Chapter 13-1 Notes Page 1

Transcription

1 Chapter 13-1 Notes Page 1 Constrained Optimization Constraints We will now consider how to maximize Sales Revenue & Contribution Margin; or minimize costs when dealing with limited resources (constraints). We will first discuss using the graphical approach to solving these constrained optimization problems. Then we will discuss how to use Excel to solve these problems. The first thing you need to do is to state your Objective Function. This is your goal (the object of your analysis). The Objective Function should state your goal (e.g., maximize or minimize something) along with a mathematical formula for the item that you are maximizing or minimizing (e.g., Revenue, Contribution Margin or Cost). The Objective Function Variables should be your alternative courses of action (e.g., the number of each type of product that you will produce). The Objective Function is: For example, assume that Karloff Bolt Company makes two types of bolts, Bolt A and Bolt B. Bolt A has a Unit Contribution Margin of 10, and the Contribution Margin per Unit for Bolt B is 12. You want to know how many units of each bolt Karloff should produce in order to maximize its total Contribution Margin. Maximize.10A and.12b where A is the number of A Bolts that you will produce and B is the number of B Bolts that you should produce. A and B are the Objective Function Variables. Without more information, Karloff would produce an infinite number of each type of bolt, and thereby make an infinite total Contribution Margin. In reality, there is a limit to the number of bolts that can be produced.

2 Chapter 13-1 Notes Page 2 Let us assume that each bolt must pass through the following three machines, and the time required on each machine differs, as shown in the table below: Machine I Machine II Machine III A Bolt.1 min.1 min.1 min B Bolt.1 min.4 min.5 min In a day, there are 240, 720, and 160 minutes available, on Machine I, Machine II and Machine III, respectively. Karloff wants to know how many of each type of bolts it should produce in a day. We now have to add these constraints to our Objective Function: Description Problem This is the Objective Function Max.10(A) +.12(B) These are the constraints: subject to: You only have 240 minutes available on Machine I..1A +.1B 240 You only have 720 minutes available on Machine II..1A +.4B 720 You only have 160 minutes available on Machine III..1A +.5B 160 You can't manufacture a negative number of bolts. A, B 0 Next, you have to graph the area of production that meets all of these constraints (the Feasible Region or Area). The graph should consist of a Cartesian axis with the Objection Function Variables as the x-axis and the y-axis. For example, the last constraint alone means that you are dealing with the top right quarter of the Cartesian axis: The wiggly line represents that the area goes out to infinity. You now want to graph the other constraints, which are inequalities. They each include the equal sign combined with either a greater than ( ) or lesser than ( ) symbol. The area covered by such an inequality consists of a line that divides Cartesian plane and

3 Chapter 13-1 Notes Page 3 one side of that line. The formula of the line that divides the Cartesian plane is the inequality formula with an equal sign (=) substituted for the inequality (, ). The way to graph such an inequality is to first graph the line that divides the Cartesian plane and then chose the side of the line that satisfies the inequality. To graph the first constraint (.1A+.1B 240), first graph the line,.1a+.1b=240. The easiest way to graph the line is identify the points where it crosses each axis. We know that B is zero on any point on the A-axis. Therefore, the point where the line crosses the A-axis has zero as the B coordinate. So, we replace B with zero in the equation and solve for A:.1A +.1B = 240.1A +.1(0) = 240.1A = 240 A = 240/.1 A = 2400 So, we know that the line crosses the A-axis at (2400, 0). We now want to see where the line crosses the B-axis. We know that A is zero at any point on the B-axis. Therefore, the point where the line crosses the B-axis has zero as the A coordinate. So, we replace A with zero in the equation and solve for B:.1A +.1B = 240.1(0) +.1B = 240.1B = 240 B = 240/.1 B = 2400 So, we know that the line crosses the B-axis at (0, 2400). With these two points, you can now graph the line.

4 Chapter 13-1 Notes Page 4 Now, we need to pick the side of the line that satisfies this inequality. The easiest way to do this is to test one point on one side of the line. You test a point by substituting the coordinates into the inequality formula, and check to see if coordinates produce a value that satisfies the inequality. If the point that is tested satisfies the inequality, then every point on the same side of the line as the point tested will satisfy the inequality. It is easiest to use the origin (0,0) to test whether the inequality is true because you are dealing with zeros as the variables. So, plug the origin into the inequality and see if the inequality is true for that point:.1a +.1B 240 Inequality.1(0) +.1(0) 240 Test the Origin True Statement So, we know that the Feasible Region for the first constraint includes the side of the line that contains the origin. The Feasible Region that satisfies the first and last constraint consists of the following: To graph the second constraint (.1A+.4B 720), we want to graph the line,.1a+.4b=720. We know that this line crosses the A-axis at (0, 7200):.1A +.4B = 720.1A +.4(0) = 720.1A = 720 A = 720/.1 A = 7200

5 Chapter 13-1 Notes Page 5 We know that this line crosses the B-axis at (1800, 0):.1A +.4B = 720.1(0) +.4B = 720.4B = 720 B = 720/.4 B = 1800 By testing the origin, we see that the side that contains the origin satisfies the inequality:.1a +.4B 720 Inequality.1(0) +.4(0) 720 Test the Origin True Statement So, we know that the Feasible Region for the second constraint includes the side of the line that contains the origin. The Feasible Region that satisfies the first, second and last constraints consists of the following: To graph the third constraint (.1A+.5B 160), we want to graph the line,.1a+.5b=160. We know that this line crosses the A-axis at (0, 1600):.1A +.5B = 160.1A +.5(0) = 160.1A = 160 A = 160/.1 A = 1600

6 Chapter 13-1 Notes Page 6 We know that this line crosses the B-axis at (320,0):.1A +.5B = 160.1(0) +.5B = 160.5B = 160 B = 160/.5 B = 320 By testing the origin, we see that the side that contains the origin satisfies the inequality: 1A +.5B 160 Inequality 1(0) +.5(0) 160 Test the Origin True Statement So, we know that the Feasible Region for the third constraint includes the side of the line that contains the origin. The Feasible Region that satisfies all of the constraints is the following: The fact that the Feasible Region no longer touches the lines for the first and second constraints tells you that the first and second constraints are not binding. The time on Machines I and II could be unlimited and it would not affect our production possibilities. Once, you have identified the area that satisfies all of the constraints, you have to decide which points within the Feasible Region maximize your Contribution Margin. The corners points of the Feasible Region are the most extreme points, and therefore, they represent the production levels that produce the most extreme Contribution Margins (e.g., the highest and lowest Contribution Margins).

7 Chapter 13-1 Notes Page 7 In order to find the highest Contribution Margin, you plug each of these corner points into the original Objective Function [.10(A)+.12(B)] and check to see which corner point produces the highest Contribution Margin: Corner.10(A)+.12(B) Contribution Margin (1600, 0).1 (1600) +.12 (0) = = $ (0, 320).1 (0) +.12 (320) = = (0, 0).1 (0) +.12 (0) = = 0 The firm will generate the highest Contribution Margin by producing 1600 A Bolts and no B Bolts. Using Excel To Solve Constrained Optimization Problems We can solve these Constrained Optimization Problems using Excel. You need to have one cell on your Excel spreadsheet represent the value of each of your two Objective Function Variables (number of A Bolts and number of B Bolts). Excel will place the optimal value in these cells after it solves the problem. For now, give A Bolts a value of 10 and give B Bolts a value of 20 so that you can tell if your other formulas are correct: Cell Label Cell Contents A Bolts 10 B Bolts 20 You also need to have another cell represent the value of the Objection Function formula using the values in your Objective Function Variable cells: Cell Label Objective Function Cell Contents =.1*[A Bolt Cell]+.12*[B Bolt Cell] For your constraints, you need to have a separate cell for each constraint. The cell will contain the value of the side of the constraint inequality formula that uses the Objective Function Variables (e.g., the time on each machine uses the number of each type of bolt produced). The side of the constraint inequality formula containing the fixed value will be supplied later: Cell Label Time on Machine I Time on Machine II Time on Machine III Cell Contents =.1*[A Bolt Cell]+.1*[B Bolt Cell] =.1*[A Bolt Cell]+.4*[B Bolt Cell] =.1*[A Bolt Cell]+.5*[B Bolt Cell]

8 Chapter 13-1 Notes Page 8 Your spreadsheet should look something like the following: Now, select Tools on the Menu Bar, and then click on Solver. (If you do not see Solver, select Add-Ins and then select Solver Add-In.) Once you select Solver, the following dialogue box will open: For Set Target Cell, click on the cell that contains the Objective Function. For Equal To, click whether you want to maximize the Objective Function or minimize it. For By Changing Cells, highlight the two cells that contain the Objective Function Variables. For Subject to the Constraints:, click on Add and the following dialogue box will open: For each constraint: (i) the left box should name the cell that contains the portion of the constraint inequality that contains the Objective Function Variables; (ii) the middle box

9 Chapter 13-1 Notes Page 9 should give the proper inequality symbol; and (iii) the right box should contain the fixed value portion of the constraint inequality. For each non-negative constraint (e.g., A O): (i) the left box should name the cell that contains the Objective Function Variable; (ii) the middle box should have >= as the inequality symbol; and (iii) the right box should contain 0 as the fixed amount. In our example, we have the following constraints: Once you have entered all of this information, click on Solve. Excel will change the values in the Objective Function Variable cells to reflect their optimal value. You can see that Excel gives the same answer that we got using the graphical method (produce 1600 A Bolts and zero B Bolts): A Solver Results dialogue box will open. Leave Keep Solver Solution checked. Also, under Reports click on Answer and Sensitivity. When you click OK, two new worksheets will be created on your spreadsheet, the Answer Report 1 sheet and the Sensitivity Report 1 sheet. The Answer Report 1 sheet contains a description of your constraints, and indicates which ones were binding : Answer Report 1 indicates that the constraints involving Machines I and II are not binding. We noted the same thing when we used the graphical approach. Notice that

10 Chapter 13-1 Notes Page 10 Excel says that restricting the B Bolts to a non-negative answer was binding. This indicates that without this constraint, Excel would have given a negative number for the number of B Bolts to be produced. A negative B Bolt value would allow you to produce more A Bolts. For example, if B could be (-1), then you could obtain an extra 30 seconds on Machine III. This would allow you to produce 5 more A Bolts and stay within the constraint:.1a+.5b 160 Constraint.1(1600) +.5(0) = 160 Optimal Value.1(1605) +.5(-1) = 160 Allowing B to be (-1) The negative one B Bolt would cost you 12 (a negative Contribution Margin) but the five A Bolts would produce an additional Contribution Margin of 50. Thus, if you could go negative, the next negative B Bolt would produce a marginal profit of 38. This potential is reported as the Reduced Gradient on Sensitivity Report 1: The Lagrange Multiplier is the Shadow Price of a minute on Machine III. It reports that you would be willing to pay up to $1 for an extra minute on Machine III. You can calculate this yourself. We know that each A Bolt takes.1 minute on Machine III. You can, therefore, produce 10 A Bolts with an extra minute on Machine III. Each A Bolt produces a Contribution Margin of 10. Thus, an extra minute on Machine III, which allows us to produce 10 A Bolts is worth $1 (10 x 10) to us.