1 Question:

Use the Linear Optimaziton method to find the most efficient way for paper recycling.

1.1 Conditions:

1.1.1 Material

  • 70 tons of newspaper
  • 50 tons of mixed paper
  • 30 tons of white office paper
  • 40 tons of cardboard

1.1.2 Production:

  • 60 newsprint pulp
  • 40 packaging paper pulp
  • 50 print stock pulp

2 Solution

2.1 Linear Model

2.1.1 Minimize the Total Cost

\[ 13x{15} + 12x{16} + 11x{25}+ 13x{26}+ 9x_{35} + 10x_{36}+ 13x_{45}+ 14x_{46}+ 5x_{57}+ 6x_{58} + 8x_{59} + 6x_{67}+ 8x_{68} +7x_{69}\]

2.1.2 Constraints:

2.1.2.1 General Constraints:

x_{ij} >= 0 for all \(ij\)

2.1.2.2 Raw Material constraints :

  • newspaper : \(x_{15}+ x_{16} \leq 70\)

  • mixed paper \(x_{25}+ x_{26} \leq 50\)

  • White office paper \(x_{35}+ x_{36} \leq 30\)

  • cardboard \(x_{45}+ x_{46} leq 40\)

2.1.2.3 Recycling Processes Constraints:

  • Recycle Method 1: \(0.9x_{15}+ 0.8x_{25}+ 0.95x_{35}+ 0.75x_{45}- x_{57} -x_{58} - x_{59} \geq 0\)

  • Recycle Method 2: \(0.85x_{16}+ 0.85x_{26}+ 0.9x_{36}+ 0.85x_{46}- x_{67} -x_{68} - x_{69} \geq 0\)

2.1.2.4 Production Paper Constraints:ΒΆ

newsprint : \(0.95x_{57}+ 0.9x_{67} \geq 60\)

packaging paper: \(0.9x_{58}+ 0.95x_{68} \geq 40\)

stock paper: \(0.9x_{59}+ 0.95x_{69} \geq 50\)

#minimize
obj_fun <- c(13,12,11,13,9,10,13,14,5,6,8,6,8,7)

#constrain
constr <- matrix (c(1,1,0,0,0,0,0,0,0,0,0,0,0,0,
                    0,0,1,1,0,0,0,0,0,0,0,0,0,0,
                    0,0,0,0,1,1,0,0,0,0,0,0,0,0,
                    0,0,0,0,0,0,1,1,0,0,0,0,0,0,
                    -0.95,0,-0.8,0,-0.95,0,-0.75,0,1,1,1,0,0,0,
                    0,-0.85,0,-85,0,-0.9,0,-0.85,0,0,0,1,1,1,
                    0,0,0,0,0,0,0,0,-0.95,0,0,-0.9,0,0,
                    0,0,0,0,0,0,0,0,0,-0.9,0,0,-0.95,0,
                    0,0,0,0,0,0,0,0,0,0,-0.9,0,0,-0.95), ncol=14, byrow=TRUE)

constr_dir <- c("<=",
                "<=",
                "<=",
                "<=",
                ">=",
                ">=",
                ">=",
                ">=",
                ">=")
                    
rhs <-c(70,50,30,40,0,0,-60,-40,-50)

2.2 Output

As we can see the Minimun Price for the Recycling is $3447.735 .

prod_sol <- lp("max", obj_fun, constr, constr_dir, rhs, compute.sens=TRUE)
prod_sol$objval
## [1] 3447.735