Question

Problem 4-05 (Algorithmic) Kilgore's Deli is a small delicatessen located near a major university. Kilgore's does a large walk-in carry-out lunch business. The deli offers two luncheon chili specials, Wimpy and Dial 911. At the beginning of the day, Kilgore needs to decide how much of each special to make (he always sells out of whatever he makes). The profit on one serving of Wimpy is $0.35, on one serving of Dial 911, $0.59. Each serving of Wimpy requires 0.45 pound of beef, 0.45 cup of onions, and 5 ounces of Kilgore's special sauce. Each serving of Dial 911 requires 0.45 pound of beef, 0.72 cup of onions ounces of Kilgore's special sauce, and 4 ounces hot sauce. Today, Kilgore has 33 pounds of beef, 47 cups of onions, 81 ounces of Kilgore's special sauce, and 48 ounces hot sauce on hand a. Develop an LP model that will tell Kilgore how many servings of Wimpy and Dial 911 to make in order to maximize his profit today. Let W #of servings of Wimpy to make D-# of servings Dial 911 to make Max s.t |(Beef) W (Onions) . (Special Sauce) + (Hot Sauce) w, D 0 b. Find an optimal solution. Round the answer for profit to the nearest cent and, if required, round the answers for W and D to one decimal place. Solution: W ,Profit c. What is the dual value for special sauce? Round your answer to the nearest cent Dual value for special sauce $ d. Increase the amount of special sauce available by 1 ounce. Give the new solution. Round the answer for profit to the nearest cent. Solution: W Profit $ Does the solution confim the answer to part (c)?

Answer 1

a) Let,

W = # of servings of Wimpy to make
D = # of servings of Dial 911 to make

Objective is to maximize profit so objective function = Max 0.35W + 0.59D

Subject to,

0.45W + 0.45D <= 33 (beef)
0.45W + 0.72D <= 47 (Onions)
5W + 1D <= 81 (Special sauce)
4D <= 48 (Hot sauce)

E,D >= 0(Non-negativity constraint)

b.

Solution W = 13.8, D = 12 Profit = $11.91

	W	D
	14	12				Objective function	11.98
Beef	0.45	0.45	11.7	<=	33
Onions	0.45	0.72	14.94	<=	47
Special Sauce	5	1	82	<=	82
Hot Sauce		4	48	<=	48

c.

Sensitivity report

Variable Cells
			Final	Reduced	Objective	Allowable	Allowable
	Cell	Name	Value	Cost	Coefficient	Increase	Decrease
	$B$3	W	13.8	0	0.35	2.6	0.35
	$C$3	D	12	0	0.59	1E+30	0.52
Constraints
			Final	Shadow	Constraint	Allowable	Allowable
	Cell	Name	Value	Price	R.H. Side	Increase	Decrease
	$D$5	Beef	11.61	0	33	1E+30	21.39
	$D$6	Onions	14.85	0	47	1E+30	32.15
	$D$7	Special Sauce	81	0.07	81	237.6666667	69
	$D$8	Hot Sauce	48	0.13	48	204.1269841	48

Dual value for Special sauce = $0.07

d.

New solution

W = 14, D = 12 and Profit = 11.98

Yes solution confirm to C) as the increase of 1 ounce is within allowable increase limit, shadow price will remain valid for increase of 1. So the profit will increase by $0.07 which has happened ($13,98-$13.91 = $0.07)