Chapter 15, Problem 2P

(a)

To determine

The linear programming problem to maximize the profit if the factors are given as below,

Resource	Regular	Premium	Supreme	Resource Availability
Raw gas	$7{\text{ m}}^{\text{3}}\text{/tonne}$	${\text{11 m}}^{\text{3}}\text{/tonne}$	${\text{15 m}}^{\text{3}}\text{/tonne}$	${\text{154 m}}^{\text{3}}\text{/week}$
Production time	$\text{10 hr/tonne}$	$\text{8 hr/tonne}$	$\text{12 hr/tonne}$	${\text{80 m}}^{\text{3}}\text{/week}$
Storage	$\text{9 tonnes}$	$\text{6 tonnes}$	$\text{5 tonnes}$
Profit	$\text{150/tonne}$	$\text{175/tonne}$	$\text{250/tonne}$

Answer to Problem 2P

Solution:

The linear programming problem to maximize the profit is,

$\text{Maximize }P=150x+175y+250z$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z\le 154\\ 10x+8y+12z\le 80\\ x\le 9\\ y\le 6\\ z\le 5\\ x,y,z\ge 0\end{array}$

Explanation of Solution

Given Information:

The factors are given as below,

Resource	Regular	Premium	Supreme	Resource Availability
Raw gas	$7{\text{ m}}^{\text{3}}\text{/tonne}$	${\text{11 m}}^{\text{3}}\text{/tonne}$	${\text{15 m}}^{\text{3}}\text{/tonne}$	${\text{154 m}}^{\text{3}}\text{/week}$
Production time	$\text{10 hr/tonne}$	$\text{8 hr/tonne}$	$\text{12 hr/tonne}$	${\text{80 m}}^{\text{3}}\text{/week}$
Storage	$\text{9 tonnes}$	$\text{6 tonnes}$	$\text{5 tonnes}$
Profit	$\text{150/tonne}$	$\text{175/tonne}$	$\text{250/tonne}$

Assume x be the amount of regular resource, y be the amount of premium resource and z be the amount of supreme resource

Therefore, total amount of raw gas is $7x+11y+15z$.

But the resource availability is ${\text{154 m}}^{\text{3}}\text{/week}$. Therefore, the material constrain is,

$7x+11y+15z\le 154$

Now, the total production time from the provided table is $10x+8y+12z$.

But the resource production time is ${\text{80 m}}^{\text{3}}\text{/week}$. Therefore, the time constrain is,

$10x+8y+12z\le 80$

Now, regular resource can store up to 9 tones. Therefore,

$x\le 9$

And, premium resource can store up to 6 tones. Therefore,

$y\le 6$

Also, supreme resource can store up to 5 tones. Therefore,

$z\le 5$

Since, the amount of resource cannot be negative. Therefore, the positivity constraints are,

$x,y,z\ge 0$

Now, the company makes profits of $\text{150/tonne}$ on regular resources, $\text{175/tonne}$ on premium resources and $\text{250/tonne}$ on supreme resources. Therefore, maximum profit is,

$\text{Maximize }P=150x+175y+250z$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z\le 154\\ 10x+8y+12z\le 80\\ x\le 9\\ y\le 6\\ z\le 5\\ x,y,z\ge 0\end{array}$

(b)

To determine

The solution of the linear programming problem,

$\text{Maximize }P=150x+175y+250z$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z\le 154\\ 10x+8y+12z\le 80\\ x\le 9\\ y\le 6\\ z\le 5\\ x,y,z\ge 0\end{array}$

By the Simplex method.

Answer to Problem 2P

Solution:

The values of variables are $x=0,y=6,z=2.6667$. The maximum $P=1716.7$.

Explanation of Solution

Given Information:

The linear programming problem,

$\text{Maximize }P=150x+175y+250z$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z\le 154\\ 10x+8y+12z\le 80\\ x\le 9\\ y\le 6\\ z\le 5\\ x,y,z\ge 0\end{array}$

Consider the provided linear programming problem,

$\text{Maximize }P=150x+175y+250z$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z\le 154\\ 10x+8y+12z\le 80\\ x\le 9\\ y\le 6\\ z\le 5\\ x,y,z\ge 0\end{array}$

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 $S1\ge 0,S2\ge 0,S3\ge 0,S4\ge 0\text{ and }S5\ge 0$.

Thus, the linear programming model would be:

$\text{Maximize }P=150x+175y+250z+0\cdot S1+0\cdot S2+0\cdot S3+0\cdot S4+0\cdot S5$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z+S1=154\\ 10x+8y+12z+S2=80\\ x+S3=9\\ y+S4=6\\ z+S5=5\end{array}$

The above linear programming models consist of three non-basic variables $\left(x,y,z\right)$ and five basic variables $\left(S1,S2,S3,S4,S5\right)$.

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

Basic	$P$	x	y	z	$S_1$	$S_2$	$S_3$	$S_4$	$S_5$	Solution	Intercept $\left(\frac{\text{solution}}{z}\right)$
$P$	1	-150	-175	-250	0	0	0	0	0	0
$S_1$	0	7	11	15	1	0	0	0	0	154	10.2667
$S_2$	0	10	8	12	0	1	0	0	0	80	6.66667
$S_3$	0	1	0	0	0	0	1	0	0	9	∞
$S_4$	0	0	1	0	0	0	0	1	0	6	∞
$S_5$	0	0	0	1	0	0	0	0	1	5	5

The negative minimum, P is $-250$ and it corresponds to variable z. So, the entering variable is z.

The minimum ratio is 5 and it corresponds to basis variable S5. So, the leaving variable is S5.

Therefore, the pivot element is 1.

Basic	$P$	x	y	z	$S_1$	$S_2$	$S_3$	$S_4$	$S_5$	Solution	Intercept $\left(\frac{\text{solution}}{y}\right)$
$P$	1	-150	-175	0	0	0	0	0	250	1250
$S_1$	0	7	11	0	1	0	0	0	-15	79	7.18182
$S_2$	0	10	8	0	0	1	0	0	-12	20	2.5
$S_3$	0	1	0	0	0	0	1	0	0	9	∞
$S_4$	0	0	1	0	0	0	0	1	0	6	6
z	0	0	0	1	0	0	0	0	1	5	∞

The negative minimum, P is $-175$ and it corresponds to variable y. So, the entering variable is y.

The minimum ratio is 2.5 and it corresponds to basis variable S2. So, the leaving variable is S2.

Therefore, the pivot element is 8.

Basic	$P$	x	y	z	$S_1$	$S_2$	$S_3$	$S_4$	$S_5$	Solution	Intercept $\left(\frac{\text{solution}}{S_5}\right)$
$P$	1	68.75	0	0	0	21.88	0	0	-12.5	1687.5
$S_1$	0	-6.75	0	0	1	-1.375	0	0	1.5	51.5	34.3333
y	0	1.25	1	0	0	0.125	0	0	-1.5	2.5	-1.66667
$S_3$	0	1	0	0	0	0	1	0	0	9	∞
$S_4$	0	-1.25	0	0	0	-0.125	0	1	1.5	3.5	2.33333
z	0	0	0	1	0	0	0	0	1	5	5

The negative minimum, P is $-12.5$ and it corresponds to variable S5. So, the entering variable is S5.

The minimum positive ratio is 2.33333 and it corresponds to basis variable S4. So, the leaving variable is S4.

Therefore, the pivot element is 1.5.

Basic	$P$	x	y	z	$S_1$	$S_2$	$S_3$	$S_4$	$S_5$	Solution
$P$	1	58.3333	0	0	0	20.83	0	8.33	0	1716.7
$S_1$	0	-5.5	0	0	1	-1.25	0	-1	0	48
y	0	0	1	0	0	0	0	1	0	6
$S_3$	0	1	0	0	0	0	1	0	0	9
$S_5$	0	-0.8333	0	0	0	-0.083	0	0.67	1	2.3333
z	0	0.83333	0	1	0	0.083	0	-0.67	0	2.6667

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

Hence, the values of variables are $x=0,y=6,z=2.6667$. The maximum $P=1716.7$.

(c)

To determine

The solution of the linear programming problem,

$\text{Maximize }P=150x+175y+250z$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z\le 154\\ 10x+8y+12z\le 80\\ x\le 9\\ y\le 6\\ z\le 5\\ x,y,z\ge 0\end{array}$

By the use of software.

Answer to Problem 2P

Solution:

The maximum profit is 1716.7 with $x=0,y=6,z=2.67$.

Explanation of Solution

Given Information:

The linear programming problem,

$\text{Maximize }P=150x+175y+250z$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z\le 154\\ 10x+8y+12z\le 80\\ x\le 9\\ y\le 6\\ z\le 5\\ x,y,z\ge 0\end{array}$

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

Step 1: Enter the coefficients of x, y and z for each constraint as below,

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

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

Step 4: Press the solve button.

The result obtained as,

Hence, the maximum value profit is 1716.7 with $x=0,y=6,z=2.67$.

(d)

To determine

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

Answer to Problem 2P

Solution:

The increasing production time will give the maximum profit.

Explanation of Solution

Given Information:

The linear programming problem,

$\text{Maximize }P=150x+175y+250z$

Subject to the constraints:

$\begin{array}{c}7x+11y+15z\le 154\\ 10x+8y+12z\le 80\\ x\le 9\\ y\le 6\\ z\le 5\\ x,y,z\ge 0\end{array}$

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 production time has a high shadow price.

Hence, the production time will give the maximum profit.