Case Study-This is for [email protected]
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Julia Food Booth Case Problem
Assignment #3
Sherika R Furlow
Strayer University
MAT540
November 25, 2012
A. Formulate and solve an L.P. model for this case.
Let, X1 = # of pizza slices,
X2 = # of hot dogs,
X3 = # of barbeque sandwiches
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=$6.00/8=$0.75
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X1 |
X2 |
X3 |
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SP |
1.50 |
1.50 |
3.00 |
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-COST |
0.75 |
0.50 |
1.50 |
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PROFIT |
0.75 |
1.00 |
1.50 |
Maximize Total profit Z = $0.75X1 + 1.00X2 +1.50X3
Constraints:
1. Budget constraint: 0.75X1+0.50X2+1.50X3<=$1500
2. Space constraint:
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=16*16=256 in-square
Space required for a slice of pizza=256/8=32 in-square approximately.
Thus, space constraint can be written as: 32X1 + 15X2 +25X3 <= 55,296 (In-square Of Oven Space)
3. At least as many slices of pizza as hot dogs and barbeque sandwiches combined: X1 ≥ X2 + X3 = X1 - X2 - X3 ≥ 0 (at least as many slices of pizza as hot dogs and barbeque sandwiches combined)
4. At least twice as many hot dogs as barbeque sandwiches X2/X3 ≥ 2 = X2 ≥2 X3 =X2 - 2 X3 ≥ 0 (at least twice as many hot dogs as barbeque sandwiches)
X1, X2, X3 >= 0 (Non negativity constraint)
Model:
Maximize Total profit Z = $0.75X1 + 1.00X2 +1.50X3
Subject to: 0.75X1+0.50X2+1.50X3<=$1500 (Budget)
32X1 + 15X2 +25X3 <= 55,296 (In-square Of Oven Space)
X1-X2 - X3>=0 (at least as many slices of pizza as hot dogs and barbeque sandwiches combined)
X2-2X3>=0 (at least twice as many hot dogs as barbeque sandwiches)
X1, X2, X3 >= 0 (Non negativity constraint)
(B). Evaluate the prospect of borrowing money before the first game.
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Food items: |
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Pizza |
Hot Dogs |
Barbecue |
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Profit per item: |
0.75 |
1.00 |
1.50 |
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Constraints: |
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Available |
Usage |
Left over |
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Budget ($) |
0.75 |
0.50 |
1.50 |
1,500 |
1,500.00 |
0 |
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Oven space (sq. in.) |
24 |
16 |
25 |
55,296 |
50,000.00 |
5296 |
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Demand |
1 |
-1 |
-1 |
0 |
- |
0 |
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Demand |
0 |
1 |
-2 |
0 |
1,250.00 |
-1250 |
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Stock |
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Pizza= |
1250 |
slices |
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Hot Dogs= |
1250 |
hot dogs |
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Barbecue= |
0 |
sandwiches |
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Profit= |
2,250.00 |
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Adjustable Cells |
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Final |
Reduced |
Objective |
Allowable |
Allowable |
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Cell |
Name |
Value |
Cost |
Coefficient |
Increase |
Decrease |
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$B$6 |
Pizza= |
1250 |
0 |
0.75 |
1 |
1 |
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$B$6 |
Hot Dogs= |
1250 |
0 |
0.5 |
1E+30 |
0.27 |
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$B$6 |
Barbecue= |
0 |
0 |
1.5 |
0.375000011 |
1E+30 |
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Constraints |
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Final |
Shadow |
Constraint |
Allowable |
Allowable |
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Cell |
Name |
Value |
Price |
R.H. Side |
Increase |
Decrease |
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$G$9 |
Demand Usage |
1250 |
0 |
0 |
1250 |
1E+30 |
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$G$7 |
Oven space (sq. in.) Usage |
50000 |
- |
55296 |
1E+30 |
5296 |
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$G$6 |
Budget ($) Usage |
1500 |
1.5 |
1500 |
158.88 |
1500 |
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$G$8 |
Demand Usage |
- |
-0.38 |
0 |
2000 |
3333.33 |
Julia would increase her profit if she borrowed some more money from a friend. The shadow price, or dual value, is $1.50 for each additional dollar that she earns. Julia can only borrow $158.88 from her friend, giving her an additional profit of $238.32.
(C). Evaluate the prospect of paying a friend $100/game to assist.
Yes, I believe Julia should hire her friend for $100 per game. In order for Julia to prepare the hot dogs and barbeque sandwiches needed in a short period of time to make her profit, she needs the additional help. Also, with her borrowing the extra $158.88 from her friend, Julia would be able to pay her friend for the time spent per game helping with the food booth.
(D). Analyze the impact of uncertainties on the model.
The biggest uncertainty in this model is demand. Although Julia may have a good idea of what people will buy and not buy during the game by talking to previous vendors who have sold at prior home games, it is not always guaranteed that this is the way it will go every home game. For this reason, if the demand changes then the solution to the linear programming will change and may affect her capability to make proceeds greater than $1000. Weather plays an important role as well. What if there is a home game in late October or early November, then most people look forward to eating chili or something warmer than a slice of pizza or a hotdog. Consequently weather has the greatest affect on the food sales and Julia’s overall achievement or failure.
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Julia Food Booth Case Problem
Assignment #3
Sherika R Furlow
Strayer University
MAT540
November 25, 2012