Mat 540 Assign 3
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
Solve the LPM
Based on the excel solution the optimum solution:
Pizza (X1) = 1250; Hotdogs(X2) = 1250 and Barbecue sandwiches (X3) = 0
Maximum value of Z = $2250
Julia should stock 1250 slices of pizza, 1250 hot dogs and no barbecue sandwiches.
Maximum Profit = $2250.
|Maximum Profit |$ 2,250.00 | |Booth Rent per game |$ (1,000.00) | |Warming Oven 600 for total of 6 home |$ (100.00) | |games 600/6 =100
(A) Formulate and solve an
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
Solve the LPM
Based on the excel solution the optimum solution:
Pizza (X1) = 1250; Hotdogs(X2) = 1250 and Barbecue sandwiches (X3) = 0
Maximum value of Z = $2250
Julia should stock 1250 slices of pizza, 1250 hot dogs and no barbecue sandwiches.
Maximum Profit = $2250.
|Maximum Profit |$ 2,250.00 | |Booth Rent per game |$ (1,000.00) | |Warming Oven 600 for total of 6 home |$ (100.00) | |games 600/6 =100
(A) Formulate and solve an