Mathematical formulation of L.P.P.

Mathematical formulation of an L.P.P. follows a fixed three-step working rule:

• Step 1 — identify the decision variables: assign symbols, typically $x,y$ (or $x_1,x_2$ for more than two

  quantities), to the unknown quantities the problem is asking to determine.

• Step 2 — identify the objective function: write the quantity to be maximized or minimized (profit, cost, distance,

  etc.) as a linear mathematical expression in terms of the decision variables just assigned.

• Step 3 — identify the constraints: express every resource limit, minimum requirement, or other restriction stated

  in the problem as a linear equation or inequation in the decision variables, and add the non-negativity constraints.

Solved Example 1 — Toy manufacturer (bicycles and tricycles). A manufacturer produces bicycles and tricycles, each

processed through Machine A (max. 120 hours/day) and Machine B (max. 180 hours/day). A bicycle needs 4 hours on A and 10

hours on B; a tricycle needs 6 hours on A and 3 hours on B. Profits are Rs.65 per bicycle and Rs.45 per tricycle.

Formulating: let $x$ = number of tricycles and $y$ = number of bicycles produced, so $x\ge0,\ y\ge0$. The total profit is

$z=45x+65y$, to be maximized. From the machine-hours table (6 hrs/tricycle + 4 hrs/bicycle on Machine A, capped at 120;

3 hrs/tricycle + 10 hrs/bicycle on Machine B, capped at 180), the remaining constraints are $6x+4y\le120$ and

$3x+10y\le180$. The formulated L.P.P. is: Maximize $z=45x+65y$ subject to $x\ge0,\ y\ge0,\ 6x+4y\le120,\ 3x+10y\le180$.

Solved Example 2 — Toy company (cutting and assembling). A company manufactures toys A and B. Toy A needs 2 minutes

cutting + 1 minute assembling; toy B needs 3 minutes cutting + 4 minutes assembling. There are 3 hours (=180 minutes)

available for cutting and 2 hours (=120 minutes) for assembling. Profit is Rs.10 per toy A and Rs.20 per toy B.

Formulating: let $x$ = toys of type A and $y$ = toys of type B, so $x\ge0,\ y\ge0$. Total profit

$P=10x+20y$, to be maximized. The cutting-time constraint is $2x+3y\le180$ and the assembling-time constraint is

$x+4y\le120$. The formulated L.P.P. is: Maximize $P=10x+20y$ subject to $x\ge0,\ y\ge0,\ 2x+3y\le180,\ x+4y\le120$.

Solved Example 3 — Horticulturist's fertilizer mix. A horticulturist wants to mix two fertilizer brands to provide at

least 15 units of potash, 20 units of nitrate, and 24 units of phosphate. Brand I (Rs.120/unit) supplies 3 units potash, 1

unit nitrate, 3 units phosphate per unit; Brand II (Rs.60/unit) supplies 1 unit potash, 5 units nitrate, 2 units

phosphate per unit. Formulating: let $x$ = units of Brand I and $y$ = units of Brand II, so $x\ge0,\ y\ge0$. The total

cost is $z=120x+60y$, to be minimized. Since each nutrient has a stated MINIMUM requirement, all three nutrient rows

become $\ge$ constraints: $3x+y\ge15$ (potash), $x+5y\ge20$ (nitrate), $3x+2y\ge24$ (phosphate). The formulated L.P.P. …