1. 1 The Dual in Linear Programming In LP the solution for the profit- maximizing combination of outputs automatically determines the input amounts that must be used in the production process.

2. 2 If the optimal output combination, suggested by the LP solution, uses up all available inputs, we say that the capacity constraints are binding.

3. 3 Under this condition, any reduction/increase in the use of inputs will reduce/increase the firm’s profits. marginal opportunity cost of using the input How much profit do we lose not having one more unit of an input?

4. 4 Finding the Opportunity Costs For every maximization problem in LP there exists a symmetrical minimization problem and vice versa The problems are called the primal problem and the dual problem

5. 5 Primal Problem and Dual Problem optimal solutions for these two problems are always the same objective of the dual problem is to find shadow prices (= dual prices) shadow prices can be used to decide whether the firm should employ an additional unit of a resource or not

6. 6 Primal Problem Gross profit: Max GP = 50X + 30Y st. 5X + 2Y  220 3X + 2Y  180 X, Y  0 Dual Problem* Total opportunity cost: Min C = 220a + 180p st.5a + 3p  50 2a + 2p  30 a, p  0 Example: Primal and Its Dual *John von Neumann proved the Duality Theorem

7. 7 Coefficient matrix of a dual problem is a transpose of the primal’s coefficient matrix a = opportunity cost of using an additional unit of labor for the assembly of PCs and printers p =opportunity cost of using an additional unit of labor for the packaging of PCs and printers X Y RHS a p RHS

8. 8 The solution for the above problem is: GP = 2800, X = 20 and Y = 60 a = 2.50 and p = 12.50

9. 9 Linear Programming: Sensitivity Analysis Postoptimality analysis Sensitivity analysis is the study of how changes in the coefficients of a linear program affect the optimal solution or in the value of right hand sides of the problem affect the optimal solution

10. 10 Sensitivity Analysis continued Using sensitivity analysis we can answer questions such as: 1.How will a change in a coefficient of the objective function affect the optimal solution? We can define a range of optimality for each objective function coefficient by changing the objective function coefficients, one at a time

11. 11 Sensitivity Analysis continued 2.How will a change in the right-hand- side value for a constraint affect the optimal solution? The feasible region may change when RHSs (one at a time) are changed and perhaps cause a change in the optimal solution to the problem

12. 12 ￭ Sensitivity analysis is important since real world problems exist in a changing environment: prices of raw materials change, product demand changes, new machinery is bought to replace old, stock prices fluctuate, employee turnover occurs, etc.  we can expect some of the coefficients to change over time Sensitivity analysis can also be used to determine which coefficients are crucial

13. 13 Graphical Sensitivity Analysis 1.Change in the slope of the objective function (change in the coefficients of the objective function) The shaded region defines the range in which the objective function coefficients can vary without changing the optimal solution for the problem: the output combination does not change but the value of the objective function naturally changes…

14. 14

15. 15 Graphical Sensitivity Analysis continued 2.Changing the right-hand-side of a constraint The shaded region is an increase in the feasible set resulting from a change in the RHS of one of the constraints: increasing the availability of one input (resource)

16. 16

17. 17 Interpretation of Computer Output We will use Solver with Excel to solve LP-problems On the spreadsheet after using Solver we can find the optimal value for the objective function and the decision variable values associated with this particular solution, as well as the right-hand-side values for the constraints

18. 18 An Example: What to Plant? A farmer owns a 100 acre farm He plans to plant at most three crops. The seed for crop A, B, and C cost $40, $20, and $30 per acre, respectively A maximum of $3,200 can be spent on seed Crop A, B, and C require 1, 2, and 1 workdays per acre, respectively, and there are a maximum of 160 workdays available If the farmer can make a profit of $100 per acre on crop A, $300 per acre on crop B, and $200 per acre on crop C, how many acres of each crop should be planted to maximize profit?

19. 19 Excel Spreadsheet with Solver

20. 20 More detailed information about the solution can be found on the Answer and Sensitivity Reports: Slack tells how much, if any, of a resource has been left unused Reduced Costs indicate how much the objective function coefficient of each decision variable would have to improve before it would be possible for that variable to assume a positive value in the optimal solution

21. 21

22. 22

23. 23 Sensitivity Analysis in the Computer Output Allowable Increase and Decrease in the “Adjustable Cells” tells the ranges in which the coefficients of the objective function can vary without changing the optimal solution i.e. the values of decision variables in optimal solution; the value of the objective function itself will naturally changes when the coefficients change

24. 24 Sensitivity Analysis in the Computer Output continued Shadow Price tells how much a one unit increase/decrease in the RHS of a constraint would increase/decrease the value of the objective function Allowable Increase and Decrease in the “Constraint R.H. sinde” tells the ranges in which the RHSs can vary without changing the basis of the optimal solution (= the set of variables with a positive value)

25. 25 Example: Resource Allocation Problem Let’s consider a manufacturing facility that produces five different products using four machines The scarce resources are the times available on the machines and the alternative activities are the individual production volumes The machine requirements in hours per unit are shown for each product in the table The unit profits are also shown in the table The facility has four machines of type 1, five of type 2, three of type 3 and seven of type 4 Each machine operates 40 hours per week The problem is to determine the optimum weekly production quantities for the products The goal is to maximize total profit

26. 26 Machine data and processing requirements (hrs./unit) Machine QuantityProduct 1Product 2Product 3Product 4Product 5 M141.21.30.70.00.5 M250.72.21.60.51.0 M330.90.71.31.00.8 M471.42.80.51.20.6 Unit profit, $ ——1825101215 The number of hours available on each machine type is 40 times the number of machines.

27. 27 Machine Availability Constraints M1 :1.2P 1 + 1.3P 2 + 0.7P 3 + 0.0P 4 + 0.5P 5 < 160 M2 :0.7P 1 + 2.2P 2 + 1.6P 3 + 0.5P 4 + 1.0P 5 < 200 M3 :0.9P 1 + 0.7P 2 + 1.3P 3 + 1.0P 4 + 0.8P 5 < 120 M4 : 1.4P 1 + 2.8P 2 + 0.5P 3 + 1.2P 4 + 0.6P 5 < 280 Nonnegativity P j > 0 for j = 1,...,5 Objective Function Max Z = 18P 1 + 25P 2 + 10P 3 + 12P 4 + 15P 5

28. 28

29. 29 Quiz 1. Explain the meaning of Reduced Cost in this context. 2. Explain the meaning of ranges in Variable Analysis. Note! What do they mean in this context? 3. Explain the meaning of Shadow Price in this context. 4. Explain the meaning of ranges in Constraint Analysis. Again in this context!

30. 30 Variable Analysis: Reduced Cost Reduced cost $13.53 means that the coefficient of P 3 (its unit profit) had to increase by this amount before it would be profitable to start producing product 3 The "reduced cost" column indicates the increase in the objective function per unit change in the value of the associated variable Reduced cost of P 3 indicates that if this variable were increased from 0 to 1, i. e. if we would produce on unit of product P3, the objective function value (profit) will decrease by $13.53 It is not surprising that the reduced cost is negative since the optimum value of P 3 is zero

31. 31 Variable Analysis: Reduced cost continued The ranges at the right of the display indicate how far the associated objective coefficient may change before the current solution values (P 1 through P 5 ) must change to maintain optimality For example, the unit profit on P1 may assume any value between 13.26 and 24.81 The "---" used for the lower limit of P 3 indicates an indefinite lower bound Since P 3 is zero at the optimum, reducing its unit profit by any amount will make it even less appropriate to produce that product

32. 32 Constraint Analysis A shadow price indicates the increase in the objective function value resulting from an unit increase of the associated constraint (resource) From the table we see that increasing the hour limit of 120 for M 3 increases the objective function value by the most ($8.96), while increasing the limit for M 4 increases the objective function value by the least ($0.36)

33. 33 Constraint Analysis: Shadow Price continued The ranges at the right of the display indicate how far the limiting value may change while keeping the same optimum basis The shadow prices remain valid within this range As an example consider M 1. For the solution, there are 160 hours of capacity for this machine. The capacity may range from 99.35 hours to 173 hours while keeping the same basis optimal. Changes above 160 cause an increase in profit of $4.82 per unit, while changes below 160 cause a reduction in profit by $4.82 per unit.