1. Linear Programming
虞台文

2. Content Examples Forms of LP Geometric View of LP
Intuitive View of the Simplex Method

3. Linear Programming
Examples

4. Managing a Production Facility
Unit
Market
Price 1 2 n
. . .
Unit
Value 1 2 3 m
. . .
Raw
Material
. . .
Stock b1 b2 b3 bm
. . .
Product
. . .

5. Managing a Production Facility
aij : number of units of the ith raw material needed
to produce one unit of the jth product.
Managing a Production Facility
xj : the number of units of the jth product produced.
Unit
Market
Price 1 2 n
. . .
Unit
Value 1 2 3 m
. . .
Raw
Material
. . .
Unit
Profit c1 c2 cn
. . .
Stock b1 b2 b3 bm
. . .
Product
. . .

6. The Resource Allocation Problem
aij : number of units of the ith raw material needed
to produce one unit of the jth product.
The Resource Allocation Problem
xj : the number of units of the jth product produced.
Unit
Value 1 2 3 m
. . .
Stock b1 b2 b3 bm
Raw
Material
Product
Profit c1 c2 cn
Market
Price 1 2 n
Maximize
Subject to

7. The Diet Problem c1 c2 c3 cm . . . b1 b2 bn . . . . . . . . .
aij : number of units of nutrient i in one unit food j.
xj : number of units of food j in the diet.
The Diet Problem
Unit
Price c1 c2 c3 cm
. . .
Required
Units b1 b2 bn
. . .
Food
. . .
Nutrient
. . .
Minimize
Subject to

8. Linear Programming
Forms of LP

9. The Linear Programming Problem
An application of linear algebra.
Its central goal is to maximize or minimize a linear function on a domain defined by linear inequalities and equations.

10. The General Form x1, …, xn : decision variables Maximize or minimize
Subject to

11. The General Form x1, …, xn : decision variables
Maximize or minimize
A linear objective function
A set of linear constraints
linear inequalities or
linear equations
Subject to

12. The Canonical Form Minimize Subject to “greater than” only
Non-negative

13. The Canonical Form Minimize Subject to
This can also be considered canonical.
The Canonical Form
“ less than” only
Minimize
Subject to
Non-negative

14. The Standard Form
Minimize
Subject to

15. Equivalence of LP Forms
How to convert among these forms?
Equivalence of LP Forms

16. Maximization  Minimization
Maximize
Minimize
E.g.,
Minimize

17. Surplus/Slack Variables
Surplus variable
Slack variable

18. Replace Unrestricted Variables
Replace each unrestricted variables with the difference of two nonnegative variables.
For example:
Let x be an unrestricted variable.

19. Example
Maximize
Minimize
Subject to
Subject to
Canonical Form

20. Example
Minimize
Subject to
Standard Form

21. Forms of LP Standard Form Canonical Form General Form
These forms are all equivalent.

22. The Standard Form

23. LP  Convex Programming
Local optimality  Global optimality
LP  Convex Programming
a convex function
a convex set
The optimal solution is at a corner of the convex polytope.

24. The Simplex Algorithm a convex function a convex set
Local optimality  Global optimality
The Simplex Algorithm
Finding optimum by moving around the corner of the convex polytope in cost descent sense.
a convex function
a convex set
basic
feasible
solution
at a corner
The optimal solution is at a corner of the convex polytope.

25. Linear Programming
Geometric View of LP

26. An Example of 2 Decision Variablesy 1 2 3 4 5 6 7 8
Maximize
Subject to x

27. An Example of 2 Decision Variables1 2 3 4 5 6 7 8 x y
Maximize
Subject to

28. An Example of 2 Decision Variables1 2 3 4 5 6 7 8 x y
Optimum solution
Maximize
Subject to *
(6, 2)

29. Nonempty feasible region
The feasible region determined by a collection of linear inequalities is the collection of points that satisfy all of the inequalities.
Feasible LP Problems 1 2 3 4 5 6 7 8 x y
Maximize
Subject to
A feasible LP problem
Nonempty feasible region

30. Infeasible LP Problemsy x 1 1 2 3 4 5 6 2
Maximize
Subject to
An infeasible LP problem
The feasible region is empty

31. An unbounded LP problem
Unbounded LP Problems y x 1 1 2 3 4 5 6 2
Maximize
Subject to
An unbounded LP problem
The feasible region is unbounded.
Some coefficients of objective function is non-negative.

32. Intuitive View of the Simplex Method
Linear Programming
Intuitive View of the Simplex Method
大同大學資工所
智慧型多媒體研究室

33. An Example Two possible ways to help maximization:
1. Increase the variables with positive coefficients.
An Example
2. Decrease the variables with negative coefficient.
Maximize
Objective Function
Subject to

34. An Example
Maximize
Subject to

35. Converting the LP into Standard Form
Maximize
Subject to

36. Basic Solutions

37. Basic Solutions
Such a basic solution is feasible if and only if all bi’s are feasible.

38. Adjust their values to increase  progressively.
How?

39. How? Two possible ways to help maximization:
1. Increase the variables with negate coefficients.
How?
2. Decrease the variables with positive coefficient.

40. Active/Inactive Variables
The variable whose corresponding column has only one nonzero entry is active; otherwise, it is inactive.
Active/Inactive Variables
Inactive
Variables
Active
Variables

41. Set Up the Initial Tableau
Active x1 x2 x3 w1 w2 w3 
Ans 2 4 3 5 1 3 5 11 8
w1 = 5
w2 = 11
w3 = 8
 = 0

42. Select the Pivot Column
Active x1 x2 x3 w1 w2 w3 
Ans 2 3 1 1 5
w1 = 5 4 1 2 1 11
w2 = 11 3 4 2 1 8
w3 = 8 5 4 3 1
 = 0
From bottom row, choose the negative number with the largest magnitude.
Its column is the pivot column.
If there are two candidates, choose either one
If all the numbers in the bottom row are zero or positive, then you are done, and the basic solution is the optimal solution.

43. Select the Pivot in the Pivot Column
Active x1 x2 x3 w1 w2 w3 
Ans 2 3 1 1 5
5/2
=30/12
w1 = 5 4 1 2 1 11
11/4
=33/12
w2 = 11 3 4 2 1 8
8/3
=32/12
w3 = 8 5 4 3 1
 = 0
The pivot must always be a positive number.
For each positive entry b in the pivot column, compute the ratio a/b, where a is the number in the rightmost column in that row. We call this a test ratio.
Of these ratios, choose the smallest one. The corresponding number b is the pivot.

44. Clear the Pivot Column Active x1 x2 x3 w1 w2 w3  Ans 2 3 1 1 5 5/25
5/2
=30/12
w1 = 5 4 1 2 1 11
11/4
=33/12
w2 = 11 3 4 2 1 8
8/3
=32/12
w3 = 8 5 4 3 1
 = 0
Active x1 x2 x3 w1 w2 w3 
Ans 2 3 1 1 5
x1 = 2.5 5 2 1 1
w2 = 1
0.5
0.5
1.5 1
0.5
w3 = 0.5
11.5
0.5
2.5 1
12.5
 =12.5

45. Select the Pivot Column
Active x1 x2 x3 w1 w2 w3 
Ans 2 3 1 1 5
x1 = 2.5 5 2 1 1
w2 = 1
0.5
0.5
1.5 1
0.5
w3 = 0.5
11.5
0.5
2.5 1
12.5
 =12.5

46. Select the Pivot in the Pivot Column
Active x1 x2 x3 w1 w2 w3 
Ans
5/1 =5 2 3 1 1 5
x1 = 2.5 5 2 1 1
w2 = 1
0.5/0.5 =1
0.5
0.5
1.5 1
0.5
w3 = 0.5
11.5
0.5
2.5 1
12.5
 =12.5

47. Clear the Pivot Column Active x1 x2 x3 w1 w2 w3  Ans 2 4 4 2 44 2 4
x1 = 2 5 2 1 1
w2 = 1
0.5
0.5
1.5 1
0.5
x3 = 1 11 1 1 1 13
 =13
Active x1 x2 x3 w1 w2 w3 
Ans
5/1 =5 2 3 1 1 5
x1 = 2.5 5 2 1 1
w2 = 1
0.5/0.5 =1
0.5
0.5
1.5 1
0.5
w3 = 0.5
11.5
0.5
2.5 1
12.5
 =12.5

48. All of them are positive.
The Optimum
Active x1 x2 x3 w1 w2 w3 
Ans 2 4 4 2 4
x1 = 2 5 2 1 1
w2 = 1
All of them are positive.
0.5
0.5
1.5 1
0.5
x3 = 1 11 1 1 1 13
 =13
The optimal solution is here

49. The Optimum
Maximize
Subject to

50. The Optimum
Maximize 13
Subject to 5 10 8

51. Summary
Step 1: Using slack/surplus variables, convert the LP problem to a system of linear equations.
Step 2: Set up the initial tableau.
Step 3: Select the pivot column.
Step 4: Select the pivot in the pivot column.
Step 5: Use the pivot to clear the pivot column in the normal manner. This gives the next tableau.
Step 6: Repeat Steps 3-5 until there are no more negative numbers in the bottom row (with the possible exception of the Answer column).