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MAT540 - Week 7 - Assignment 3_Julia's Food Booth

# MAT540 - Week 7 - Assignment 3_Julia's Food Booth -...

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Assignment 3: Julia’s Food Booth Case Problem 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 calculations is broken down to the equations. Let, X1 =No of pizza slices, X2 =No of hot dogs, X3 = 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. Therefore, Objective function for the model can be written as: Maximize Total profit Z = \$0.75X1 + 1.05X2 +1.35X3 Subject to constraints: 1

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Assignment 3: Julia’s Food Booth Case Problem \$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>=0 X1, X2, X3 >= 0 Final Model: Maximize Total profit Z = \$0.75X1 + 1.05X2 +1.35X3 Subject to: \$0.75X1 + .0.45X2 + 0.90X3 <= 1,500 (Budget)
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