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LP model Using
Excel & R
Product mix problem - Beaver Creek Pottery
Company
How many bowls and mugs should be
produced to maximize profits given labor and
materials constraints?
ï‚— Step 1: define decision variables
ï‚— Let x1=number of bowls to produce/day
ï‚— x2= number of mugs to produce/day
ï‚— Step 2: define the objective function
ï‚— maximize Z = $40x1 + 50x2
ï‚— where Z= profit per day
ï‚— Step 3: state all the resource constraints
ï‚—
ï‚— constraint 1) 1x1 + 2x2 <= 40 hours of labor ( resource
ï‚— constraint 2) 4x1 + 3x2 <= 120 pounds of clay
(resource
ï‚— Step 4: define non-negativity constraints
ï‚— x1>=0; x2 >= 0
ï‚— Complete Linear Programming Model:
ï‚— maximize Z=$40x1 + 50x2
ï‚— subject to
ï‚— 1x1 + 2x2 <= 40
ï‚— 4x2 + 3x2 <= 120
ï‚— x1, x2 >= 0
Maximize Z = $40x1 + $50x2
subject to:
1x1 + 2x2 <= 40
4x1 + 3x2 <=120
x1, x2 >= 0
Maximize Z = $100x1 + $50x2
subject to:
1x1 + 2x2 <= 40
4x1 + 3x2 <= 120
x1, x2 >= 0
Maximize Z = $40x1 + $100x2
subject to:
1x1 + 2x2 <= 40
4x1 + 3x2 <= 120
x1, x2 >= 0
Final presentation
Final presentation
Final presentation
Final presentation
Final presentation
Using R

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Final presentation

  • 2. Product mix problem - Beaver Creek Pottery Company How many bowls and mugs should be produced to maximize profits given labor and materials constraints?
  • 3. ï‚— Step 1: define decision variables ï‚— Let x1=number of bowls to produce/day ï‚— x2= number of mugs to produce/day ï‚— Step 2: define the objective function ï‚— maximize Z = $40x1 + 50x2 ï‚— where Z= profit per day ï‚— Step 3: state all the resource constraints ï‚— ï‚— constraint 1) 1x1 + 2x2 <= 40 hours of labor ( resource ï‚— constraint 2) 4x1 + 3x2 <= 120 pounds of clay (resource ï‚— Step 4: define non-negativity constraints ï‚— x1>=0; x2 >= 0 ï‚— Complete Linear Programming Model: ï‚— maximize Z=$40x1 + 50x2 ï‚— subject to ï‚— 1x1 + 2x2 <= 40 ï‚— 4x2 + 3x2 <= 120 ï‚— x1, x2 >= 0
  • 4. Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 <= 40 4x1 + 3x2 <=120 x1, x2 >= 0
  • 5. Maximize Z = $100x1 + $50x2 subject to: 1x1 + 2x2 <= 40 4x1 + 3x2 <= 120 x1, x2 >= 0
  • 6. Maximize Z = $40x1 + $100x2 subject to: 1x1 + 2x2 <= 40 4x1 + 3x2 <= 120 x1, x2 >= 0