Paper/ Subject Code 86001/ Operation Research
TYBMS SEM 6
Operation Research
(Q.P. April 2019 with Solution)
1) April 2019 Q.P. with Solution (PDF)
2) November 2019 Q.P. with Solution (PDF)
4) April 2023 Q.P. with Solution (PDF)
Please check
whether you have got the right question paper.
Note:
1. All questions
are compulsory. (Subject to Internal Choice)
2. Figures to the
right indicate full marks
3. Use of
non-programmable calculator is allowed and mobile phones are not allowed.
4. Normal
distribution table is printed on the last page for reference. Support your
answers with diagrams / illustrations, wherever necessary
6. Graph papers
will be supplied on request.
____________________________________________
Q1.A) Multiple choice questions (Attempt Any 8) (8)
i) ln linear programming, unbounded solution means ____________.
a) infeasible
solution b) Degenerate solution
c) Infinite
solutions d) Unique solution
Ans: c) Infinite solutions
ii) If M + N-1 = Number of allocations in transportation, it means _________.
(Where 'M' is
number of rows and N' is number of columns)
a) There is no
degeneracy b) Problem is unbalanced
c) Problem is
degenerate d) Solution is optimal
Ans: a) There is no degeneracy
ijü) Floats for critical activities will be always _________________.
a) One b) Zero
c) Highest d) Same ás duration of the activity
Ans: b) Zero
iv) The total time required to complete all the jobs-in a job sequencing problem is known as __________,
a) Idle time b) Processing time
c)Elapsed time d) Processing order
Ans: c)Elapsed time
v) In linear programming, _____________ represents mathematical equation of the limitations imposed by the problem.
a) Objective
function b) Decision variable
c)Redundancy d) Constraints
Ans: d) Constraints
vi) If in an assignment problem, number of rows is not equal to number of columns then ____________.
a) Problem is
degenerate b) Problem is
‘unbalanced
c) It is a
maximization problem d)
Optimal solution is not possible
Ans: b) Problem is ‘unbalanced
vii) The maximum time in which an activity will be completed assuming all possible delays and postponements is termed as __________.
a) Optimistic time
b) Most likely time
c) Pessimistic time d) Expected time
Ans: c) Pessimistic time
viii) The various alternatives or courses of actions available to each player in a game are called as __________.
a) Saddle points b) Strategies
c) Pay off d) 'n' player game
Ans: b) Strategies
ix) In simplex, a maximization problem is optimal when all Delta J, i.e., Cj-Zj values are
a) Either zero or
positive b) Either zero or negative
c) Only positive d) Only negative
Ans: b) Either zero or negative
x) Which of the following considers difference between two least costs for each row and column while finding initial basic feasible solution in transportation?
a) North west
corner rule b) Least
cost method
c) Vogel's
Approximation method d) Row minima
method
Ans: b) Least cost method
Q1.B) True or
false (Attempt Any 7) (7)
i) Probability of a project completing in its expected time (Te) will be always 100%
Ans: False
ii) lf saddle point is available in a game, it is called as pure strategy game.
Ans: True
iii) Slack represents unutilized resources.
Ans: True
iv) If in a transportation problem, number of rows is not equal to number of columns, then the problem is unbalanced.
Ans: False
v) lf we introduce an unnecessary dummy activity, the error is termed as redundancy.
Ans: True
vi) Job sequencing problems are solved to ensure that, both, the total time to complete all jobs and idle time of each machine are maximum.
Ans: False
vii) When more than one optimal solution is possible in a linear programming problem, it is termed as 'unique solution'.
Ans: False
viii) Regret matrix is made to convert a maximization problem into minimization problem in assignment.
Ans: True
ix) Critical path method (CPM) considers the three-time estimates: most likely, optimistic and pessimistic time estimates.
Ans: False
x) In solving a job sequencing problem, it is assumed that all jobs require the same sequence of operations.
Ans: False
Q.2 A) A company produces 2 products A and B: x1 and x2, are the quantities manufactured of Products A and B respectively. The following objective function along with constraints is given to you:
Max Z = 8x1 + 16x2,
Subject to
constraints:
x1 + x2
< 200
x2 <
125
x1 + 2x2
< 300
x1 , x2 > 0
Find how many units of Product A and Product B should be produced by the company so that the profit is maximized. Ls it a case of multiple optimal solutions? Use graphical method to solve the LPP.
Video Solution
Solution:
Max Z = 8x1
+ 16x2,
Subject to
constraints:
x1 + x2
< 200 ---- (i)
x2 <
125 ---- (ii)
x1 + 2x2
< 300 ---- (iii)
x1 , x2 > 0
x1 + x2
< 200 ---- (i)
x1 + x2
= 200
Let X1
= 0
0 + x2 =
200
Therefore x2 = 200 (x1, x2) = (0, 200)
x1 + x2
= 200
Let X2
= 0
x1 + 0 = 200
Therefore x1 = 200 (x1, x2) = (200, 0)
x2 < 125 ---- (ii)
x2 = 125
Let x2 =
0
x2 = 125 (x1, x2) = (0, 125)
x1 + 2x2
< 300 ---- (iii)
x1 + 2x2
= 300
Let x1 =
0
0 + 2x2 = 300
2x2
= 300
x2 = 300 / 2 Therefore x2 = 150 (x1, x2) = (0, 150)
x1 + 2x2
= 300
Let x2 =
0
x1 + 2(0)
= 300
x1 =
300 (x1, x2)
= (300, 0)
Corner Point Method:
|
Vertex |
Co-ordinate |
Max Z = 8x1
+ 16x2, |
|
O (0,0) |
x1 =
0, x2 = 0 |
8(0) + 16(0) = 0 |
|
A (0, 125) |
x1 =
0, x2 = 125 |
8(0) + 16(125) =
2000 |
|
B (50, 125) |
x1 = 50, x2
= 125 |
8(50) + 16(125)
= 2400 |
|
C (100, 100) |
x1 = 100, x2
= 100 |
8(100) + 16(100) = 2400 |
|
D (200, 0) |
x1 = 200, x2
= 0 |
8(200) + 16(0) = 1600 |
Optimal Profit Maximization is Rs. 2,400
Q2.B) You are given the per unit cost of transporting goods from 3 factories to 4 customers. The 3 factories A B and C have capacity to supply 500,300 and 200 units respectively. The 4 customers P, Q, R and S Require 180,150,350 and 320 units respectively.
|
Customers Factory |
P |
Q |
R |
S |
|
A |
12 |
10 |
12 |
13 |
|
B |
12 |
11 |
8 |
14 |
|
C |
6 |
16 |
11 |
7 |
i) You are
required to find the Initial Basic Feasible Solution using Vogel's
Approximation Method
(5)
ii) Find the total
cost of transportation schedule obtained using VAM (2)
Video Solution
Solution:
Total Demand =
1000
Total Supply =
1000
Total Demand =
Total Supply
Cost data is given
Minimization
Problem
RIM Condition:
M + N – 1 = No. of
Allocation
3 + 4 – 1 = 6
6 = 6
Therefore, RIM
Condition is satisfied.
Optimal Solution
is Non-Degeneracy
(ii) Total Cost of
Transportation :
= (150 x 10) + (230 x 12) + (120 x
13) + (180 x 7) + (120 x 8) + (200 x 7)
= Rs. 9, 440
OR
Q.2 C) You are given the
following details for a project consisting of 8 activities:
|
ACTIVITY |
NODE |
DURATION
(days) |
|
A |
1-2 |
4 |
|
B |
1-3 |
6 |
|
C |
1-5 |
13 |
|
D |
2-3 |
5 |
|
E |
2-4 |
20 |
|
F |
4-6 |
10 |
|
G |
3-6 |
6 |
|
H |
5-6 |
16 |
(i) Draw the network diagram and identify the critical path. (3)
(ii) Find earliest start time, earliest finish time, latest start time and latest finish time for each activity. (4)
(iii) Find free float for activity B. (1)
Solution: [Video Explanation]
(i)
Network Diagram
Paths in Network:
(1) A-E-F = 4 + 20 + 10 = 34 days
(2) A-D-G = 4 + 5 + 6 = 15 days
(3) B-G = 6 + 6 = 12 days
(4) C- H = 13 + 16 = 29
Projection Completion Time : 34 days
(ii)
|
Activity |
EST
= Ei |
EFT
= Ei – t |
LST
= Lj – t |
LFT
= Lj |
|
A |
0 |
0 + 4 = 4 |
4 – 4 = 0 |
4 |
|
B |
0 |
0 + 6 = 6 |
28 – 6 = 22 |
28 |
|
C |
0 |
0 + 13 = 13 |
18 – 13 = 5 |
18 |
|
D |
4 |
4 + 5 = 9 |
28 – 5 = 23 |
28 |
|
E |
4 |
4 + 20 = 24 |
24 – 20 = 4 |
24 |
|
F |
24 |
24 + 10 = 34 |
34 – 10 = 24 |
34 |
|
G |
9 |
9 + 6 = 15 |
34 – 6 = 28 |
34 |
|
H |
13 |
13 + 16 = 29 |
34 – 16 = 18 |
34 |
(ii)
Free Float for Activity B:
= Ej – Ei – t
= 9 – 0 – 6
= 15
Q.2 D) There are 6 jobs to be performed in a factory and each would go through 2 machines A and B in the order AB. The processing time (in hours) is given for each job in each machine.
|
Job |
Machine A |
Machine B |
|
I |
7 |
3 |
|
II |
4 |
8 |
|
III |
2 |
6 |
|
IV |
5 |
6 |
|
V |
9 |
4 |
|
VI |
8 |
1 |
(i) Determine the
sequence of performing the jobs that would minimize the total time of
completing all the jobs: (2)
(ii) Find the
elapsed time. (3)
(iii) Find idle
time for both the machines. (2)
Solution : [Video Explanation]
(i)
Optimal Sequence
A B
|
III |
II |
IV |
V |
I |
VI |
(ii)
Elapsed Time Calculation :
|
Job |
Machinery A |
Machinery B |
||
|
In |
Out |
In |
Out |
|
|
III |
0 |
0 + 2= 2 |
2 |
2 + 6 = 8 |
|
II |
2 |
2 + 4 = 6 |
8 |
8 + 8 = 16 |
|
IV |
6 |
6 + 5 = 11 |
16 |
16 + 6 = 22 |
|
V |
11 |
11 + 9 = 20 |
22 |
22 + 4 = 26 |
|
I |
20 |
20 + 7 = 27 |
27 |
27 + 3 = 30 |
|
VI |
27 |
27 + 8 = 35 |
35 |
35 + 1 = 36* |
(iii) Idle time for Machinery A = 36 – 35 = 1 Hrs
Idle
time for Machinery A = 36 – 28 = 8 Hrs
Q.3. B) You are given the following details for a project with 8 activities.
|
ACTIVITY |
NODE |
Optimistic
time days |
Most likely
days |
Pessimistic
time days |
|
A |
1-2 |
4 |
6 |
8 |
|
B |
2-3 |
5 |
7 |
15 |
|
C |
2-4 |
4 |
8 |
12 |
|
D |
3-5 |
10 |
18 |
26 |
|
E |
4-6 |
8 |
9 |
16 |
|
F |
5-7 |
4 |
8 |
12 |
|
G |
6-7 |
1 |
2 |
3 |
|
H |
7-8 |
6 |
7 |
8 |
(i) Draw the network diagram. (3)
(ii) Find the expected time of project completion along with
standard deviation. (2)
(iii) What is the probability of the project completing in
55 days? (2)
Ans: [Video Explanation]
|
ACTIVITY |
NODE |
Expected
Time te = [a+ 4m+ b / 6] |
SD [b-a /
6] |
Variance (S.D)2 |
|
A |
1-2 |
6 |
4/6 |
16/36 |
|
B |
2-3 |
8 |
10/6 |
100/36 |
|
C |
2-4 |
8 |
8/6 |
64/36 |
|
D |
3-5 |
18 |
16/6 |
256/36 |
|
E |
4-6 |
10 |
8/6 |
64/36 |
|
F |
5-7 |
8 |
8/6 |
64/36 |
|
G |
6-7 |
2 |
2/6 |
4/36 |
|
H |
7-8 |
7 |
2/6 |
4/36 |
Network Diagram:
Paths in the Network:
1) 1-2-3-5-7-8
= 47 days
2) 1-2-4-6-7-8
= 33 days
Expected Project Completion Time : 47 days
Variance of critical path = 16/36 + 100/36 + 256/36 +
64/36 + 4/36 = 440/36 =12.22
SD of critical path = √variance
SD = √ 12.22 =
3.495 = 3.5 days
Time = 55 days = te
Z = t – CP / SD
= 55 – 47 / 3.5 =
2.29
Table value = 0.4890
P(55 days) = 0.5 + 0.4890 = 0.9890 =
98.90%
OR
Q.3. C) You are given information about the cost (in Rs. Thousands) of performing different jobs by different persons. PI cannot perform J3. P3 cannot perform J4
|
|
|
Job |
||||
|
P E R S O N |
|
J1 |
J2 |
J3 |
J4 |
J5 |
|
P1 |
27 |
18 |
X |
20 |
21 |
|
|
P2 |
31 |
24 |
21 |
12 |
17 |
|
|
P3 |
20 |
17 |
20 |
X |
16 |
|
|
P4 |
22 |
28 |
20 |
16 |
27 |
|
(i)Obtain optimal assignmnt and find cost of such
assignment.
(i)Is it a case of alternative optimal solution?
Ans:
Prohibited Assignment = X (infinity)
The problem is unbalanced.
Taking P5 as Dummy
Cost is given.
Hence Minimization Problem.
|
Person |
Job |
||||
|
|
J1 |
J2 |
J3 |
J4 |
J5 |
|
P1 |
27 |
18 |
X |
20 |
21 |
|
P2 |
31 |
24 |
21 |
12 |
17 |
|
P3 |
20 |
17 |
20 |
X |
16 |
|
P4 |
22 |
28 |
20 |
16 |
27 |
|
P5 |
0 |
0 |
0 |
0 |
0 |
Row Minimization
|
Person |
Job |
||||
|
|
J1 |
J2 |
J3 |
J4 |
J5 |
|
P1 |
27 |
18 |
X |
20 |
21 |
|
P2 |
31 |
24 |
21 |
12 |
17 |
|
P3 |
20 |
17 |
20 |
X |
16 |
|
P4 |
22 |
28 |
20 |
16 |
27 |
|
P5 |
0 |
0 |
0 |
0 |
0 |
Column Minimization Not Required.
Min. No. of Lines = 4
Size of Matrix = 5 x 5
Therefore Not optimal
Change = Min. open Value = 4
|
Person |
Job |
Cost (in Rs. 000) |
|
P1 |
J2 |
18 |
|
P2 |
J4 |
12 |
|
P3 |
J5 |
16 |
|
P4 |
J3 |
20 |
|
P5 |
J1 |
0 |
|
|
Total |
66 |
Optimal Cost = 66,000
Q3.D) Two firms, Lacko textiles and Rayon textiles have 3 strategies each to select from. The 3 strategies are no advertisement, using moderate advertising and using heavy advertising. You are given the pay off matrix from view point of Lacko textiles, showing its market share under several combinations of strategies:
( (i)Find the saddle point and value of game.
(ii)Comment on the strategy to be selected by both
the companies.
Value of the Game = 50 x 10000 = 5,00,000
Q. 4 A) you are given a solution for a transportation cost problem. Figures in each cell represent per unit transportation cost. Figures in circle within each cell represent number of units allocated for transportation. X, Y and Z. are the 3 factories and A, B, C and D are the 4 Customers.
(i) You are required to check the above solution for
optimality.
(ii) If it is not optimal, use modified distribution method
to obtain optimal solution.
(iii) Find optimal transportation cost.
Ans: [Video Explanation]
No negative ∆, Hence Optimal Solution.
(iii) Optimal transportation Cost:
= (200 x 7) + (120 x 18) + (100 x 15) + (280 x 7) + (180 x
11) + (12 x 5)
= RS. 9,600
B) You are given the following information, for a project
with 8 activities:
The cost of completing the eight activities in normal time
is Rs.6,500.
Indirect cost is Rs 160 per day. The contract includes a
penalty of Rs. 100 per day for every day of delay more than 17 days.
|
Node |
Normal
Duration (days) |
Crash cost
per day (Rs) |
Maximum
possible crash time |
|
1-2 |
6 |
80 |
2 |
|
1-3 |
8 |
90 |
4 |
|
1-4 |
5 |
30 |
2 |
|
2-4 |
3 |
- |
0 |
|
2-5 |
5 |
40 |
2 |
|
3-6 |
12 |
200 |
4 |
|
4-6 |
8 |
50 |
3 |
|
5-6 |
6 |
- |
0 |
(i (ii) Crash the projèct duration to find the total
'cost of completing the project in 17 days (4)
Q.4. C) A company produces 2 products x and xX) using three
resources S1, S2 and S. Product x1 gives profit of Rs.30 per unit and product
x2 gives profÃt of Rs.40 per unit. The 3 resources Si. S and S; are available
to the extent of 200 units, 600 units and 500 units respectiveiy The following
objective function and constraints are given to you:
Max Z = 30x1 + 40x2
Subject to constraints;
x1 + 2x2 < 200
8x1 + 5x2 < 600
3x1 + 4x2 < 500
X1 >0; x2 >0
You are given the following simplex solution to the above
problem:
|
|
|
Cj à |
30 |
40 |
0 |
0 |
0 |
|
C |
X |
B |
X1 |
X2 |
S1 |
S2 |
S3 |
|
40 |
X2 |
100 |
½ |
1 |
½ |
0 |
0 |
|
0 |
S2 |
100 |
11/2 |
0 |
(-) 5/2 |
1 |
0 |
|
0 |
S3 |
100 |
1 |
0 |
(-) 2 |
0 |
1 |
|
|
|
Zj -Ã |
20 |
40 |
20 |
0 |
0 |
i) Check if the above solution is optimal or not.
ii) lf it is not optimal, find optimal solution.
B) With reference to the optimal simplex table in the above
problem obtained by you, answer the following:
iii) Find the optimal product mix and optimal profit
iv) Which resources are scarce and which are, unutilized?
v) Is it a case of alternative solution? Justify your answer
vi) What are the shadow prices of the resources? Justify
Ans: [Video Explanation]
Testing the given solution for optimality ;
Positive ∆
Hence, solution not optimal.
(ii) Revising the Solution for
Optimality:
No positive ∆
Therefore, Optimal Solution
(iii) Optimal Product Mix;
No. of units of X1 = 200/11 = 18.18
No. of units of X1 = 1000/11 = 90.90
Optimal Profit:
Max Z = [30 x 200/11] + [40 x
1000/11]
=
Rs. 46,000 /11
=
Rs. 4,181.81
(iv) Abundant Resource:
S3 = 900/11 ----- in optimal
solution
Therefore Unutilized Capacity of
S3 = 900 / 11
S3 is Abundant resource.
S1 and S2 are non-basis variable
in optimal solution.
S1 = 0
S2 = 0
Unutilized Capacity of S1 and S2 are 0
(i)
∆ of
non basis variable :
∆ of
S1 = - 170 / 11
∆ of
S2 = -20/11
No
zero ∆ value.
Hence,
this is a case of Unique Solution.
There
is no alternate optimal solution.
(ii)
Zj
value of slack variables:
|
Slack variable |
Zj value |
Shadow Price |
|
S1 |
170/11 |
Rs. 15.45 / unit |
|
S2 |
20/11 |
Rs. 1.89 / unit |
|
S3 |
0 |
Rs. 0 / unit |
Q.5 A) Explain the concepts : Total float, free float, Independent float and Interfering float. (8)
Ans:
(a) Total float:
lt is the amount of time by which the completion of an activity can be delayed beyond the earliest possible finishing time without affecting total project completion time.
Total Float = (LST - EST) or (LFT - EFT)
.:: Total float = LST - EST -[;-t]-E Total Float = [Lj- E -t
(b) Free float: It is the amount of time by which the completion of an activity can be delayed beyond the earliest finishing time without affecting the earliest possible start of a subsequent activity.
Free float = Earliest starting time of subsequent activity - Earliest finishing time of activity ij = Ej- [E, + t;] . Free Float = Ej- E-tj
(c) Independent float: It is the amount of time by which the start of an activity can be delayed without affecting the earliest start of a subsequent activity, assuming that the preceding activity has finished at its latest finishing time. .
:. Independent float = Earliest starting time of subsequent activity - Latest finishing time of preceding activity - ij
:. Independent Float = [Ej-Li]-tij;
(d) Interfering float: It is that part of total float which reduces the float of a subsequent activity. It is the amount of time by which the earliest possible start of a subsequent activity will be delayed if activity ÷ finishes on latest finishing time.
:. Interfering float = Latest finishing time of activity ij - Earliest starting time of subsequent activity. :. Interfering Float= Lj-Ej
B) Discuss any 5 areas where techniques of operation Research can be applied. (7)
Ans:
OR techniques are applied to a variety of business problems. Some examples are:
(A) Production management:
(1) To calculate loss of time due to waiting time, queuing time etc.
(2) To decide optimum allocation of jobs and optimum sequence in which jobs should be sequenced.
(B) Personnel management:
(1) To study labour turnover. (2) To do human resource planning. (3) To decide number of personnel required to be kept on standby in case of demand for higher manpower.
(C) Inventory management: (1) To study economic lot size to be ordered.
(D) Marketing management: (1) To decide optimal product mix for maximum profit. (2) Media selection for advertising for maximum reach. (3) Sales forecasting.
(E) Transportation management: (1) To determine transportation schedule for minimum cost or minimum time.
(F) Project management (1) To identify critical and non-critical activities of a project (2) To determine minimum project completion time. (3) To determine optimal project cost.
(G) Financial management: project. (1) To decide investment portfolio to maximize return on investment. of such nature that it is impossible to find a feasible solution that can satis1y a the constraints.
OR
c) Answer any 3 of the following : (15)
i) Explain the terms: Redundant constraints and infeasible solution in linear programming.
Ans: A redundant constraint is the one which does not affect the optimal solution. Even if that constraint is not considered, we can still obtain the solution to the problem. Since the constraint is not required for obtaining the optimal solution, it is called redundant constraint cost.
Following sketch explains redundancy (redundant constraint):
There is no common feasible region for line AB and line CD
Hence, solution is infeasible.
A redundarnt constraint is the one which does not affect the optimal
solution. Even if that constraint is not considered, we can still obtain the
solution to the problem. Since the constraint is not required for obtaining the
optimal solution, it is called redundant constraint. cost.
Following sketch explains redundancy (redundant constraint):
The feasible region for the above problem is OABC. The 3rd constraint does
not affect the feasible region. Hence, the constraint [X1 S 16] is redundant
constraint.
ii) what do you means by alternative optimal solution in transportation? How do you find that alternative solution?
Ans:
Alternate optimal solution or multiple optimal solutions mean there are two
sets of solutions which provide the same optimal cost or optimal profit:
(a) In the optimal solution to a transportation problem, if there is an
empty cell with zero A value, it means there is alternate optimal solution.
(b) To find alternate optimal solution, we should construct a closed loop
from the empty cell with zero A. The new table we obtain after looping is the alternate
optimal solution.
iii) Explain the time coat trade off in project crashing.
Ans:
The execution of a project involves two types of costs, direct cost and
indirect cost.
(a) Direct costs: These include costs of materials, machinery, tools,
manhours etc. When we estimate the duration of various activities in the
project, it is assumed that normal amount of labour (manhours) & machines
will be required to complete these activities. The estimation of direct costs
is done based on the normal amount of resource required. This estimation gives
us minimum possible direct cost required to complete the project. When we want
to complete the project in a shorter period than critical path, we will need to
employ more resources. Hence, direct cost will increase.
(b) Indirect costs: These include rent, overheads, administrative costs etc.
Indirect costs vary with time. They are expressed on per day (or per week etc.)
basis. Hence, when we shorten the project completion time, total indirect cost
reduces.
Total P'roject Cost = Direct Cost + Indirect Cost
iV) Discuss the signification of theory of game. Briefly discuss the terms. Player and pay off.
Ans: Every one
business party. The situation (also called competitive situation) involves more
parties involved could be corporate. individuals, small firms or A number of
outcome alternatives are available to each party for decision making. The (i.e.
profit or loss) of the business decision does not depend on the alternative
selected by one party alone, but on the interaction between the decisions of
all the parties involved in the situation. Each party will try to select the
alternative which maximizes gains for it, at the cost of its competitors.
"Games Theory" deals with such business situations or problems where
multiple parties are involved in the conflict through interaction of their
alternatives or decisions.
Players: The various participants or decision markers in the game are called
"Players". A game must have minimum two players (or competitors). A
game having two competitors opposing each other is called 'two person game' and
a game having more than 2 players is called 'n-person game'.
Payoff: Payoff is the outcome of the interaction of selected
strategies of opponents in the game. Positive payoff is gain and negative
payoff is loss. In a 'two person game', when one party gains, the other party
loses.
v) Distinguished between PERT and CPM.
|
CPM |
PERT |
|
In CPM there is only one time given for each activity. |
In the PERT there are three time estimates. Optimistic time, Most likely
time and Pessimistic time. From these, we find Expected time for completion of an activity (te) |
|
After drawing the network we find critical path, which gives us project
completion time. |
We draw the network using Expected time (te) and then we find critical
path, which gives us expected project completion time |
|
There is no probability information
in CPM |
Probability is associated with Project completion time (Te). The
probability of completing the project in that duration is 50% (0.5). In other words, the probability that the project will be completed in
expected project completion time is 50%. |
|
There is no standard deviation and variance for activities. We assume that
all activities will be completed in specified time. |
We find standard deviation and variance of critical activities and from
that we find variance and standard deviation of the critical path. |
|
As there is no probability, there is no relation with normal distribution
table |
We can find "z" value and from "z" value the
probability of completing the project in specified time duration. |
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