The document summarizes the results of solving a linear programming problem with 2 variables and 6 constraints. The optimal solution was found at the first step with an objective value of 660. The values of the variables X1 and X2 at the optimal solution are 70 and 90, respectively. Ranges for the objective coefficients and right-hand side values are provided where the basis remains unchanged.

1. Rows= 6 Vars= 2 No. integer vars= 0 ( all are linear)
Nonzeros= 12 Constraint nonz= 7( 5 are +- 1) Density=0.667
Smallest and largest elements in absolute value= 1.00000 250.000
No. < : 3 No. =: 0 No. > : 2, Obj=MAX, GUBs<= 2
Single cols= 0
Optimal solution found at step: 1
Objective value: 660.0000
Variable Value Reduced Cost
X1 70.00000 0.0000000E+00
X2 90.00000 0.0000000E+00
Row Slack or Surplus Dual Price
1 660.0000 1.000000
2 0.0000000E+00 0.3333333
3 0.0000000E+00 2.333333
4 30.00000 0.0000000E+00
5 70.00000 0.0000000E+00
6 90.00000 0.0000000E+00
Variable ValueReducedCost
X1 70.00000 0.0000000E+00
X2 90.00000 0.0000000E+00
Row Slack or Surplus Dual Price
1 660.0000 1.000000

2. 2 0.0000000E+00 0.3333333
3 0.0000000E+00 2.333333
4 30.00000 0.0000000E+00
5 70.00000 0.0000000E+00
6 90.00000 0.0000000E+00
Ranges in which the basis is unchanged:
Objective Coefficient Ranges
Current Allowable Allowable
Variable Coefficient Increase Decrease
X1 3.000000 7.000000 0.5000000
X2 5.000000 1.000000 3.500000
Righthand Side Ranges
Row Current Allowable Allowable
RHS IncreaseDecrease
2 230.0000 270.0000 90.00000
3 250.0000 45.00000 135.0000
4 120.0000 INFINITY 30.00000
5 0.0 70.00000 INFINITY
6 0.0 90.00000 INFINITY
1]max=3*x1+5*x2;
2]2*x1+x2<=230;
3]x1+2*x2<=250;

3. 4]x2<=120;
5]x1>=0;
6]x2>=0;
7]
http://es.calameo.com/read/00114496451c6f9fe7a29