Chapter 6.8, Problem 11P

Explanation of Solution

a.

Solving LP using LINDO:

LP OPTIMUM FOUND AT STEP 4

OBJECTIVE FUNCTION VALUE

1)      3000.000

VARIABLE	VALUE	REDUCED COST
X1	200.000000	0.000000
X2	150.000000	0.000000
X3	0.000000	3.000000
X4	0.000000	0.500000

ROW	SLACK OR SURPLUS	DUAL PRICES
2)	0.000000	0.000000
3)	0.000000	3.500000
4)	0.000000	0.000000
5)	0.000000	4.500000

NUMBER OF ITERATIONS=4

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES

VARIABLE	CURRENT COEF	ALLOWABLE INCREASE	ALLOWABLE DECREASE
X1	9.000000	7.000000	1.000000
X2	8.000000	INFINITY	3...

Explanation of Solution

b.

Examples of Oddities 1-3:

Oddity 1: Row 3 has an $\text{AI}$ and $\text{AD}$ of 0.

Oddity 2: ${\text{x}}_{\text{4}}$ is a non-basic variable with Reduced Cost of $\text{.5}$. It turns out that if one increase ${\text{c}}_{\text{4}}$ by $\text{.6}$, we still have ${\text{x}}_{\text{4}}\text{=0}$