Julia’s Food Booth
NAME
Strayer University
MAT 540
LECTURER
DATE

Julia’s Food Booth
Formulate and solve an L.P. model for this case.
Julia Robertson is trying to advance her education and financing her education, to do so she started a food booth outside the Tech stadium selling Slices of cheese pizza, hot dogs, and barbecue sandwiches. This covers the items X1, X2, and X3 respectively. The costs of these items are $4.50, $0.50 and $1.00 respectively. As one knows the objective is to maximize Julia Robertson’s profit. The method of doing this is by evaluating subtracted costs from the selling price. Things to consider are constraints and the cost of everything involved. Being that cost always seems the most important this paper will start there.
Profit on pizza slice = $1.50 - $0.75 = $ 0.75
Profit on hot dog = $1.50 – 0.45 = $1.05
Profit on sandwich = $2.25 - $0.90 = $1.35
The overall profit would be the following:
Z=0.75 X1+ 1.05 X2 + 1.35 X3
Constraints:
Budget Constraints would go as follows:
0.75X1 + 0.45X2 + 0.90X3 <=$1500
Space Constraints:
Space available = 3 x 4 x 16 = 192 sq. feet = 192 x 12 x 12 =27648 sq. inches
The oven will be refilled during half time.
Thus total space available = 27648 x 2 = 55296
Space required for pizza = 14 x 14 = 196 sq. inches
Space required for pizza slice = 196/ 8 = 24.5 sq. inches
Total Oven Space required: 24.5 X1 + 16 X2 + 25 X3
Constraint: 24.5 X1 + 16 X2 + 25 X3 ≤ 55296 sq. Inches
Another area to consider is the demand or potential demand for each item and what the availability is. Julia can sell at least as many slices of pizza as she is able to sell hot dogs and sandwiches represented by:
X2 ≥2 X3 X2 - 2 X3 ≥ 0
LP programming model :
Maximize Z = 0.75 X1 + 1.05 X2 + 1.35 X3
Subject to: 0.75 X1 + 0.45 X2 + 0.90 X3 ≤ 1500 24.5 X1 + 16 X2 + 25 X3 ≤ 55296 X1 - X2 - X3 ≥ 0 X2 - 2 X3 ≥ 0 X1≥ 0, X2≥ 0 and X3 ≥0
Excel Worksheet (excel attached) | | | | | | | | | | | | | | | | Cell | Name | Original Value | Final Value | | | | | $H$21 | Profit= | 2250 | 2250 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Cell | Name | Original Value | Final Value | Integer | | | | $H$18 | Pizza= | 1250 | 1250 | Contin | | | | $H$19 | Hot Dogs= | 1250 | 1250 | Contin | | | | $H$20 | Barbecue= | 0 | 0 | Contin | | | | | | | | | | | | | | | | | | | | | | | | | | | | Cell | Name | Cell Value | Formula | Status | Slack | | | $M$15 | Demand Usage | 0 | $M$15>=$L$15 | Binding | 0 | | | $M$13 | Budget Usage | 1500 | $M$13<=$L$13 | Binding | 0 | | | $M$14 | Oven space (sq. inches) Usage | 50625 | $M$14<=$L$14 | Not Binding | 4671 | | | $M$16 | Demand Usage | 1250 | $M$16>=$L$16 | Not Binding | 1250 | | | $H$18 | Pizza= | 1250 | $H$18>=0 | Not Binding | 1250 | | | $H$19 | Hot Dogs= | 1250 | $H$19>=0 | Not Binding | 1250 | | | $H$20 | Barbecue= | 0 | $H$20>=0 | Binding | 0 | | | | | | | | | | |

From the table above, we get the optimum solution as follows:
X1(Pizza) = 1250; X2(hot dog) = 1250 and X3(Barbecue) = 0 and Maximum value of Z (profit) = $2250
Base on this and the demand usage, Julia should stock 1250 slices of pizza and 1250 hot dogs. She should not stock barbecue sandwiches.
Maximum Profit expected is $2250.
Lease cost for the booth per game = $1000
Lease cost for the oven per game = $100
Net profit after both expenses above = $2250 – $1100 = $1150
This means it is worth leasing the booth because the of the $1150 profit after deducting all expenses which will be more than the actual needed amount by $150.

Evaluate the prospect of borrowing money before the first game.
Excel Worksheet for sensitivity report.( excel attached) | | | | | | | | | | Final | Reduced | Objective | Allowable | Allowable | | Cell | Name | Value | Cost | Coefficient | Increase | Decrease | | $H$18 | Pizza= | 1250 | 0 | 0.75 | 1 | 1 | | $H$19 | Hot Dogs= | 1250 | 0 | 1.05 | 1E+30 | 0.272727273 | | $H$20 | Barbecue= | 0 | -0.375 | 1.35 | 0.375 | 1E+30 | | | | | | | | | | | | | | | | | | | | Final | Shadow | Constraint | Allowable | Allowable | | Cell | Name | Value | Price | R.H. Side | Increase | Decrease | | $M$15 | Demand Usage | 0 | -0.375 | 0 | 2000 | 3333.333333 | | $M$13 | Budget Usage | 1500 | 1.5 | 1500 | 158.8 | 1500 | | $M$14 | Oven space (sq. inches) Usage | 50625 | 0 | 55296 | 1E+30 | 4671 | | $M$16 | Demand Usage | 1250 | 0 | 0 | 1250 | 1E+30 | | | | | | | | | |

The above report shows that, the Shadow price for the budget usage constraint is 1.5 and the allowable increase is 158.8 which can be translated as: For each dollar added to the budget, can increase a profit of $1.5 and therefore the maximum allowable increase is 158.8. The full sum Julia can borrow from her associates is $158.8 which will be an additional profit of 158.8 x 1.5 = $ 238.20.
Evaluate the prospect of paying a friend $100/game to assist.
If Julia is willing to hire a friend to help at the price of $100 per game, her net profit per game will now be $1150-$100 = $1050. It is still worth renting the booth. Knowing this Julia should hire her friend to help her accomplish her task within the short period of time. Also Julia can decide to borrow more money and expand her stock because of the additional help.
Analyze the impact of uncertainties on the model.
The main factor though uncertain is demand. She has no doubt that she will sell everything she has in stock based on her customer base. If she is wrong then the formula processed in this paper will be unfounded and wrong. So to go through all the areas (A), the net profit that can be earned is $1150. This is more than her $1000 target. This means the most she can lose at this point is $150. When you look at (C), in order to complete this goal she will need to hire an associate at $100 dollars per game to complete 1250 hotdogs. Knowing this Julia may want to lower the number of hotdogs she will produce in order to avoid hiring an associate. This should take some of the uncertainty out of the equation when it comes to demand. The best thing to do is reduces the number of hot dogs to 1108 this will reduce the profit by $149.10 which is still above the $1000 target.
Also another uncertainty Julia can face is the weather, because the weather is unpredictable this can affect the attendance especially if is it a rainy day resulting in lower sale. Also the mood of the attendees can affect the sales for example if the home team is losing, they the customer might decide to go home and not cater to buy for any food in the standium.

Reference
Taylor, B. M. (2010). Introduction to management science (10th ed.). Upper Saddle River, NJ: Pearson/Prentice Hall