...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 |Value |Cost |Coefficient |Increase |Decrease ...
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...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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...Julia’s Food Booth A.Formulate and solve an L.P. model for this case. Decision Variables | X1 | # of Pizzas | X2 | # of Hotdogs | X3 | # of BBQ Sandwiches | Objective Function | | X1 | X2 | X3 | Selling Price | $ 1.50 | $ 1.50 | $ 2.25 | Cost | $ 0.75 | $ 0.45 | $ 0.90 | Profit | $ 0.75 | $ 1.05 | $ 1.35 | Maximize total profit Z=$0.75X1+$1.05X2+$1.35X3 Budget Constraints: $0.75X1+$0.45X2+$0.90X3<=$1,500 Space Constraints: 3x4x16= 196 Sq. Ft., 192x12x12=27,648 In Sq. Oven will be filled at the beginning and halftime of the game: 2x27, 648=55,296 In. Sq. Space for Pizza: 14x14=196 In. Sq., 196/8(Slices) =24.5 In. Sq., Round down to 24 Space in oven constraint: 24X1+16X2+25X3<=55,296 She can sell at least as many pizzas as hotdogs and BBQ sandwiches: X1>=X2+X3 She can sell twice as many hotdogs as BBQ sandwiches: X2/X3>=2, X2-2X3>=0 X1, X2, X3 >=0 Linear Program: | | | | | Maximize total profit Z=$0.75X1+$1.05X2+$1.35X3 | Subject to the following constraints: | | | 1 | $0.75X1+$0.45X2+$0.90X3<=$1,500 | | 2 | 24X1+16X2+25X3<=55,296 inch squared | | 3 | X1-X2-X3>=0 | | | | 4 | X2-2X3>=0 | | | | | 5 | X1, X2, X3 >=0 | | | | Linear Program Solved: Maximize Total Profit Z=$0.75(1,250) +$1.05(1,250) +$1.35(0) Maximize Total Profit Z=$2,250 Constraints: 1. $0.75(1,250)+$0.45(1,250)+$0.90(0)<=$1,500 $1,500<=$1...
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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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...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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...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 $1,500 on hand to purchase and prepare food for the first home game. A constraint must be formed for the budget. Cost of each item and money available is what is known, so it is easy to form the...
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...|Case 3 - Julia's Food Booth | | | | | | | | | | | | | | | | |Constraints: | | | |Available |Usage |Left over | | |Budget ($) |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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...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 Kenneth W. Dayton Strayer University 11/25/2011 Math 540 Quantitative Reasoning Julia Robertson is a senior at Tech University and wants to find a way to make extra money to finance her final year at school. She is considering leasing a food booth outside the Tech stadium during home football games. Julia knows the games sell out and the crowd eats a lot of food. Julia has spoken with Ken a consultant and has agreed to investigate her leasing a food booth. Ken will look at the following items to determine if it is a good decision. The first item is creating an L.P. model (linear progression) to help understand the constraints and how to maximize her profit. Julia was asking if she should borrow money from a friend so this will also be evaluated. Julia wants to hire friend to help her with the booth and she is uncertain about the impact of the model. By looking at the L.P. model Ken can determine her best course of action and if she should lease the booth. The first step is to create the L.P. model to determine an objective function. The objective function ken determines to use is Max Z= .75p+1.05h+1.35b which can be done by looking at the information available such as items being sold, how much those items cost Julia to buy, how much she plans to sell each item for, looking at how one item can sell as much as or more than another item. Julia wants to sell Pizza, Hot Dogs, and Bar-B-Q sandwiches. Julia is going to have a Pizza company deliver a 14 inch...
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...creative genius who changed the culinary world with her energetic personality and fine cooking skills. She is a fine example of the gusto-olfactory intelligence with special accents of visual spatial and interpersonal intelligence. There were many chefs that could fall in this intelligence category, including escoffier, beard, and farmer. Choosing Julia Child wasn’t hard, she is the women who broke new ground in the world of cooking. She empowered women to cook and different style of cooking. A gusto olfactory intelligent person is able to create a specific taste, to create their own additions to recipes. Julia as always The Begin Of Her Career As A Chef. 6 have a love for cooking food. After moving to Paris, the love develop even more. So while going to school, she decided that she wanted to make a cookbook. Having a goal of adapting sophisticated French cuisine for mainstream Americans. Julia earned a $750 advance for the work. However, the book was 734 pages long, it weighs 3-lb. The book was in September 1961, under the title mastering the art of French cooking. The book was being considered groundbreaking. It was remained the bestselling cooking for five straight years after its publication. Which was really amazing, and considering Julia was a female. Which in the late 1900, a lot of female wasn’t publicating cookbook’s (Biography.com, 2017). Julia Child was the face of French cooking. Julia has groundbreaking...
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...Running Head: Julia’s Food Booth Strayer University April 2012 Julia is a senior at Tech, and a small entrepreneur. She wants to lease a food booth outside the Tech stadium for the home football games, so she can make profit to finance a final year. Tech sells out every home game, and the one thing Julia knows from attending the every games, is that everyone eats a lot of food. She has 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. Then, she had a great idea, 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. A. Formulate a linear programming model for this case X1= the number of slices of pizza X2=the number of hot dog X3=the number of sandwiches 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=$6/8=$0.75 For hot dog= 1.50-0.45=1.05 For sandwishes=2.25-0.90=1.35 Maximize; Z = $0.75x1 + $1.05x2 + $1.35x3 Subject to: $0.75x1 + $0.45x2 + $0.90x3 = 2.0 X1, X2, X3 >= 0 Constraints; Cost $0.75x1 + $0.45x2 + $0.90x3 =2X3 A)- Yes, she...
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