Chapter 16, Problem 5P

A chemical plant makes three major products on a weekly basis. Each of these products requires a certain quantity of raw chemical and different production times, and yields different profits. Thepertinentin formation is in Table P16.5. Note that there is sufficient warehouse space at the plant to store a total of 450 kg/week.

TABLE P16.5

	Product 1	Product 2	Product 3	Resource Availability
Raw chemical	$7kg/kg$	$5kg/kg$	13kg/kg	$3000kg$
Production time	$0.005hr/kg$	$0.1hr/kg$	0.2hr/kg	$55hr/week$
Product	$\$30/kg$	$\$30/kg$	$\$35/kg$

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

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

(c) Solve the problem with a software package.

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

To determine

A linear programming problem for a chemical plant that makes three products on a weekly basis. The data is given below:

$\begin{array}{ccccc}& \text{Product 1}& \text{Product 2}& \text{Product 3}& \text{Resource Availability}\\ \text{Raw Chemical}& 7\text{ kg/kg}& 5\text{ kg/kg}& 13\text{ kg/kg}& 3000\text{ kg}\\ \text{Production Time}& \text{0}\text{.05 hr/kg}& \text{0}\text{.1 hr/kg}& 0.2\text{ hr/kg}& \text{55 hr/week}\\ \text{Profit}& \text{\$30/kg}& \$30\text{/kg}& \$35\text{/kg}& \end{array}$

Answer to Problem 5P

Solution:

The Linear Programming formulation is given as:

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

Explanation of Solution

Given Information:

Following data is given for a chemical plant that makes three products on a weekly basis:

$\begin{array}{ccccc}& \text{Product 1}& \text{Product 2}& \text{Product 3}& \text{Resource Availability}\\ \text{Raw Chemical}& 7\text{ kg/kg}& 5\text{ kg/kg}& 13\text{ kg/kg}& 3000\text{ kg}\\ \text{Production Time}& \text{0}\text{.05 hr/kg}& \text{0}\text{.1 hr/kg}& 0.2\text{ hr/kg}& \text{55 hr/week}\\ \text{Profit}& \text{\$30/kg}& \$30\text{/kg}& \$35\text{/kg}& \end{array}$

Also, the warehouse can store a total of $450\text{ kg/week}$.

Let $x_1,x_2,x_3$ be respectively the weights needed of Product 1, Product 2 and Product 3 to maximize profit and satisfy the provided conditions.

The Linear Programming Model can be set up as follows:

As Profit is to be maximized, the objective function is $\text{Maximize }C=30x_1+30x_2+35x_3$.

The constraints are:

The raw chemical constraint is $7x_1+5x_2+13x_3\le 3000$.

As the total production time must be equal to or less than $55\text{ hr}$. Thus, the time constraint is $0.05x_1+0.1x_2+0.2x_3\le 55$.

The storage available is $450\text{ kg/week}$. Thus, the storage constraint is $x_1+x_2+x_3\le 450$.

Also, the weights can never be negative. Thus, positivity constraint is $x_1\ge 0,x_2\ge 0,x_3\ge 0$

Hence, the Linear Programming formulation is given as:

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

To determine

To calculate: The solution of the linear programming problem given below:

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

Answer to Problem 5P

Solution:

The values of variables are $x_1=0,x_2=356.25,x_3=93.75$. The maximum $C=13968.75$.

Explanation of Solution

Given Information:

A linear programming problem,

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

Calculation:

Consider the provided linear programming problem,

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

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:

$\text{Maximize }C=30x_1+30x_2+35x_3-0\cdot S1+0\cdot S2+0\cdot S3$

$\begin{array}{c}7x_1+5x_2+13x_3+S1=3000\\ 0.05x_1+0.1x_2+0.2x_3+S2=55\\ x_1+x_2+x_3+S3=450\end{array}$

$x_1\ge 0,x_2\ge 0,x_3\ge 0,S1\ge 0,S2\ge 0\text{ and }S3\ge 0$

The above linear programming models consist of three non-basic variables $\left(x_1,x_2,x_3\right)$ and three basic variables $\left(S1,S2,S3\right)$.

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

$\begin{array}{cccccccccc}\text{Basis}& C& x_1& x_2& x_3& S1& S2& S3& \text{Solution}& \text{Minimum Ratio}\\ P& 1& -30& -30& -35& 0& 0& 0& 0& \\ S1& 0& 7& 5& 13& 1& 0& 0& 3000& 230.7692\\ S2& 0& 0.05& 0.1& 0.2& 0& 1& 0& 55& 275\\ S3& 0& 1& 1& 1& 0& 0& 1& 450& 450\end{array}$

The negative minimum, P is $-35$ and it corresponds to variable $x_3$. So, the entering variable is $x_3$

The minimum ratio is 230.7692 and it corresponds to basis variable S1. So, the leaving variable is S1.

Therefore, the pivot element is 13.

$\begin{array}{cccccccccc}\text{Basis}& C& x_1& x_2& x_3& S1& S2& S3& \text{Solution}& \text{Minimum Ratio}\\ P& 1& -11.1538& -16.5385& 0& 2.69231& 0& 0& 8076.923& \\ x_3& 0& 0.53846& 0.38462& 1& 0.07692& 0& 0& 230.7692& 600\\ S2& 0& -0.05769& 0.02308& 0& -0.01538& 1& 0& 8.846154& 383.3333\\ S3& 0& 0.46154& 0.61539& 0& -0.07692& 0& 1& 219.2308& 356.25\end{array}$

The negative minimum, P is $-16.5385$ and it corresponds to variable $x_2$. So, the entering variable is $x_2$.

The minimum ratio is 356.25 and it corresponds to basis variable S3. So, the leaving variable is S3.

Therefore, the pivot element is 0.61539.

$\begin{array}{cccccccccc}\text{Basis}& C& x_1& x_2& x_3& S1& S2& S3& \text{Solution}& \text{Minimum Ratio}\\ P& 1& 1.25& 0& 0& 0.625& 0& 26.875& 13968.75& \\ x_3& 0& 0.25& 0& 1& 0.125& 0& -0.625& 93.75& 750\\ S2& 0& -0.075& 0& 0& -0.0125& 1& -0.0375& 0.625& -50\\ x_2& 0& 0.75& 1& 0& -0.125& 0& 1.625& 356.25& -2580\end{array}$

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

Hence, the values of variables are $x_1=0,x_2=356.25,x_3=93.75$. The maximum $C=13968.75$.

To determine

To calculate: The solution of the linear programming problem given below using software package:

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

Answer to Problem 5P

Solution:

The values of variables are $x_1=0,x_2=356.25,x_3=93.75$. The maximum $C=13968.75$.

Explanation of Solution

Given Information:

A linear programming problem,

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

Calculation:

Consider the provided linear programming problem,

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

The solution can be obtained using Excel.

Set up the values and use the formula as shown below:

The values obtained are:

Now press Solver under Data tab and enter the constraints and objective as shown below:

The resulting solution is:

Hence, the values of variables are $x_1=0,x_2=356.25,x_3=93.75$. The maximum $C=13968.75$.

To determine

The factor among increasing raw material, production time or storage that will rise profits the most for the linear programming problem given below:

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

Answer to Problem 5P

Solution:

Increasing storage will result in the most profits.

Explanation of Solution

Given Information:

A linear programming problem,

$\text{Maximize }C=30x_1+30x_2+35x_3$

Subject to Constraints:

$7x_1+5x_2+13x_3\le 3000$

$0.05x_1+0.1x_2+0.2x_3\le 55$

$x_1+x_2+x_3\le 450$

$x_1\ge 0,x_2\ge 0,x_3\ge 0$

Calculation:

Open the Excel sheet of part (c), then Press Solver under Data and select Simplex LP as a solving method as shown below:

Press solve then select Sensitivity in reports as shown below:

The sensitivity report obtained is:

The high shadow price for storage implies that increasing storage will result in the most profits.