B) Evaluate the prospect of borrowing money before the first game.

Yes, 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. The upper limit given in the model is $1,658.88, which means that Julia can only borrow $158.88 from her friend, giving her an additional profit of $238.32.

C) 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
A major uncertainty that could play a factor in Julia’s analysis in weather. Weather is always un predictable and it could be sunny one day and raining the next. If the weather is rainy, there may not be as big of a crowd as there is on a nice day. The weather might also be too cold or too hot and game patrons may not want to eat before and during half time. Julia has to reach her goal of at least $1,000 per game so that she can pay for the booth each home...

...Julia’sFoodBooth
Julia Robertson is a senior at Tech, and she's investigating different ways to finance her final year at school. She is considering leasing a foodbooth outside the Tech stadium at home football games. Tech sells out every home game, and Julia knows, from attending the games herself, that everyone eats a lot of food. She has to pay $1,000 per game for a booth, and thebooths are not very large. Vendors can sell either food or drinks on Tech property, but not both. Only the Tech athletic department concession stands can sell both inside the stadium. She thinks slices of cheese pizza, hot dogs, and barbecue sandwiches are the most popular food items among fans and so these are the items she would sell.
Most food items are sold during the hour before the game starts and during half time; thus it will not be possible for Julia to prepare the food while she is selling it. She must prepare the food ahead of time and then store it in a warming oven. For $600 she can lease a warming oven for the six-game home season. The oven has 16 shelves, and each shelf is 3 feet by 4 feet. She plans to fill the oven with the three food items before the game and then again before half time.
Julia has negotiated with a local pizza delivery company to deliver 14-inch cheese pizzas...

...A. Formulate a linear programming model for Julia that will help you to advise her if she should lease the booth.
Let, X1 =No. of pizza slices,
X2 =No. of hot dogs,
X3 = No. 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=$4.5/6=$0.75
| X1 | X2 | X3 |
SP | $1.50 | $1.60 | $2.25 |
-Cost | 0.75 | $0.50 | $1.00 |
Profit | $0.75 | $1.10 | $1.25 |
Maximize Total profit Z = $0.75X1 + 1.10X2 +1.25X3
* Constraints:
1. Budget constraint:
0.75X1+0.50X2+1.00X3<=$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=14*14=196 in-square
Space required for a slice of pizza=196/6=32.667in-square approximately.
Thus, space constraint can be written as:
33X1 + 16X2 +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 (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>= 2.0 (at least twice as many hot dogs as barbeque sandwiches)
This constraint can be rewritten as:
X2-2X3>=0
X1,...

...00 |$B$10>=0 |Not Binding |1250.00 |
| |$C$10 |X2 |1250.00 |$C$10>=0 |Not Binding |1250.00 |
| |$D$10 |X3 |0.00 |$D$10>=0 |Binding |0.00 |
From the table, we get the optimum solution as follows:
X1 = 1250; X2 = 1250 and X3 = 0 and Maximum value of Z = $2250
Julia should stock 1250 slices of pizza and 1250 numbers of Hot dogs. She need not stock sandwiches.
Maximum Profit that can be expected is $2250.
Lease cost for the booth per game = $1000
Lease cost for the oven per game = $100
Net profit after all the expenses = 2250 – 1100 = $1150
Now it is clear that as per the strategy it is worth leasing the booth.
B) The sensitivity report of the solution is given below
| | | | | | | |
| | |Final |Reduced |Objective |Allowable |Allowable |
|Cell |Name |Value |Cost |Coefficient |Increase |Decrease |
|$B$10 |X1 |1250.00 |0.00 |0.75 |1...

...Running Head: JULIA’SFOODBOOTH
Assignment #3: Case Problem "Julia'sFoodBooth"
Mat540
Quantitative Methods
August 22, 2012
Julia’sFoodBooth
(A) Formulate and solve a L.P. model for this case.
Variables:
Pizza - X1 $1.33 $1.50 14 inches
Hot Dogs - X2 $0.45 $1.50 16 square inches
Barbecue - X3 $0.90 $2.25 25 square inches
Maximize Z= $0.75x1, 1.05x2, 1.35x3
Subject to:
$0.75x1 + $0.45x2 + $0.90x3 ≤ $1,500
24x1 + 16x2 + 25x3 ≤ 55,296 in2 of oven space
x1 ≥ x2 + x3 (changed to –x1 + x2 + x3 ≤ 0 for constraint)
x2/x3 ≥ 2 (changed to –x2 +2x3 ≤ 0 for constraint)
x1, x2, x3 ≥ 0
Solution: x1 = 1,250 pizza slices
x2= 1,250 hot dogs
x3= 0 barbecue sandwiches
Z= $2,250
Solution:
Variable | Status | Value |
X1 | Basic | 1250 |
X2 | Basic | 1250 |
X3 | NONBasic | 0 |
slack 1 | NONBasic | 0 |
slack 2 | Basic | 5296.0 |
slack 3 | NONBasic | 0 |
slack 4 | Basic | 1250 |
Optimal Value (Z) | | 2250 |
B) Evaluate the prospect of borrowing money before the first game.
Yes, Julia would increase her profit if she borrowed money from a friend before the first game to purchase more ingredients. Her outcome would be an increase in profit. The shadow price, or dual value, is $1.50 for each additional dollar that she earns. ...

...Julia’sFoodBooth Case Problem
Assignment 3
Max Z =Profit1x1+ Profit2x2+ Profit3x3
A - Formulation of the LP model
x1 - number of pizza slice
x2 - number of hot dogs
x3 - number of barbecue sandwiches
Constraints
Cost
Maximum fund available for food = $1500
Cost per pizza $6 ÷08 (slices) = $0.75
Cost for a hot dog = $0.45
Cost for a barbecue sandwich = $0.90
Constraint: 0.75x1+0.45x2+0.90x3 ≤1500
Oven space
Space available 16.3.4.2 = 384ft^2
384.144=55296 in ^2
Space required for pizza: 14.14 = 196 ^2 inches
Space for slice of pizza; 196 ÷8 = 24.50 in ^2
Space for hot dog: 16 in ^2
Space for barbecue = 25 in ^2
Constraint 24.50x1+16x2+25x3 ≤55296
Julia can sell at least as many slice of pizza (x1) as hot dogs (x2) and Barbecue sandwiches (x3) combined.
x1-x2-x3≥0
Julia can sell at least twice as many hot dogs as Barbecue sandwiches
+x2-2x3≥0
Non negative constraint
x1,x2,x3≥0
Objective Function
| SELL | COST | PROFIT |
Pizza slice (x1) | $1.50 | $0.75 | $0.75 |
Hot dog (x2) | $1.50 | $0.45 | $1.05 |
Barbecue Sandwich (x3) | $2.25 | $0.90 | $1.35 |
Profit = Sell - Cost
Max Z=0.75x1+1.05x2+1.35x3
LPP Model:
Maximize Z = 0.75 X1 + 1.05 X2 + 1.35 X3
Subject to 24.5 X1 + 16 X2 + 25 X3 ≤ 55296
0.75 X1 + 0.45 X2 + 0.90 X3 ≤ 1500
X1 -...

...Assignment #3: Julia’sFoodBooth
Quantitative Methods 540
Buddy L. Bruner, Ph.D.
Shirley Foster
11/25/2012
Assignment 3: Case problem “Julia’sFoodBooth” Page 1
A. Julia Robertson is making an allowance for renting a foodbooth at her school. She is seeking ways to finance her last year and believed that a foodbooth outside her school’s stadium would be ideal. Her goal is to earn the most money possible thus increasing her earnings. In this case problem, she decided to sell pizza, hotdogs and BBQ sandwiches. The following LP model illustrates the maximum net profit and constraints that will determine whether or not to least the boot.
Variables:
X1 – Pizza Slices
X2 – Hot Dogs
X3 – Barbeque Sandwiches
Subject to:
$0.75x1 + $0.45x2 + $0.90x3 ≤ $1,500
24x1 + 16x2 + 25x3 ≤ 55,296 in2 of oven space
X1 ≥ x2 + x3 (changed to –x1 + x2 + x3 ≤ 0 for constraint)
X1, X2, X3 ≥ 0
Solution:
Variable | Status | Value |
X1 | Basic | 1250 |
Assignment 3 Case problem “Julia’sFoodBooth” Page 2
X2 | Basic | 1250 |
X3 | NON Basic | 0 |
Slack 1 | NON Basic | 0 |
Slack 2 | Basic | 5296.0 |
Slack 3 | NON Basic | 0 |
Slack 4 | Basic | 1250 |
Optimal Value (Z) | | 2250 |
Built on the above LP model, Julia is estimated...

... Together
Constraint 4 H - 2B >= 0 selling at least twice as many hot dogs as barbecue . Sandwiches
p,H,B >= 0 . si>= 0
Based on QM for windows l solution the optimum solution:
Pizza (p) = 1250; Hotdogs(H) = 1250 and Barbecue sandwiches (B) = 0
Maximum value of Z = $2250
Julia should stock 1250 slices of pizza, 1250 hot dogs and no barbecue sandwiches.
Maximum Profit = $2250.
|Maximum Profit 2250
Booth Rent per game 1000
Warming Oven 600 for total of 6 games 600/6 =100 for 1 game
2250-1000-100=1150 is the profit after paying all the expenses.
*I think she should lease the booth
B) the amount of Borrowing money that will increase her profit is defined by the upper limit value of the cost constrain. .0001=<C1<= 1638.4. this means that she can only borrow 1638.4-1500= 138.5 so she can increase her profit from 2250 to 2457 and reduce her slack values from S2 = 4671 to S2= .0019. Any money mere than that will not increase the profit bur it will cause increasing in the value of slack variables
C) I thing that hiring her friend for $100 she will still have a profit above 1000 (1150-100=1050) so she can do that.
D) One uncertainty is the weather because...