x 4 = 40 +2x 5 +6x x 6 x 1 = 10 2x x 6 x 3 = 20 +x 5 x x 6 z = 540 3x 5 x 2 3x 6 x 4 x 5 x 6 x x

Transcription

1 MATH 4 A Sensitivity Analysis Example from lectures The following examples have been sometimes given in lectures and so the fractions are rather unpleasant for testing purposes. Note that each question is imagined to be independent; the changes are not intended to be cumulative. We wish to consider a trucking firm who can choose to purchase three types of trucks subject to three constraints on capital, space and number of drivers. We use variable x i to denote the number of trucks of type i to be purchased. Truck Truck 2 Truck availability capital units $K space trucks drivers 6 6 people net profit Now setting x i = number of trucks of type i to be produced we have the LP: We get the final dictionary: max 2x +2x 2 +2x 24x +4x 2 +46x 2 x +x 2 +x x +6x 2 +6x x, x 2, x a) Give B, appropriately labelled: x 4 = 4 +2x +6x x 6 x = 2x + x 6 x = 2 +x x 2 x 6 z = 4 x x 2 x 6 B = x x 2 x b) Give the marginal values associated with capital,space and drivers: capital:, space:, drivers: i.e. extra capital worth so not helpful, extra space worth, extra drivers worth. But this is only true on the margin, for large changes, these values are unlikely to be valid (although they provide an upper bound). You d be willing to pay units for a parking spot and sell a parking spot for units. c) Give values for c, c 2, c so {x 4, x, x } still yields an optimal basis: In this case we wish c T N c T BB A N : ( x 2 x x 6 c 2 ) ( x 4 x x c c ) x x 2 x = ( x 2 x x 6 c 2 c c 2c c c ) x 2 x x 6 x 4 4 x x 6 6

2 d) For what range on c is the current basis {x 4, x, x } still optimal? c T N c T BB A N = [, 2 2c, c 7). We deduce that 2 c 2. Note that in this range the optimal solution is x = and x = 2 and so the profit would be c e) What is the optimal solution if c =? We create the updated dictionary and try pivoting to optimality. We believe we are close to optimal since we have made a small change. x 4 = 4 +2x +6x x 6 x = 2x + x 6 x = 2 +x x 2 x 6 z = +x x 2 x 6 x enters and x leaves x 4 = x = x = 2 z = x 2 x 2 x 6 We have a new optimal solution x = 2. The profit hasn t been worked out but you can since it is 2 2 = 2, a slight reduction. f) What is the optimal solution if c = 22? x 4 = 4 +2x +6x x 6 x = 2x + x 6 x = 2 +x x 2 x 6 z = 2x x 2 + x 6 x 6 enters and x leaves x 4 = 48 x = x 6 = 6 z = 22x 2x 2 x We have a new optimal solution x =. The profit hasn t been worked out but you can since it is 22 = 66. We have made truck valuable so we now buy lots of it. g) For what range on c 2 is the current basis optimal. We simply need c 2 c B B A 2. Now c B B = [,, ]. So we find that for c 2 2 the the current basis (and solution) remain optimal. h) What if we introduce a new truck (we ll use variable x 7 ) with requirements of 4 capital, parking spot, crew and value 22. Sounds like a winner compared with truck. We may imagine adding x 7 and the associated date to our original problem and compute 4 c 7 c B B A 7 = c 7 [ ] = 4. Thus we would like to buy truck 7. We compute B A 7 = x x 2 x 4 = 6 2

3 We insert this in our dictionary: x 4 = 4 +2x +6x x 6 6 x 7 x = 2x + x 6 x 7 x = 2 +x x 2 x 6 2 x 7 z = 4 x x 2 x 6 +4x 7 Here x 7 enters and x 4 leaves but there is more pivoting to be done so I ll ignore this case. Note that since e we will have increased z above 4 then we know that track 7 will be purchased. i) Describe a profitable truck. It is one in which times the space requirement and times the labour requirement is less than the profit. j) Determine the value of the new solution when we decrease capital by, increase space by and decrease labour by. Given our marginal values we expect the profit to change by i b i y i = ( ) + + ( ) =. But this prediction works if the current basis remains feasible. Check x x 2 + = + x 2 x x 2 x + = Thus we gently increase trucks of type (to take advantage of increased parking and handle decreased labour) while decreasing type. The changes b, b 2, b can vary quite a bit and still have B b. k) For what values of b 2 is the current basis still optimal? Thus from which we deduce B b + b b b 2 2 b 2 b 2 2 i.e. 2 b 2. The marginal value of is valid in this interval. What happens if B (b + b)? l) What is the optimal solution if b = and b 2 =? We compute 6 4 B b + = We would have the dictionary x x 2 x = 8 x 4 = 8 +2x +6x x 6 x = 2x + x 6 x = +x x 2 x 6 z = 4 x x 2 x 6

4 Recall that if c T N c T BB A N T, and so c T BB is a feasible dual solution of objective function value c T BB b. If we do not have a primal feasible solution because B (b + b), we still have a dual feasible solution and so we seek to find a dual optimal solution (and so a primal optimal solution) by minimizing the dual objective function while preserving feasibility. Following the Dual Simplex method we have x 4 leave and seek the largest λ so that ( x x 2 x 6 ) + λ ( ) T. We choose λ = /6 and so x 2 enters resulting in the following dictionary. 4 x 2 = x + x 6 4 x 9 6 x = 2x + x 6 x = + 4x x x 9 6 z = 6 2 8x x 6 4 6x 9 6 One more pivot required. We choose x to leave and then seek the largest λ so that ( x x 4 x 6 ) ( ) λ T. There is a tie with λ = 2 and so either x or x 6 enters but following an extended form of Anstee s rule we choose x to enter. We obtain the new dictionary (or at least the parts we need) x 2 = 2 2 x = 2 x = 2 z = 2x x 2 4 Thus we get an optimal solution (2, 2/2,,, /2, ) with z = (we make more money as predicted) with new marginal values of /2 for capital, for space, and for labour. Our tie for the entering variable results in a degeneracy in the dual. m) What is the optimal solution if we add the constraint x 2 + x? First note that adding a constraint can only decrease the value of the objective function (it may even make the LP ifeasible). The answer here is easy. Our current solution x = and x = 2 with z = 4 remains optimal. n) What is the optimal solution if we add the constraint x 2 + x 22? We start by noting that our current optimal solution is not feasible. So we have to do something. We add a new slack variable x 7 = x 2 + x 22. We can t introduce this directly into our dictionary because it contains a basic variable but we can express x in terms of non-basic variables and obtain We have the dictionary x 7 = x 2 + x 22 = x 2 + (2 + x x 2 6 x 6) 22 = 2 + x 6 x 6 x 4 = 4 +2x +6x x 6 x = 2x + x 6 x = 2 +x x 2 x 6 x 7 = 2 +x 6 x 6 z = 4 x x 2 x 6 We follow the dual simplex algorithm, attempting to decrease z value while maintaining dual feasibility by having x 7 leave the basis. ( x x 2 x 6 ) + λ ( ) 6 T. We can take λ = and choose x as the entering

5 variable. Our new dictionary is: x 4 = 44 x = 6 x = 22 x = 2 z = 4 x 7 x 2 4x 6 We have a new optimal solution x = 6 and x = 22 (thus we satisfy the constraint x 2 + x 22) with z = 4 (a reduction in profitability from the added constraint). The marginal values are fro capital, for space, 4 for labour and for the new constraint. We assume the new constraint is written as x 2 x 22 and so increasing 22 to 2 results in less of a restriction and so an expectation of increased profit of units. When LINDO is given the constraint explicitly as x 2 + x 22 it will return a dual price of so that if you increase 22 to 2 then you expect profit to drop by. After n questions perhaps you don t need anymore but we discussed in class two questions: o) How do we delete a variable? Two suggestions were offered. If for example we wished to remove x we could set c =, making x unprofitable and our sensitivity techniques would drive it to. We could add a constraint x = (or x ) and proceed as above. If we wished to remove x 2, since it is non-basic we could just delete it from further consideration. p) How do we delete a constraint? If for example we wished to remove x + x 2 + x we could simply change the right hand side to some large number (say ) essentially eliminating the constraint. Or we could add a variable to the constraint x + x 2 + x x 7 where c 7 =. If the constraint is currently non-binding, such as capital, then essentially the constraint has been eliminated. q) How do we change an entry in A? This is more difficult but still possible. For the column of a nonbasic variable this is reasonable (try it!). In a test environment, only one pivot suffices to get you to optimality but this is unrealistic and for some changes it may be advisable to start from scratch.