 # 1Introduction to Linear ProgrammingLesson 2 Introduction to Linear Programming.

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1Introduction to Linear ProgrammingLesson 2 Introduction to Linear Programming

2 Lesson 2  A Linear Programming model seeks to maximize or minimize a linear function, subject to a set of linear constraints.

3Introduction to Linear ProgrammingLesson 2  The linear model consists of the following components:  A set of decision variables.  An objective function.  A set of constraints.

4Introduction to Linear ProgrammingLesson 2  Many problems lend themselves to linear programming formulations. programming formulations.  Many problems can be approximated by linear models.  The output generated by linear programs provides useful “what-if” information.

5Introduction to Linear ProgrammingLesson 2  Galaxy manufactures two toy models:  Space Ray.  Zapper.

6Introduction to Linear ProgrammingLesson 2  Resources are limited to  1000 pounds of special plastic.  40 hours of production time per week.

7Introduction to Linear ProgrammingLesson 2  Marketing requirement  Total production cannot exceed 700 dozens.  Number of dozens of Space Rays cannot exceed number of dozens of Zappers by more than 350.

8Introduction to Linear ProgrammingLesson 2  Technological input  Space Rays requires 2 pounds of plastic and 3 minutes of labor per dozen.  Zappers requires 1 pound of plastic and 4 minutes of labor per dozen.

9Introduction to Linear ProgrammingLesson 2  Current production plan calls for:  Producing as much as possible of the more profitable product, Space Ray (\$8 profit per dozen).  Use resources left over to produce Zappers (\$5 profit per dozen).

10Introduction to Linear ProgrammingLesson 2 The current production plan consists of: Space Rays = 450 dozens Space Rays = 450 dozens Zapper = 100 dozens Zapper = 100 dozens Profit = 4100 dollars per week Profit = 4100 dollars per week

12Introduction to Linear ProgrammingLesson 2 A Linear Programming Model can provide an intelligent can provide an intelligent solution to this problem solution to this problem

13Introduction to Linear ProgrammingLesson 2  Decisions variables:  X1 = Production level of Space Rays (in dozens per week).  X2 = Production level of Zappers (in dozens per week).  Objective Function:  Weekly profit, to be maximized

14Introduction to Linear ProgrammingLesson 2 Max 8X 1 + 5X 2 (Weekly profit) subject to 2X 1 + 1X 2 < = 1000 (Plastic) 3X 1 + 4X 2 < = 2400 (Production Time) X 1 + X 2 < = 700 (Total production) X 1 + X 2 < = 700 (Total production) X 1 - X 2 < = 350 (Mix) X 1 - X 2 < = 350 (Mix) X j > = 0, j = 1,2 (Nonnegativity) X j > = 0, j = 1,2 (Nonnegativity)

15Introduction to Linear ProgrammingLesson 2 The set of all points that satisfy all the constraints of the model is called FEASIBLE REGION

16Introduction to Linear ProgrammingLesson 2 Using a graphical presentation Using a graphical presentation we can represent all the constraints, we can represent all the constraints, the objective function, and the three the objective function, and the three types of feasible points. types of feasible points.

1000 500 The Plastic constraint Feasible The plastic constraint: 2X1+X2<=1000 X2 Infeasible Production Time 3X1+4X2<=2400 Total production constraint: X1+X2<=700 600 700 Production mix constraint: X1-X2<=350 There are three types of feasible points Interior points. Boundary points.Extreme points. X1 700 800

Solving Graphically for an Optimal Solution

Recall the feasible Region 600 800 1200 400600800 X2 X1 We now demonstrate the search for an optimal solution: Start at some arbitrary profit, say profit = \$2,000... Start at some arbitrary profit, say profit = \$2,000... Profit = \$ 000 2, Then increase the profit, if possible... 3, 4,...and continue until it becomes infeasible Profit =\$4360

600 800 1200 400600800 X2 X1 Let’s take a closer look at the optimal point the optimal point Feasible region Feasible region Infeasible

21Introduction to Linear ProgrammingLesson 2 Space Rays = 320 dozens Zappers = 360 dozens Zappers = 360 dozens Profit = \$4360 Profit = \$4360 Why is this production schedule is able to attain better profit than the original, considering that there was no increase in any input resources?

22Introduction to Linear ProgrammingLesson 2  Extreme Points and Optimal Solutions  If a linear programming problem has an optimal solution, an extreme point is optimal.  However, not all extreme points are optimal  Multiple Optimal Solutions  For multiple optimal solutions to exist, the objective function must be parallel to a part of the feasible region.

23Introduction to Linear ProgrammingLesson 2  The optimal solution will remain unchanged as long as  An objective function coefficient lies within its range of optimality  There are no changes in any other input parameters.

24Introduction to Linear ProgrammingLesson 2  The value of the objective function will change if the coefficient multiplies a variable whose value is nonzero.

600 800 1200 400600800 X2 X1 The effects of changes in an objective function coefficient on the optimal solution Max 8x1 + 5x2 Max 4x1 + 5x2 Max 3.75x1 + 5x2 Max 2x1 + 5x2

600 800 1200 400600800 X2 X1 The effects of changes in an objective function coefficients on the optimal solution Max8x1 + 5x2 Max 3.75x1 + 5x2 Max8x1 + 5x2 Max 3.75 x1 + 5x2 Max 10 x1 + 5x2 3.75 10 Range of optimality

27Introduction to Linear ProgrammingLesson 2