Julia's Food Booth Case Problem
Julia’s Food Booth 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 - 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 | $
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 | $