Chapter 15, Problem 1P

A company makes two types of products, A and B. These products are produced during a 40-hr work week and then shipped out at the end of the week. They require 20 and 5 kg of raw material per kg of product, respectively, and the company has access to 9500 kg of raw material per week. Only one product can be created at a time with production times for each of 0.04 and 0.12 hr, respectively. The plant can only store 550 kg of total product per week. Finally, the company makes profits of $45 and $20 on each unit of A and B, respectively. Each unit of product is equivalent to a kg.

(a) Set up the linear programming problem to maximize profit.

(b) Solve the linear programming problem graphically.

(c) Solve the linear programming problem with the simplex method.

(d) Solve the problem with a software package.

(e) Evaluate which of the following options will raise profits the most: increasing raw material, storage, or production time.

To determine

The linear programming problem to maximize the profit if the company makes profits of $45 on each unit of A and $20 on each unit of B.

Answer to Problem 1P

Solution:

The linear programming problem to maximize the profit is,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

Explanation of Solution

Given Information:

Product A and B are produced during a 40-hr of work week.

Company has access to 9500 kg of raw material per week.

Company required 20 kg of raw material per kg of product A and 5 kg of raw material per kg of product B.

One product of A can be created in 0.04 hour and one product of B can be created in 0.12 hour.

Plant can store 550 kg of total product per week.

And, the company makes profits of $45 on each unit of A and $20 on each unit of B.

Assume $x_a$ be the amount of product A produced weekly and $x_b$ be the amount of product B produced weekly.

Therefore, total number of products is $x_a+x_b$.

But plant can store 550 kg of total product per week. Therefore, the storage constrain is,

$x_a+x_b\le 550$

Company required 20 kg of raw material per kg of product A and 5 kg of raw material per kg of product B. Therefore, total raw material per week is,

$\text{Raw material }=20x_a+5x_b$

But the company has access to 9500 kg of raw material per week. Therefore, raw material constraint is,

$20x_a+5x_b\le 9500$

One product of A can be created in 0.04 hour and one product of B can be created in 0.12 hour. Therefore, the total production time is,

$\text{Production time}=0.04x_a+0.12x_b$

Since, product A and B are produced during a 40-hr of work week. Therefore, the production time constraint is,

$0.04x_a+0.12x_b\le 40$

Since, the amount of product cannot be negative. Therefore, the positivity constraint is,

$x_a,x_b\ge 0$

Now, the company makes profits of $45 on each unit of A and $20 on each unit of B. Therefore, maximum profit is,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

To determine

To calculate: The solution of the linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

By the graphical method.

Answer to Problem 1P

Solution:

The maximum value of P is approximately $22,250$ at $x_a=450\text{ and }{x}_b=100$.

Explanation of Solution

Given Information:

The linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

Formula used:

The equation of a straight line is,

$y=mx+C$

Where, m is the slope and C is the intercept of y.

Calculation:

Consider the linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

Reformulate the constraints to straight lines by replacing inequality by equal sign and solving for $x_b$ as below,

$\begin{array}{c}{x}_b=1900-4x_a\\ x_b=333.33-0.33x_a\\ x_b=550-x_a\\ x_b=\left(\frac{1}{20}\right)P-2.25x_a\end{array}$

The above equations are the equation of straight lines and represent the constraints.

The value of P in the objective function $x_b=\left(\frac{1}{20}\right)P-2.25x_a$ is increased till it reaches the highest value that obeys all constraints.

Plot all the straight lines.

The graph obtained is,

Hence, the maximum value of P is approximately $22,250$ at $x_a=450\text{ and }{x}_b=100$.

To determine

The solution of the linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

By the Simplex method.

Answer to Problem 1P

Solution:

The values of variables are $x_a=450,x_b=100$. The maximum $P=22250$.

Explanation of Solution

Given Information:

The linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

Consider the provided linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

First convert the above problem to standard form by adding slack variables.

As the constraints are subjected to less than condition, non- negative slack variables are added to reach equality.

Let the slack variables be $S1\ge 0,S2\ge 0\text{ and }S3\ge 0$.

Thus, the linear programming model would be:

$\text{Maximize }P=45x_a+20x_b+0\cdot S1+0\cdot S2+0\cdot S3$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b+S1=9500\\ 0.04x_a+0.12x_b+S2=40\\ x_a+x_b+S3=550\\ x_a,x_b,S1,S2,S3\ge 0\end{array}$

The above linear programming models consist of three non-basic variables $\left(x_a,x_b\right)$ and three basic variables $\left(S1,S2,S3\right)$.

Now the apply the Simplex method and solve the above problem as:

Basic	$P$	$x_a$	$x_b$	$S_1$	$S_2$	$S_3$	Solution	Intercept
$P$	1	-45	-20	0	0	0	0
$S_1$	0	20	5	1	0	0	9500	475
$S_2$	0	0.04	0.12	0	1	0	40	1000
$S_3$	0	1	1	0	0	1	550	550

The negative minimum, P is $-45$ and it corresponds to variable $x_a$. So, the entering variable is $x_a$.

The minimum ratio is 475 and it corresponds to basis variable S 1. So, the leaving variable is S 1.

Therefore, the pivot element is 20.

Basic	$P$	$x_a$	$x_b$	$S_1$	$S_2$	$S_3$	Solution	Intercept
$P$	1	0	-8.75	2.25	0	0	21375
$x_a$	0	1	0.25	0.05	0	0	475	1900
$S_2$	0	0	0.11	-0.002	1	0	21	190.9090
$S_3$	0	0	0.75	-0.05	0	1	75	100

The negative minimum, P is $-8.75$ and it corresponds to variable $x_b$. So, the entering variable is $x_b$.

The minimum ratio is 100 and it corresponds to basis variable S 3. So, the leaving variable is S 3.

Therefore, the pivot element is 0.75.

Basic	$P$	$x_a$	$x_b$	$S_1$	$S_2$	$S_3$	Solution	Intercept
$P$	1	0	0	1.6667	0	11.6667	22250
$x_a$	0	1	0	0.0667	0	-0.3333	450	1900
$S_2$	0	0	0	0.0053	1	-0.1467	10	190.9090
$S_3$	0	0	1	-0.0667	0	1.3333	100	100

Since $P\ge 0$, optimal solution is obtained.

Hence, the values of variables are $x_a=450,x_b=100$. The maximum $P=22250$.

To determine

The solution of the linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

By the use of software.

Answer to Problem 1P

Solution:

The maximum profitis $22,250$ with $x_a=450\text{ and }{x}_b=100$.

Explanation of Solution

Given Information:

The linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

Use excel solver as below, to solve the linear programming,

Step 1: Enter the coefficients of $x_a$ and $x_b$ for each constraint as below,

Step 2: Use formulas in column D to find total are as below,

Step 3: click on Solver button under the Data Ribbon. Set the values in pertinent cells of Solver dialogue box as below:

Step 4: Press the solve button.

The result obtained as,

Hence, the maximum value profitis $22,250$ with $x_a=450\text{ and }{x}_b=100$.

To determine

The constraint among increasing raw material, storage or production time that gives the maximum profit.

Answer to Problem 1P

Solution:

The storage will give the maximum profit.

Explanation of Solution

Given Information:

The linear programming problem,

$\text{Maximize }P=45x_a+20x_b$

Subject to the constraints:

$\begin{array}{c}20x_a+5x_b\le 9500\\ 0.04x_a+0.12x_b\le 40\\ x_a+x_b\le 550\\ x_a,x_b\ge 0\end{array}$

To obtain the maximum profit, the shadow price should be high.

Use excel as below to find the shadow price by generating the sensitivity report,

Follow same steps up to the step 4 of part (d) then select the report as sensitivity as below,

The sensitivity report for the linear programming problem is as follows:

From the above sensitivity report, it is observed that the storage has a high shadow price.

Hence, the storage will give the maximum profit.