F,G
6
10
14
10
16/9
26
36
26
36
0
I
E
4
5
6
5
1/9
22
27
28
33
6
J
I
3
4
5
4
1/9
27
31
33
37
6
K
H
1
1
1
1
0
36
37
36
37
0
µ = ∑𝑇𝑒 = 37
σ = 𝜎 2 = 2.028
Parameters of project duration
E
C
I
J
A
F
B
D
G
K
H
Contd. with solution



Project completion = Te = 37 weeks
σ = 2.028 weeks
Prob. of completing in 40 weeks
40−37
= 2.028 =

=Z

area between mean and Z= 1.48 under normal
curve is .4306, and the required probability is = 0.5 +
0.4306 = 0.9306
To be 95% sure in project completion calculate the
value of X corresponding to which an area equal to
95% to the left of it .
Use normal area table Z value corresponding to the
area of .45 (area between μ and X) is 1.64


1.48
Case Example-IV
Activity
Predec
essor
Time Estimation (weeks)
To
Tm
Tp
Te
A
1-2
3
15
8
8
2
B
1-3
4
8
6
6
2/3
C
1-4
6
20
12
12
7/3
D
2-4
6
6
6
6
0
E
2-6
5
9
7
7
2/3
F
3-4
6
15
9
9
3/2
G
3-5
4
12
6
6
6
H
4-7
10
12
11
11
11
I
5-7
4
8
6
6
6
J
5-8
10
10
10
10
10
K
6-7
10
18
14
14
14
L
7-8
2
6
3
3
3
𝜎
Critical path:
1-2-6-7-8 (A E K L)
Expected Duration (Te) of the
Project = 8+7+14_3 =32 weeks
Variance 𝜎 2 = 22 + 2/32 + 4/32 +
2/32 = 20/3 weeks
𝜎 = 20/3 = 2.582 weeks
Prob. of completing in 38
weeks:
Contd….
Chapter Summary






Crashing implies attempting to reduce the duration of the
project by employing additional resources
Obtain per unit incremental cost also called cost slope for each
activity
Step by step reduce the activity times on critical path till we are
able to crash the project duration to desired level or till we get
the minimum project cost
In PERT analysis, use three time estimates for each activity,
and on the basis of expected activity time (Te) critical path is
determined to get the expected project duration.
Critical activities variances yield variance of the project
completion time.
Probabilities of project completion are calculated by
considering the project completion time to be normally
distributed with parameters of mean and standard deviation
Thank you
All THE BEST