Chapter 6.4, Problem 6P

Explanation of Solution

Optimal solution of the diet problem:

• Number of iterations= 5
• The ranges in which the basis is unchanged:

Variable	Current coefficient	OBJ coefficient ranges allowable increase	Allowable decrease
BR	50.000000	Infinity	27.500000
IC	20.000000	18.333334	5.000000
CLA	30.000000	10.000000	30.000000
PC	80.000000	Infinity	50.000000

Row	Current RHS	Right-hand side ranges allowable increase	Allowable decrease
2	500.000000	250.000000	Infinity
3	6.000000	4.000000	2.857143
4	10.000000	Infinity	4.000000
5	8.000000	5.000000	Infinity

• As brownies and pineapple cheesecake have no zero reduced costs, so case 2 is applied to the states where,
  • “At least one variable whose objective function coefficient is changed has a reduced cost of zero”.
• In case 2 for each constraint j the ratio r_j is defined as follows:
  • If $\Delta b_j\ge 0$, then $r_j=\frac{\Delta b_j}{I_j}$
  • If $\Delta b_j\le 0$, then $r_j=\frac{-\Delta b_j}{D_j}$
• Here,
  • $\Delta b_j=$ Change in original objective function coefficient (c_j) for x_j
  • $I_j=$ Maximum allowable increase in c_j from LINDO output.
  • $D_j=$ Maximum allowable decrease in c_j from LINDO output.
• The 100% rule for objective function states that,
  • “If $\sum r_j\le 1$, then the current basis remains optimal”
  • Also,
  • “If $\sum r_j>1$, then the current basis may or may not be optimal”.
• If the current basis remains optimal, the value of the decision variables remains unchanged, but the optimal z-value may change.
• Now, the price of a bottle of soda decreases to $15\cancel{c}$ and the price of a piece of a cheesecake decreases to $60\cancel{c}$