Mat 540 Assignment 3 Julias Food Booth

MAT540 Assignment #3: Case Problem "Julia's Food Booth"

6/6/13

Problem summary and definitions of quantities:

Booth cost $1,000 / game

No. games 6

Oven cost $600/season = $100/game

Oven size, 3’ x 4’ x 16 = 27648 in2

Hot dog size, 16 in2

BBQ size, 25 in2

Pizza size, π(14/2)2/8 ≈ 20 in2

Pizza cost, $6 / 8 = $0.75 / slice

Hot dog cost, $0.45 / dog

BBQ cost, $0.90 / sandwich

Pizza retail, $1.50 / slice

Hot dog retail, $1.50 / dog

BBQ retail, $225 / sandwich

Cash available $1,500

Required profit $1,000

a) Formulate Linear Programming Model:

Decision Variables:

x1 = pizza slices

x2 = hot dogs

x3 = barbeque sandwiches

The Model is for the first home game,

Maximize: Z= $0.75 x1 + 1.05x2 + 1.35x3 ← Objective Function

Subject To the Constraints,

$0.75 x1 + 0.45x2 + 0.90x3 ≤ 1500

24 x1 + 16x2 + 25x3 ≤ 55296 in2 of oven space

X1 ≥ x2 + x3

X2/x3 ≥ 2.0 or x2 ≥ 2x3

X1, x2, x3 ≥ 0

Solution:

X1 = 1250

X2 = 1250

X3 = 0

Z = $2250

Oven cost $600/season = $100/game. In this scenario, she sells 1,250 each of hot dogs pizza slices, but does not sell any BBQ sandwiches. The profit is $1,150. Given this is above her $1,000 target for net profit, I would recommend she go forward with the plan.

A “tricky” aspect of the model formulation is $1500 used to purchase the ingredients. Since the objective function reflects net profit, the $1500 is recouped and can be used for the next home game to purchase food ingredients; thus its not necessary for Julia to use any of her $1500 profit to buy ingredients for the next game.

b) Yes, she would increase her profit; the dual value is $1.50 for each additional dollar. The upper limit of sensitivity range for budget is $1658.88, so she should only borrow approx. $158.

Her additional profit would be $238.32 or a total profit of $2488.32.

c) If we add an additional fixed cost of $100, the optimal solution from (a) is still...