How to Formulate the given Linear programming program

Question

Chemlabs produce the domestic cleaning solutions $A$ and $B$ by processing the raw materials for $I$ & $II$. The processing of $1$ unit of raw material $I$ costs $\text{Rs. }80$ and produces $0.5$ units of solution $A$ and $0.5$ units of solution $B$. Moreover, the processing of $1$ unit of raw material $II$ costs $\text{Rs. }50$ & produces $0.4$ units of solution $A$ and $0.6$ units of solution $B$. The daily demand for solution $A$ lies between $10$ and $15$ units and that for solution $B$ lies between $12$ & $20$ units. Formulate the L.P.P.

Answer:

Let $x$ and $y$ be no. of units of raw material $I$ and $II$ to be processed respectively.

The objective function is $$\min z = 80x +50y.$$

Since the demand for solution $A$ lies between $10$ and $15$ units,

$$ 10 \le 0.5 x + 0.4x \le 15 \tag{1}$$

Since the demand for solution $B$ lies between $12$ and $20$ units,

$$ 12 \le 0.5 y+ 0.6 y \le 20 \tag{2}$$

Total LPP is $$\min z = 80x+50y $$ subject to

$$ 10 \le 0.9x \le 15$$ $$ 12 \le 1.1y \le 20$$ $$x, y \ge 0.$$

Is this formulation correct? Or I misunderstood the problem.

Answer 1

If I represent the given constraints using two tables, the formulation for this cost minimization problem will be clear.

\begin{array}{crrrr} \text{raw material} & \text{cost per unit} & \text{product $A$} & \text{product $B$} & \text{number of units processed} \\ I & \text{Rs. 80} & 0.5 & 0.5 & x \\ II & \text{Rs. 50} & 0.4 & 0.6 & y \end{array}

From the above table, we can see that the amount of solutions $A$ and $B$ produced are $0.5x + 0.4y$ and $0.5x + 0.6y$ respectively.

\begin{array}{crrrr} \text{product} & \text{number of units produced} & \text{min. daily demand} & \text{max. daily demand} \\ A & 0.5x + 0.4y & 10 & 15 \\ B & 0.5x + 0.6y & 12 & 20 \end{array}

Hence, the LPP should be formulated as

$$\min z= 80x + 50y$$ subject to \begin{align} 0.5x + 0.4y &\ge 10 \\ 0.5x + 0.4y &\le 15 \\ 0.5x + 0.6y &\ge 12 \\ 0.5x + 0.6y &\le 20. \end{align}