We will now construct the dual to Example : Primal Problem:
Interpretation of Dual Constraints
The first inequality states, that the time to produce product A on machine 1 (two hours) times the opportunity cost per hour of using machine 1 (u1) plus the time to produce product A on machine 2 (one-half hour) times the cost per hour of using machine 2 (u2) is greater than or equal to $16. The $16 is the net profit of a unit of A (see the profit function of the primal). Thus, the opportunity cost of producing A is going to be either equal to the net profit (in which case A will be produced) or greater than the net profit (in which case no units of A will be produced and the resources will be used elsewhere to attain higher profit).
The interpretation of the second constraint is similar. The total cost per unit of producing product B is 0.5u1, (cost of using machine 1) plus 1.5u2 (cost of using machine 2). The total cost per unit is equal to or greater than $6, where $6 is the net increase in profit per unit of product B. Thus the cost of producing B is going to be either equal to the net profit per unit of B (in which case B will be produced) or greater than the net profit (in which case no units of B will be produced).
It should be noted that the form of solution does not allow the opportunity costs of producing either product to be less than the incremental profit of the product. This is reasonable, since the value of the machine-hours is measured by the profit they can produce. To have the total costs less than the profit would imply that we should produce more units of the product; but if the product is produced to the limit of productive capacity, the costs of the last unit will be equal to the profit. The only time the costs will be greater than the incremental profit will be when it is not desirable to produce any units of the product. Remember that these are opportunity, not accounting, costs.
The constraint equations, expanded to include slack variables and artificial variables, are:
The large coefficient M is assigned to u5 and u6, in the expanded cost equation to drive these two variables from the solution; this is done because u5 and u6 are artificial variables. The final simplex solution to this dual LP see Dual Solution.