FM Lesson LP2

8/13/2019 FM Lesson LP2

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the Further Mathematics network

Lesson Notes Decision 1

Lesson 7 Linear Programming

Formulating a problem as a linear programming problem

First: Identify the variables about which a decision is to be made. These aresometimes called the decision variables. For example, if your problem is to decidehow many chairs to make and how many tables to make to maximise profitLet x = no. of chairsLet y = no. of tables.

If your problem is to work out how many grams of wheat germ and how many gramsof oat flour there should be in a new food product to meet nutrition requirements andminimise cost then:Let x = no. of grams of wheat germLet y = no. of grams of oat flour.

Next: Decide what the objective function is in terms of x and y (this is the value youare trying to maximise or minimise) and what the constraints are as inequalities(although sometimes equalities) involving x and y . Be careful to use unitsconsistently. Sometimes you could be given some units in metres and others incentimetres. Choose one type of units and convert everything into those units.

ExampleA clothing retailer needs to order at least200 jackets to satisfy demand over the nextsales period. He stocks two types of jacket which cost him 10 pounds and 30 poundsto purchase. He sells them at 20 pounds and 50 pounds respectively. He has 2700pounds to spend on jackets.

The cheaper jackets are bulky and each need 20cm of hanging space. The expensive jackets need only 10 cm each. He has 40m of hanging space.

Formulation as a linear program

The decision is about how many of two types of jacket need to be ordered.Let x = no. of cheaper jackets orderedLet y = no. of expensive jackets orderedThe profit, P, given by selling all of these, is P = 10 x + 20 y , since the profit made on acheaper jacket is 10 pounds and the profit made on an expensive one is 20 pounds.The constraints are

1. "needs to order at least 200" giving 200 x y+ 2. "cost him 10 pounds and 20 pounds" and "has 2700 pounds to spend" giving

10 30 2700 x y+ 3. "20cm of hanging space" and "10cm" and "has 40m of hanging space" giving

0.2 0.1 40 x y+


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SolutionDrawing the line representing a constraint.As an example, take the constraint 0.2 0.1 40 x y+ from the example over the page.

The initial aim is to draw the line 0.2 0.1 40 x y+ =

. We know this is a straight line soit's enough to find two points on the line and join them. When x = 0, y = 400 andwhen y = 0, x = 200. So the points (0, 400) and (200, 0) are on the line. Then shadeout the unacceptable region. To find the unacceptable region just test appoint out tosee if it satisfies the constraint or not. For example in this case, (10, 10) clearlysatisfies the constraint and so is in the acceptable region.

The feasible regionOnce you have drawn all the constraints, the feasible region is the intersection of theacceptable regions for all of them.

x

y

350 50

350

50

x

y

35050

350

50

FeasibleRegion


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Finding the solutionThe solution of the problem will be at one of the vertices of the feasible region. Youwill need to solve simultaneous equations to find the coordinates of these vertices.Then each vertex must be checked to find the best. For example, in the above we havea feasible region as in the diagram below. The coordinates of point A are found bysolving 200 x y+ = and 10 30 2700 x y+ = simultaneously. The solutions are x = 165and y = 35. So the point is (165, 35) and the profit at that point is P = 10 x + 20 y = 1650+ 700 = 2350.

A (165, 35) => P = 2350B (186, 28) => P = 2420C (200, 0) => P = 200

So B is the optimal point. Produce 186 cheap jackets and 28 expensive ones.

Integer ProgrammingIf th solution to a problem has to have integer values, then points with integer valuecoordinates close to the optimal point should be checked. This is likely to reveal theoptimal integer solution but is not guaranteed to. You should be careful to check thatsuck points actually are in the feasible region. Do this by substituting their coordinatesinto the constraints.

More Questions

1. A builder can build either luxury houses or standard houses ona plot of land. Planning regulations prevent the builder frombuilding more than 30 houses altogether, and he wants to build atleast 5 luxury houses and at least 10 standard houses. Eachluxury house requires 300 m 2 of land, and each standard houserequires 10 m 2 of land. The total Area of the plot is 6500 m 2.

A B

C


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2. A company manufactures two types of container, eachrequiring the same amount of material. The first type ofcontainer requires 4 seconds on a cutting machine and 3 secondson a sewing machine.

The second type of container requires 2 seconds on the cuttingmachine and 7 seconds on the sewing machine. Each machine isavailable for 1 hour. Thre first type of container gives a profit of40p. The second type gives a profit of 30p. How many of eachtype should be made to maximize profit?

3. A car parl with total usable area 300 m 2 is to have spaces

marked out for small cars and large cars. A small car space hasarea 10 m 2 and a large car space has area 12 m 2. The ratio ofsmall cars to large cars parked at any one time is estimated to bebetween 2:3 and 2:1. Find the number if spaces if each type thatshould be provided so as to maximise the number of cars that canbe parked.

FM Lesson LP2

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