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from the companion CD - Chapter 2 of the book:Balakrishnan, Render, and Stair,
“Managerial Decision Modeling with Spreadsheets”, 2nd ed., Prentice-Hall, 2007
http://www.stmartin.edu/
Rev. 2.3 by M. Miccio on January 20, 2015
Fundamental Theorem of Linear Programming
2
(stated here in two variables)A linear expression ax + by, defined over a closed bounded convex set S whose sides are line segments, takes on its maximum value at a vertex of S and its minimum value at a vertex of S. If S is unbounded, there may or may not be an optimum value, but if there is, it occurs at a vertex.
Example LP Model Formulation:The Product Mix Problem
Decision: How much to make of > 2 products?
Objective: Maximize profit
Constraints: Limited resources
4
Example: Flair Furniture Co.
Two products: Chairs and Tables
Decision: How many of each to make this month?
Objective: Maximize profit
5
Flair Furniture Co. DataTables
(per table)
Chairs(per chair)
Hours Available
Profit Contribution
$7 $5
Carpentry 3 hrs 4 hrs 2400
Painting 2 hrs 1 hr 1000
Other Limitations:• Make no more than 450 chairs• Make at least 100 tables
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Decision Variables:
T = Num. of tables to make
C = Num. of chairs to make
Objective Function: Maximize Profit
Maximize $7 T + $5 C
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Constraints:
• Have 2400 hours of carpentry time available
3 T + 4 C < 2400 (hours)
• Have 1000 hours of painting time available
2 T + 1 C < 1000 (hours)
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More Constraints:• Make no more than 450 chairs
C < 450 (num. chairs)
• Make at least 100 tables T > 100 (num. tables)
Nonnegativity:Cannot make a negative number of chairs or tables
T > 0C > 0
9
Model Summary
z = 7T + 5C max! (profit)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
C < 450 (max # chairs)
T > 100 (min # tables)
T, C > 0 (nonnegativity)
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Graphical MethodGraphing an LP model helps provide insight into LP models and their solutions:A straight line is plotted in place of each disequationA convex and bounded set (hopefully) is generatedAn ideal line, that is a family of parallel lines, is drawn to represent the objective functionThe optimum is found at the interception of the ideal line with a vertex
While this can only be done in two dimensions, the same properties apply to all LP models and solutions.
11
LP-2D.avi
Carpentry
Constraint Line
3T + 4C = 2400
Intercepts
(T = 0, C = 600)
(T = 800, C = 0)
0 800 T
C
600
0
Feasible
< 2400 hrs
Infeasible
> 2400 hrs
3T + 4C = 2400
12
Painting
Constraint Line
2T + 1C = 1000
Intercepts
(T = 0, C = 1000)
(T = 500, C = 0)
0 500 800 T
C1000
600
0
2T + 1C = 1000
13
0 100 500 800 T
C1000
600
450
0
Max Chair Line
C = 450
Min Table Line
T = 100
Feasible
Region
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0 100 200 300 400 500 T
C
500
400
300
200
100
0
Objective Function Line
z = 7T + 5C Profit
7T + 5C = $2,100
7T + 5C = $4,040
Optimal Point(T = 320, C = 360)7T + 5C
= $2,800
15
0 100 200 300 400 500 T
C
500
400
300
200
100
0
Additional Constraint
Need at least 75 more chairs than tables
C > T + 75
or
C – T > 75
T = 320C = 360
No longer feasible
New optimal point
T = 300, C = 375
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LP-2D conclusion
• Optimal Solution: The corner point with the best objective function value is optimal
17
Extreme Point theorem : Any LP problem with a nonempty bounded feasible region has an optimal solution; moreover, an optimal solution can always be found at an (or at least one) Corner Point (extreme point) of the problem's feasible region.
LP-2D with MatLab®
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LINEAR PROGRAMMING WITH MATLABcourse by Edward Neuman Department of Mathematics
Southern Illinois University at Carbondale
Function drawfr
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function drawfr(c, A, rel, b) % Graphs of the feasible region and the line level % of the LP problem with two legitimate variables % % min (max)z = c*x % Subject to Ax <= b (or Ax >= b), % x >= 0 % Enter a sequence of instructions like these into the COMMAND WINDOW:% c=[1 2];% A=[-1 3; 1 1; 1 -1; 1 3; 2 1];% rel='<<<>>';% b=[10; 6; 2; 6; 4];% NB: % b must be a COLUMN vector
% components of b vector can % indifferently be <0 or >0
It is possible to read the vertex coordinates in the Figure window by activating in the menu bar:Tools >>> Data Cursor
Function drawfrCautions in its use
20
function drawfr(c, A, rel, b)
% min (max)z = c*x % Subject to Ax <= b (or Ax >= b), % x >= 0
Negative value of a resourcedrawfr accepts one or more negative resource in the vector b
Unbounded Feasible Regiondrawfr is currently unable of drawing an unbounded region
Equality constraintsA constraint of the typeai1 x1 + ai2 x2 +.....+ aij .xj+.....+ ain xn = bi
must be transformed in 2 constraints of the typeai1 x1 + ai2 x2 +.....+ aij .xj+.....+ ain xn ≤ bi’