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IME634: Management Decision Analysis

IME634: Management Decision Analysis. Raghu Nandan SENGUPTA IME Department Indian Institute of Technology Kanpur, INDIA. Operations Research (OR) OR techniques Linear programming Integer linear programming Dynamic programming Nonlinear programming Network programming.

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IME634: Management Decision Analysis

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  1. IME634: Management Decision Analysis Raghu Nandan SENGUPTA IME Department Indian Institute of Technology Kanpur, INDIA

  2. Operations Research (OR) • OR techniques • Linear programming • Integer linear programming • Dynamic programming • Nonlinear programming • Network programming RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 • 2x1 + x2 ≤ 8 • x1 + 3x2 ≤ 8 • x1, x2 ≥ 0 RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  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) RNSengupta,IME Dept.,IIT Kapur,INDIA

  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. RNSengupta,IME Dept.,IIT Kapur,INDIA

  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. 10000) • Constraints: 6x1 + 4x2 ≤ 24 x1 + 2x2 ≤ 6 -x1 + x2 ≤ 1 x2 ≤ 2 x1 , x2 ≥ 0 RNSengupta,IME Dept.,IIT Kapur,INDIA

  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) RNSengupta,IME Dept.,IIT Kapur,INDIA

  11. JOBCO Problem Maximize z = 3x1 +2x2 Subject to 2x1 + x2 ≤ 8 x1 + 3x2 ≤ 8 x1, x2 ≥ 0 RNSengupta,IME Dept.,IIT Kapur,INDIA

  12. Optimal Solution to the JOBCO Problem (x1 = 3.2, x2 = 1.6), z= 12.8 RNSengupta,IME Dept.,IIT Kapur,INDIA

  13. 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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  15. 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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  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? RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  19. Simplex Table Objective function z row  z-3x1-2x2-0s1-0s2=0 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) RNSengupta,IME Dept.,IIT Kapur,INDIA

  20. Second iteration Third iteration RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 x1 + 3x2 ≤ 8 x1 ≥ 10 Plot and verify!!! RNSengupta,IME Dept.,IIT Kapur,INDIA

  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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  23. 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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  24. Simplex Method A company which produces 2 different types of glass products has 3 factories, and the data for the same is as given Capacity per unit production rate Capacity available Product 1 2 Plant 1 1 0 4 2 0 2 12 3 3 2 18 Unit profit (USD) 3 5 RNSengupta,IME Dept.,IIT Kapur,INDIA

  25. Simplex Method The model formulation is as Max Z=3x1+5x2 s.t: x1 <= 4 2x2<=12 3x1+2x2<=18 x1, x2>=0 RNSengupta,IME Dept.,IIT Kapur,INDIA

  26. Simplex Method Introduce slack variables (red coloured) such that equality can be brought into the picture. Thus we have Max Z=3x1+5x2+0x3+0x4+0x5 s.t: x1+x3= 4 2x2+x4=12 3x1+2x2+x5=18 x1, x2, x3, x4, x5,>=0 RNSengupta,IME Dept.,IIT Kapur,INDIA

  27. Initial tableau Basic Eq # Z x1 x2x3x4x5 RHS Variable Z 0 1 -3 -5 0 0 0 0 x3 1 0 1 0 1 0 0 4 x4 2 0 0 2 0 1 0 12 x5 3 0 3 2 0 0 1 18 • You main task is to replace the basic variables with the actual real variables or decision variables which for your case is x1 and x2. • As this is an maximization problem then make a decision which one will be the entering variables such that it has the largest non-negative coefficient • Name the column corresponding to this entering variable as pivot variable, hence we will choose x2. WHY? RNSengupta,IME Dept.,IIT Kapur,INDIA

  28. Changing tableau Basic Eq # Z x1 x2x3x4x5 RHS Variable Z 0 1 -3 -5 0 0 0 0 x3 1 0 1 0 1 0 0 4 x4 2 0 0 2 0 1 0 12 x5 3 0 3 2 0 0 1 18 • Now if a variable enters, one has to leave. We now have to decide which variable leaves. • Calculate the ratios of 4/0 (non defined, hence left out), 12/2=6, 18/2=9. • Choose the one for which the ratio is the smallest. WHY? • The column marked is the pivot column corresponding to x2. RNSengupta,IME Dept.,IIT Kapur,INDIA

  29. Changing tableau Basic Eq # Z x1 x2x3x4x5 RHS Variable Z 0 1 -3 -5 0 0 0 0 x3 1 0 1 0 1 0 0 4 x4 2 0 0 2 0 1 0 12 x5 3 0 3 2 0 0 1 18 • Choose the row corresponding to this as the pivot row and the element common to pivot column and pivot row is the pivot number. • Hence x4 leaves and x2 enters and 2 is the pivot number. RNSengupta,IME Dept.,IIT Kapur,INDIA

  30. Changing tableau Basic Eq # Z x1 x2x3x4x5 RHS Variable Z 0 1 -3 0 0 5/2 0 30 x3 1 0 1 0 1 0 0 4 x2 2 0 0 1 0 1/2 0 6 x5 3 0 3 0 0 -1 1 6 • How are the new rows formed: (New row) = (old row) – (pivot column coefficient) X (new pivot row) • Eq # 0 has changed as per the following calculation to (-3 -5 0 0 0 0)-(-5)X(0 1 0 ½ 0 6)=(-3 0 0 5/2 0 30) • Eq # 1 has changed as per the following calculation to (1 0 1 0 0 4)-(0)X(0 1 0 ½ 0 6)=(1 0 1 0 0 4), i.e., it does not change as we have 0 • Eq # 2 has changed as per the following calculation to ½(0 2 0 1 0 12)=(0 1 0 ½ 0 6) • Eq # 3 has changed as per the following calculation to (3 2 0 0 1 18)-(2)X(0 1 0 ½ 0 6)=(3 0 0 -1 1 6) RNSengupta,IME Dept.,IIT Kapur,INDIA

  31. Changing tableau Basic Eq # Z x1 x2x3x4x5 RHS Variable Z 0 1 -3 0 0 5/2 0 30 x3 1 0 1 0 1 0 0 4 x2 2 0 0 1 0 1/2 0 6 x5 3 0 3 0 0 -1 1 6 RNSengupta,IME Dept.,IIT Kapur,INDIA

  32. Changing tableau Basic Eq # Z x1 x2x3x4x5 RHS Variable Z 0 1 0 0 0 3/2 1 36 x3 1 0 0 0 1 1/3 -1/3 2 x2 2 0 0 1 0 1/2 0 6 x1 3 0 1 0 0 -1/3 1/3 2 • How are the new rows formed: (New row) = (old row) – (pivot column coefficient) X (new pivot row) • Eq # 0 has changed as per the following calculation to (-3 0 0 5/2 0 30)-(-3)X(1 0 0 -1/3 1/3 2)=(0 0 0 3/2 1 36) • Eq # 1 has changed as per the following calculation to (1 0 1 0 0 4)-(1)X(1 0 0 -1/3 1/3 2)=(0 0 1 1/3 -1/3 2) • Eq # 2 has changed as per the following calculation to (0 1 0 ½ 0 6)-(0)X(1 0 0 -1/3 1/3 2)=(0 1 0 ½ 0 6) • Eq # 3 has changed as per the following calculation to 1/3(3 0 0 -1 1 6)=(1 0 0 -1/3 1/3 2) RNSengupta,IME Dept.,IIT Kapur,INDIA

  33. Final solution • x=(x1, x2, x3, x4, x5)=(2,6,2,0,0) and maximum Z is 36 • What are the constraints now • X1+x3=4, hence slack is x2=2 • 2x2+x4=12, hence slack is x3=0 • 3x1+2x2+x5=18, hence slack is x5=0 • Which would be better? To have all slacks as zeros or optimize RNSengupta,IME Dept.,IIT Kapur,INDIA

  34. 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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  35. 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 RNSengupta,IME Dept.,IIT Kapur,INDIA

  36. Rules for Constructing the Dual Problem RNSengupta,IME Dept.,IIT Kapur,INDIA

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