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

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

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

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

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

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

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)

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. 10000) 5

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. 10000) Constraints: 6x1 + 4x2 ≤ 24 x1 + 2x2 ≤ 6 -x1 + x2 ≤ 1 x2 ≤ 2 x1 , x2 ≥ 0

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)

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

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

JOBCO: An Increase in the Machine Available Hours Rate of revenue change due to increase in machine capacity by 1 hour = (14.2-12.8)/1=$1.4, is the shadow price or dual price Similar analysis will yield the shadow price of M2 = $0.2

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

Optimality Range: Revenue Component 1/3 ≤ c1/c2 ≤ 2/1 (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

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?

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

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

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)

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

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 x1 ≥ 10 Plot and verify!!!

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

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

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 ≤

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

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