Chapter 3, Problem 25RP

Explanation of Solution

Formulation of LP:

Let,

${\text{x}}_{\text{1}}$= Gallons (in 1000s) of wine-1 produced annually.

${\text{x}}_2$= Gallons (in 1000s) of wine-2 produced annually.

${\text{x}}_3$= Gallons (in 1000s) of wine-3 produced annually.

${\text{x}}_4$= Gallons (in 1000s) of wine-4 produced annually.

Objective function:

Maximize profit: $6\times {\text{x}}_1+12\times {\text{x}}_2+20\times {\text{x}}_3+30\times {\text{x}}_4$

Inventory constraint:

On average one can store only 50000 Gallons in the inventory as wine-1 needs only $1/3$ times of year one can produce wine 3 times in a year, one can produce wine-2 only once in a year, wine-3 only once in 2 years, wine-4 only once in 3.333years.

That is,

$\frac{1}{3}{x}_1+x_2+2x_3+3.333x_4\le 50$

If the one person is producing only wine-1, then in all, he/she can produce $50000\times 3$ gallons of wine-1 in a year; similarly, one can produce maximum 50000 gallons in a year of only wine-2 $1/2\times 50000$ gallons of only wine-3 and $(1/3.333)\times 50000$ gallons of only wine-4.

Total production constraint:

At most, 100000 gallons of wine can be produced each year as per low.

$x_1+x_2+x_3+x_4\le 100$

Labour hour’s constraint:

Total labour hours available are, 12000 hours,

$0.2x_1+0.3x_2+0.3x_3+0.5x_4\le 12$

Tank usage hour’s constraint:

Total tank usage hours available are 32000 hours

$0.5x_1+0.5x_2+x_3+1.5x_4\le 32$

Non-negativity constraints:

$\begin{array}{l}{x}_1\ge 0\\ x_2\ge 0\\ x_3\ge 0\\ x_4\ge 0\end{array}$

Here, the unit of decision variable in 1000’s of gallons, one can divide the right hand of the inequality constraints with 1000 to make it computationally easier for the solver to solve