Linear Optimization

These two examples are ones we covered in class. One minimization of cost of production of to products and one more complex maximization on a split for a mutual fund.

Libraries Needed

Load these libraries into your environment. Remember you will like need to remove the hashtag in front of install.packages(“lpSolve”) to run the code in r

# Install the lpSolve package if you haven't
#install.packages("lpSolve")

# Load the package
library(lpSolve)
Warning: package 'lpSolve' was built under R version 4.3.3

Min Example

Step 1 — Mathematical Formulation of the Linear Programming Model

Decision Variables

Let

[

\[\begin{aligned} A &= \text{number of gallons of product A to produce} \\ B &= \text{number of gallons of product B to produce} \end{aligned}\]

]

Objective Function

Minimize the total production cost:

\[ Z = 2A + 3B \]

Subject to the Constraints

  1. Customer demand for product A:
    \[A \ge 125 \]

  2. Minimum combined production requirement:
    \[ A + B \ge 350 \]

  3. Processing time limitation (600 hours):
    \[ 2A + B \le 600 \]

  4. Non-negativity constraints:
    \[A,B \ge 0 \]

This code will walk you through the set up of the minimization progment and then the constraints

Minimization

# Objective coefficients (for 2A + 3B)
objective <- c(2, 3)

Constraints

# Constraint matrix
constraints <- matrix(c(
  1, 0,   # A >= 125
  1, 1,   # A + B >= 350
  2, 1    # 2A + B <= 600
), nrow = 3, byrow = TRUE)

# Direction of constraints
direction <- c(">=", ">=", "<=")

# Right-hand side of constraints
rhs <- c(125, 350, 600)

Linear Formula

# Solve the LP
solution <- lp(direction = "min", objective.in = objective, const.mat = constraints, const.dir = direction,const.rhs = rhs)

Answer

# Display results
solution$objval       # Minimum total cost
[1] 800
solution$solution      # Optimal values for A and B
[1] 250 100

Max Example

Step 1 — Welte Mutual Funds

Decision Variables

Let

[

\[\begin{aligned} x_1 &= \text{dollars invested in Atlantic Oil} \\ x_2 &= \text{dollars invested in Pacific Oil} \\ x_3 &= \text{dollars invested in Midwest Steel} \\ x_4 &= \text{dollars invested in Huber Steel} \\ x_5 &= \text{dollars invested in Government Bonds} \end{aligned}\]

]

Objective Function

Maximize the total projected return:

\[ \text{Maximize } Z = 0.073x_1 + 0.103x_2 + 0.064x_3 + 0.075x_4 + 0.045x_5 \]

Subject to the Constraints

  1. Total investment (available funds):
    \[ x_1 + x_2 + x_3 + x_4 + x_5 = 100{,}000\]

  2. Oil industry limit:
    \[ x_1 + x_2 \le 50{,}000 \]

  3. Steel industry limit:
    \[ x_3 + x_4 \le 50{,}000 \]

  4. Government bonds requirement (≥ 25 % of steel investment):
    \[ x_5 \ge 0.25(x_3 + x_4) \] or equivalently \[ -x_5 + 0.25x_3 + 0.25x_4 \le 0 \]

  5. Pacific Oil limit (≤ 60 % of total oil investment):
    \[ x_2 \le 0.60(x_1 + x_2) \]
    or equivalently \[ -0.6x_1 + 0.4x_2 \le 0 \]

  6. Non-negativity:
    \[ x_1,,x_2,,x_3,,x_4,,x_5 \ge 0 \]

Maximizations

# Objective coefficients
objective <- c(0.073, 0.103, 0.064, 0.075, 0.045)

Constraints

# Constraint matrix
constraints <- matrix(c(
  # x1  x2   x3   x4   x5
   1,   1,   1,   1,   1,      # total investment = 100000
   1,   1,   0,   0,   0,      # oil ≤ 50000
   0,   0,   1,   1,   0,      # steel ≤ 50000
   0,   0,   0.25, 0.25, -1,   # bond ≥ 0.25*(steel)
  -0.6, 0.4,  0,   0,   0      # Pacific Oil ≤ 0.6*(oil)
), nrow = 5, byrow = TRUE)

# Direction of constraints
directions <- c("=", "<=", "<=", "<=", "<=")

# Right-hand side of constraints
rhs <- c(100000, 50000, 50000, 0, 0)

Linear Formula

# Solve the LP
solution <- lp(direction = "max",
               objective.in = objective,
               const.mat = constraints,
               const.dir = directions,
               const.rhs = rhs)

Answer

# Results
solution$status        # 0 = optimal
[1] 0
solution$objval        # Maximum return
[1] 8000
solution$solution      # Optimal investment in x1-x5
[1] 20000 30000     0 40000 10000