Question

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 food booth 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 the booths 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 twice each game—2 hours before the game and right after the opening kickoff. Each pizza will cost her $6 and will include 8 slices. She estimates it will cost her $0.45 for each hot dog and $0.90 for each barbecue sandwich if she makes the barbecue herself the night before. She measured a hot dog and found it takes up about 16 square inches of space, whereas a barbecue sandwich takes up about 25 square inches. She plans to sell a slice of pizza and a hot dog for $1.50 apiece and a barbecue sandwich for $2.25. She has $1,500 in cash available to purchase and prepare the food items for the first home game; for the remaining five games she will purchase her ingredients with money she has made from the previous game.

Julia has talked to some students and vendors who have sold food at previous football games at Tech as well as at other universities. From this she has discovered that she can expect to sell at least as many slices of pizza as hot dogs and barbecue sandwiches combined. She also anticipates that she will probably sell at least twice as many hot dogs as barbecue sandwiches. She believes that she will sell everything she can stock and develop a customer base for the season if she follows these general guidelines for demand.

If Julia clears at least $1,000 in profit for each game after paying all her expenses, she believes it will be worth leasing the booth.

Formulate and solve a linear programming model for Julia that will help you advise her if she should lease the booth.

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)?

When Julia looked at the solution in (A), she realized that it would be physically difficult for her to prepare all the hot dogs and barbecue sandwiches indicated in this solution. She believes she can hire a friend of hers to help her for $100 per game. Based on the results in (A) and (B), is this something you think she could reasonably do and should do?

Julia seems to be basing her analysis on the assumption that everything will go as she plans. What are some of the uncertain factors in the model that could go wrong and adversely affect Julia’s analysis? Given these uncertainties and the results in (A), (B), and (C), what do you recommend that Julia do?

Answer 1

Answer:

By using given data,

Julia, through this business, must want to maximize her profit. So this is a maximization problem. With given selling prices per item, the revenues will be calculated. Different cost items are also given. Subtracting total cost from the revenue will yield the profit. We have to maximize that.

The resources and constraints involved, qualitatively can be described as:

• The maximum space that she has, based on the overall space of the oven
• The overall numbers of each individual items are not independent (“she can expect to sell at least as many slices of pizza as hot dogs and barbecue sandwiches combined”, “she will probably sell at least twice as many hot dogs as barbecue sandwiches”)
• Minimum profit after each game should be enough to sustain for the next game
• For the first game, maximum amount available for purchase of food items is $1,500
• Minimum profit after each game is 1000

Formulations:

Let P_i = No. of pizza for the i-th game

      H_i = No. of hot dogs for the i-th game

      S_i = No. of sandwiches for the i-th game                               i = 1,2,3,4,5,6

Space available in the oven = 16*3*4 = 192 square feet

We will take double of this i.e. 384 as twice a time it can be emptied for a game.

Space taken by one pizza = 3.14*(7²) = 153.86 square inch = 1.068 square feet

Space taken by one hot dog = 16 sq.inches = 0.111 square feet

Space taken by one sandwich = 25 sq.in = 0.174 sqft

Game1

Payment to Tech - $ 1000

Payment for the oven - $ 600

Cost for food purchase/ cooking – 6(P1) + 0.45(H1) + 0.90(S1) <= 1500 ----------------------------------------1

Considering the total space:

1.068(P1) + 0.111(H1) + 0.174(S1) <= 382--------------------------------------------------------------------------------2

Revenue = 1.5*8(P1) + 1.5(H1) + 2.25(S1)

So, Profit would be

12(P1) + 1.5(H1) + 2.25(S1) – [ 6(P1) + 0.45(H1) + 0.90(S1) ] – 1000 – 600

= 6(P1) + 1.05(H1) + 1.35(S1) – 1600

6(P1) + 1.05(H1) + 1.35(S1) – 1600 >= 1000 -----------------------------------------------------------------------------3

Game2

Payment to Tech - $ 1000

Cost for food purchase/ cooking – 6(P2) + 0.45(H2) + 0.90(S2)

Now, as she must sustain from the previous profit only,

6(P2) + 0.45(H2) + 0.90(S2) <= 6(P1) + 1.05(H1) + 1.35(S1) – 1600

Or,

6(P2) + 0.45(H2) + 0.90(S2) - 6(P1) - 1.05(H1) - 1.35(S1) – 1600 <= 0 ---------------------------------------------4

Considering the space:

1.068(P2) + 0.111(H2) + 0.174(S2) <= 382--------------------------------------------------------------------------------5

Revenue = 1.5*8(P2) + 1.5(H2) + 2.25(S2)

So, Profit would be

12(P2) + 1.5(H2) + 2.25(S2) – [ 6(P2) + 0.45(H2) + 0.90(S2) ] - 1000

= 6(P2) + 1.05(H2) + 1.35(S2) – 1000

6(P2) + 1.05(H2) + 1.35(S2) – 1000 >= 1000 -------------------------------------------------------------------------------6

Game3

Payment to Tech - $ 1000

Cost for food purchase/ cooking – 6(P3) + 0.45(H3) + 0.90(S3)

Now, as she must sustain from the previous profit only,

6(P3) + 0.45(H3) + 0.90(S3) <= 6(P2) + 1.05(H2) + 1.35(S2) – 1000

Or,

6(P3) + 0.45(H3) + 0.90(S3) - 6(P2) - 1.05(H2) - 1.35(S2) – 1000 <= 0 ---------------------------------------------7

Considering the space:

1.068(P3) + 0.111(H3) + 0.174(S3) <= 382--------------------------------------------------------------------------------8

Revenue = 1.5*8(P3) + 1.5(H3) + 2.25(S3)

So, Profit would be

12(P3) + 1.5(H3) + 2.25(S3) – [ 6(P3) + 0.45(H3) + 0.90(S3) ] - 1000

= 6(P3) + 1.05(H3) + 1.35(S3) – 1000

6(P3) + 1.05(H3) + 1.35(S3) – 1000 >= 1000 -----------------------------------------------------------------------------9

Game’j’    for j = 4, 5, 6

Payment to Tech - $ 1000

Cost for food purchase/ cooking – 6(P_j) + 0.45(H_j) + 0.90(S_j)

Now, as she must sustain from the previous profit only,

6(P_j) + 0.45(H_j) + 0.90(S_j) <= 6(P_j-1) + 1.05(H_j-1) + 1.35(S_j-1) – 1000

Or,

6(P_j) + 0.45(H_j) + 0.90(S_j) - 6(P_j-1) - 1.05(H_j-1) - 1.35(S_j-1) – 1000 <= 0 --------------------------------10, 13 and 16

Considering the space:

1.068(P_j) + 0.111(H_j) + 0.174(S_j) <= 382---------------------------------------------------------------------11, 14 and 17

Revenue = 1.5*8(P_j) + 1.5(H_j) + 2.25(S_j)

So, Profit would be

12(P_j) + 1.5(H_j) + 2.25(S_j) – [ 6(P_j) + 0.45(H_j) + 0.90(S_j) ] - 1000

= 6(P_j) + 1.05(H_j) + 1.35(S_j) – 1000

6(P_j) + 1.05(H_j) + 1.35(S_j) – 1000 >= 1000 ------------------------------------------------------------------12, 15 and 18

P_i >= 0

H_i >= 0

S_i >= 0                              for i = 1, 2, 3, 4, 5, 6

Demand constraint

H_i + S_i >= (P_i /8)

H_i >= 2*S_i                        for i = 1, 2, 3, 4, 5, 6

Objective Function (max.Z) = 6(P1+P2+….+P6) + 1.05(H1+H2+…..+H6) + 1.35(S1+S2+…+S6) - 6600