Linear Programming:Model Constraints, Ingredients Mixing

<< Linear Programming:Formulation of the Linear Programming Problem, Decision Variables

Linear Programming:VITAMIN CONTRIBUTION, Decision Variables >>

Operations Research (MTH601)

Formulate this problem as a linear programming model by defining each component of the model separately and

then combining the components into a single model.

Decision Variables

The decision confronting management in this problem is how many plates and mugs to produce. As such,

there are two decision variables that represent the number of plates and mugs to be produced on a daily basis. The

quantities to be produced can be represented symbolically as,

X1 = the number of plates to produce

X2 = the number of mugs to produce

The Objective Function

The objective of the company is to Maximize total profit. The company's profit is the sum of the individual

profits gained from each plate and mug. As such, profits from plates is determine by multiplying the unit profit for

each plate, Rs. 4, by the number of plates produced, X1. Likewise, profit derived from mugs is the unit profit of a

mug, Rs. 5, multiplied by the number of mugs produced, X2. Thus, total profit, Z, can be expressed mathematically

Maximize Z = 4X1 + 5X2

where

Z = total profit per day

Rs 4X1 = profit from plates

Rs 5X2 = profit from mugs

By placing the term Maximize in front of the profit function, the relationship expresses the objective of the

firm to Maximize total profit.

Model Constraints

This problem has two resources used for production, which are limited, labor and clay. Production of plates

and mugs require both labor and clay. For each plate produce, one hour of labor is required. Therefore, the labor

used for the production of plates is 1X1 hours. Similarly, each mug requires two hours of labor; the labor used for

the production of mugs is 2X2 hours. Thus, the labor used by the company is the sum of the individual amounts of

labor used for each product.

1X1 + 2X2

However, the amount of labor represented "1X1 + 2X2" is limited to 40 hrs per day, thus, the complete labor

constraint is

Operations Research (MTH601)

1X1 + 2X2 < 40 hours

The "less than or equal to ( < )" inequality is employed instead of an equality ( = ) because the forty hours

of labor is a maximum limitation that can be used, but not an amount that must be used, but not an amount that must

be used. This allows the company more flexibility in that it is not restricted to use the 40 hours exactly, but whatever

amount necessary to Maximize profit up to and including forty hours. This means that the possibility of "idle or

excess capacity" (i.e., the amount under forty hours not used) exists.

The constraint for pottery clay is formulated in the same way as the labor constraint. Since each plate

requires four pounds of clay, the amount of clay used daily for the production of plates is 4X1 pounds, and since

each mug requires three pounds of clay, the amount of clay used for mugs daily is 3X2. Given that amount of clay

available for production each day is 120 pounds, the material constraint can be formulated as

4X1 + 3X2 < 120 pounds

A final restriction is that the number of plates and mugs produced be either zero or a positive value, since it

would be impossible to produce negative items. These restrictions are referred to as nonnegative constraints and are

expressed mathematically as

X1 > 0, X2 > 0

The complete linear programming model for this problem can now be summarized as

Maximize

Z = Rs. 4X1 + 5X2

Subject to

1X1 + 2X2 < 40

4X1 + 3X2 < 120

X1, X2 > 0

The solution of this model will result in numerical values for X1 and X2, which will maximize total profit,

Z. As one possible solution, consider X1 = 5 plates and X2 = 10 mugs. First we will substitute this hypothetical

solution into each of the constraints in order to make sure that the solution does not require more resources than the

constraints show are available.

1(5) + 2(10) < 40

25 < 40

4(5) + 3(10) < 120

50 < 120

and

4(5) + 3(10) < 120

50 < 120

Thus, neither one of the constraints is violated by this hypothetical solution. As such, we say the solution is

feasible (i.e., it is possible). Substituting these solution values in the objective function gives Z = 4(5) + 5(10) = Rs.

70. However, the maximum profit.

Now consider a solution of X1 = 10 plates and X2 = 20 mugs, This would result in a profit of

Z = Rs. 4(10) + 5 (20) = 40 + 100 = Rs. 140

Operations Research (MTH601)

While this is certainly a better solution in terms of profit, it is also infeasible (i.e., not possible) because it

violates the resource constraint or labor:

1(10) + 2(20) < 40

50 < 40

Thus, the solution to this problem must both Maximize profit and not violate the constraints. The actual

solution to this model which achieves this objective is X1 = 24 plates and X2 = 8 mugs, with a corresponding profit

of Rs. 136.

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