Food Booth

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Julia’s Food Booth. Parts A thru C.
Please provide linear programming model, graphical solution, sensitivity report, and answers to questions A thru C. (Problem on page 2)

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A) Formulate and solve a linear programming model for Julia that will help you advise her if she should lease the booth. Let, X1 =No of pizza slices,
X2 =No of hot dogs,
X3 = barbeque sandwiches
Formulation:
1. Calculating Objective function co-efficients:
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.

Thus, Objective function for the model can be written as: Maximize Total profit Z = $0.75X1 + 1.05X2 +1.35X3
Subject to constraints:
$0.75X1 + .0.45X2 + 0.90X3 = 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 =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) | | | | | | |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$28 |X3 |0 |0 | | | | | | | | | | | | | | | | | | | |Constraints | | | | | | |Cell |Name |Cell Value |Formula |Status |Slack | | |$E$30 |Budget |1500 |$E$30=$G$33...
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