This presentation includes LPP, Queuing theory and decision making theory and which is all describe detailed with relevant examples

2. LINEAR PROGRAMMING PROBLEM

3.  LPP – simultaneous equation / in equality condition
linear function / object function.
i) Optimize linear function.
ii) Subjective to some condition or restriction.
iii) Non – negativity restrictions
max z =x1+x2

4.  A general LPP includes a set of simultaneous linear
equations which represent the conditions of the
problem and a linear constrains which express the
objective linear problem.
 The linear functions which is the optimized is called
the objective function and the condition of the problem
express the simultaneous linear equations (or) the
qualities are referred as constrains.

5. A general linear programming problem can be started as
follows:
Max z = 5x1 + 10x2
St = 1x1 +2x2 < 100
2x1 +3x2 < 150
x1 + x3 > 0

6.  LPP includes:
i) Optimize the linear
ii) Subject to some constrains / restrictions
iii) Non – regativity restrictions / 0,1,2,3……
a) The LPP is applicable when a concern producers 2
(or) more qualities and each quantity shall on 2 (or)
more process.
b) A large no. of decision problem faced by the
business managers involved allocation of resources
to varies activities with the objectives of increasing
profit or decreasing (or) both.

7. c) The decision problem becomes complicated when no.
of resources are required to be allocated and there are
several activities to perform.
d) The decision problems can be formulated solved as a
mathematical programming problem.
example:
Rule of thumb even a experienced manager, in all
likely hood may not provide the right answer in such
cases.

8. e) Mathematical programming involved optimization of
a certain functions called the objective function, subject
too certain constrains.
Example:
A manager may faced with the problem of
decisioning appropriate product mix of the 4 product
with the profitability of product along with their
requirements of raw materials, labour extra…. His
problem can be formulated as a mathematical
programming problem taking the objective funtions as
the maximization of product obtainable from the mix
the various constrain.

9. The availability of raw material, labour, supply market
and so on….
The method of mathematical programming can be
divided into 3 groups:
I) Linear
II) Integer
III) Non – Linear Programming

10. Problem 1:
Max z = 6x1 + 11x2
St = 2x1 + x2 < 104
x1 + 2x2 < 76
x1 , x2 > 0
Mathematical Conversion:
Max z = 6x1 + 11x2 + x3 + x4
St = 2x1 + x2 + x3 = 104
x1 + 2x2 + + x4 = 76
x1 , x2 = 0

11. INITIAL TABLE:
B CB XB X1 X2 X4 X5 XB/X2
X3 0 104 2 1 1 0
104/1
=104
X4 0 76 1 2 0 1
76/2=
38
(min)
Z 6 11 0 0
CB 0 0 0 0
Z-CB 6 11
(max)
0 0
2

12. WORKINGS:
76 1 2 0 1 / 2
38 1/2 1 0 1/2 - 2 (2ND ROW)
38 ½ 1 0 ½ * 1
104 2 1 1 0
38 ½ 1 0 ½
66 3/2 0 1 -1/2 ( 1ST ROW)
I SIMPLEX TABLE:
B CB XB X1 X2 X3 X4 XB/X1
X3 0 66 3/2 0 1 -1/2 66*2/3=4
4(MIN)
X2 11 38 ½ 1 0 ½ 38*2/1=7
6
Z 6 11 0 0
CB 11/2 11 0 -11/2
CB-Z
½(MAX)
0 0 -11/2
3/
2

13. WORKINGS:
66 3/2 0 1 ½ / 3/2
44 1 0 2/3 -1/3 * ½ (I ST ROW)
22 ½ 0 1/3 -1/3
38 ½ 1 0 ½ (-)
22 ½ 0 1/3 -1/3
16 0 1 -1/3 2/3 (IIND ROW)
II SIMPLEX TABLE:
B CB XB X1 X2 X3 X4
X1 6 44 1 0 2/3 -1/3
X2 11 16 0 1 -1/3 2/3
Z 6 11 0 0
CB 6 11 -1/3 28/3
Z-CB 0 0 -1/3 -28/3

14. THE OPTIMUM SOLUTION IS
X1 = 44
X2 = 16
MAXZ = 6X1 + 11X2
= (6*44) + ( 11*16)
= 264 + 176
MAXZ = 440
Problem 2:
Solve the following LPP problem by applying simplex method.
MAX z = 5x1 + 3x2
St = 5x1 + 5x2 < 15
= 5x1 + 2x2 < 10
x1 , x2 > 0

15. SOLUTION:
MAX z = 5x1 + 3x2 + x3 + x4
St = 3x1 + 5x2 + x3 =15
= 5x1 + 2x2 + x4 = 10
x1 , x2 = 0
INITIAL TABLE:
B CB XB X1 X2 X3 X4 XB/X1
X3 0 15 3 5 1 0 15/3=5
X4 0 10 5 2 0 1 15/5=3
(MIN)
Z 5 3 0 0
CB 0 0 0 0
CB-Z 5(MAX) 3 0 0
5

16. WORKINGS:
10 5 2 0 1 / 5
2 1 2/5 0 1/5 (2ND ROW) * 3
15 3 5 1 0 (-)
6 3 6/5 0 -3/5
9 0 19/5 1 -3/5
I- SIMPLEX TABLE:
B CB XB X1 X2 X3 X4 XB/X3
X3 0 9 0 19/5 1 -3/5 9*5/19=
2.3(MIN)
X1 5 2 1 2/5 0 1/5 2*5/2=5
Z 5 3 0 0
CB 5 2 0 1
Z-CB 0 1(MAX) 0 0
19/5

17. WORKINGS:
9 0 10/5 1 -3/5 / 19/5
45/19 0 1 5/19 -3/19 (1ST ROW) * 2/5
2/1 1 2/5 0 1/5 (-)
18/19 0 2/5 2/19 -6/95
20/19 1 0 -2/19 5/19
II SIMPLEX TABLE:
B CB XB X1 X2 X3 X4
X2 3 45/19 0 0 5/19 -3/19
X1 5 20/19 1 1 -2/19 5/19
Z 5 3 0 0
CB 5 3 -5/19 -16/19
Z-CB 0 0 -5/19 -16/19

18. THE OPTIMUM SOLUTION IS:
X1 = 45/19
X2 = 20/19
MAX Z = 5X1 + 3X2
(5*45/19) + (3*20/19)
225/19 + 60/19
285/19
MAXZ = 15
Problems 3:
SOLVE THE PROBLEM BY APPLING THE SIMPLEX METHOD:
MAXZ = 30X1 + 23X2 + 29X3
St = 6X1 + 5X2 + 3X3 < 26
4X1 + 2X2 + 5X3 < 7
X1 , X2 , X3 > 0

19. SOLUTION:
MAXZ = 30X1 + 23X2 + 29X3 + X4 + X5
St = 6X1 + 5X2 + 3X3 + X4
4X1 + 2X2 + 5X3 + + X5
X1 , X2 , X3 = 0
INITIAL TABLE:
B CB XB X1 X2 X3 X4 XB/X5
X4 0 26 6 5 3 1 26/6=
4.3
X5 0 7 2 5 0 7/4=
1.75(MIN)
Z 30 23 29 0
CB 0 0 0 0
Z-CB 30(MAX) 23 29 0
4

20. WORKINGS:
7 4 2 5 0 1 / 4
7/4 1 2/4 5/4 0 1/4 (2ND ROW) * 6
26 6 5 3 1 0 (-)
21/2 6 6/2 15/2 0 -3/2
31/2 0 4/2 -9/2 1 -3/2 (1ST ROW)
I SIMPLEX TABLE:
B CB XB X1 X2 X3 X4 X5 XB/X2
X4 0 31/2 0 2 -9/2 1 -3/2 3/2*2
=62
X1 30 7/4 1 2/4 5/4 0 1/4 7/4*2/4
=7/2(MIN)
Z 30 23 29 0 0
CB 30 15 75/2 0 -15/2
Z-CB 0 8(MAX) -17/2 0 -15/2
2/4

21. WORKINGS:
7/4 1 2/4 5/4 0 1/4 / 2/4
7/2 2 1 5/2 0 ½(2ND ROW) * 2
31/2 0 2 -9/2 1 -3/2 (-)
7 4 2 5 0 1
17/2 -4 0 19/2 1 -5/2
II SIMPLEX TABLE:
B CB XB X1 X2 X3 X4 X5
X4 0 17/2 -4 0 19/2 1 -5/2
X2 23 7/2 2 1 5/2 0 1/2
Z 30 23 29 0 0
CB 46 23 115/2 0 -23/2
Z-CB -16 0 -57/2 0 -23/2

22. THE OPTIMUM SOLUTION IS:
MAXZ = 30X1 + 23X2 + 29X3
(30*0) + (23*7/2) + (29*0)
0 + 161/2 + 0
MAXZ = 161/2
Problem 4:
Use the simplex to solve the following LPP:
MAXZ = 4X1 + 10X2
St = 2X1 + X2 < 50
= 2X1 + 5X2 < 100
= 2X1 + 3X2 < 90
X1 , X2 > 0

23. WORKINGS:
MAXZ = 4X1 + 10X2 + X3 + X4 + X5
St = 2X1 + X2 + X3
= 2X1 + 5X2 + + X4
2X1 + 3X2 + X5
X1 , X2 = 0
INITIAL TABLE:
B CB XB X1 X2 X3 X4 X5 XB/X2
X3 0 50 2 1 1 0 0 50
X4 0 100 2 5 0 1 0 100/5=
20(MIN)
X5 0 90 2 3 0 0 1 90/3=
30
Z 4 10 0 0 0
CJ 0 0 0 0 0
Z-CJ 4 10(MAX) 0 0 0
5

24. WORKINGS:
100 2 3 0 1 0 / 5
20 2/5 1 0 1/5 0(2nd ROW) * 1
50 2 1 1 0 0 (-)
20 2/5 1 0 1/5 0
30 8/5 0 1 3/5 0 (1ST ROW)
20 2/5 1 0 1/5 0 * 3
90 2 3 0 0 1 (-)
60 6/5 3 0 3/5 0
30 4/5 0 0 -3/5 1 (3RD ROW)
I SIMPLEX TABLE:
B CB XB X1 X2 X3 X4 X5
X3 0 30 8/5 0 1 1/5 0
X2 10 20 2/5 1 0 1/5 0
X5 0 30 4/5 0 0 -3/5 1
Z 4 10 0 0 0
Cj 4 10 0 -2 0
Z-Cj 0 0 0 -2 0

25. THE OPTIMUM SOLUTION IS:
MAXZ = 4X1 + 10X2
X1 = 30
X2 = 20
X3 = 30
= (0*30) + (10*20) + (0*30)
= 0 + 200 + 0
MAXZ = 200
Problems 5:
SOLVE THE FOLLOWING PROBLEM BY APPLYING SIMPLEX
METHOD:
MAXZ = 5X1 + 7X2
St = X1 + X2 < 4
= 3X1 + 8X2 < 24
= 10X1 + 7X2 < 35
X1 , X2 > 0

26. SOLUTION:
MAXZ = 5X1 + 7X2 + X3 + X4 + X5
St = X1 + X2 + X3 = 4
3X1 + 8X2 + X4 = 24
10X1 + 7X2 + X5 = 35
X1 , X2 = 0
INITIAL TABLE:
B CB XB X1 X2 X3 X4 X4 XB/X
1
X3 0 4 1 1 1 0 0 4
X4 0 24 3 0 1 0 24/8=3
X5 0 35 10 7 0 0 1 35/7=5
Z 5 7 0 0 0
Cj 0 0 0 0 0
Z-Cj 5 7 0 0 0

27. WORKINGS:
24 3 8 0 1 0 / 8
3 3/8 1 0 1/8 0 (2 ROW) * 1
4 1 1 1 0 0 (-)
3 3/8 1 0 1/8 0
1 5/8 0 1 -1/8 0 (1 ROW)
3 3/8 1 0 1/8 0
35 10 7 0 0 1 (-)
21 21/8 7 0 -7/8 0
14 59/8 0 0 -7/8 1 (3 ROW)
I SIMPLEX TABLE:
B CB XB X1 X2 X3 X4 X5 XB/X1
X3 0 1 5/8 0 1 -1/8 0 5/8(MIN)
X2 7 3 3/8 1 0 ½ 0 8
X5 0 14 59/8 0 0 -7/8 1 812/8
Z 5 7 0 0 0
Cj 21/8 7 0 7/8 0
Z-Cj 19/8(MAX) 0 0 -7/8 0
5/
8

28. WORKINGS:
1 5/8 0 1 -1/8 0 / 5/8
8/5 1 0 8/5 -1/5 0 (1 row) * 3/8
3 3/8 1 0 1 /8 0 (-)
3/5 3/8 0 3/5 -3/5 0
12/5 0 1 -3/5 29/40 0 (2 row)
8/5 1 0 8/5 7/5 0 * 59/8
14 59/8 0 0 -7/8 1 (-)
59/5 59/8 0 -59/5 59 0
11/5 0 0 -59/5 -479/5 1 (3 row)

29. Problem 6:
SOLVE THE SIMPLEX METHOD BY LPP PROBLEM:
MINZ = X1 - 3X2 + 2X3
St = 3X1 - X2 + 2X3 < 7
-2X1 + 4X2 < 12
-4X1 + 3X2 + 8X3 < 10
X1 , X2 < 0
METHAMETICAL CONVERTION:
MINZ = -X1 + 3X2 - 2X3 + X4 + X5 + X6
St = 3X1 + X2 + 2X3 + X4 = 7
-2X1 + 4X2 + + X5 = 12
-4X1 + 3X2 + 8X3 + X6 = 10
X1 , X2 = 0

30. INITIAL TABLE:
WORKINGS:
12 -2 4 0 0 1 0 / 4
3 -1/2 1 0 0 ¼ 0 (2 ROW) * -1
7 3 -1 2 1 0 0 (-)
-3 -1/2 1 0 0 -1/4 0
10 5/2 0 2 1 -1/4 0 ( 1 ROW)
B CB XB X1 X2 X3 X4 X5 X6 XB/X2
X3 0 7 3 -1 2 1 0 0 7/-1=-7
X4 0 12 -2 0 0 1 0 12/4=3
X5 0 10 -4 3 8 0 0 1 10/3=3.33
Z -1 3 -2 0 0 0
Cj 0 0 0 0 0 0
Z-Cj -1 3 -2 0 0 0

31. 3 -1/2 1 0 0 ¼ 0 * 3
10 -4 3 8 0 0 1 (-)
9 -1/2 13 0 0 -3/4 0
1 -5/2 0 8 0 -3/4 1 (3 ROW)
I SIMPLEX TABLE:
B CB XB X1 X2 X3 X4 X5 X6 XB/X1
X4 0 10 5/2 0 2 1 -1/4 0 10/(5/2)
=4(MIN)
X2 3 3 -1/2 1 0 0 ¼ 0 3/(1/2)
=6
X6 0 1 -5/2 0 8 0 -3/4 1 1/(-5/2)
=2/-5
Z -1 3 -2 0 0 0
Cj 3/2 3 0 0 -3/4 0
Z-Cj 5/2*(MA
X)
0 2 0 -3/4 0
5/
2

32. WORKINGS:
10 5/2 0 2 1 -1/4 0 / 5/2
4 1 0 4/5 2/5 -1/10 0(1 ROW) * -1/2
3 -1/2 1 0 0 ¼ 0 (-)
-2 -1/2 0 -2/5 1/5 -1/20 0
5 0 1 -2/5 1/5 6/20 0 (2 ROW)
4 1 0 4/5 2/5 1/10 0 * -5/2
1 -5/2 0 8 0 -3/4 1
-10 -5/2 0 -2 -1 -1/4 0
-11 0 0 10 1 -2/4 1

33. II SIMPLEX TABLE:
THE OPTIMUM SOLUTION IS:
X1 = -1 , X2 = 3 , X3 = 2
(-1*4) + (3*5) * (0*-1)
-4 + 15
= 11
B CB XB X1 X2 X3 X4 X5 X6
X1 -1 4 1 0 4/5 2/5 1/10 0
X2 3 5 0 1 2/5 1/5 6/20 0
X6 0 -11 0 0 10 1 2/4 1
Z -1 3 -2 0 0 0
Cj -1 3 2 1 1 0
Z-Cj -2 0 -4 -1 -1 0

50. Thank you