# Linear Programming II: Minimization 2006 Samuel L. Baker Assignment 11 is on page 16.

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1 LINEAR PROGRAMMING II 1 Linear Programming II: Minimization 2006 Samuel L. Baker Assignment 11 is on page 16. Introduction A minimization problem minimizes the value of the objective function rather than maximizing it. Minimization problems generally involve finding the least-cost way to meet a set of requirements. Classic example -- feeding farm animals. Animals need: 14 units of nutrient A, 12 units of nutrient B, and 18 units of nutrient C. Learning Objective 1: Recognize problems that linear programming can handle. Linear programming lets you optimize an objective function subject to some constraints. The objective function and constraints are all linear. Two feed grains are available, X and Y. A bag of X has 2 units of A, 1 unit of B, and 1 unit of C. A bag of Y has 1 unit of A, 1 unit of B, and 3 units of C. A bag of X costs \$2. A bag of Y costs \$4. Minimize the cost of meeting the nutrient requirements. To solve, express the problem in equation form: Cost = 2X + 4Y objective function to be minimized Constraints: 2X + 1Y \$ 14 nutrient A requirement 1X + 1Y \$ 12 nutrient B requirement 1X + 3Y \$ 18 nutrient C requirement 8 8 Read vertically to see how much of each nutrient is in each grain. X \$ 0, Y \$ 0 non-negativity Learning objective 2: Know the elements of a linear programming problem -- what you need to calculate a solution. The elements are (1) an objective function that shows the cost or profit depending on what choices you make, (2) constraint inequalities that show the limits of what you can do, and (3) non-negativity restrictions, because you cannot turn outputs back into inputs.

2 LINEAR PROGRAMMING II 2 Graph method of solution Graph the constraints as equalities, like before. The constraints are now \$ rather than #, so the feasible area is everything to the right and above all of the constraint lines. You want to find the lowest cost point of this area. It will be a corner. You superimpose isocost lines that you get from the objective function. When you find an isocost line that just nicks one corner of the area, that corner is the solution. Feasible region in minimization problem Isocost lines and optimum Learning objective 3: Understand the principles that the computer uses to solve a linear programming problem. The computer uses the simplex method to systematically trace around the feasible area (the simplex) in the figure above. It goes from one corner to the next until it finds the best value for the objective function. Learning objective 4a: What linear programming problems have no solution? Those that have no feasible area. This means it's impossible to satisfy all the constraints at once. Learning objective 4b: What difference does linearity make? If the constraints are not linear, the feasible area has curved edges. The simplex method doesn't work because you can't be sure that the solution is at a corner. The solution may be in the middle of a curved edge. If the objective function is not linear, the solution may not even be on an edge. It may be in the interior of the feasible area.

3 LINEAR PROGRAMMING II 3 The simplex method of solution for minimization problems The simplex method gets more complicated when you have constraints that are >. The method can't start at the origin (with X = 0, Y = 0, etc.) because that is not a feasible solution. To find a starting place, we invent some new variables in addition to the slack variables. The new variables are called "artificial" and are conventionally labelled with A's. Artificial variables allow us to automatically come up with a feasible solution to start the simplex method. In this case, the starting feasible solution is A 1 =14, A 2 =12, A 3 =18, and all the rest (X, Y, and the S s) =0. The equations now look like this. 2X + 1Y - 1S 1 + 0S 2 + 0S 3 + 1A 1 + 0A 2 + 0A 3 = 14 1X + 1Y + 0S 1-1S 2 + 0S 3 + 0A 1 + 1A 2 + 0A 3 = 12 1X + 3Y + 0S 1 + 0S 2-1S 3 + 0A 1 + 0A 2 + 1A 3 = 18 Each constraint equation has one slack variable (S) and one artificial variable (A) with non-zero coefficients. The slack variable in each equation has a -1 coefficient because in a minimization problem we have to meet or exceed each constraint. This means we have to subtract the slack variable rather than add it as we did when we were maximizing the objective function. The objective function, which we want to minimize, is expanded to: 2X + 4Y + 0S 1 + 0S 2 + 0S 3 + MA 1 + MA 2 + MA 3 = Cost M stands for a mammoth number, meaning a very large unit cost for the A's, large enough to be sure that none of the A's will be in the final solution. In matrix form we have: X Y S 1 S 2 S 3 A 1 A 2 A M -M -M 0 The first thing we do is to get rid of those -M's that are at the bottom of the three columns for the A's. To do this, multiply each of the constraint rows by M and add, element by element, to the last row M-2 5M-4 -M -M -M M The columns for the A's now have one 1 and the rest 0's. The A's, and only them, are in the initial basic feasible solution. The solution is A 1 = 14, A 2 = 12, A 3 = 18, all X's and S's = 0. Cost is 44M, a monstrously mammoth number that we want to reduce. We then proceed with the simplex method as before. The pivot column is the second, because it has the biggest number (5M-4) in the bottom row.

4 LINEAR PROGRAMMING II M-2 5M-4 -M -M -M M To find the pivot row, we divide each number in the last column by the corresponding number in the second column, to get 14/1=14, 12/1=1, and 18/3=6. 6 is the smallest so the third row is the pivot row. The pivot element is the 3. The next step is to use that third row to eliminate the other elements in the second column. We multiply the row by 1/3 and subtract that from the first row. We do the same for the second row. For the bottom row we multiply the third row by (4/3-5M/3) and subtract that from the fourth row. For our new third row, we divide the original third row by 3. We get: X Y S 1 S 2 S 3 A 1 A 2 A M M -M M M+24 This puts Y in the solution, along with A1 and A3. The next iteration adds X to the solution. The iteration after that adds S 1. The final tableau looks like this: X Y S 1 S 2 S 3 A 1 A 2 A M 1-M 1-M 30 This is the final tableau because all the numbers in the bottom row are 0 or negative. Notice that the only M's left are under the A's. The non-zero variables in the solution are shown by the columns that have one 1 and the rest 0's. The solution amounts are in the right column, and match up with X, Y, and S 1 according to where the 1 in each column is. The solution is: X = 9, Y = 3, S 1 = 7. S2 and S3 are not in the final solution, so S 2 = S 3 = 0. The total cost is \$30. The shadow prices for the requirements are in the bottom row, in columns for S2 and S3. Both S 2 and S 3 have a shadow price of -\$1. These mean that it would cost you \$1 if you had to provide one more unit of either nutrient B or nutrient C. Having to provide one more unit of nutrient A costs nothing because you are already providing 7 extra units of A.

5 LINEAR PROGRAMMING II 5 Minimizing with Excel Fortunately, you don t have to do the simplex method by hand anymore. You can get Excel or Quattro Pro to do it for you. The challenge becomes setting the problem up properly. Learning objective 5: Be able to solve small linear programming problems yourself. The tricky part is setting up the objective function and the constraints. Write them on paper (without the S s and A s). Then let your spreadsheet crunch them for you. Here is the minimization example, as set up in Excel: Column B is for Grain X. Column C is for Grain Y. Column D has formulas for total cost and total amounts of nutrients provided. If you type the formula in D3 first, and use \$ signs in the right places, you can copy D3 to D6:D8. Here s the same spreadsheet, showing the formulas in the D column.

6 LINEAR PROGRAMMING II 6 Select Tools and then Solver... from Excel s menu. If you are re-using a spreadsheet from an earlier assignment, the safest way to start is to click on the Reset All button. This clears all the old settings and cell choices, preventing some elusive bugs. Then set the Target Cell to D3 and the Changing Cells to B2:C2. On the Equal To: line, check Min. If you don t do this, your solution will be unbounded. Solver will fly off looking for the upper-right corner of your boundless feasible region. Click on the Add button to add constraints. The nutrient amounts our grain provides, in D6:D8, must be greater than or equal to the required nutrients in E6:E8. Be sure to change the relation in the center to >=. Then click on this box s Add button to save this constraint and get ready to add another one. This time, it s the non-negativity constraints. B2 and C2 must be >= 0. Click on OK once this box is filled in as shown, because we re done adding constraints.

7 LINEAR PROGRAMMING II 7 Here s what the Solver Parameters box should look like when you re ready to solve. Click on Solve. Done! A Solver Results dialog box pops up in front of your spreadsheet. In the Reports section of the Solver Results box, select the Answer and Sensitivity Reports. Then click OK. You will then see that Excel has filled in the best solution, as shown to the right. This says that we satisfy the requirements for the least possible cost with 9 bags of X and 3 bags of Y. The cost is \$30.

9 LINEAR PROGRAMMING II 9 One edge of the feasible region is now parallel to the iso-cost lines. The diagram to the right shows how the lowest iso-cost line that touches the feasible region touches it along a whole line segment, not just at one corner. Excel s solver does not directly tell you that there are multiple solutions. Instead, it seems to pick one solution arbitrarily. When I solved this with Excel, I got an arbitrary value near the middle of this line segment. This was a true solution, but it was only one of many. Any combination of X and Y that sum to 12 from (2,10) to (9,3) is an optimum. I solved again in Excel, this time first clicking the Options button and checking the Assume Linear Model box. Excel came back with the (9,3) solution, which is at one end of the line segment of optimal solutions. The senstivity report, shown here, does have a subtle indication that something is up: The Shadow Price for nutrient C is \$0, even though there is no slack in nutrient C, the amount of C provided being equal to the amount required. The \$0 shadow price for C means that the C constraint is not really active at the (9,3) solution. Ordinarily in a linear programming problem, the number of active constraints (constraints with positive shadow prices) equals the number of variables. We have two variables, X and Y, so ordinarily we would have two active constraints. The fact that we have only one active constraint indicates that we have parallelism, and that the solution displayed is one among many.

12 LINEAR PROGRAMMING II 12 To get the logic behind this spreadsheet, consider row 6, for example. It's the constraint for Monday. The total number of people on the job on Monday equals the number of people who work Monday-Friday, plus Thursday-Monday, Friday-Tuesday, Saturday-Wednesday, and Sunday-Thursday. Row 6 has 1's in those columns. People who work Tuesday-Saturday and Wednesday-Sunday are not working Monday, so Row 6 has 0's in those columns. Cell I6's SUMPRODUCT formula calculates the total number of people working Monday. It must be greater than or equal to what is in J6. An alternative way to set up the problem is: The formulas in Column E do what the first layout does with 0's and 1's. The idea behind them is, for example, that the number of staff working Monday equals total number of people who started their 5-day work week on Monday, Thursday, Friday, Saturday, and Sunday. The constraints are that each of the formulas in the E column has to be greater than or equal to the corresponding Need amount in the F

13 LINEAR PROGRAMMING II 13 column. Both methods also need the non-negativity constraints, that the cells containing the amounts to hire are all >= 0. B. Transportation You are distributing pharmaceuticals from two locations. Location A has 100 units. Location B has 250 units. You have customers at locations C, D, and E, who want 50, 120, and 180 units respectively. Minimize cost, based on these unit transportation costs: From: A B To: C \$5 \$3 D \$10 \$4 E \$12 \$9 Here is a spreadsheet layout for this transportation problem: The key insight is that the variables are the six possible routes, that is, the six possible combinations of the two sources and the three destinations. The constraints are a mixture of >= and <=. Be careful when you enter these to be sure that they are all in the right directions. The constraints are: H5:H6 <= I5:I6 H7:H9 >= I7:I9 B2:G2 >= 0 If Solver gives you an error message that the solution is unbounded, check the directions of your > and < signs. Also check that you did include the non-negativity constraint, that the amounts are all >= 0.

14 LINEAR PROGRAMMING II 14 The dual illustrated Every linear programming problem has a "dual" -- a restatement of the problem that switches the roles of the constraints and the activities. The original problem is called the "primal." Here is the desks and tables example, showing the primal and the dual. The numbers are the same in the primal and the dual, but the columns in one are the rows in the other, the constraint constants in one are the coefficients of the objective function in the other, and the directions of the inequalities are switched. In the primal version of the desks and tables problem we maximize the objective function. We minimize the objective function in the dual. Primal: The constraints in the primal are: D6:D8 <= E6:E8 B2:C2 >= 0. The objective is to maximize D3. Dual: The constraints in the dual are: E5:E6 >= F5:F6 B2:D2 >= 0. The objective is to minimize E3. In the dual you are doing some accounting. You are allocating the profit value of each finished product (desk or table) to the inputs used to make it. These inputs' accounting values tell you how much you should be willing to pay if you can buy more inputs. These accounting values are the shadow prices of the inputs. Because they come from the Dual, they are sometimes called dual values. The E5 >= F5 constraint in the dual says that you must choose values for the three inputs such that the total value of the inputs into a desk is at least as great as \$3, the value of a desk. The E6 >= F6 constraint in the dual says that the values you put on those three inputs must also be such that the total value of the inputs that go into a table is at least as big as \$4, the value of a table. The dual's objective function says that you want to chose values that minimize the total value of your initial stock of available resources. A mind-bender, isn't it? The solutions are on the next page.

15 LINEAR PROGRAMMING II 15 The primal s solution: The primal s Solver Sensitivity Report The dual s solution: The dual s Solver Sensitivity Report The primal s solution amounts (4 and 6) are the shadow prices in the dual. Likewise, the dual s solution values (0,.667, and 1) are the shadow prices in the primal. The simplex method solves the primal and the dual simultaneously. This is how Excel gets the numbers it puts in the Sensitivity Report. Your turn:

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