Minimizing Problems and Problems 

 Where Zero Production Is Not Allowed 



The preceding example was a case where zero production was 

 allowed under the given constraints, and the objective function was to 

 be maximized. A common L.P. problem is one where zero production 

 is not allowed by the constraints and where the objective function is to 

 be minimized. 



Minimizing a positive objective function is nothing more than max- 

 imizing a negative objective function. If an objective function, Z = 

 SXj -{- 4X2, is to be minimized, it is the same as maximizing — Z r= 

 — 8X^ — 4X->. In this case the bottom row of the beginning matrix would 

 contain: 



(X,) (Xo) (KPo)) 



— Z = *" 



indicating that the objective solution could be reduced by the amounts 

 shown for X^ and X2. Reducing by a negative amount, however, means 

 adding to or increasing the objective solution, which would defeat the 

 minimizing objective. This objective solution can be reduced further 

 only if a positive value appears in the bottom row, which leaves the pre- 

 viously stated rule (see page 3, Pivot Column) unchanged. The ob- 

 jective solution will naturally be a negative, but with a simple sign 

 change the objective solution becomes an acceptable value. For ex- 

 ample, if — Z = — 85, then Z = 85. 



This minimizing case is directly related to the case where zero ac- 

 tivity »is not allowed. The first case requires the second, for minimizing 

 a problem where zero activity is allowed leads to the trivial solution of 

 zero, no activity. 



In the example problem, suppose that the second constraint were 



changed to: 



.5X + .25X2 ^ 3 



(greater than, or equal to). In order to make this into an equality a 

 slack variable must be subtracted from the left hand side of the in- 

 equality sign: .5X1 -f .25X2 — S2 = 3. When this equality is restated 

 in terms of the positive slack variable, a negative sign appears in the 



last term: 



02 ^^^^ .5Xj -|- .25X2 — o. 



