5. Linear Programming Objectives: 1.Problem formulation 2.Solving an LP problem graphically 3.Bounded regions and corner points Refs: B&Z 5.2.

See handout for description. Example 1 Our first task is to formulate a mathematical model of this problem. In this example, we are asked to determine the number of ads to place in two different newspapers. So let x be the number of ads placed in Yarra News y be the number of ads placed in Sun City Paper The variables x and y are called decision variables.

Our objective is to maximize the number of people who see our ad. The exposure depends on the number of people who read each newspaper. For each ad we place in Yarra News, we can guarantee that 50,000 people see it. As the number of ads placed in Yarra News is represented by x we see that 50,000x people see our ad.

For each ad we place in Sun City Paper we can guarantee that 20,000 people see it. As the number of ads placed in Sun City Paper is represented by y we can see that 20,000y people see this ad. So our exposure statement can be written as P=50,000x+20,000y This is called the objective function.

Obviously, making x and y as large as possible will increase the value of P. But there are bounds (or constraints) on the values of x and y. The first constraint concerns the total amount of money which the company can spend on advertising. We are told that there is an upper limit of $9,000. So we need to formulate an equation which represents the cost of advertising. 1. Each ad in Yarra News will cost $300. Since x represents the number of ads we place in Yarra News, we can represent this part of our cost by 300x cost per ad number of ads total cost of ads in Yarra News

2. Similarly, each ad in Sun City Paper will cost us $100. Since y represents the number of ads we place in Sun City Paper, we can represent this component of cost by 100y cost per ad number of ads total cost of ads in Sun City Paper So in total, the cost of advertising is 300x + 100y.

To satisfy the conditions of our problem, we must ensure that 300x + 100y ≤ 9,000. The second constraint concerns the number of ads we place each month. This can be described as x + y ≥ 30 number of ads in Y.N. number of ads in S.C.P. total number of ads not less than 30.

A final obvious (but necessary) constraint is that the number of ads we place in each paper cannot be negative. So x ≥ 0,y ≥ 0. These are callednon-negativity constraints. We are now ready to give a mathematical description of our problem. maximizeP=50,000x + 20,000y Subject to 300x + 100y ≤ 9,000 x + y ≥ 30 x ≥ 0 y ≥ 0

To solve such a problem we must firstly determine which pairs of points (x, y) satisfy all of the above constraints. x + y = x + 100y = 9,000 y x Any point in this region will satisfy all of our constraints. x = 0 y = 0

This region is called the feasible region and any point in it is called a feasible point. A feasible region is either bounded (which means it can be enclosed within a circle) or unbounded (it cannot be enclosed within a circle). Each point (x,y) in the feasible region will give us a value for P. Our task is to determine which point, or points, gives the maximum value of P on the feasible region. Let’s sketch the line 50,000x + 20,000y = P for various values of P. **remember this is our objective function

1. P = 800,000 50,000x + 20,000y = 800, P = 1,200,000 50,000x + 20,000y = 1,200, P = 1,600,000 50,000x + 20,000y = 1,600, P = 2,000,000 50,000x + 20,000y = 2,000,000 y x 30 90

Notice that we can write the equation P = 50,000x + 20,000y So we see that changing the value of P does not change the slope of the lines - they are parallel. Also note that as P increases the y-intercept increases. as y = P - 50,000x 20,000

Notice that the line 50,000x + 20,000y = 2,000,000 does not intersect the feasible region. Since increasing P will move the line further from the feasible region, we can see that our solution must have P< 2,000,000. The line 50,000x + 20,000y = 1,600,000 does intersect The feasible region - any point in the feasible region which is on this line will give P = 1,600,000. Since there are feasible points to the right of this line we conclude that we can find an (x, y) which will give us a value of P > 1,600,000.

So now we have 1,600,000 < P* < 2,000,000 but how do we find P* - the maximum value of P ? Place a ruler on the line 50,000x + 20,000y = P (any P will do) and slide it along, parallel to this line (in the direction of increasing P) until you are just touching the edge of the feasible region. Draw the line. What has happened?

y x (x, y) = (0, 90) This line determines the maximum value of P and intersects The feasible region at one point only. To determine this value of P we simply evaluate the objective function at the point of intersection (x, y).

Point of intersection. Easy in this case because it is on the axis. x = 0,y = 90. At this point P = 50,000x + 20,000y = 50,000(0) + 20,000(90) =1,800,000. So the optimal solution is (x, y) = (0,90) and max P = 1,800,000.

You may now do Questions 1, 2 and 3 Example Sheet 2 From the Orange Book.