Introduction

Optimization of resources (cost and time) is required in every aspect of our lives. We need the optimization because we have limited time and cost resources, and we need to take maximum out of them. From manufacturing to resolving supply chain issues, every aspect of the business world today requires optimization to stay competitive.

Linear programming offers the most easiest way to do optimization as it simplifies the constraints and helps to reach a viable solution to a complex problem. In this article, we will solve some of the linear programming problems through graphing method.

 

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Exercise 1

A transport company has two types of trucks, Type A and Type B. Type A has a refrigerated capacity of 20 m^# and a non-refrigerated capacity of 40 m^3 while Type B has the same overall volume with equal sections for refrigerated and non-refrigerated stock. A grocer needs to hire trucks for the transport of 3,000 m^3 of refrigerated stock and 4, 000 m^# of non-refrigerated stock. The cost per kilometer of a Type A is $30, and $40 for Type B. How many trucks of each type should the grocer rent to achieve the minimum total cost?

Exercise 2

A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers available. The rental cost for a large bus is $800 and $600 for the small bus. Calculate how many buses of each type should be used for the trip for the least possible cost.

Exercise 3

A store wants to liquidate 200 of its shirts and 100 pairs of pants from last season. They have decided to put together two offers, A and B. Offer A is a package of one shirt and a pair of pants which will sell for $30. Offer B is a package of three shirts and a pair of pants, which will sell for $50. The store does not want to sell less than 20 packages of Offer A and less than 10 of Offer B. How many packages of each do they have to sell to maximize the money generated from the promotion?

 

 

Solution of exercise 1

A transport company has two types of trucks, Type A and Type B. Type A has a refrigerated capacity of 20 m^3 and a non-refrigerated capacity of 40 m^3 while Type B has the same overall volume with equal sections for refrigerated and non-refrigerated stock. A grocer needs to hire trucks for the transport of 3000 m ^3 of refrigerated stock and 4,000 m^3 of non-refrigerated stock. The cost per kilometer of a Type A is $30, and $40 for Type B. How many trucks of each type should the grocer rent to achieve the minimum total cost?

a) Choose the unknowns.

x = Type A trucks

y = Type B trucks

b) Write the objective function.

f(x,y) = 30x + 40y

c) Write the constraints as a system of inequalities.

40x + 30y \geq 4000

20x + 30y \geq 3000

x \geq 0

y \geq 0

d) Find the set of feasible solutions that graphically represent the constraints.

Example 1 - part d

e) Calculate the coordinates of the vertices from the compound of feasible solutions.

Example 1 - part e

f) Calculate the value of the objective function at each of the vertices to determine which of them has the maximum or minimum values.

f(0, 400/3) = 30 \cdot 0 + 40 \cdot \frac {400}{3}  = 5,333.33^ 2

f(150,0) = 30 \cdot 150 + 40 \cdot 0 = 4500

As x and y must be natural numbers round the value of y.

f(50,67) = 30 \cdot 50 + 40 \cdot 67 = 4180

By default, we see what takes the value x to y = 66 in the equation 20x + 30y = 3000 \cdot x = 51 which it is within the feasible solutions.

f(51,66) = 30 \cdot 51 + 40 \cdot 66 = 4170

The minimum cost is $4170. To achieve this 51 trucks of Type A and 66 trucks of Type B are needed.

 

Solution of exercise 2

A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and 8 buses of 40 seats, but only has 9 drivers available. The rental cost for a large bus is $800 and $600 for the small bus. Calculate how many buses of each type should be used for the trip for the least possible cost.

a)Choose the unknowns.

x = small buses

y = big buses

b) Write the objective function.

f(x,y) = 600x + 800y

c) Write the constraints as a system of inequalities.

40x + 50y \geq 400

x + y \leq 9

x \geq 0

y \geq 0

d)  Find the set of feasible solutions that graphically represent the constraints.

Example 2 - part d

e)  Calculate the coordinates of the vertices from the compound of feasible solutions.

Example 2 - part e

f)  Calculate the value of the objective function at each of the vertices to determine which of them has the maximum or minimum values.

 

f(0,8) = 600 \cdot 0 + 800 \cdot 8 = 6400

(0,9) = 600 \cdot 0 + 800\cdot 9 = 72000

f(5,4) = 600 \cdot 5 + 800 \cdot 4 = 6200

Hence, the minimum cost is $6200. This is achieved with 4 large and 5 small buses.

We substituted the points (0,9), (0,8), and (5,4) in the equation to determine the minimum cost. However, you can tell this by directly looking at the graph. The coordinate (5,4) comes under the feasible region and is the minimum point of it.

 

Solution of exercise 3

A store wants to liquidate 200 of its shirts and 100 pairs of pants from last season. They have decided to put together two offers, A and B. Offer A is a package of one shirt and a pair of pants which will sell for $30. Offer B is a package of three shirts and a pair of pants, which will sell for $50. The store does not want to sell less than 20 packages of Offer A and less than 10 of Offer B. How many packages of each do they have to sell to maximize the money generated from the promotion?

a) Choose the unknowns.

x = number of packages of Offer A

y = number of packages of Offer B

b) Write the objective function.

f(x,y) = 30x + 50y

c) Write the constraints as a system of inequalities.

x + y \leq 100x + 3y \leq 200

x \geq 20

y \geq 10

d) Find the set of feasible solutions that graphically represent the constraints.

Example 3 - part d

e)  Calculate the coordinates of the vertices from the compound of feasible solutions.

Example 3 - part e

f) Calculate the value of the objective function at each of the vertices to determine which of them has the maximum or minimum values.

f(x,y) = 20 \cdot 20 + 50 \cdot 10 = 1100

f(x,y) = 30 \cdot 90 + 50 \cdot 10 = 3200

f(x,y) = 30 \cdot 20 + 50 \cdot 60 = 3600

f(x,y) = 30 \cdot 50 + 50 \cdot 50 = 4000

50 packages of each offer generates a maximum amount of $4000 in sales.

 

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.