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Blend: A model for mineral blending Description Several ores are blended to a final product that must
have a certain quality ('grade'). We wish to determine
the quantity of every ore to be used in the blend with
the objective to maximize the total profit (calculated
as sales revenues - raw material cost). Further explanation of this example: 'Xpress Python Reference Manual'
Source Files By clicking on a file name, a preview is opened at the bottom of this page.
Data Files blend.py
'''*******************************************************
* Python Example Problems *
* *
* file blend.py *
* Example for the use of the Python language *
* (Blending problem) *
* *
* Reading data from file. *
* *
* (c) 2018-2024 Fair Isaac Corporation *
*******************************************************'''
from __future__ import print_function
import xpress as xp
from Data.blend_data import COST, AVAIL, GRADE
p = xp.problem()
ROres = range(2)
REV = 125 # Unit revenue of product
MINGRADE = 4 # Min permitted grade of product
MAXGRADE = 5 # Max permitted grade of product
x = [p.addVariable(ub=AVAIL[o]) for o in ROres]
# Objective: maximize total profit
p.setObjective(xp.Sum((REV - COST[o]) * x[o] for o in ROres),
sense=xp.maximize)
# Lower and upper bounds on ore quality
p.addConstraint(xp.Sum((GRADE[o] - MINGRADE) * x[o] for o in ROres) >= 0)
p.addConstraint(xp.Sum((MAXGRADE - GRADE[o]) * x[o] for o in ROres) >= 0)
p.optimize()
# Print out the solution
print("Solution:\n Objective:", p.getObjVal())
for o in ROres:
print(" x(", o, "): ", p.getSolution(x[o]))
print("Grade: ", sum(GRADE[o] * p.getSolution(x[o]) for o in ROres)
/ sum(p.getSolution(x[o]) for o in ROres),
" [min,max]: [", MINGRADE, ",", MAXGRADE, "]")
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| © Copyright 2024 Fair Isaac Corporation. |
