Lecture 7: Sensitivity Analysis. Reading: Chapter 9

Transcription

1 Lecture 7: Sensitivity Analysis Reading: Chapter 9 1

2 Sensitivity Analysis: Answering the What If Questions After you have solved a given LP and found an optimal solution: 1. What is the effect of a change in one or more parameters on the optimal tableau? current solution? 2. If the solution is no longer optimal, what needs to be done in order to find a new optimal solution? 3. What happens when we add a new variable? a new constraint? 4. What if we want to test changes in parameters over a range of possible parameter values? 2

3 Problem Parameters and the Optimal Tableau Suppose we start with a canonical max form tableau z x 1 x 2... x n s 1 s 2... s m rhs 0 a 11 a a 1n b 1 0 a 21 a a 2n b a m1 a m2... a mn b m 1 c 1 c 2... c n and we pivot (using whatever method) to optimal tableau basis z x 1 x 2... x n s 1 s 2... s m rhs x B1 0 a 11 a a 1n s 11 s s 1m b 1 x B2 0 a 21 a a 2n s 21 s s 2m b x Bm 0 a m1 a m2... a mn s m1 s m2... s mm b m z 1 c 1 c 2... c n y 1 y 2... y m z 0 with b 0 and c 0. We start by identifying the basis matrix B = [A B1,..., A Bm ], the basic costs c B = [c B1,..., c Bm ], the inverse matrix S = B 1, and the dual variables y = c B S. 3

4 Example: Woody s Problem Initial tableau z x 1 x 2 x 3 s 1 s 2 s 3 rhs Optimal tableau basis z x 1 x 2 x 3 s 1 s 2 s 3 rhs s / x /4 0 2/3 2 x / z / We can read off from the the slack columns S = 15/ /4 0 2/3 1/2 0 1 and note that y = ( 5/2, 0, 5 ) c B = ( c 5, c 2, c 1 ) = ( 0, 60, 35 ) 4

5 The Key Equations Recall the matrix equations describing the final tableau derived in Lecture 4: In formulae: b = S b (1) A = S A (2) c = c c B S A z 0 = c y A (3) = c c B A (3 ) = c B S b = y b (4) = c B b (4 ) (1) b i = s i1b 1 + s i2b s imb m i = 1,..., m (2) a ij = s i1a 1j + s i2a 2j s ima mj i = 1,..., m, j = 1,..., n (3) c j = c j y1a 1j y2a 2j... yma mj j = 1,..., n (3 ) = c j c B1 a 1j c B2 a 2j... c Bm a mj j = 1,..., n (4) z0 = y1b 1 + y2b ymb m (4 ) = c B1 b 1 + c B2 b c Bm b m 5

6 Dual Variables as Shadow Prices The dual variables serve an important role as a mechanism for pricing the parameters of the LP. In particular, The dual variable for the i th constraint represents the shadow price for that constraint, and is the amount the objective function value will increase per unit increase of b i. That is, y i = z/ b i. Example: Suppose Woody has the opportunity to buy additional amounts of either mahogany or pine at $2 per linear foot. Which (if any) of these opportunities should he take advantage of? Answer: The shadow price of pine is 5/2, which means that each additional unit of pine obtained will result in an increase of the in Woody s profits by $2.50. Thus buying pine at $2 per linear foot results in a profit of $.50 for each additional linear foot purchased. Mahogany has $0 shadow price (why?). Thus there is no change in the objective per additional units of pine, and so there is no price attractive enough to convince Woody to buy more mahogany. 6

7 Changes in a Resource Suppose we want to change the values of one or more b k values. Effect on the optimal tableau: b k only appears in Equations (1) and (4), i.e. only the right-handside column (b and z 0) is affected Effect on the current solution: All basic solution values, as well as the objective function value, may change. Dual solution remains the same (although its objective function value may change). Effect on feasibility: Primal solution may become infeasible. Dual solution remains dual feasible. Reoptimization method: Dual Simplex Method. 7

8 Example Suppose Woody wants to change the amount of pine he uses each day, that is, he is changing b 1 currently 120 to an amount b 1 + = New tableau and solution values: s 2 b1 15/ x B = x 2 = b2 = 1/4 0 2/3 x 1 b3 1/2 0 1 = /4 2 1/ /2 z 0 = ( 5/2, 0, 5 ) = / In particular, for every four linear feet of pine Woody buys, he will make two more chairs and one less table (and still no desks), with an additional profit of $10. There is a limit here, however; if, say, Woody buys 12 more linear feet of pine the RHS values will be 75, -1, and 18 (and z 0 = 570). In this case, we would have to reoptimize by performing a dual simplex pivot in Row 2, in particular, with x 4 replacing x 2 in the basis. 8

9 Changes in Nonbasic Cost Coefficients Suppose that we want to change the costs c k for one or more variables that are nonbasic in the optimal tableau. Effect on the optimal tableau: c k only appears in the Equation (3) and only for j = k, i.e. only the entry c k is affected. Effect on the current solution: Primal and dual solution values and objective function value remain the same. Effect on feasibility: Primal solution remains feasible, but dual solution may become infeasible. Reoptimization method: Primal Simplex Method 9

10 Example Suppose that Woody wants to change the price for desks so that the current value of c 3 = 75 changes to an amount New value of c 3: c 3 = (75 + ) ( 5/2, 0, 5 ) = 10 + Reduced costs and ranging for optimality: The reduced cost of any nonbasic variable in the optimal tableau represents the minimum increase allowed for the profit of that variable before it enters the basis. (For a min problem, this represents the minimum decrease in that cost.) Any less than that increase will not affect the optimal solution for either the primal or dual problem. In our example, if the profits for desks stays below 75+ = $85 then there will be no change in the optimal solution values for the problem. If profits for desks exceed $85 then a pivot would be called for in column 3, with desks (x 3 ) replacing chairs (x 2 ) in the basis. 10

11 Changes in Basic Cost Coefficients Suppose we want to analyze a change in the cost c k of a basic variable in the optimal tableau. Effect on the optimal tableau: c k appears in Equation (3 ) and (4 ), and affects the entire objective function row (z 0 and all c j) Effect on the current solution: Primal solution remains the same (although its objective function may change) and dual solution values may change. Effect on feasibility: Primal solution remains feasible, dual solution may become infeasible. Reoptimization method: primal simplex method 11

12 Example Suppose that Woody wants to analyze a change in the price of chairs from its current value of c 1 = 35 to an amount (Negative of the) new tableau values: Using (3 ), (4 ) with c B = (c 5, c 2, c 1 ) = (0, 60, 35 + ) we get c = ( 35 +, 60, 75, 0, 0, 0 ) ( 0, 60, 35 + ) / /4 0 2/ /2 0 1 = ( 0, 0, 10 +, 5/2 1/2, 0, 5 + ) z 0 = ( 0, 60, 35 + ) = Thus if c 1 increases by = 5 the optimal solution value would not change, although its objective function now becomes 600, and the dual solution becomes ỹ = (5, 0, 0). If profits decrease by, say, = 10, though, then c 4 becomes 5 (corresponding tableau value is 5) and so x 4 enters the basis, replacing x 5. 12

13 Changes in Nonbasic Production Coefficients Suppose we want to analyze a change in the value a kl of a nonbasic variable in the optimal tableau. Effect on the optimal tableau: a kl appears in Equations (2) and (3), and affects all values in Column l (and nothing else). Effect on the current solution: Primal and dual solution values and objective function value remain the same. Effect on feasibility: Primal solution remains feasible, but dual solution may become infeasible. Reoptimization method: Primal Simplex Method 13

14 Example Suppose that Woody wants to decrease the number of carpenter-hours spent on making desks from its current value of a 33 = 9 to value a 33 = 9, to see when it becomes profitable to make desks. New tableau values: ã 13 15/ ã 23 = 1/4 0 2/3 ã 33 1/2 0 1 c 3 = 75 ( 5/2, 0, 5 ) = /3 1 + = Value at which Woody would start making desks: Since the only change in the objective row occurs at c 3, then the tableau and hence current solution will remain optimal as long as c 3 = 10 5 remains nonnegative. In this case we need 10 5 < 0, i.e. > 2 (a 33 < 7) before it becomes preferable to pivot on Column 3, that is, make desks. We would then make primal simplex method pivot to reoptimize. 14

15 Adding a Variable Suppose Woody wants to consider producing executive desks, with each executive desk requiring 12 linear feet of pine, 30 linear feet of mahogany, and 15 carpenter-hours, with a profit of $175. Effect on the optimal tableau: A new column (using x 4 = number of executive desks) is added, using Equations (2) and (3), as if the Revised Simplex Method were being applied. Remaining values remain unchanged. Effect on the current solution: Solution value and objective function value remain the same. New tableau values: ã 14 ã 24 ã 34 = 15/ /4 0 2/3 1/2 0 1 c 4 = 175 ( 5/2, 0, 5 ) = = 70 c 4 = 70, so Woody should produce executive desks. Reoptimization Method: Primal Simplex Method. In this case the first pivot would be in Row 2, that is, Woody would replace table production with executive desk production. 15

16 Changes in Basic Production Coefficients Suppose we want to analyze a change in the value a kl of a basic variable x l = x Br in the optimal tableau. Effect on the optimal tableau: Again, a kl appears in Equations (2) and (3), and affects values in Column l. Unfortunately, Column l is a basic column, and hence the resulting tableau may stop being basic. Remedy: After substituting in the new column values, we pivot on the (r, l) th entry. The tableau then becomes basic again, but all tableau entries can change. Tableau can become primal infeasible, dual infeasible, or both. Method of reoptimization: Depends upon tableau status. If the tableau happens to remain primal feasible apply the Phase II Simplex Method; if the tableau happens to remain dual feasible apply the Dual Simplex Method. If it becomes both primal and dual infeasible, then we would have to apply some Phase I method. 16

17 Example Suppose that Woody wants to analyze a change in the amount of carpenter-hours spent on making chairs (a 31 ) from 3 to 2. Then the values in Column 1 become ã 11 ã 21 ã 31 = 15/ /4 0 2/3 1/2 0 1 c 1 = 35 ( 5/2, 0, 5 ) = = /3 2 We replace Column 1 by the above values, and then pivot on entry (3, 1), to get new tableau basis z x 1 x 2 x 3 s 1 s 2 s 3 rhs s / x /3 1/12 0 1/3 6 x /2 1/4 0 1/2 6 z /2 15/4 0 5/2 570 In this case the tableau remains dual feasible, and so we apply the dual simplex method to reoptimize, pivoting on entry (1, 6). This again produces an optimal tableau. 17

18 Adding a Constraint Suppose we want to add a constraint of the type a m+1,1 x 1 + a m+1,2 x a m+1,n x n b m+1 1. Enter the corresponding equality constraint (including the slack variable) into the current tableau as the (m + 1) st row. 2. Cost out the current basis by subtracting a m+1,bi (Row i) from Row (m + 1). Equivalently, use the formulae à m+1, = A m+1 A m+1,b A S m+1, = A m+1,b S bm+1 = b m+1 A m+1,b b where A m+1,b = (A m+1,b1,..., A m+1,bm ). 3. If b m+1 < 0, continue to apply the Dual Simplex Method to reoptimize. 18

19 Example Woody wants to limit himself to 10 pieces of furniture each day, that is, he wants to add the constraint x 1 + x 2 + x 3 10 to his tableau. He can add this row directly to the tableau and cost out the basic variables, or use the formulae: Ã 4, = A 4 (A 45, A 42, A 41 )A = ( 1, 1, 1 ) ( 0, 1, 1 ) = ( 0, 0, 0 ) S 4, = (A 45, A 42, A 41 )S = ( 0, 1, 1 ) = ( 1/4, 0, 1/3 ) 15/ /4 0 2/3 1/ b4 = 10 ( 0, 1, 1 ) = 4. 19

20 This gives the following tableau: basis z x 1 x 2 x 3 s 1 s 2 s 3 s 4 rhs s / x /4 0 2/3 0 2 x / s /4 0 1/3 1 4 z / Two dual simplex pivots produce optimal tableau basis z x 1 x 2 x 3 s 1 s 2 s 3 s 4 rhs x /10 1/2 3/2 3 x /5 2/3 2 0 x /10 1/6 1/2 7 s / z / so that Woody should produce 7 chairs and 3 desks. 20

21 Other Types of Inequalities constraints: Negate the constraint and treat it as a constraint. Equality constraints: Add this constraint just as you would a constraint, but without the slack variable. Then do a dual simplex pivot on the added row (ignore the sign of the RHS) to determine a basic variable for that tableau. Then continue with the Dual Simplex Method. Example: Suppose Woody wanted to make exactly 10 pieces of furniture. The initial tableau will look like basis z x 1 x 2 x 3 s 1 s 2 s 3 rhs s / x /4 0 2/3 2 x / ? /4 0 1/3 4 z / Pivot on the s 1 column to put s 1 into the basis in Row 4. One more pivot gives the same optimal tableau and solution as in the case, only without the s 4 column. 21

22 Summary of Sensitivity Analysis parameter relevant parts of tableau method of changed equation(s) affected reoptimization b i (1) & (4) entire r.h.s. column dual simplex nonbasic c j (3) a 0j only primal simplex basic c j (3 ) & (4 ) entire objective row primal simplex nonbasic a ij (2) & (3) entire j th column primal simplex basic a ij (2) & (3) all entries (depends upon + pivot type of tableau) added column (2) & (3) new column primal simplex added constraint cost out new row dual simplex 22