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 - X2 - X3 ≥ 0
X2 - 2 X3 ≥ 0
X1≥ 0, X2≥ 0 and X3 ≥0
Solve the LPM -answer in QM for Windows solution
Based on the QM for Windows solution the optimum solution:
Pizza (X1) = 1250; Hotdog s(X2) = 1250 and Barbecue sandwiches (X3) = 0 Optimal solution value Z = $2250
Julia should stock 1250 slices of pizza, 1250 hot dogs and no barbecue sandwiches. Maximum Profit = $2250.
Maximum Profit| $ 2,250.00|
Booth Rent per game| $ (1,000.00)|
Warming Oven 600 for total of 6 home games 600/6 =100| $ (100.00)| Profit for the 1st Game| $ 1,150.00...

...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...

...MAT 540
Case Assignment #3
Julia’sFoodBooth
A. Formulate and solve a linear programming model for Julia that will help you advice her if she should lease the booth.
JULIA’S PRODUCT STORAGE
VARIABLES COST SELLING PRICE NEEDED
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
$0.75x1 + .0.45x2 + 0.90x3 = 1,500
24x1 + 16x2 + 25x3 < 55,296 in2 of oven space.
x1 > x2 + x3
>2.0
x1, x2, x3 > 0
(A) Solution: x1 = 1,250 pizza slices
x2= 1,250 hot dogs
x3= 0 barbecue sandwiches
Z= $2,250
B. If Julia were to borrow some more money from a friend before the first game to purchase more ingredients, could she increase her profit? If so, how much should she borrow and how much additional profit would she make? What factor constrains her from borrowing even more money than this amount (indicated in your answer to the previous question?)
| |Final |Reduced |Objective |Allowable |Allowable |
|Name |Value |Cost |Coefficient |Increase |Decrease |
|X1 |1250.00 |0.00 |0.75 |1 |1.00000001 |...

...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,...

...(A) Formulate and solve an L.P. model for this case
Variable Food Cooking Area
x1 Pizza Slice 24in sq
x2 Hot Dogs 16in sq
x3 BBQ Sandwiches 25in sq
*The oven space required for a pizza slice is calculated by dividing the total area arequired for a whole pizza by the number of slices in a pizza 14 x 14 = 196 in2, by 8, or approximately 24 in2 per slice. The total space available is the dimension of a shelf, 36 in. x 48 in. = 1,728 in2, multiplied by 16 shelves, 27,648 in2, which is multiplied by 2, the times before kickoff and halftime the oven will be filled = 55,296 in2.
Maximize Z = $0.75x1 + 1.05x2 + 1.35x3
Subject to:
$.75x1 + $.45x2 + $.90x3≤1500
24x1 + 16x2 + 25x3 ≤ 55296in sq of oven space
x1 ≥ x2 + x3
x2/x3 ≥ 2
x1,x2,x3 ≥ 0
Solution:
X1 = 1250 slices of pizza
X2 = 1250 hotdogs
X3 = 0 BBQ sandwiches
Julia would profit $2250. Her lease per game for the tent is $1000.00 and $100.00 for the warming oven. This means she still clears $1150 which is more than her $1000 minimum profit to open the concession stand.
(B) Evaluate the prospect of borrowing money before the first game.
Julia should borrow money based on the given scenario. She could increase her profit if she borrows money from a friend. The shadow price, or dual value, is $1.50 for each extra dollar that she earns. The upper limit given in the model is $1,658.88. This means that she would max out her profit at $1,658.88 of spending. Since she...

... 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...

...Julia’sFoodBooth
Strayer University
Quantitative Methods – MAT 540/ Spring 2012
Dr. Buddy Bruner
May 19, 2012
A) Formulate and solve an L.P. model for this case.
[pic]
[pic]
B) Evaluate the prospect of borrowing money before the first game.
After observing the ranging chart calculations indicate that the upper bound in the budget is equal to 1638.4. The original value of the budget is 1500. If you subtract the 1500 from the 1638.4 it will leave 138.4 which indicate room for Julie to borrow. This is a good thing for Julie because for every 138.4 she borrows dual value will increase 1.5. She will also have enough funds to purchase more sandwiches. After she pays any funds back that she has borrowed then she will earn a profit of at least $69.20 which is half of the $138.40 borrowed.
C) Evaluate the prospect of paying a friend $100/game to assist.
The linear programming results show that Julie will have a profit of $2250 minus the $1000 for booth rent for the month which will leave $1150. She only wants to make a profit of $1000 therefore she has at least $150 to keep $50 to purchase more sandwiches and $100 to pay a friend. After the purchase of more sandwiches Julie made need some additional help. Therefore it will be feasible to hire a friend for extra help which...