Optimization Methods. LI Xiao-lei email@example.com http://www.csce.sdu.edu.cn ftp://22.214.171.124:112/upload. Optimization Tree. Introduction to Operations Research. Encyclopedia of Mathematics Optimization Theory See Operations Research
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OPERATIONS RESEARCH:Mathematical Programming(THIRD EDITION)
WAYNE L. WINSTON, 2003
Anything about Operations Research, Management Science, Decision Sciences, Mathematical Programming, Logistics Research.
Giapetto’s Woodcarving, Inc., manufactures two types of wooden toys: soldiers and trains.
Requires two types of skilled labor: carpentry and finishing.
A Soldier sells for $27 and uses $10 worth of raw materials, each costs variable labor and overhead by $14. requires 2 hours of finishing labor and 1 hour of carpentry labor.
A train sells for $21 and uses $9 worth of raw materials, each cost variable labor and overhead by $10. requires 1 hour of finishing labor and 1 hour of carpentry labor.
Each week, Giapetto can obtain all the needed raw material but only 100finishing hours and 80 carpentry hours. Demand for trains is unlimited, but at most 40 soldiers are bought.
Formulate a mathematical model to maximize Giapetto’s weekly profit (revenues -costs).
Giapetto must decide how many soldiers and trains should be manufactured each week. With this end, we define
x1=number of soldiers produced each week.
x2=number of trains produced each week.
The decision maker wants to maximize (usually revenue or profit) or minimize (usually cost) some function of the decision variables. The function is called the objective function.
=weekly revenues from soldiers + weekly revenues from trains
=(dollars/soldier) (soldiers/week) + (dollars/train) (trains/week)
Weekly raw material costs=10x1+9x2
Other weekly variable costs=14x1+10x2
Then Giapetto wants to maximize,
We use the variable z to denote the objective function value of LP. Giapetto’s objective function is,
The coefficient of a variable in the objective function is called the objective function coefficient of the variable.
The value of x1 and x2 are limited by the following three restrictions.
Constraint 1 Each week, no more than 100 hours of finishing time may be used.
Constraint 2 Each week, no more than 80 hours of carpentry time may be used.
Constraint 3 Because of limited demand, at most 40 soldiers should be produced each week.
Total finishing hrs./Week
=(finishing hrs./soldier) (soldiers made/week) +(finishing hrs./train) (trains made/week)
Constraint 1 may be expressed by,
Note: For a constraint to be reasonable, all terms in the constraint must have the same units.
Total carpentry hrs./week
=(carpentry hrs./soldier) (soldiers/week) + (carpentry hrs./train) (trains/week)
The constraint 2 may be written as,
If a decision variable xican only assume nonnegative values, we add the sign restrictionxi≥0
If a decision variable xiallowed to assume both positive and negative values, we say that xiis unrestricted in sign.
For the Giaprtto problem,
x1≥0 and x2≥0
s.t. 2x1 + x2≤100
x1 + x2≤80
s.t. (subject to) means that the values of the decision variables must satisfy all the constraints and all the sign restrictions.
A function f(x1,x2,…,xn) of x1,x2,…,xn is a linear function if and only if for some set of constants c1,c2,…,cn, f(x1,x2,…,xn) = c1x1,+c2x2+…+cnxn
For any linear function f(x1,x2,…,xn) and any number b, the inequalities f(x1,x2,…,xn)≤b and f(x1,x2,…,xn)≥b are linear inequalities.
A linear programming problem (LP) is an optimization problem for which we do the following:
The feasible region for an LP is the set of all points satisfying all the LP’s constraints and all the LP’s sign restrictions.
For a maximization problem, an optimal solution to an LP is a point in the feasible region with the largest objective function value. Similarly, for a minimization problem, an optimal solution is a point in the feasible region with the smallest objective function value.
The feasible region for the Giapetto problem is the set of all points (x1.x2) satisfying all the inequalities,
The optimal solution will be the point in the feasible region with the largest value of objective function z=3x1+2x2.
A set of points S is a convex set if the line segment joining any pair of points in S is wholly constrained in S.
Any LP that has an optimal solution has an extreme point that is optimal. This reduces the set of points yielding an optimal solution from the entire feasible region (infinite) to the set of extreme points (finite).
max z= 3x1+ 2x2
max z= 3x1+ 2x2
max z=2x1 - x2
s.t. x1 - x2≤1