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-- |
-- Module : Documentation.SBV.Examples.Optimization.Production
-- Copyright : (c) Levent Erkok
-- License : BSD3
-- Maintainer: erkokl@gmail.com
-- Stability : experimental
--
-- Solves a simple linear optimization problem
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module Documentation.SBV.Examples.Optimization.Production where
import Data.SBV
-- | Taken from
--
-- A company makes two products (X and Y) using two machines (A and B).
--
-- - Each unit of X that is produced requires 50 minutes processing time on machine
-- A and 30 minutes processing time on machine B.
--
-- - Each unit of Y that is produced requires 24 minutes processing time on machine
-- A and 33 minutes processing time on machine B.
--
-- - At the start of the current week there are 30 units of X and 90 units of Y in stock.
-- Available processing time on machine A is forecast to be 40 hours and on machine B is
-- forecast to be 35 hours.
--
-- - The demand for X in the current week is forecast to be 75 units and for Y is forecast
-- to be 95 units.
--
-- - Company policy is to maximise the combined sum of the units of X and the units of Y
-- in stock at the end of the week.
--
-- How much of each product should we make in the current week?
--
-- We have:
--
-- >>> optimize Lexicographic production
-- Optimal model:
-- X = 45 :: Integer
-- Y = 6 :: Integer
-- stock = 1 :: Integer
--
-- That is, we should produce 45 X's and 6 Y's, with the final maximum stock of just 1 expected!
production :: Goal
production = do x <- sInteger "X" -- Units of X produced
y <- sInteger "Y" -- Units of X produced
-- Amount of time on machine A and B
let timeA = 50 * x + 24 * y
timeB = 30 * x + 33 * y
constrain $ timeA .<= 40 * 60
constrain $ timeB .<= 35 * 60
-- Amount of product we'll end up with
let finalX = x + 30
finalY = y + 90
-- Make sure the demands are met:
constrain $ finalX .>= 75
constrain $ finalY .>= 95
-- Policy: Maximize the final stock
maximize "stock" $ (finalX - 75) + (finalY - 95)