Mat540 Julia's Food Booth Case Problem
Formulation of the LPP
Suppose Julia stock X1 numbers of pizza slices, X2 numbers of hot dogs and X2 numbers of sandwiches.
Constraints:
1. On the 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 during half time. Thus total space available = 27648 x 2 = 55296 Space required for pizza = 14 x 14 = 196 sq. inches Space required for pizza slice = 196/ 8 = 24.5 sq. inches Total Space required: 24.5 X1 + 16 X2 + 25 X3 Constraint: 24.5 X1 + 16 X2 + 25 X3 ≤ 55296
2. On the cost: Maximum fund available for the purchase = $1500 Cost per pizza slice = 6/8 = $0.75 Funds required: 0.75 X1 + 0.45 X2 + 0.90 X3 Constraint: 0.75 X1 + 0.45 X2 + 0.90 X3 ≤ 1500
3. On demand: She can sell at least as many slices of pizza as hot dogs and barbecue sandwiches combined Constraint: X1 ≥ X2 + X3 [pic]X1 - X2 - X3 ≥ 0
She can sell at least twice as many hot dogs as barbecue sandwiches X2/X3 ≥ 2 (at least twice as many hot dogs as barbeque sandwiches) X2 ≥2 X3 [pic] X2 - 2 X3 ≥ 0
Objective Function (Profit):
Profit on pizza slice = $1.50 - $0.75 = $ 0.75
Profit on hot dog = $1.50 – 0.45 = $1.05
Profit on sandwich = $2.25 - $0.90 = $1.35
Profit function: Z = 0.75 X1 + 1.05 X2 + 1.35 X3
A) 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 Solution Solution of the LPP by using Excel Solver gives the following Report
|Target Cell (Max) | | | | | |
| | | | | | |
| |Cell |Name
Suppose Julia stock X1 numbers of pizza slices, X2 numbers of hot dogs and X2 numbers of sandwiches.
Constraints:
1. On the 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 during half time. Thus total space available = 27648 x 2 = 55296 Space required for pizza = 14 x 14 = 196 sq. inches Space required for pizza slice = 196/ 8 = 24.5 sq. inches Total Space required: 24.5 X1 + 16 X2 + 25 X3 Constraint: 24.5 X1 + 16 X2 + 25 X3 ≤ 55296
2. On the cost: Maximum fund available for the purchase = $1500 Cost per pizza slice = 6/8 = $0.75 Funds required: 0.75 X1 + 0.45 X2 + 0.90 X3 Constraint: 0.75 X1 + 0.45 X2 + 0.90 X3 ≤ 1500
3. On demand: She can sell at least as many slices of pizza as hot dogs and barbecue sandwiches combined Constraint: X1 ≥ X2 + X3 [pic]X1 - X2 - X3 ≥ 0
She can sell at least twice as many hot dogs as barbecue sandwiches X2/X3 ≥ 2 (at least twice as many hot dogs as barbeque sandwiches) X2 ≥2 X3 [pic] X2 - 2 X3 ≥ 0
Objective Function (Profit):
Profit on pizza slice = $1.50 - $0.75 = $ 0.75
Profit on hot dog = $1.50 – 0.45 = $1.05
Profit on sandwich = $2.25 - $0.90 = $1.35
Profit function: Z = 0.75 X1 + 1.05 X2 + 1.35 X3
A) 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 Solution Solution of the LPP by using Excel Solver gives the following Report
|Target Cell (Max) | | | | | |
| | | | | | |
| |Cell |Name