An Example Problem 



Given: the following inequality constraints: ^ 



2X1 + 3X2 ^ 24 

 .5X1 4- .25X0 ^ 3 

 Xi ^5 



and objective function: 



Z = 3X1 + 2X2 (max.) 



The inequality constraints are then changed into equalities by add- 

 ing or subtracting slack variables. 



2X1 + 3X2 + Si = 24 



.5X1 + .25X2 + S2 = 3 



Xi + Sh = 5 



In order to fit these equality constraints into this modified simplex 

 technique, they are restated in terms of the positive slack variables, and 

 set up in matrix form. 



Si = 



52 = 



53 = 



Si = 



52 = 



53 = 

 z = 



2X1 — 3X2 + 24 

 .5X1 — .25X2 + 3 



+ 5 



Restated 

 Constraints 



Xi 

 Xi 



X2 



KPoi 



Constraint 

 Matrix 



The box in the lower right-hand corner of the matrix is the oh- 

 jective solution. The initial solution is zero because, as the table is read, 

 there are 24 units of Sj, 3 units of So, and 5 units of S^ in the solution. 

 In other words, none of the resources have yet been used; therefore, 

 the objective solution equals zero. However, the first two figures in 

 the bottom row (objective function) indicate that the objective solu- 

 tion can be increased by three for each unit of Xj brought into solu- 

 tion and by 2 for each unit of X2 brought into solution. 



1 The other constraints generally assumed in LP problems are that the values 

 Xi, X2, Si, S2j etc., are non-negative; that is, the solution cannot contain negative 

 amounts of an activity. These constraints are assumed in the algorithm presented, 

 and are explicitly stated in the requirements from which the rules are developed 

 (see p. 22, requirement A). 



