[M]-Integer-Optimisation.md

Integer Optimisation

Sample Task 1

Maximise profits with the given constraints.

Maximise total profit using decision variables X1 = number of apples to sell, X2 = whether to sell one orange or not	Profit = 11X1 + 12X2
Budget Constraint	0.1X1 + 0.2X2 ≤ 12
Space Constraint	1.1X1 + 1.2X2 ≤ 34
Contract Constraint (X1)	X1 ≥ 5
Non-negativity Constraint (X1)	X1 ≥ 0
Non-negativity Constraint (X2)	X2 ≥ 0
Integer Constraint (X1)	X1 is an integer
Binding Constraint (X2)	X2 is binary

Preparation Code

# Functions
library(lpSolve)

Actual Code

1. Run the integer optimisation model.

# Define parameters
objective.fn <- c(11, 22)

const.mat <- (matrix(
  c(0.1, 0.2,
    1.1, 1.2,
    1, 0), # R assumes non-negativity, so the non-negativity constraint does not need to be specified
  ncol = 2, # Number of decision variables
  byrow = TRUE))

const.dir <- c("<=","<=",">=")

const.rhs <- c(12,34,5)

# Run the model
lp.solution <- lp("max", objective.fn, const.mat, const.dir, const.rhs,
                  int.vec = 1, binary.vec = 2, # For other questions ----- eg int.vec = c(1,2,3) / all.int = TRUE
                  compute.sens = FALSE)

lp.solution
lp.solution$solution

2. If required, obtain shadow prices.

lp.solution$duals

3. If required, conduct sensitivity analysis.

lp.solution$sens.coef.to # Upper bounds to the coefficients
lp.solution$sens.coef.from # Lower bounds to the coefficients

Sample Task 2

Minimise costs with the given constraints.

Minimise total cost using decision variables X1 = number of apples to sell, X2 = whether to sell one orange or not	Cost = 11X1 + 12X2
Budget Constraint	0.1X1 + 0.2X2 ≤ 12
Space Constraint	1.1X1 + 1.2X2 ≤ 34
Contract Constraint (X1)	X1 ≥ 5
Non-negativity Constraint (X1)	X1 ≥ 0
Non-negativity Constraint (X2)	X2 ≥ 0
Integer Constraint (X1)	X1 is an integer
Binding Constraint (X2)	X2 is binary

Preparation Code

# Functions
library(lpSolve)

Actual Code

1. Run the integer optimisation model.

# Define parameters
objective.fn <- c(11, 22)

const.mat <- (matrix(
  c(0.1, 0.2,
    1.1, 1.2,
    1, 0), # R assumes non-negativity, so the non-negativity constraint does not need to be specified
  ncol = 2, # Number of decision variables
  byrow = TRUE))

const.dir <- c("<=","<=",">=")

const.rhs <- c(12,34,5)

# Run the model
lp.solution <- lp("min", objective.fn, const.mat, const.dir, const.rhs,
                  int.vec = 1, binary.vec = 2, # For other questions ----- eg int.vec = c(1,2,3) / all.int = TRUE
                  compute.sens = FALSE)

lp.solution
lp.solution$solution

2. If required, obtain shadow prices.

lp.solution$duals

3. If required, conduct sensitivity analysis.

lp.solution$sens.coef.to # Upper bounds to the coefficients
lp.solution$sens.coef.from # Lower bounds to the coefficients