Linear Programming Quiz Solution
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This document discusses solving a linear programming problem to maximize profit given two decision variables (number of newspaper and social media ads) and two constraints (budget and work hours). It presents the decision variables, objective function to maximize profit, constraints, graphing the feasible region, identifying the extreme points, and using simultaneous equations and the multiplication method to find the optimal solution where the constraints intersect at (76.92, 9.23). This maximizes profit of $326 given the budget of $240 and 100 work hours available.
Linear Programming Quiz Solution
- 1. LP Quizzes 4 to 8 Professor Ed Dansereau
- 2. Decision Variables The two decision variables are the number of social media and newspaper ads. Decision variables represent what management control. We can decide how many of each to run. X1 = Newspaper X2 = Social Media All other expressions (OF and Constraints are defined in terms of decision variables).
- 3. Objective Function Objective: Maximize Profit Objective Function: How much profit from each of our decision variables. $4 profit from Newspapers $2 profit from Social Media So Zmax = 4X1 + 2X2
- 4. Constraints Constraints are limitations, we can only dream of having unlimited resources such as time and money. For Budget Constraint We have a maximum of 240 dollars, we could use 200 dollars but can not spend 250. Our limitation is less than or equal to 240 dollars. Each newspaper ad consumes $3 and each social media $1 Budget → 3X1 + 1X1 <= 240
- 5. Constraints continued Work-Hour constraint A maximum of 100 hours are available. Each Newspaper consumes 1 hour and each social media 2.5 hours. X1 + 2.5X2 <= 100
- 6. Graph Each constraint is Linear - a straight line. To graph we take advantage of the straight lines and that the axis of any variable equals zero. The horizontal axis is X1 and the Vertical is X2. Budget X1 → set X2 = 0, X1 + 2.5(0) <= 100, solve for X1, X1 = 100 X2 → set X1 = 0, 1(0) + 2.5X2 = 100, X2 = 40 Work Hours
- 7. Graph
- 8. Feasible Region Feasible Region - the area limited by the constraints. Since our constraints are less than or equal to, the feasible region is area below both lines. Note that for most of the graph the binding constraint is the blue line, after the point (76.9, 9.2) the binding constraint is the red line. All LP problems assume greater than zero, so the axis represents the other boundaries (you would not do a project if your profits were negative). Feasible Region
- 9. Extreme Points Extreme Points are the points at the edges of the feasible region. With two constraints there are four extreme points, there can be more if there are additional constraints. Our extreme points are (the labels are arbitrary) A: (0,0) B: (0,40) C: (76.9, 9.2) Extreme Points
- 10. Optimal Solution Any point within the feasible area is a solution and will provide some measure of profit or at least break-even The Optimal Solution is the point that provides the maximum profit given our constraints. In Linear Programming it will always be one of the extreme points regardless if the objective is to maximizing profit or minimizing costs. Put Coordinates of extreme points into Objective Function to determine Maximum Profit or Minimum Cost (Max profit in our example). Zmax = 4X1 + 2X2 Za (0,0) = 4(0)+ 2(0) = 0 Zb (0,40) = 4(0) +2(40) =40 Zc (76.9, 9.2) = 4(76.9) + 2(9.2) = 326Optimal Solution
- 11. Binding and Slack A BINDING constraints limits the optimal solution. Slack is extra capacity. For example the Budget constraint has slack up until it intersects with the Work Hour constraint. After that point, the Budget constraint is binding.
- 12. Finding the intersection of two lines There are three ways to determine the intersection of two lines: 1. Estimate, take your best guess by looking at graph, for our example I would guess (77,10). 2. Simultaneous Solution 3. Mathematical Method
- 13. Simultaneous Solution 1. Write out the two linear lines, the constraint equations 2. Solve for one variable in terms of the other for first equation 3. Substitute solution into second equation, which will result a solution for one variable 4. Solve for second variable
- 14. Simultaneous Solution 1. Write out two linear lines a. Work Hours → X1 + 2.5X2 <= 100 b. Budget → 3X1 + X2 <= 240 2. Solve for one variable in terms of the other for first equation a. X1 + 2.5X2 = 100, so X1 = 100 - 2.5X2 b. Solve for any variable in either equation. I choose Work Hours and the variable X1 because the math was easy. With a coefficient of 1, (240-2.5X2)/1 = (240-2.5X2). You do not even need a calculator for that!
- 15. Simultaneous Solution 3. Substitute solution into second equation, which will result a solution for one variable a. Budget → 3X1 + X2 <= 240 b. Work Hours → X1 = 100 - 2.5X2 c. Sub work hours into Budget → 3(100-2.5X2) + X2 = 240 i. 300 - 7.5X2 +x2 =240 ii. -6.5X2 = 240 - 300
- 16. Simultaneous Solution 4. Solve for second variable X2 = 9.27 3X1 + X2 = 240 (either equation will work) 3X1 + 9.27 = 240 X1 = (240 - 9.27) / 3 X1 = 76.92, solution for second variable
- 17. Multiplication Method In this method we try to eliminate one variable but multiplying by an inverse (negative number). 1. Look at your two linear lines (constraints) and choose a variable to eliminate. I look for a number that is easy to multiple. 2. Multiple the other equation by the inverse. 3. Add two equations together and solve for first variable. 4. Solve for second variable
- 18. Multiplication Method 1. Look at your two linear lines (constraints) and choose a variable to eliminate a. Budget → 3X1 + 1X2 <= 240 b. Hours → 1X1 + 2.5X2 <= 100 c. The X1 in budget is three times the X1 in hours, so I choose to multiple the Work Hours constraint by inverse, or simply -3. 2. Multiple the other equation by the inverse a. Hours → -3(1X1 +2.5X2 = 100),
- 19. Multiplication Method 3. Add two equations together and solve for first variable. 3X1 + 1X2 = 240 + -3X1 - 7.5X2 = -300 = 0X1 - 6.5X2 = -60 X2 = (-60)/(-6.5) X2 = 9.23 (Does this number look familiar?)
- 20. Multiplication Method 4. Solve for second variable X2 = 9.23 Budget → 3X1 + 1X2 <= 240, you may choose either equation 3X1 + 9.23 = 240 X1 = (240 -9.23) / 3 X1 = 76.92
- 21. Finding the intersection of two lines Both the Simultaneous Solution and Mathematical Method produce the same solution for X1 and X2 (76.92, 9.23). These coordinates are the point on the graph where the two lines intersect. Yes, you need to learn both methods. We will use the Mathematical Method in the next section.