1. MBA 651 Quantitative Methods for Decision Making
Operations Research
MBA 651
Quantitative Methods for Decision Making

2. Operations Research (OR)
OR techniques
Linear programming
Integer linear programming
Dynamic programming
Nonlinear programming
Network programming

3. Phases of an OR study Identifying the management problem
Development of mathematical model
Solving the model
Validation of the model
Implementation of the model

4. How much to produce?
JOBCO produces two products on two machines. A unit of product 1 requires 2 hours on machine1 and 1 hour on machine 2. For product 2, a unit requires 1 hour on machine 1 and 3 hours on machine 2. The revenues per unit of products 1 and 2 are $3 and $2, respectively. The total daily processing time available for each machine is 8 hours. JOBCO wants the optimum mix of Products 1 and 2 that maximizes their profit

5. Mathematical Modeling
Decision variables
Quantity of product 1/day: x1
Quantity of product 2/day: x2
Objective
Revenue/unit of product 1: $ 3
Revenue/unit of product 2: $ 2
Maximize the revenue: z = 3x1 + 2x2
Constraints
Hours required per unit of
Product 1
Product 2
Available hours of machine/day
Machine 1 2 1 8
Machine 2 3
2x1 + x2 ≤ 8
x1 + 3x2 ≤ 8
x1, x2 ≥ 0

6. Linear Programming (LP)
JOBCO problem in LP: maximize z = 3x1 + 2x2
Properties of LP
Proportionality
Additivity
Divisibility
Certainty
Subject to 2x1 + x2 ≤ 8
x1 + 3x2 ≤ 8
x1, x2 ≥ 0

7. Violations of the Properties of LP
Proportionality:
Economies of scale
Marketing effort
Additivity
Complementing products
Certainty
Stochastic business environment (price and cost may vary in future)
Divisibility
Integer decision variables (people to be employed)

8. Can You Help Asian Paints….?
Asian paints produces both interior and exterior paints from two raw materials. The following table shows the basic data. A market survey indicates that, the daily demand for interior paint cannot exceed that of the exterior paint by more than 1 ton. Also, the maximum demand of interior paint is 2 tons. Asian paints want the optimal product mix.
Tons of raw materials/ton of
Maximum daily availability (tons)
Exterior paint
Interior paint
Raw material 1 6 4 24
Raw material 2 1 2
Profit per ton (Rs ) 5

9. LP Formulation: Asian Paints
Decision variables:
x1 = tons of exterior paint produced daily
x2 = tons of interior paint produced daily
Objective: Maximize z = 5x1 + 4x2 (Rs )
Constraints: 6x1 + 4x2 ≤ 24
x1 + 2x2 ≤ 6
-x1 + x2 ≤ 1
x2 ≤ 2
x1 , x2 ≥ 0

10. Graphical Method for LP model
Plot the constraints
Identify the region satisfying the constraints (feasible solution)
Plot the objective function
Identify the direction of increase
Identify the maximum value of objective function under the given constraints
(optimal solution)

11. JOBCO Problem Maximize z = 3x1 +2x2 Subject to 2x1 + x2 ≤ 8

12. Optimal Solution to the JOBCO Problem
(x1 = 3.2, x2 = 1.6), z= 12.8

13. JOBCO: An Increase in the Machine Available Hours
Rate of revenue change due to increase in machine capacity by 1 hour
= ( )/1=$1.4, is the shadow price or dual price
Similar analysis will yield the shadow price of M2 = $0.2

14. Feasibility range of Machines
Range of the limit of Machine 1 capacity for which the shadow (dual) price remains the same:
Value of M1 corresponding to (0, 8) = 16
Value of M1 corresponding to (2.67, 0)= 2.67
Hence, range= 2.67 ≤ M 1Capacity ≤ 16
For M2, 4 ≤ M 2Capacity ≤ 24

15. Optimality Range: Revenue Component
1/3 ≤ c1/c2 ≤ 2/ (1)
Range of c1: fix c2 and evaluate (1)
0.667 ≤ c1 ≤ 4
Range of c2: fix c1 and evaluate (1)
1.5 ≤ c2 ≤ 9

16. Some Insights
If JOBCO can increase the capacity of both machines, which machine should be preferred first?
Is it advised to increase the capacities of Machines 1 and 2 at the cost of $1/hr?
If the capacity of the machine is increased from 8 hrs. to 13 hrs., how will it impact the revenue?
Suppose that the unit revenues for product 1 and 2 are changed to $3.5 and $2.5, will the current optimum remain same?

17. Minimization : JOBCO Problem
Objective
Cost/unit of product 1: $ 2
Cost/unit of product 2: $ 1
Minimize w = 2x1 +x2
Subject to
2x1 + x2 ≤ 8
x1 + 3x2 ≤ 8
2x1 + x2 ≥ 5
x1 + 3x2 ≥ 3
x1, x2 ≥ 0
Hours required per unit of
Product 1
Product 2
Minimum hours of machine/day
Available hours of machine/day
Machine 1 2 1 5 8
Machine 2 3

18. Simplex Method: JOBCO Problem
Maximize z = 3x1 + 2x2
Subject to 2x1 + x2 ≤ 8
x1 + 3x2 ≤ 8
x1, x2 ≥ 0
Objective : Maximize z= 3x1 + 2x2 + 0s1+ 0s2
Constraint equations
2x1 + x2 + s1=8
x1 + 3x2 + s2=8
s1 and s2 are the slack variables ≥ 0
x1, x2, s1, s2 ≥ 0

19. Simplex Table Objective function z row  z-3x1-2x2-0s1-0s2=0
Basic z x1 x2 s1 s2
Solution
Minimum ratio 1 -3 -2 2 8
8/2 3
8/1
Leaving
(feasibility condition)
Entering (optimality condition)
Maximization: entering variable = with most negative coefficient
Leaving: minimum non negative ratio
3. For pivot row (s1 row in this table)
new pivot row= current pivot row/pivot element ( )
4. For other rows
new row= current row- (row’s pivot column coefficient ×
new pivot row)

20. Second iteration
Basic z x1 x2 s1 s2
Solution
Minimum ratio 1
-1/2
3/2 12
1/2 4 8
5/2
8/5
Third iteration
Basic z x1 x2 s1 s2
Solution 1
14/10
1/5
128/10
3/5
-1/5
16/5
2/5
8/5

21. Special Cases 1. Alternate optima---Many solutions
Maximize z = 4x1 + 2x2
2x1 + x2 ≤ 8
x1 + 3x2 ≤ 8
x1, x2 ≥ 0
2. Unbounded solution---no bound on solution
Subject to 2x1 + x2 ≥ 8
x1 + 3x2 ≥ 8
3. Infeasible solution---no solution
Subject to 2x1 + x2 ≤ 8
x ≥ 10
Plot and verify!!!

22. Minimization Problem: JOBCO
Minimize w = 2x1 + x2 + 0s1 +0s2+0s3 +0s4
Subject to
2x1 + x2 + s1 = 8
x1 + 3x2 + s2 = 8
2x1 + x2 – s3 = 5
x1 + 3x2 – s4 = 3
x1, x2 , s1, s2, s3, s4 ≥ 0
s3 and s4 are the surplus variables

23. Minimization Problem: JOBCO: Simplex Table---Big M Method
Minimize w = 2x1 + x2 + 0s1 +0s2+0s3 +0s4+MA1+MA2
Subject to
2x1 + x2 + s = 8
x1 + 3x2 + s = 8
2x1 + x2 – s3 +A1= 5
x1 + 3x2 – s4 +A2 = 3
x1, x2 , s1, s2, s3, s4 , A1 , A2 ≥ 0
s1, s = slack variables
s3, s4 = surplus variables
A1 , A2 = Artificial variables

24. Primal and Dual Problems
Primal problem: JOBCO
Maximize z = 3x1 + 2x2
Subject to 2x1 + x2 ≤ 8
x1 + 3x2 ≤ 8
x1, x2 ≥ 0
Basic rules for constructing the dual problem
Dual problem
Objective
Constraint
Variable sign
Primal : Maximization
Minimization ≥
unrestricted
Primal: Minimization
Maximization ≤

25. Primal and Dual Problems
Primal problem Maximize z = 3x1 + 2x2 + 0s1+ 0s2 2x1 + x2 + s1+ 0s2 = 8 y1 x1 + 3x2 + 0s1+ s2 = 8 y2 s1 and s2 are the slack variables ≥ 0 x1, x2, s1, s2 ≥ 0 Dual Problem Minimize w = 8y1+8y2 2y1 + y2 ≥ 3 y1 + 3y2 ≥ 2 y1 + 0y2 ≥ 0 and y1 unrestricted  y1 ≥ 0 0y1 + y2 ≥ 0 and y2 unrestricted  y2 ≥ 0 Solving the dual problem will give y1 = 14/10, and y2 = 1/5, which are the dual price or the worth of the resources and objective w = 128/10. Hence, at optimality, z = w.
Dual variables

26. Rules for Constructing the Dual Problem
Primal: Maximization
Dual: Minimization
Constraints
Variables ≤
≥ 0 ≥
≤ 0 =
unrestricted