In the earlier linear optimization problem, we got the solution as X1 (product 1) = 24 and X2 (product 2) = 16. The interesting thing to notice here is that both are whole numbers; you can’t make and sell fractional products! It may only be the case in some cases. Here is an example.
A factory makes three products: chairs, tables and sofas.
One chair makes $150 profit, requires 7 hours to produce, and 3 m2 of space to store
One table makes $250 profit, requires 9 hours to produce, and 6 m2 of space to store
One sofa makes $300 profit, requires 11 hours to produce, and 7 m2 of space to store
Now, the constraints. The resources available for the factory are only sufficient to cover up to 80 hours a week. The warehouse has an area of 40 m3, and based on historical data, a maximum of 4 chairs sell in a week. Estimate an optimised production plan.
Here is an R code and the results.
f.obj <- c(150, 250, 300)
f.con <- matrix (c(7, 9, 11, 3, 6, 7, 4, 0, 0), nrow = 3, byrow = TRUE)
f.dir <- c("<=", "<=", "<=")
f.rhs <- c(80, 40, 8)
results <- lp ("max", f.obj, f.con, f.dir, f.rhs)
results
results$solutionSuccess: the objective function is 1757.143
[1] 2.000000 0.000000 4.857143While making two chairs a week seems reasonable, it doesn’t make any sense to make 4.85 sofas and expect someone to buy a 0.85 item. At the same time, rounding off to 5 will end up violating the two constraints (on resource and space).
So, we will specify integer constraints to variables 1, 2 and 3 by adding ‘int.vec = ‘ in the options.
f.obj <- c(150, 250, 300)
f.con <- matrix (c(7, 9, 11, 3, 6, 7, 4, 0, 0), nrow = 3, byrow = TRUE)
f.dir <- c("<=", "<=", "<=")
f.rhs <- c(80, 40, 8)
results <- lp ("max", f.obj, f.con, f.dir, f.rhs, int.vec = c(1,2,3))
results
results$solutionSuccess: the objective function is 1750
[1] 2 1 4Reference
Linear Optimization: Desmond C. Ong
