Linear Programming Problems, Duality and Zero sum Games

1. Linear Programming Problem, Duality
and Game Theory
(February 21{26, 2015)
Dr. Purnima Pandit
Assistant Professor
Department of Applied Mathematics
Faculty of Technology and Engg.
The Maharaja Sayajirao University of Baroda

2. Different Kinds of
Optimization

3. Elements
 Variables
 Objective function
 Constraints
Obtain values of the variables
that optimizes the objective function
( satisfying the constraints )

4. • Linear programming (LP) problems are
optimization problems where the objective
function and the constraints of the problem are
all linear.
• Many practical problems in operations research
can be expressed as linear programming
problems.
• A lot of work is generated on the research of
specialized algorithms for the solutions of
specific LP problems.
• In mathematical optimization theory, the
simplex algorithm of George Dantzig is the
fundamental technique for numerical solution of
the LP problem.
Introduction to LPP

5. Constrained Optimization
 If objective function and constraints all
linear, this is “linear programming”
 Observation: minimum must lie at corner
of region formed by constraints
 Simplex method: move from vertex to
vertex, minimizing objective function

6. LP Model Formulation
 Decision variables
 mathematical symbols representing levels of activity of an
operation
 Objective function
 a linear relationship reflecting the objective of an operation
 most frequent objective of business firms is to maximize
profit
 most frequent objective of individual operational units (such
as a production or packaging department) is to minimize cost
 Constraint
 a linear relationship representing a restriction on decision
making

7. LP Model Formulation (cont.)
Max/min z = c1x1 + c2x2 + ... + cnxn
subject to:
a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
:
am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
xj = decision variables
bi = constraint levels
cj = objective function coefficients
aij = constraint coefficients

8. MCQ
1. Which of the following is not correct about LPP?
(a) All constraints must be linear relationship.
(b) Objective function must be linear
(c) All the constraints and decision variables must be of either ≤ or ≥ type.
(d) All decision variables must be non-negative.
2. A constraint in LPP restricts
(a) value of objective function
(b) value of decision variable.
(c) use of available resource
(d) uncertainity of optimum value.
3. Which of the following is correct?
(a) LP takes into consideration the effect of time and uncertainity
(b) An LPP can have only two decision variables.
(c) Decision variables in an LPP may be more or less than the number of constraints
(d) LP deals with problems involving only a single objective.

9. LP Model: Example
Labor Clay Revenue
PRODUCT (hr/unit) (lb/unit) ($/unit)
Bowl 1 4 40
Mug 2 3 50
There are 40 hours of labor and 120 pounds of clay
available each day
Decision variables
x1 = number of bowls to produce
x2 = number of mugs to produce
RESOURCE REQUIREMENTS

10. LP Formulation: Example
Maximize Z = $40 x1 + 50 x2
Subject to
x1 + 2x2 40 hr (labor constraint)
4x1 + 3x2 120 lb (clay constraint)
x1 , x2 0
Solution is x1 = 24 bowls x2 = 8 mugs
Revenue = $1,360

11. Graphical Solution Method
1. Plot model constraint on a set of coordinates
in a plane
2. Plot objective function to find the point on
boundary of this space that maximizes (or
minimizes) value of objective function
 2 dimensional problems
 Observation: Optimum must lie at corner
of region formed by constraints

12. Graphical Solution Method
1. Plot model constraint on a set of coordinates
in a plane
2. Identify the feasible solution space on the
graph where all constraints are satisfied
simultaneously
3. Plot objective function to find the point on
boundary of this space that maximizes (or
minimizes) value of objective function

13. Computing Optimal Values:
x1 +2x2 = 40
4x1 +3x2 = 120
4x1 +8x2 = 160
-4x1-3x2 = -120
5x2 = 40
x2 = 8
x1 +2(8) = 40
x1 = 24
4 x1 + 3 x2 120 lb
x1 + 2 x2 40 hr
40 –
30 –
20 –
10 –
0 – |
10
|
20
|
30
|
40
x1
x2
Z = $50(24) + $50(8) = $1,360
24
8

14. Extreme Corner Points
x1 = 224 bowls
x2 = 8 mugs
Z = $1,360 x1 = 30 bowls
x2 = 0 mugs
Z = $1,200
x1 = 0 bowls
x2 = 20 mugs
Z = $1,000
A
B
C|
20
|
30
|
40
|
10 x1
x2
40 –
30 –
20 –
10 –
0 –

15. ISO – Profit:
4 x1 + 3 x2 120 lb
x1 + 2 x2 40 hr
Area common to
both constraints
50 –
40 –
30 –
20 –
10 –
0 – |
10
|
60
|
50
|
20
|
30
|
40 x1
x2

16. 4x1 + 3x2 120 lb
x1 + 2x2 40 hr
40 –
30 –
20 –
10 –
0 –
B
|
10
|
20
|
30
|
40 x1
x2
C
A
Z = 70x1 + 20x2
Optimal point:
x1 = 30 bowls
x2 = 0 mugs
Z = $2,100
Objective Function

17. Minimization Problem
CHEMICAL CONTRIBUTION
Brand Nitrogen (lb/bag) Phosphate (lb/bag)
Gro-plus 2 4
Crop-fast 4 3
Minimize Z = $6x1 + $3x2
subject to
2x1 + 4x2  16 lb of nitrogen
4x1 + 3x2  24 lb of phosphate
x1, x2  0

18. 14 –
12 –
10 –
8 –
6 –
4 –
2 –
0 – |
2
|
4
|
6
|
8
|
10
|
12
|
14 x1
x2
A
B
C
Graphical Solution
x1 = 0 bags of Gro-plus
x2 = 8 bags of Crop-fast
Z = $24
Z = 6x1 + 3x2

19. 19
Two type of Constraints
Binding and Nonbinding constraints:
A constraint is binding if the left-hand and right-
hand side of the constraint are equal when the
optimal values of the decision variables are
substituted into the constraint.
A constraint is nonbinding if the left-hand side
and the right-hand side of the constraint are
unequal when the optimal values of the decision
variables are substituted into the constraint.

20. 20
Special Cases of Graphical Solution
• Some LPs have an infinite number of solutions
(alternative or multiple optimal solutions).
• Some LPs have no feasible solution (infeasible
LPs).
• Some LPs are unbounded: There are points in
the feasible region with arbitrarily large (in a
maximization problem) z-values.

21. 21
Alternative or Multiple Solutions
Any point (solution)
falling on line segment
AE will yield an optimal
solution with the same
objective value
X1
X2
10 20 30 40
1020304050
Feasible Region
F
50
60
z = 60
z = 100
z = 120
A
B
C
D
E

22. 22
No feasible solution
X1
X2
10 20 30 40
1020304050
No Feasible Region
50
60
x1 >= 0
x2 >=0
No feasible region exists
Some LPs have no
solution. Consider
the following
formulation:

23. 23
Unbounded LP
X11 2 3 4
1
2
3
4
X2
5
6
5 6
A
B
C
Feasible Region
z = 4
z = 6
D
The constraints are
satisfied by all points
bounded by the x2 axis
and on or above AB and
CD.
There are points in the
feasible region which
will produce arbitrarily
large z-values
(unbounded LP).

24. MCQ
1.A feasible solution to an LPP
(a) must satisfy all the problem's constraints simultaneously
(b) must be a corner point of the feasible region.
(c) need not satisfy all the constraints, only some of them.
(d) must optimize the value of the objective function.
2. An iso-profit line represents
(a) An infinite number of solutions all of which yield the same profit.
(b) an infinite numbe of optimum solutions.
(c) an infinite number of solutions all of which uses the same resources.
(d) a boundary of feasible region.
3. If an iso-profit line yielding the optimum solution coincided with a constraint
line, then
(a) the soultion is unbounded.
(b) the solution is infeasible.
(c) the constraint which coincides is redundant.
(d) none of above.

25. MCQ
4. Using graphic method, the optimum solution of the LPP of
Maximizing Z = 10x+15y
s.t.
2x+y  26, x+2y  28 y-x  5 and x  0 and y  0
is obtained as:
(a) x=8 and y =10
(b) x=6 and y =1
(c) x=6 and y =10
(d)x=8 and y =8

26. Standard form
 Standard form is a basic way of describing a LP
problem.
 It consists of 3 parts:
A linear function to be maximized
maximize c1x1 + c2x2 + … + cnxn
Problem constraints
subject to a11x1 + a12x2 + … + a1nxn < b1
a21x1 + a22x2 + … + a2nxn < b2
…
am1x1 + am2x2 + … + amnxn < bm
Non-negative variables
x1> 0, x2 > 0,... , xn > 0

27. Step 0 – Obtain Canonical Form
General Simplex LP model:
min (or max) z =  ci xi
s.t.
A x = b
x  0
In order to get and maintain this form, use
 slack, if x  b, then x + slack = b
 surplus, if x  b, then x - surplus = b
 artificial variables (sometimes need to be added to
ensure all variables  0 )
IMPORTANT: Simplex only deals with equalities

28. Simplex method
Solve the following problem using the simplex
method
Maximize Z = 3X1+ 5X2
Subject to
X1  4
2 X2  12
3X1 +2X2  18
X1 , X2  0

29. Simplex method
Standard form
Maximize Z,
Subject to
Z - 3X1- 5X2 = 0
X1 + S1 = 4
2 X2 + S2 = 12
3X1 +2X2 + S3 = 18
X1 , X2, S1, S2, S3  0
Sometimes it is called
the augmented form of
the problem because the
original form has been
augmented by some
supplementary variables
needed to apply the
simplex method

30. Definitions
A basic solution is an augmented corner point solution.
A basic solution has the following properties:
1. Each variable is designated as either a non-basic variable or a
basic variable.
2. The number of basic variables equals the number of functional
constraints. Therefore, the number of non-basic variables
equals the total number of variables minus the number of
functional constraints.
3. The non-basic variables are set equal to zero.
4. The values of the basic variables are obtained as simultaneous
solution of the system of equations (functional constraints in
augmented form). The set of basic variables are called “basis”
5. If the basic variables satisfy the non-negativity constraints, the
basic solution is a Basic Feasible (BF) solution.

31. Initial tableau
Basic
variable
X1 X2 S1 S2 S3 RHS
S1 1 0 1 0 0 4
S2 0 2 0 1 0 12
S3 3 2 0 0 1 18
Z -3 -5 0 0 0 0
Pivot column
Pivot row
Pivot
number
Entering
variable
Leaving
variable

32. Optimality test
By investigating the last row of the initial
tableau, we find that there are some
negative numbers. Therefore, the current
solution is not optimal

33. Basic
variable
X1 X2 S1 S2 S3 RHS
S1 1 0 1 0 0 4
X2 0 1 0 1/2 0 6
S3 3 0 0 -1 1 6
Z -3 0 0 5/2 0 30
The most negative
value; therefore, X1 is
the entering variable
The smallest ratio is 6/3
=2; therefore, S3 is the
leaving variable
This solution is not optimal, since there is a negative numbers in the last
row

34. Apply the same rules we will obtain this solution:
Basic
variable
X1 X2 S1 S2 S3 RHS
S1 0 0 1 1/3 -1/3 2
X2 0 1 0 1/2 0 6
X1 1 0 0 -1/3 1/3 2
Z 0 0 0 3/2 1 36
This solution is optimal; since there is no negative solution in the last row: basic
variables are X1 = 2, X2 = 6 and S1 = 2; the non-basic variables are S2 = S3 = 0
Z = 36

35. Special cases of linear
programming
Infeasible solution
Multiple solution (infinitely many solution)
Unbounded solution
Degenerated solution

36. Notes on the Simplex tableau
1. In any Simplex tableau, the intersection of any basic variable with itself is always one
and the rest of the column is zeroes.
2. In any simplex tableau, the objective function row (Z row) is always in terms of the
nonbasic variables. This means that under any basic variable (in any tableau) there is
a zero in the Z row. For the non basic there is no condition ( it can take any value in
this row).
3. If there is a zero under one or more nonbasic variables in the last tableau (optimal
solution tableau), then there is a multiple optimal solution.
4. When determining the leaving variable of any tableau, if there is no positive ratio (all the
entries in the pivot column are negative and zeroes), then the solution is unbounded.
5. If there is a tie (more than one variables have the same most negative or positive) in
determining the entering variable, choose any variable to be the entering one.
6. If there is a tie in determining the leaving variable, choose any one to be the leaving
variable. In this case a zero will appear in RHS column; therefore, a “cycle” will occur,
this means that the value of the objective function will be the same for several
iterations.
7. A Solution that has a basic variable with zero value is called a “degenerate solution”.
8. If there is no Artificial variables in the problem, there is no room for “infeasible solution”

37. Example Problem
Maximize Z = 5x1 + 2x2 + x3
subject to
x1 + 3x2 - x3 ≤ 6,
x2 + x3 ≤ 4,
3x1 + x2 ≤ 7,
x1, x2, x3 ≥ 0.

38. Duality

39. LP: Dual Formation
 Formulation of the Dual from the Prime:
 Standardize constraint set:
 Max problem - all to (≤) type /Min problem, all to (≥).
 Replace equality constraint with two inequality ones.
 Transforming standardized Prime to Dual following rules.
 No. of Variables in Dual = No. of Constraints in Prime
 No. of Constraints in Dual = No. of Variables in Prime
 Cj in Prime → bi in Dual / bi in Prime → Cj in Dual
 aij in Prime → aji in Dual

40. jallfor0X
iallforbXa..
Xc
j
i
j
jij
j
jj



ts
Max
iallfor0
jallforcas.t.
b
i
j
i
iji
i
ii



Y
Y
YMin
Primal
Dual
The Primal-Dual Relationship

41. Continued…..

45. S-
Game Theory

46. Rules, Strategies, Payoffs, and
Equilibrium
A game is a contest involving two or more
decision makers, each of whom wants to win
A two-person game involves two parties (X
and Y)
 A zero-sum game means that the sum of losses
for one player must equal the sum of gains for the
other. Thus, the overall sum is zero

47. Minimax Criterion
 The value of a game is the average or expected game outcome if the game is
played an infinite number of times
 A saddle point indicates that each player has a pure strategy i.e., the strategy is
followed no matter what the opponent does

48. Pure Strategy - Minimax Criterion
Player Y’s
Strategies
Minimum Row
Number
Y1 Y2
Player X’s
strategies
X1 10 6 6
X2 -12 2 -12
Maximum
Column Number
10 6

49. Mixed Strategy Game
 When there is no saddle point:
 The most common way to solve a mixed strategy is to use the expected gain or loss
approach
 A player plays each strategy a particular percentage of the time so that the expected
value of the game does not depend upon what the opponent does
Y1
P
Y2
1-P
Expected Gain
X1
Q
4 2 4P+2(1-P)
X2
1-Q
1 10 1P+10(1-p)
4Q+1(1-Q) 2Q+10(1-q)

50. Solving for P & Q
4P+2(1-P) = 1P+10(1-P)
or: P = 8/11 and 1-p = 3/11
Expected payoff:
1P+10(1-P)
=1(8/11)+10(3/11)
EPX= 3.46
4Q+1(1-Q)=2Q+10(1-q)
or: Q=9/11 and 1-Q = 2/11
Expected payoff:
EPY=3.46

51. Example
Using the solution procedure for a mixed
strategy game, solve the following game

52. Example
This game can be solved by setting up the
mixed strategy table and developing the
appropriate equations:

53. Example

54. Analytical Method
A 2x2 game without saddle point can be solved using
following formula.

55. Dominance
A strategy can be eliminated if all its game’s
outcomes are the same or worse than the
corresponding outcomes of another strategy
A strategy for a player is said to be dominated
if the player can always do as well or better
playing another strategy

56. Domination
S-56
Y1 Y2
X1 4 3
X2 2 20
X3 1 1
Initial Game
X3 is a dominated strategy as player X can
always do better with X1 or X2
Y1 Y2
X1 4 3
X2 2 20
Revised Game

57. Example
Solve using graphical method
57

58. Solution
58

59. Cont..
V = 66/13
SA = (4/13, 9 /13)
SB = (0, 10/13, 3 /13

60. Example
Solve by graphical method

61. Solution

62. Cont..
V = 3/9 = 1/3
SA = (0, 5 /9, 4/9, 0)
SB = (3/9, 6 /9)