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Analyzing the Profitability of Leasing a Booth for Food Sales at a Sports Event: A Linear Programming Approach

This paper uses linear programming to assess the profitability of leasing a booth for food sales at a sports event.

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    Assignment 3 : Julia’s Food Booth Case Problem 1 Analyzing the Profitability of Leasing a Booth for Food Sales at a Sports Event: A Linear Programming Approach A. Formulate and solve a linear programming model for Julia that will help you to advise her if she should lease the booth. By Formulating the model for the first home game , the profit function and constraints and c alculations is broken down to the equations. Let, X 1 =No of pizza slices, X2 =No of hot dogs, X 3 = barbeque sandwiches Formulation: 1. Calculating Objective function co - efficient : The objective is to Maximize total profit. Profit is calculated for each variable by subtracting cost from the selling price. • For Pizza slice, Cost/slice=$6/8=$0.75 X1 X2 X3 SP $ 1.50 $ 1.50 $ 2.25 - Cost $ 0.75 $ 0.45 $ 0.90 Profit $ 0.75 $ 1.05 $ 1.35 • Total space available=3*4*16=192 sq feet =192*12*12=27,648 in - square The oven will be refilled during half time. Thus, the total space available=2*27,648= 55,296 in - square • Space required for a pizza=14*14=196 in - square Space required for a slice of pizza=196/8=24 in - square approximately.

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    Assignment 3 : Julia’s Food Booth Case Problem 2 T herefore, Objective function for the model can be written as: Maximize Total profit Z = $0.75 X 1 + 1.05 X 2 +1.35 X 3 Subject to constraints: $0.75 X 1 + .0.45 X 2 + 0.90 X 3 <= 1,500 (Budget constraint) 24 X 1 + 16 X 2 +25 X 3 <= 55,296 (Inch square Of Oven Space) X 1>=X2 + X 3 (at least as many slices of pizza as hot dogs and barbeque sandwiches combined) X2/X3>= 2.0 (at least twice as many hot dogs as barbeque sandwiches) This constraint can be rewritten as: X2 - 2X3>=0 X 1, X 2, X 3 >= 0 Final Model: Maximize Total profit Z = $0.75 X 1 + 1.05 X 2 +1.35 X 3 Subject to: $0.75 X 1 + .0.45 X 2 + 0.90 X 3 <= 1,500 (Budget) 24 X 1 + 16 X 2 +25 X 3 <= 55,296 (In - square Of Oven Space) X 1 - X2 - X 3>=0 (at least as many slices of pizza as hot dogs and barbeque sandwiches combined) X2 - 2X3>=0 (at least twice as many hot dogs as barbeque sandwiches) X 1, X 2, X 3 >= 0 (Non negativity constraint) Solution: Following table shows the Excel solver solution for the model: Target Cell (Max) Cell Name Original Value Final Value $E$28 $ 2,250.00 $ 2,250.00 Adjustable Cells Cell Name Original Value Final Value $B$28 X1 1250 1250 $C$28 X2 1250 1250 $D$28 X3 0 0

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