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ISyE 6669, Deterministic Optimization Homework #4, Due never (but you’ll need to know how to do these problems in order to do well on the first midterm)
- In class, we described the Dual Revised Simplex algorithm for maximization problems: start with a dual feasible basis, and iteratively move to a better dual feasible basis until we find one that is also primal feasible – this will be optimal. Primal: Dual: Maximize cx Minimize ub st Ax = b st uA c x 0
- Start with a dual feasible basis B
- If the basis is also primal-feasible (if xB = B-1b 0) then stop, we have the optimal solution. Otherwise, a) Find a primal-infeasible variable (a variable that is negative) to leave the basis. Call this row i. b) Pick a variable to enter the basis, making sure that we retain dual feasibility. In this case, dual feasibility means uA c. Since u = cBB-1, we need all ck – cBB-1ak 0 for dual feasibility. The rate-of- change for any reduced cost (value of the dual excess variable) is (B-1ak)i, so to find the entering variable that retains dual feasibility we want to choose from among all variables with negative values of (B-1ak)i and pick the one with the smallest ratio (ck – cBB-1ak)/(B- (^1) a k)i. Call this column j. So, the variable in column j will enter the basis, replacing the ith basic variable. To update B-1, create E as in primal revised simplex, and then B-1new = EB-1old. (a) That’s the dual revised simplex algorithm for a primal maximization problem. For this homework problem, I want you to describe the dual revised simplex algorithm for a primal minimization problem. Do not do what the book does (multiply the objective function by –1 and call it a max problem) – instead, think about what we’re doing in each step, and how each step would (or would not) be different for a primal minimization problem. [You should be able to do this problem now.]
- Start with a dual feasible basis B.
- If the basis is also primal feasible (if xB = B-1b 0) then stop, we have the ) then stop, we have the optimal solution. Otherwise, a) Find a primal-infeasible variable (a variable that is negative) to leave the basis. Call this row i.
b) Pick a variable to enter the basis, making sure that we retain dual feasibility. Until this step, nothing we did involved the primal objective function, so it didn’t matter whether we were maximizing or minimizing – it was all the same. But now, there’s a difference. Let’s look at our primal and dual problems: P: minimize cx D: maximize ub st Ax = b st uA c x 0) then stop, we have the Notice that, because we now have a primal minimization problem, the dual constraints are constrains (instead of constraints for the dual of a primal maximization problem). So, for dual feasibility we need uA c. Since u = cBB-1, dual feasiblity is cBB-1A c, or, for each dual constraint j, cj – cBB-1aj 0) then stop, we have the (in other worse, all primal reduced costs should be positive). We will be limited by variables xj for which (B-1aj)i < 0) then stop, we have the ; for positive values, the reduced cost will rise and will always remain positive. Therefore, we need to choose the variable which has the smallest positive value of (cj – cBB-1aj)/(-B-1aj)i. This will be the entering variable. So, the variable in column j will enter the basis, replacing the ith basic variable. To update B-1, create E as in primal revised simplex, and then B-1new = EB-1old. (b) Suppose the primal problem is a minimization problem as in part (a), and all of the costs c are positive (c > 0). Then, u = 0 is a dual feasible solution to the dual, so if we knew a basis for which all u = 0, it would be a good starting basis for dual simplex. Can you find such a basis? [You probably won’t be able to do this problem until after class on Monday, 9/24.] In fact, it may not be possible to find such a basis. Recall from class that, by complementary slackness, for every primal basic variable xj, the jth dual constraint must be satisfied at equality. If, for example, all c > 0) then stop, we have the , then none of the dual constraints uA c will be satisfied at equality when u = 0) then stop, we have the. This means that u = 0) then stop, we have the , although a feasible solution to the dual, is not a corner point of the dual feasible region – instead, it is somewhere inside the feasible region.
- From the last homework, find the optimal solution to problem 1 part (c) – the optimal basis. In part (d,ii) we found that when the right-hand side of the machine 2 constraint changed from 24 to 48, we had to start at the beginning and re-optimize the problem.
We have a negative value for a basic variable, so the solution is no longer optimal. However, it is still dual feasible, so we can use dual revised simplex to solve the problem. ITERATION 1 There is only one negative basic variable, so x 1 will leave the basis. x 1 is the fourth basic variable. Now, let’s check the reduced costs and the values of (B-1aj) 4 (because the fourth basic variable is leaving the basis). Reduced costs for nonbasic variables: It will be easier to calculate cBB-1^ once, instead of repeating the calculation for every nonbasic variable. [ -1/30) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 1 1 / 3 ] [ 1 / 3 -1 0) then stop, we have the 0) then stop, we have the 662 / 3 ] cBB-1^ = [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 40) then stop, we have the 0) then stop, we have the 0) then stop, we have the 10) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] [ 1 / 3 0) then stop, we have the -1 0) then stop, we have the 166^2 / 3 ] = [ 13^1 / 3 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the -333^1 / 3 ] [ 1 /30) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the -1/ 3 ] [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 1 ] e 1 : ce1 – cBB-1ae1 = 0) then stop, we have the – -13^1 / 3 = 13^1 / 3 s 5 : cs5 – cBB-1as5 = 0) then stop, we have the – -333^1 / 3 = 333^1 / 3 (Actually, we knew these from our final iteration of primal revised simplex in the previous homework.) Rates of change: e 1 : (B-1ae1) 4 = -1/30) then stop, we have the 0) then stop, we have the s 5 : (B-1as5) 4 = -1/ 3 Ratios: e 1 : (40) then stop, we have the / 3 ) / (^1 /30) then stop, we have the 0) then stop, we have the ) = 40) then stop, we have the 0) then stop, we have the 0) then stop, we have the s 5 : (10) then stop, we have the 0) then stop, we have the 0) then stop, we have the / 3 ) / (^1 / 3 ) = 10) then stop, we have the 0) then stop, we have the 0) then stop, we have the So s 5 will enter the basis, because it has the smallest ratio. [ 1 / 3 ] [ 66^2 / 3 ] B-1as5 = [ 166^2 / 3 ] [ -1/ 3 ] [ 1 ]
So to get the special column, we divide each entry but the fourth by 1 / 3 (because the leaving variable was the fourth one) and the fourth entry is just 1/(-1/ 3 ). [ 1 0) then stop, we have the 0) then stop, we have the 1 0) then stop, we have the ] [ 0) then stop, we have the 1 0) then stop, we have the 20) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] E = [ 0) then stop, we have the 0) then stop, we have the 1 50) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the -3 0) then stop, we have the ] [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 3 1 ] Now we calculate the new value of B-1, as in primal revised simplex. As we had before, B-1new = EB-1old. [ 1 0) then stop, we have the 0) then stop, we have the 1 0) then stop, we have the ] [ -1/30) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 1 1 / 3 ] [ 0) then stop, we have the 1 0) then stop, we have the 20) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] [ 1 / 3 -1 0) then stop, we have the 0) then stop, we have the 662 / 3 ] B-1^ = [ 0) then stop, we have the 0) then stop, we have the 1 50) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] [ 1 / 3 0) then stop, we have the -1 0) then stop, we have the 166^2 / 3 ] [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the -3 0) then stop, we have the ] [ 1 /30) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the -1/ 3 ] [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 3 1 ] [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 1 ] [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 1 0) then stop, we have the ] [ 1 -1 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] = [ 2 0) then stop, we have the -1 0) then stop, we have the 0) then stop, we have the ] [ -1/10) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 1 ] [ 1 /10) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] Now, let’s check the new values of the basic variables {s 4 ,e 2 ,e 3 ,s 5 ,x 2 }: [ 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 1 0) then stop, we have the ] [ 30) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] [ 24 ] [ 1 -1 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] [ 50) then stop, we have the 0) then stop, we have the ] [ 250) then stop, we have the 0) then stop, we have the ] xB = B-1b = [ 2 0) then stop, we have the -1 0) then stop, we have the 0) then stop, we have the ] [ 20) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] = [ 40) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] [ -1/10) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 1 ] [ 24 ] [ 18 ] [ 1 /10) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the ] [ 48 ] [ 30) then stop, we have the ] All of the basic variables are nonnegative, so the solution is optimal.
- Long John Jarvis the pirate discovered a vast treasure hidden on a deserted island. There were limitless amounts of gold and silver coins, jewel-encrusted plates, and ivory statues, all of which were completely unguarded and could be taken at no risk. Unfortunately, Long John’s ship had been sunk, and he had only a small rowboat in which to put his treasure. The rowboat could hold up to 1000 pounds (plus the pirate himself) and a volume of 200 cubic feet. The table below gives the value, size, and weight of each type of treasure: Value (dollars) Size (cubic feet) Weight (pounds)
Let usize and uweight be the dual variables associated with the primal size and weight constraints. The dual is then Minimize 20) then stop, we have the 0) then stop, we have the usize + 10) then stop, we have the 0) then stop, we have the 0) then stop, we have the uweight Subject to 0) then stop, we have the .0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 145 usize + 0) then stop, we have the .177 uweight 20) then stop, we have the 0) then stop, we have the (gold)
- then stop, we have the .0) then stop, we have the 0) then stop, we have the 0) then stop, we have the 145 usize + 0) then stop, we have the .0) then stop, we have the 96 uweight 10) then stop, we have the 0) then stop, we have the (silver)
- then stop, we have the .0) then stop, we have the 4 usize + uweight 40) then stop, we have the 0) then stop, we have the 0) then stop, we have the (plates)
- then stop, we have the .75 usize + 20) then stop, we have the uweight 10) then stop, we have the 0) then stop, we have the 0) then stop, we have the 0) then stop, we have the (statues) usize , uweight 0) then stop, we have the We can solve this problem graphically (see attached sheet). Recall that the primal shadow prices are the optimal dual variables, and the dual shadow prices are the optimal primal variables. So, if we can find the dual shadow prices from our graphical solution, we’ll know the optimal solution to Long John’s original problem. When we look at our graphical solution, we see that three of the constraints, the gold, silver, and statue constraints, are not binding at the optimal solution. Therefore, their shadow prices are zero; Long John should take zero gold coins, zero silver coins, and zero ivory statues. The optimal dual solution is at the intersection of the plate constraint and the non-negativity constraint for the variable usize. Therefore, to calculate the shadow price of the plate constraint, we can add one to the plate constraint’s right-hand side and solve for the intersection of those two constraints: usize = 0) then stop, we have the (non-negativity for usize )
- then stop, we have the .0) then stop, we have the 4 usize + uweight = 40) then stop, we have the 0) then stop, we have the 1 (plates) This intersection is at usize = 0) then stop, we have the , uweight = 40) then stop, we have the 0) then stop, we have the 1. Thus usize is the same as in the optimal solution and uweight has increased by 1, so the objective has increased by 0) then stop, we have the times the objective coefficient of usize , plus 1 times the objective coefficient of uweight , or 10) then stop, we have the 0) then stop, we have the 0) then stop, we have the. Therefore, the shadow price of the plate constraint is 10) then stop, we have the 0) then stop, we have the 0) then stop, we have the , and Long John should take nothing but 10) then stop, we have the 0) then stop, we have the 0) then stop, we have the jewel-encrusted plates, for a total worth of $4,0) then stop, we have the 0) then stop, we have the 0) then stop, we have the ,0) then stop, we have the 0) then stop, we have the 0) then stop, we have the.
The feasible region is to the upper right of all the constraints, and the optimal solution is (0) then stop, we have the ,40) then stop, we have the 0) then stop, we have the 0) then stop, we have the ). 0 500 1000 1500 2000 2500 3000 3500 4000 4500 0 200000 400000 600000 800000 1000000 1200000 1400000 1600000 u u
