SlideShare una empresa de Scribd logo
1 de 12
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?
 Step 1: define decision variables
 Let x1=number of bowls to produce/day
 x2= number of mugs to produce/day
 Step 2: define the objective function
 maximize Z = $40x1 + 50x2
 where Z= profit per day
 Step 3: state all the resource constraints

 constraint 1) 1x1 + 2x2 <= 40 hours of labor ( resource
 constraint 2) 4x1 + 3x2 <= 120 pounds of clay
(resource
 Step 4: define non-negativity constraints
 x1>=0; x2 >= 0
 Complete Linear Programming Model:
 maximize Z=$40x1 + 50x2
 subject to
 1x1 + 2x2 <= 40
 4x2 + 3x2 <= 120
 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
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.  Step 1: define decision variables  Let x1=number of bowls to produce/day  x2= number of mugs to produce/day  Step 2: define the objective function  maximize Z = $40x1 + 50x2  where Z= profit per day  Step 3: state all the resource constraints   constraint 1) 1x1 + 2x2 <= 40 hours of labor ( resource  constraint 2) 4x1 + 3x2 <= 120 pounds of clay (resource  Step 4: define non-negativity constraints  x1>=0; x2 >= 0  Complete Linear Programming Model:  maximize Z=$40x1 + 50x2  subject to  1x1 + 2x2 <= 40  4x2 + 3x2 <= 120  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
  • 7.
  • 8.
  • 9.
  • 10.
  • 11.