...Case Problem Julia's Food Booth Assignment #3 Case Problem Julia's Food Booth A: Formulation of the LP Model X1(Pizza), X2(hotdogs), X3(barbecue sandwiches) Constraints: Cost: Maximum fund available for the purchase = $1500 Cost per pizza slice = $6 (get 8 slices) =6/8 = $0.75 Cost for a hotdog = $.45 Cost for a barbecue sandwich = $.90 Constraint: 0.75X1 + 0.45X2+ 0.90(X3) ≤ 1500 Oven space: Space available = 3 x 4 x 16 = 192 sq. feet = 192 x 12 x 12 =27648 sq. inches The oven will be refilled before half time- 27648 x 2 = 55296 Space required for pizza = 14 x 14 = 196 sq. inches Space required for pizza slice = 196/ 8 = 24.50 sq. inches Space required for a hotdog=16 Space required for a barbecue sandwich = 25 Constraint: 24.50 (X1) + 16 (X2) + 25 (X3) ≤ 55296 Constraint: Julia can sell at least as many slices of pizza(X1) as hot dogs(x2) and barbecue sandwiches (X3) combined Constraint: X1 ≥ X2 + X3 = X1 - X2 - X3 ≥ 0 Julia can sell at least twice as many hot dogs as barbecue sandwiches X2/X3 ≥ 2 = X2 ≥2 X3 =X2 - 2 X3 ≥ 0 X1, X2, X3 >= 0 (Non negativity constraint) Objective Function (Maximize Profit): Profit =Sell- Cost Profit function: Z = 0.75 X1 + 1.05 X2 + 1.35 X3 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 ...
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...Assignment 3 Julia’s food booth A).Formulate and solve a linear programming model for Julia that will help you advise her if she should lease the booth. X1=number of cheese pizza slices X2=number of hot dogs X3=number of BBQ sandwiches Julia Food Booth | | | | | | | | | | | | | | | | Products | Pizza X1 | Hot Dog X2 | Barb Sand X3 | | | | | Profit per unit | $0.75 | $1.05 | $1.35 | Resources | Constraints | | | | Avail | Usage | Left Over | | Budget (i) | $0.75 | $0.45 | $0.90 | 1500 | 1500 | 0 | | Oven Space (ii) | 24 | 16 | 25 | 55,296 | 50000 | 5296 | | Demand (iii) | 1 | -1 | -1 | 0 | -2.3E-13 | 2.27E-13 | | Demand (iv) | 0 | 1 | -2 | 0 | 1250 | -1250 | | | | | | | | | | | | | | | | | | Production | | | | | | | | Pizza | 1250 | | | | | | | Hotdogs | 1250 | | | | | | | Barbecue | 0 | | | | | | | Profit | 2250 | | | | | | | B). If Julia were to borrow 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? Yes, Julia could increase her profit if she borrowed from a friend. The shadow price is 1.50 for each additional dollar that she earns. This was found by looking at the sensitivity analysis report from the computer solution output. The shadow...
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...Assignment #3: Case Problem "Julia's Food Booth" Complete the "Julia's Food Booth" case problem on page 109 of the text. Address each of the issues A - D according the instructions given. (A) Formulate and solve an L.P. model for this case. See Excel worksheet. (B) Evaluate the prospect of borrowing money before the first game. I would suggest that Julia consider borrowing money before the first game to open up her food booth. According to the first constraint, she is subject to a $1,500 budget with a potential to make a profit of $2,250 if she were to sell all her pizza and hot dogs. This result yields a profit of $750 or 50%. Even if no sales were made, the potential is high, considering the opportunity. Plus, I am sure that a small initial investment is not detrimental to her personal funds, to where if things did not go as planned, she could recover the funds. (C) Evaluate the prospect of paying a friend $100/game to assist. I would suggest that Julia consult a friend for $100/game to assist in her food booth. After running a break-even analysis (see Excel), holding all things constant, where she only sells pizza and hot dogs, she would have to sell 67 slices of pizza and 48 hotdogs to break-even after paying her friend $100. This does not seem too farfetched, considering her maximum sales, given these constraints, is 1,250 slices of pizza and 1,250 hot dogs, only about 5% and 4% of maximum sales, respectively. On top of that, Julia may need the help to meet...
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...Running head: JULIA’S FOOD BOOTH 1 Case Study Julia’s Food Booth Susan Dawson Strayer University JULIA’S FOOD BOOTH 2 Case Study: Julia’s Food Booth In this case study, we are asked to determine if Julia Robertson, a senior at Tech, should finance her final year at Tech in part by opening a food booth outside the stadium at home football games. We are also asked to determine if she should borrow more money, hire preparation assistance and what could go wrong. Julia is considering opening a food booth outside the stadium at six (6) home football games at a booth rental charge of $1,000.00 per game. Due to existing regulations, she knows she can only sell food or drinks, not both. She has determined the most popular food items are slices of pizza, hot dogs and barbecue sandwiches and that most food sales occur during the hour before the game and during half-time. She has been able to secure a deal with a pizza delivery company to deliver 14” pizzas, 8 slices each, for $6.00 and, because of demand, she must prepare the other food in advance and keep it warm ahead of sales. She found that she can rent a warming oven with 16 shelves, each 3’ x 4’, for a seasonal cost of $600.00 or $100.00 per game. Julia has $1,500.00 available to purchase food for the first game. Thereafter...
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...Julia's Food Booth Part A (Formulate) | | | | | Step 1: | Define the decision variables: | | | | | | x1 | = | How many hot dogs to produce to maximize profit | | | x2 | = | How many BBQ Sandwiches to produce to maximize profit | | x3 | = | How many Cheese Pizza slices to sell to maximize profit | | | | | | | | | | | Step 2: | Define the objective function. | | | | | | Maximize the profits of Julia's booth. | | | | | | | | | | | | | | | | | | Cost | Selling Price | Profit | | | | | Profit of Pizza = | $ 0.75 | $ 1.50 | $ 0.75 | | | | | Profit of BBQ = | $ 0.90 | $ 2.25 | $ 1.35 | | | | | Profit of Hot Dog = | $ 0.45 | $ 1.50 | $ 1.05 | | | | | | | | | | | | | | Maximize Z = | $1.05x1 | + | $1.35x2 | + | $.75x3 | | | | | | | | | | | Step 3: | Define the constraints: | | | | | | Size of the shelves in the warming oven (space is a constraint). | | | | | | | | | | | | | | Budget Constraint = | $0.45x1 + $0.90x2 + $0.75x3 <=$1,500 | | | | | | | | | | | | | Space Constraint = | Total Space in Oven = | 192 | Sq Ft | | | | | | | | 27648 | in sq | | | | | She is refilling at half time = | 55296 | in sq | | | | | | | | | | | | | | Space required for a pizza = | 196 | in sq | | | | | for a slice of pizza...
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...The three products/variables in this problem that must be considered for purchase are: x1: Pizza Slices x2: Hot Dogs x3: Barbeque Sandwiches The objective is for Julia to maximize profits. Julia’s goal is to earn a profit of at least $1,000.00 after each game. Profit = Sell – Cost Profit Function: Z = 0.75(X1) + 1.05(X2) + 1.35(X3) Constraints and Cost: The maximum amount of funds available for purchase is $1500.00 Cost per pizza slice = $0.75 because Julia purchases each pizza for $6.00 and there are 8 slices per pizza. Cost per hot dog = $0.45 Cost per sandwich = $0.90 LPP Model: Maximize Profit: Z= $0.75x1 + $0.45x2 + $0.90x3 < $1,500 Subject to 24x1 + 16x2 + 25x3 < 55,296 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 Solve the LPM: Pizza(X1) = 1,250; Hotdogs(X2) = 1,250 and Barbecue sandwiches(X3) = 0 Maximum value of Z = $2,250 It would be in Julia’s best interest to stock 1,250 slices of pizza, 1, 250 hot dogs and no barbecue sandwiches as it will yield the maximum profit of $2,250.00 (B) Evaluate the prospect of borrowing money before the first game. I do assert that Julia should borrow money from her friend to increase her profits. The shadow price is $1.50 for each additional dollar Julia earns. The upper limit in the model that is given is $1,658.88. This means that Julia can borrow $158.88 from her friend, which will help her yield...
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...Assignment #3: Case Problem “Julia’s Food Booth” La-Tia Jackson MAT 540 Dr. Albert Yin May 26, 2013 1. Formulate and solve an L.P. model for this case. There are three products or variables in this problem that we must consider for purchase. X1 = number of pizza slices Julia should purchase X2 = number of hotdogs Julia should purchase X3 = number of barbecue sandwiches Julia should purchase. Julia decided to have a food booth in order to make some money. Her goal is to maximize the profit that she can get from selling the hotdogs, pizza, and barbeque sandwiches that she plans on selling. The first thing to do is find the profit that Julia will make per Item. To find that per Item price, the cost of the item will be subtracted from the selling price. Pizza: Julia can buy a pizza that contains 8 slices for $6. That means each slice of pizza will cost her $0.75. She plans to sell each piece for 1.50. $1.50-$0.75= $0.75 profit Hot dog: $1.50 - $0.45 = $1.05 profit Barbecue Sandwiches: $2.25 - $0.90 = $1.35 profit The objective function can now be written since we have found the potential profit of each food item. The objective of this function is to maximize Z (profit). Z= $0.75x1 + $1.05x2 + $1.35X3 Budget is the one thing that has to be taken into consideration. Julia has $1,500 on hand to purchase and prepare food for the first home game. A constraint must be formed for the budget. Cost of each item and money available is what is known, so it is easy to form the...
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...Case Problem” Julia’s Food Booth” Julia’s objective is to maximize total profit; the total profit is the summary of the individual profits gained from each pizza slice, hot dog, and barbeque sandwich. I am going to estimate the profit for each item as difference between total revenue and total cost of the item. Total revenue Total cost Profit Pizza slice $1.50 $6/8 slices = $0.75 $0.75 Hot dog $1.50 $0.45 $1.05 Barbeque sandwich $2.25 $0.90 $1.35 x1 - number of pizza slices x2 - number of hot dogs x3 - number of barbeque sandwiches The profit function: maximize Z = $0.75* x1 + $1.05 * x2 + $1.35 * x3 There Z = total profit per game. The oven size constraint is formulated as 16 shelves 3*4 feet each which will be filled twice for each game, so the maximum possible size of the stove is: (((12*3)*(12*4))*16)*2 = 55,296 square inches. Estimate for each item in regards to the space: Pizza slice (14*14)/8 = 24.5 square inches (keeping in mind that she will keep whole pizza warm) Hot dog ...
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...|Case 3 - Julia's Food Booth | | | | | | | | | | | | | | | | |Constraints: | | | |Available |Usage |Left over | | |Budget ($) |0.75 |0.45 |0.90 | 1,500 | 1,500.00 |0 | | |Oven space (sq. in.) |24 |16 |25 | 55,296 | 50,000.00 |5296 | | |Demand |1 |-1 |-1 |0 | - |0 | | |Demand |0 |1 |-2 |0 | 1,250.00 |-1250 | | | | | | | | | | | |Adjustable Cells | | | | | | | | | |Final |Reduced |Objective |Allowable |Allowable | | |Cell |Name ...
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...Assignment #3: Case Problem "Julia's Food Booth" Complete the "Julia's Food Booth" case problem on page 109 of the text. Address each of the issues A - D according the instructions given. (A) Formulate and solve an L.P. model for this case. There are three products or variables in this problem that we must consider for purchase. X1 = number of pizza slices Julia should purchase X2 = number of hotdogs Julia should purchase X3 = number of barbecue sandwiches Julia should purchase. The reason why Julia is having a booth is to make some money. She wants to maximize her profit that she can get from selling the hotdogs, pizza, and barbeque sandwiches. The first thing to do is find the profit that Julia will make per Item. To find that per Item price, the cost of the Item will be subtracted from the selling price. Pizza: Julia can buy a pizza that contains 8 slices for 6$. That means each slice of pizza will cost her $0.75. She plans to sell each piece for 1.50. $1.50-$0.75= $0.75 profit Hot dog: $1.50 - $0.45 = $1.05 profit Barbecue Sandwiches: $2.25 - $0.90 = $1.35 profit The objective function can now be written since we have found the potential profit of each food item. The objective of this function is to maximize Z(profit). Z= $0.75x1 + $1.05x2 + $1.35X3 The Budget is one thing that has to be taken into consideration. Julia has $1,500 on hand to purchase and prepare food for the first home game. A constraint must be...
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... In the case study regarding Julia’s food booth a solution needs to be determined as to if investing in a booth to sell food during Tech football games is a sound investment. Julia is considering leasing a booth to sell food outside of Tech stadium at home football games to generate revenue in order to finance her final year at the school. After all of her expenses are paid Julia wants to bring in a profit of $ 1,000 per game. If she can accomplish this then she feels that it will be worth the money to invest in a booth. The fixed cost associated with leasing the booth would include $1,000 per game for booth rental and $600 to lease a warming oven for the six home football games. Julia’s has $1,500 available cash, which she will use to purchase and prepare the food for the first game. For the remaining home games she plans on using funds generated from the previous game to purchase her ingredients. In order to help Julia determine if she should lease a booth a linear programming model should be formulated and solved. The first step in setting up the model is to determine the variables. The variables for this case include: x1 =Pizza, x2 = Hotdogs, and x3= BBQ sandwiches. These variables are the three food choices Julia wishes to sell at her booth. Julia’s objective function in this case is to maximize profit (maximize Z). Julia’s objective function in this case equates to Maximize Z = .75x1 + 1.05x2 + 1.35x3 =0. Within operating the booth there are some constraints that...
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...This paper will display findings of data constructed to aid Julia in deciding if she should open a food booth during the football season at Tech High School. The team at Tech High plays 6 home games during the regular season, during which she would be responsible for providing food for the booth, rent for the cost of the booth, and an oven to prepare the food. Julia concluded that there are 2 constraints that she must consider: 1) the cost of the food & 2) the size of the oven. In addition she believes that she should be able to sell at least the same amount of hot dogs and bbq sandwiches to pizza, and also double the amount of hot dogs to bbq sandwiches. Considering these constraints the linear programming model shows Julia what her profit should be, it will also help Julia decide if she should take out a loan prior to the game opener to successfully operate the booth. It will help here conclude the amount of product she will need to sell to maximize her profit. After reviewing the model, Julia’s will have to decide 3 different situations that could impact her business. First should she take out a loan, and why? Second, Should she hire someone to help and why? And third what uncertainties could exist and how will they affect the business? Linear Programming Model for Julia’s Food Booth DECISION VARIABLES Hot dogs = x1 Pizza= x2 Barbeque Sandwiches=x3 OBJECTIVE FUNCTION: Maximize Z= $1.05x1+$0.75x2+$1.35x3 Where Z= Profit from selling hot dogs, pizza and barbeque...
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...After formulating and solving a linear programming model for Julia’s Food Booth, I feel that Julia should lease the food booth at the Tech stadium football games. Yes. Julia should lease the booth because she would potentially make a profit of $1150 ($2250-$1100) each game; this is more than what she needs to run the booth. In addition, because she has cash on hand of $1500 to purchase and prepare the food items for the first game, Julia would not have to use any of the money that she makes to buy ingredients for the next game. She could still use the cash that she has on hand to buy ingredients for the next game. Next, if Julia were to borrow some more money from a friend before the first game to purchase more ingredients, she could increase her profit. Julia should borrow $138.40, so that she can make an additional profit of $207.60. An allowable increase in the ‘budget’ constrains her from borrowing more money than this. Because the shadow price for the budget constraint is 1.50 and the allowable increase is 138.40, the budget can go up to 1638.40 and the profit would increase by 1.50 for every dollar that is added to the budget. Also, Julia should hire a friend of hers to help her for $100 per game. It would be physically difficult and probably too much for her to prepare all of the hot dogs and BBQs before the game and during half time. If she borrows the extra $138.40, she can use a portion of this money to pay her friend without losing out on any money. ...
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...Julia’s Food Booth MAT 540 February 22, 2014 Julia’s Food Booth Case Problem (A) Formulate and solve an LP model. Variables: X1 = pizza slices, X2 = hot dogs, and X3 = barbeque sandwiches Maximize Z = ($0.75 X1) + ($1.05 X2) + ($1.35 X3) Subject to: $0.75x1+ $0.45x2 + $0.90x3 ≤ $1,500 24x1 + 16x2 +25x3 ≤ 55.296in of oven space X1 ≥ x2 + x3 (change to –x1 + x2 + x3 ≤ 0 for constraint) X2/x3 ≥ 0 Solution: X1 = 1250 pizza slices X2 = 1250 hot dogs X3 = 0 barbeque sandwiches Z = $2,250 (B) Evaluate the prospect of borrowing money before the first game. Yes, I do believe Julia would increase her profit if she borrowed money. The shadow price is $1.50 for each additional dollar she earns. The upper limit in the model that is given is $1,658.88. This means that Julia can borrow $158.88 from her friend, which gives her an extra profit of $238.32 or a total profit of $2,488.32. (C) Evaluate the prospect of paying a friend $100/game to assist. According to the information presented in (A) and (B), I do believe Julia should hire her friend for $100 per game. It would be difficult for Julia to prepare all of the food needed within the amount of time to reach her goal, so she will need the additional help. If she is borrowing extra money from another friend, she would be able to pay the help for the time spent helping at the game because the $158.88 she borrowed will allow her to do so. (D) Analyze the impact of uncertainties on the model. An impact of...
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...MAT 540 WK 4 Homework Chapter 15 MAT 540 WK 5 Midterm Exam MAT 540 WK 6 Homework Chapter 2 MAT 540 WK 6 Quiz 3 Chapter 2 MAT 540 WK 7 Assignment 3 Case Problem – Julia’s Food Booth MAT 540 WK 7 Homework Chapter 3 MAT 540 WK 8 Homework Chapter 4 MAT 540 WK 8 Quiz 4 Chapter 4 MAT 540 WK 9 Homework Chapter 5 MAT 540 WK 9 Quiz 5 Chapter 9 MAT 540 WK 10 Assignment 4 Case Problem – Stateline Shipping and Transport Company MAT 540 WK 10 Homework Chapter 6 MAT 540 WK 11 Final Exam Course Home Work aims to provide quality study notes and tutorials to the students of MAT 540 Entire Course Latest Strayer in order to ace their studies. MAT 540 ENTIRE COURSE LATEST STRAYER To purchase this visit following link: https://coursehomework.com/product/mat-540-entire-course-latest-strayer/ Contact us at: HELP@COURSEHOMEWORK.COM MAT 540 ENTIRE COURSE LATEST STRAYER MAT 540 WK 1 Homework Chapter 1,11 MAT 540 WK 1 Quiz 1 Chapter 1,11 MAT 540 WK 2 Homework Chapter 12 MAT 540 WK 2 Quiz 2 Chapter 11,12 MAT 540 WK 3 Assignment 1 – JET Copies Case Problem MAT 540 WK 3 Homework Chapter 14 MAT 540 WK 4 Assignment 2 – Internet Field Trip MAT 540 WK 4 Homework Chapter 15 MAT 540 WK 5 Midterm Exam MAT 540 WK 6 Homework Chapter 2 MAT 540 WK 6 Quiz 3 Chapter 2 MAT 540 WK 7 Assignment 3 Case Problem – Julia’s Food Booth MAT 540 WK 7 Homework Chapter 3 MAT 540 WK 8 Homework...
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