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Analyzing the Profitability of Leasing a Booth for Food Sales at a Sports Event: A Linear Programming Approach - Document preview page 1

Analyzing the Profitability of Leasing a Booth for Food Sales at a Sports Event: A Linear Programming Approach - Page 1

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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.

Leo Campbell
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Analyzing the Profitability of Leasing a Booth for Food Sales at a Sports Event: A Linear Programming Approach - Page 1 preview imageAssignment3:Julia’s Food BoothCase Problem1Analyzing the Profitability of Leasing a Booth for Food Sales at a SportsEvent: A Linear Programming ApproachA. Formulate and solve a linear programming model for Julia that will help you to adviseher if she should lease the booth.ByFormulatingthe model for the first home game,theprofitfunction and constraints andcalculationsis broken down to theequations.Let,X1 =No of pizza slices,X2 =No of hot dogs,X3 = barbeque sandwichesFormulation:1. Calculating Objective function co-efficient:The objective is to Maximize total profit. Profit is calculated for each variable by subtracting cost fromthe selling price.For Pizza slice, Cost/slice=$6/8=$0.75X1X2X3SP$1.50$1.50$2.25-Cost$0.75$0.45$0.90Profit$0.75$1.05$1.35Total space available=3*4*16=192 sq feet=192*12*12=27,648 in-squareThe oven will be refilled during half time.Thus, the total space available=2*27,648= 55,296 in-squareSpace required for a pizza=14*14=196 in-squareSpace required for a slice of pizza=196/8=24 in-square approximately.
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Analyzing the Profitability of Leasing a Booth for Food Sales at a Sports Event: A Linear Programming Approach - Page 3 preview imageAssignment3:Julia’s Food BoothCase Problem2Therefore, Objective function for the model can be written as:Maximize Total profitZ= $0.75X1 + 1.05X2 +1.35X3Subject to constraints:$0.75X1 + .0.45X2 + 0.90X3 <= 1,500 (Budget constraint)24X1 + 16X2 +25X3 <= 55,296 (Inch square Of Oven Space)X1>=X2 +X3 (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>=0X1,X2,X3 >= 0Final Model:Maximize Total profitZ= $0.75X1 + 1.05X2 +1.35X3Subject to:$0.75X1 + .0.45X2 + 0.90X3 <= 1,500 (Budget)24X1 + 16X2 +25X3 <= 55,296 (In-square Of Oven Space)X1-X2-X3>=0 (at least as many slices of pizza as hot dogs and barbeque sandwiches combined)X2-2X3>=0 (at least twice as many hot dogs as barbeque sandwiches)X1,X2,X3 >= 0 (Non negativity constraint)Solution:Following table shows the Excel solver solution for the model:Target Cell (Max)CellNameOriginal ValueFinal Value$E$28$2,250.00$2,250.00Adjustable CellsCellNameOriginal ValueFinal Value$B$28X112501250$C$28X212501250$D$28X300
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