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LINEAR PROGRAMMING
Linear programming is a mathematical technique designed to aid managers in allocating scarce resources (such as labor, capital, or energy) among competing activities. It reflects, in the form of a model, the organization's attempt to achieve some objective (frequently, maximizing profit contribution, maximizing rate of return, minimizing costs) in view of limited or constrained resources (available capital or labor, service levels, available machine time, budgets).
The linear programming technique can be said to have a linear objective function that is to be optimized (either maximized or minimized) subject to linear equality or inequality constraints and sign restrictions on the variables. The term linear describes the proportionate relationship of two or more variables. Thus, a given change in one variable will always cause a resulting proportional change in another variable.
Some areas in which linear programming have been applied will be helpful in setting the climate for learning about this important technique.
(i) A company produces agricultural fertilizers. It is interested in minimizing costs while meeting certain specified levels of nitrogen, phosphate, and potash by blending together a number of raw materials.
(ii) An investor wants to maximize his or her rate of return by investing in stocks and bonds. The investor can set specific conditions that have to be met including availability of capital.
(iii) A company wants the best possible advertising exposure among a number of national magazines, and radio and television commercials within its available capital requirements.
(iv) An oil refinery blends several raw gasoline and additives to meet a car manufacturer's specifications while still maximizing its profits.
(v) A city wants to maximize the daytime use of recreational properties being proposed for purchase with a limited capital available.
This technique, called linear programming (L.P), is solved in a step-by-step manner called iterations. Each step of the procedure is an attempt to improve on the solution until the "best answer" is obtained or until it is shown that no feasible answer exists.
FORMULATION OF THE LINEAR PROGRAMMING PROBLEM
To formulate a real-life problem as a linear program is an art in itself. To aid you in this task, it is helpful to isolate the essential elements of the problem as a means of asking what the clients wants and what information can be gained from the data that has been provided.
The first step in formulating a problem is to set forth the objective called the objective function.
A second element of a problem is that there are certain constraints on the company's ability to maximize the total contribution. These constraints are:
(1) quantity of raw materials available, (2) the level of demand for the products, and (3) The equipment productive capacity.
A further element that must be considered in the problem is the time period being used. The duration may be either long term or short term. Although time is an important element, it is one that has flexibility so that the time horizon may be changed as long as the restrictions are compatible with the periods under consideration.
The last element is that every product has a likelihood of being made. These products are the dependent or decision variables. Of course, the likelihood of a variable's being in the answer may change with the price or contribution values (usually profit and the nature of the restraints). Yet, at this point there is nothing to indicate that differing chances of occurrence exists for the possibility of making each of the products.
The first stage of solving linear programming problems is to set forth the problem in a mathematical form by defining the variables and the resulting constraints. Generally, the relationship is fairly simple using only elementary algebraic notation. The relationships can be seen by first identifying the decision variables. To aid in using algebraic notation, the decision variables can be represented by symbols such as X, Y, Z.
Next, we must build the objective function. If the goal is to maximize profit, we identify our objective function as
Maximize total profit or Minimize total loss (cost).
Then we write problem constraints.
These steps are now illustrated by taking some examples.
Example 1: Product Mix
The Regal China Company produces two products daily plates and mugs. The company has limited amounts of two resources used in the production of these products clay and labor. Given these limited resources, the company desires to know how many plates to produce each day, in order to Maximize profit. The two products have the following resource requirements for production and profit per item produced (i.e., the model parameters).
Product Labor (hours/unit)
Clay (lbs./unit)
Profit (Rs./unit) Plate 1 4 4 Mug 2 3 5
There are 40 hours of labour and 120 pounds of clay available each day for production.
1X 1 + 2X 2 < 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. 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 4X 1 pounds, and since each mug requires three pounds of clay, the amount of clay used for mugs daily is 3X 2. Given that amount of clay available for production each day is 120 pounds, the material constraint can be formulated as
4X 1 + 3X 2 < 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
X 1 > 0, X 2 > 0
The complete linear programming model for this problem can now be summarized as
Maximize Z = Rs. 4X 1 + 5X 2 Subject to 1X 1 + 2X 2 < 40 4X 1 + 3X 2 < X 1 , X 2 > 0
The solution of this model will result in numerical values for X 1 and X 2 , which will maximize total profit, Z. As one possible solution, consider X 1 = 5 plates and X 2 = 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
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.
- However, the maximum profit.
Now consider a solution of X 1 = 10 plates and X 2 = 20 mugs, this would result in a profit of
Z = Rs. 4(10) + 5 (20) = 40 + 100 = Rs. 140
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 X 1 = 24 plates and X 2 = 8 mugs, with a corresponding profit of Rs. 136.
