ASSIGNMENT 3
(Operation Research)
Maximum Marks: 50
Question 1
Consider the following problem
Maximize
z  4 x1  5 x2  3x3
Subject to
x1  x2  2 x3  20
15 x1  6 x2  5 x3  50
x1  3x2  5 x3  30
and
x1  0, x2  0, x3  0
Work through the simplex method step by step to demonstrate that this problem does not posses any
feasible solutions.
Solution:
Introducing surplus, artificial and slack variables we get the following equations.
x1  x2  2 x3  Sx 4  Rx5  20
15 x1  6 x2  5 x3  Sx6
 50
x1  3x2  5 x3  Sx7
 30
Phase 1:
So in phase one we have to minimize.
min z  Rx5
  x1  x2  2 x3  Sx 4  20
Iteration I:
Iteration II:
Iteration III:
Since in phase I min z  0 , so the problem has not feasible solution.
Note: Here this problem is solved using two phase method. This problem can also be solved using any other
type of simplex method, like big M method etc.
Question 2
The Build-Em-Fast Company has agreed to supply its best customer with three widgets during
each of the next 3 weeks; even through producing them will require some overtime work. The relevant
production data are as follows:
Maximum
Maximum
Production Cost
Production,
Production,
Per Unit,
Week
Regular Time
Overtime
Regular Time
2
2
$300
1
3
2
$500
2
1
2
$400
3
The cost per unit produced with overtime for each week is $100 more than for regular time. The
cost of storage is $50 per unit for each week it is stored. There is already an inventory of two widgets in hand
currently, but the company does not want to retain any widgets in inventory after the 3 weeks.
Management wants to know how many units should be produced in each week to minimize the
total cost of meeting the delivery schedule.
(a) Formulate this problem as a transportation problem by constructing the appropriate
parameter table.
(b) Obtain an optimal solution.
Solution:
Formulation and solution:
Demand
Week 1
Week 2
Week 3
Dummy Supply
R.D
Week 1
Regular
Week1
Overtime
Week2
Regular
Week2
Overtime
Week3
Regular
Week3
overtime
Demand
300
350
400
0
400
450
500
0
M
500
550
0
M
600
650
0
M
M
400
0
M
3
M
3
400
3
0
7
3
1
300
3
400
3
500
3
4
0
600
3
400
3
500
16
Week 1
Regular
Week1
Overtime
Week2
Regular
Week2
Overtime
Week3
Regular
Week3
overtime
Demand
100
100
0
0
Week 1
Week 2
Week 3
Dummy
300
350
400
0
400
450
500
0
M
500
550
0
M
M
400
0
M
3
M
3
400
3
0
4
3
R.D
Supply
1
300
3
400
3
0
500
3
400
3
500
16
1
Week 1
Regular
Week1
Overtime
Week2
Regular
Week2
Overtime
Week3
Regular
Week3
overtime
Demand
100
100
0
0
Week 1
Week 2
Week 3
Dummy
300
350
400
R.D
Supply
1
300
3
400
3
400
0
400
450
500
0
M
M
400
0
M
3
M
3
400
3
1
3
0
1
2
500
16
0
100
100
0
0
Week 1
Week 2
Week 3
Dummy
Week 1
1
Regular
300
Week1
Overtime
400
Week2
Regular
Week2
Overtime
1
350
0
50
400
3
450
R.D
Supply
500
50
Week3
Regular
Week3
overtime
Demand
3
M
M
400
2
M
3
M
3
400
3
2
100
100
0
Week 1
Week 2
Week 3
Week 1
Regular
Week1
2
Overtime
400
Week2
Regular
Week2
Overtime
Week3
Regular
M
Week3
overtime
M
Demand
2
16
Dummy
3
450
500
M
400
M
3
400
3
R.D
Supply
1
50
3
2
16
0
0
Week 1
Week 1
Regular
Week1
Overtime
Week2
Regular
Week2
Overtime
Week3
Regular
Week3
overtime
Demand
Week 2
Week 3
Dummy
1
1
450
500
M
400
R.D
Supply
0
50
3
2
M
3
400
3
16
2
0
Week 1
Week 1
Regular
Week 2
Week 3
Dummy
Supply
R.D
Week1
Overtime
Week2
Regular
Week2
Overtime
Week3
Regular
Week3
overtime
Demand
3
M
3
0
400
2
M
2
400
3
16
0
Week 1
Week 2
Week 1
Regular
Week1
Overtime
Week2
Regular
Week2
Overtime
Week3
Regular
Week3
overtime
Demand
Week 3
Dummy
Supply
2
M
2
16
Initial total cost= 3 0  3 0 1 0 1 300  2  400 1 450  3 400
=2750
Moving toward optimal solution;
Week 1
Week 1
1
Regular
300
Week1
2
Overtime
400
Week2
Regular
M
Week2
Overtime
M
Week3
Regular
M
Week3
overtime
M
Week 2
Demand
Week 3
Dummy
Supply
1
350
400
0
450
500
0
500
550
600
650
0
M
400
0
M
400
1
3
3
3
0
3
3
3
3
1
3
0
R.D
Demand
3
Basic variable
x11
x21
x22
x34
x44
x53
x64
also
Non basic var ible
3
3
7
(u,v)equation
u1+v1=300
u2+v1=400
u2+v2=450
u3+v4=0
u4+v4=0
u5+v3=0
u6+v4=0
16
solution
u1=0, v1=300
u2=100
v2=350
u3=0, v4=0
u4=0
v3=0, u5=400
u6=0
xij
ui  v j  cij
x12
u1  v2  c12  0
x13
u1  v3  c13  400
x14
u1  v4  c14  0
x23
u2  v3  c23  400
x24
u2  v4  c24  100  dummy
x32
u3  v2  c32  150
x33
u3  v3  c33  550
x42
u4  v2  c42  250
x43
u4  v3  c43  650
x54
u5  v4  c54  400  dummy
x63
u6  v3  c63  500
Hence the optimal solution is
 1 300  2  400  1 450  3  400
 2750
Question 3
Consider the transportation problem having the following parameter table.
1
2
Destination
3
4
5
6
Supply
5
6
7
4
3
13
10
22
29
18
0
14
13
16
21
M
0
3
0
M
11
6
0
Source
18
9
19
23
11
0
30
24
34
36
28
0
3
5
4
5
6
2
Demands
Use each of the following criteria an initial BF solutions Compare the values of the objective
function for these solutions.
(a) Northwest Corner Rule
(b) Vogel’s approximation method.
1
2
3
4
5
Solution:
(a)
Northwest Corner Rule
1
1
3
2
Destination
3
4
5
6
22
29
18
0
5
2
0
16
21
M
0
6
3
0
6
0
7
2
0
0
4
3
0
0
2
0
3
2
0
2
13
10
3
2
14
3
13
X
3
3
0
5
M
2
11
1
4
18
9
3
19
23
11
1
5
Demand
Supply
30
3
0
24
5
3
0
34
4
1
0
2
36
5
0
28
6
4
1
0
25
Hence
Total Cost
 3 13  2 10  3 13  3 16  5 11  2  6  119  311  1 28  2  0
 293
(b)
Vogal Approximation Method:
1
2
3
4
5
6
1
13
10
22
29
18
0
14
13
16
21
M
0
3
0
M
11
6
0
2
3
4
18
0
19
23
11
Demand
24
5
34
4
36
5
28
6
0
2
C.D
10
9
3
10
5
0
0
1
2
3
4
5
6
13
10
22
29
18
5
10
6
13
7
3
4
9
3
30
3
1
R.D
0
2
5
Supply
1
24
25
Supply
R.D
5
3
2
14
13
16
21
M
3
0
M
11
6
18
0
19
23
11
Demand
30
3
24
5
34
4
36
5
28
6
C.D
10
1
9
3
10
5
0
1
2
3
4
5
6
3
4
4
6
3
7
3
4
5
0
4
1
1
13
10
22
29
18
14
13
16
21
M
3
0
M
11
2
6
3
11
25
Supply
R.D
5
3
6
3
7
1
3
6
4
5
4
1
Demand
30
3
24
1
34
4
C.D
10
10
6
36
5
10
1
13
10
22
13
16
21
0
M
11
1
3
3
25
0
12
5
3
6
1
29
2
14
28
6
1
0
3
4
5
Demand
30
3
24
1
34
4
36
5
C.D
10
0
10
6
10
1
6
1
3
25
5
13
22
29
2
9
2
14
16
21
Demand
30
3
34
4
36
5
C.D
0
1
6
8
6
2
1
4
3
4
5
1
25
7
2
22
29
5
2
6
16
21
Demand
34
4
36
5
C.D
6
0
8
1
5
3
4
5
2
1
1
25
2
22
2
1
1
0
16
3
4
5
1
34
4
Demand
25
3
1
2
2
22
2
3
0
4
5
1
34
3
Demand
25
1
1
2
3
4
1
5
1
34
1
Demand
0
25
0
Hence,
Total cost= 2  0  4  0  6  6 1 0  313  5  21 116  2  22 1 34
=274
Question 4
The coach of an age group swim team needs to assign swimmers to a 200-yard medley relay
team to send to the junior Olympics. Since most of his best swimmers are very fast in more then one
stroke, it is not clear which swimmer should be assigned to each of the four strokes. The five fastest
swimmers and the best times (in seconds) they have achieved in each of the strokes (for 50 yards) are:
Stock
Carl
Chris
David
Tony
Ken
Backstroke
Breaststroke
Butterfly
Freestyle
37.7
43.4
33.3
29.2
32.9
33.1
28.5
26.4
33.8
42.2
38.9
29.6
37.8
34.7
30.4
28.5
35.4
41.8
33.6
31.1
The coach wishes to determine how to assign four swimmers to the four different strokes to
minimize the sum of the corresponding best times.
(a) Formulate this problem as an assignment problem.
(b) Obtain as optimal solution.
Solution:
Formulation and Solution:
JOB
Stock
Backstroke
Breaststroke
Butterfly
Freestyle
Dummy
Carl
37.7
43.4
33.3
29.2
0
Chris
32.9
33.1
28.5
26.4
0
MEN
David
33.8
42.2
38.9
29.6
0
Stock
Backstroke
Breaststroke
Butterfly
Freestyle
Dummy
Carl
4.8
10.3
4.8
2.8
0
Chris
0
0
0
0
0
David
0.9
9.1
10.4
3.2
0
Tony
4.9
1.6
1.9
2.1
0
Ken
2.5
8.7
5.1
4.7
0
Stock
Carl
Chris
David
Tony
Ken
3.9
9.4
3.9
1.9
0
0
0
0
0
0.9
0
8.2
9.5
2.3
0
4
0.7
1.0
1.2
Carl
Chris
David
Tony
3.9
8.7
3.9
1.9
0
0.7
0
0
0
1.6
0
7.5
8.8
1.6
0
4
0
0.3
0.5
0
Carl
Chris
David
Tony
3.9
8.7
2.9
0.9
0
1.0
0.3
0
0
1.9
0
7.5
8.5
1.3
0
4
0
0
0.2
0
Carl
Chris
David
Tony
0
7.5
8.5
1.3
0.9
4
0
0
0.2
0.9
Backstroke
Breaststroke
Butterfly
Freestyle
Dummy
Stock
Backstroke
Breaststroke
Butterfly
Freestyle
Dummy
Stock
Backstroke
Breaststroke
Butterfly
Freestyle
Dummy
Stock
Backstroke
Breaststroke
Butterfly
Freestyle
Dummy
Hence Assign:
3.0
7.8
2.0
0
0
David 
1.0
0.3
0
0
2.8
Backstroke
Tony
37.8
34.7
30.4
28.5
0
Ken
35.4
41.8
33.6
31.1
0
0
1.6
7.8
4.2
3.8
0
Ken
1.6
7.1
3.5
3.1
0
Ken
1.6
7.1
3.2
2.8
0
Ken
0.7
6.2
2.3
1.9
0
Tony 
Breaststroke
Chris 
Butterfly
Carl
 Freestyle
Sum of time = 33.8+34.7+28.5+29.2
= 126.2 secs
Question 5
Consider the following network, where each number along a link represents the actual distance
between the pair of nodes connected by that link. The objective is to find the shortest path from origin to
destination.
f 2*  A  11
A
5
f3*  D   6
9
D
7
Origin
6
O
6
B
T Destination
8
E
7
6
7
f3*  E   7
C
f2*  C   13
(a) What are the stages and states for the dynamic programming formulation of this problem?
(b) Use dynamic programming to solve this problem by manually constructing the usual tables
for n = 3, n = 2, and n = 1.
Solution:
(a)
In this problem we have three stages.
Stage 1:
A
O
B
C
where O, A, B and C are states.
Stage 2:
A
D
B
E
C
where A, B, C, D and E are states.
Stage 3:
D
T
E
Where D, E and T are states.
(b)
For n  3
D
E
T
f*
X 3*
6
7
6
7
T
T
D
E
f*
X 2*
11
13
-
15
13
11
13
13
D
D
E
For n  2
A
B
C
For n  1
O
A
B
C
f*
X 1*
20
19
20
19
B
Conclusion:
O
B
D
T