Question

A pharmaceutical company has 100 kg of A, 180 kg of B and 120 kg of C ingredients available

per month. The company can use these materials to make three basic pharmaceutical products

namely 5-10-5, 5-5-10 and 20-5-10, where the numbers in each case represent the percentage

of weight of A, B and C, respectively in each of the products. Further, in each of the products,

the remaining ingredient (apart from the ingredients A, B and C) is an inert ingredient. The

cost of these raw materials is as follows:

Ingredient Cost per kg (Rs)

A 80

B 20

C 50

Inert ingredient 20

The selling price of these products are Rs 40.5, Rs 43 and 45 per kg, respectively. There is a

capacity restriction of the company for product 5-10-5, because of which the company cannot

produce more than 30 kg per month. Determine how much of each of the products the company

should produce in order to maximize its monthly profits. Solve the problem using Simplex

method.

Answer 1

Let P1, P2 and P3 be the three products to be manufactured. The data of the problem can

then be summarized in a table as follows:

Cost of P1 $=5\%\times80+10\%\times20+5\%\times50+80\%\times20=4+2+2.50+16=$ Rs 24.50 per kg

Cost of P2 $=5\%\times 80+5\%\times20+10\%\times50+80\%\times 20=4+1+5+16=$ Rs 26 per kg

Cost of P3 $=20\%\times80+5\%\times20+10\%\times50+65\%\times20=16+1+5+13=$ Rs 35 per kg

Let $x_{1},x_{2}$ and $x_{3}$ be the quantities to be manufactured (in kg) of P1, P2 and P3, respectively . The Linear Progression (LP) problem can then be formulated as;

Maximize (net profit) Z= (Selling price – Cost price)$\times$ (Quantity of product)

= (40.50 – 24.50) $\times$ 1 + (43 – 26) $\times$ 2 + (45 – 35) $\times$ 3 = 16$\times$ 1 + 17$\times$ 2 + 10

subject to the constraints

$\frac{1}{20}x_{1}+\frac{1}{20}x_{2}+\frac{1}{20}x_{3}\leq100$ or $x_{1}+x_{2}+x_{3}\leq2,000$

$\frac{1}{10}x_{1}+\frac{1}{20}x_{2}+\frac{1}{20}x_{3}\leq180$ or $2x_{1}+x_{2}+x_{3}\leq3,600$

$\frac{1}{20}x_{1}+\frac{1}{10}x_{2}+\frac{1}{10}x_{3}\leq120$ or $x_{1}+2x_{2}+2x_{3}\leq2,400$

$x_{1}\leq30$

and $x_{1},x_{2},x_{3}\ge0$

Standard form : We Introduce slack variables $S_{1}, S_{2}$ and $S_{3}$ to convert the given LP model into its standard form as follows:

Maximize Z = $16x_{1}+17x_{2}+10x_{3}+Os_{1}+Os_{2}+Os_{3}+Os_{4}$

subject to the constraints;

$x_{1}+x_{2}+4x_{3}+s_{1}=2,000$

$2x_{1}+x_{2}+x_{3}+s_{2}=3,600$

$x_{1}+2x_{2}+2x_{3}+s_{3}=2,400$

$x_{1}+s_{4}=30$ and

$x_{1},x_{2},x_{3},s_{1},s_{2},s_{3},s_{4}\ge0$

Solving using simplex method:

An initial basic feasible solution is obtained by setting $x_{1},x_{2},x_{3}=0$ . Thus, the initial solution shown in the table below is: $s_{1}=2,000, s_{2}=3,600, s_{3}=2,400, s_{4}=30$ and Max $Z=0$

Since $c_{2}-z_{2}=17$ in $x_{2}$ - column is the largest positive value, we apply the following row operations

in order to get a new improved solution by entering variable $x_{2}$ into the basis and removing variable $s_{3}$

from the basis.

R3 (new) $\to$ R3 (old) $\div$ 2 ( key element) R1 (new) $\to$ R1 (old) - R3 (new)

R2 (new) $\to$ R2 (old) – R3 (new)

The new solution is shown in the table below.

The solution shown in this table is not optimal because $c_{1}-z_{1}>0$ in $x_{1}$ - column. Thus, applying the

following row operations to get a new improved solution by entering variable $x_{1}$ into the basis and removing the variable $s_{4}$ from the basis, we get

R4 (new)$\to$ R4 (old) $\div$ 1 ( key element); R1 (new) $\to$ R1 (old) – ($\frac{1}{2}$ ) R4 (new)

R2 (new) $\to$ R2 (old) – ($\frac{3}{2}$ ) R4 (new); R3 (new) $\to$ R3 (old) – ($\frac{1}{2}$ ) R4 (new)

The new solution is shown in the table below.

Since all $c_{j}​−z_{j}​<0$ corresponding to non-basic variables columns, the current solution is an optimal

solution. Thus, the company must manufacture,$x_{1}$ = 30 kg of P1, $x_{2}$ = 1,185 kg of P2 and $x_{3}$ = 0 kg of

P3 in order to obtain the maximum net profit of Rs 20,625