1. Optimization Linear Programming and Simplex Method

2. Optimization Methods One-Dimensional Unconstrained Optimization
Golden-Section Search
Quadratic Interpolation
Newton's Method
Multi-Dimensional Unconstrained Optimization
Non-gradient or direct methods
Gradient methods
Linear Programming (Constrained)
Graphical Solution
Simplex Method

3. Standard Form Linear Programming Problem
Basic linear programming (LP) problem consists of two major parts:
The objective function
A set of constraints
For maximization problem, the objective function is generally expressed as
cj= payoff of each unit of the jth activity that is undertaken
xj= magnitude of the jth activity
Z= total payoff due to the total number of activities

4. Standard Form Linear Programming Problem
The constraints can be represented generally as
where
aij= amount of the ith resource that is consumed for each unit of the jth activity
bi = amount of the ith resource that is available
The general second type of constraints specifies that all activities must have a positive value. i.e,

5. Problem
How many tonnes of regular and premium gas to produce in order to maximize weekly profit?
Products
Resource
Regular
Premium
Resource Availability
Raw gas
7 m3/tonne
11 m3/tonne
77 m3/week
Production time
10 hr/tonne
8 hr/tonne
80 hr/week
Storage
9 tonne
6 tonne
Profit
$150/tonne
$175/tonne

6. Developing LP Formulation
Let x1 = 1 tonne of regular gas
x2 = 1 tonne of premium gas
Maximize Z = 150x x2
subject to
7x1 + 11x2 ≤ 77 (material constraint) (1)
10x1 + 8x2 ≤ 80 (time constraint) (2)
x1 ≤ 9 (regular gas storage) (3)
x2 ≤ 6 (premium gas storage) (4)
x1, x2 ≥ 0 (positivity) (5, 6)

7. Graphical Solution
Feasible solution space (the shaded area) contains all points that obey the constraints or points that represent the feasible solutions.
Points located outside of this area are known as infeasible solution.

8. Graphical Solution
We can adjust the line representing the objective function Z = 150x x2 to locate the optimum point.
One of the corner points (points where two or more lines intersect) will be an optimum point.
Corner points can be feasible or infeasible

9. Graphical Solution Some insights:
Increasing storages ( constraints (3) and (4)) does not improve profit.
Raising either materials or production time can improve profit.

10. Possible outcomes that can generally be obtained in a LP problem
Unique solution. The maximum objective function intersects a single point.
Alternate solutions. Problem has an infinite number of optima corresponding to a line segment.
No feasible solution.
Unbounded problems. Problem is under-constrained and therefore open-ended.

11. Unique solution – represented by a corner point
Alternate solutions – represented by points on one of the constraint line
(b) No feasible solution
(c) Unbounded problems

12. Key properties of LP that drives the design of Simplex Method (An algorithm for solving LP problems)
The optimum point is always at a feasible corner point. (Why?)
We only need to check the corner points.
If a corner point feasible solution has an objective function value that is better that or equal to all adjacent corner point feasible solutions, then it is optimal.
We don't necessarily need to search all corner points.

13. Key properties of LP that drives the design of Simplex Method
There are finite number of corner point feasible solutions.
Any method that checks only corner points will terminate eventually.

14. Basic idea behind the Simplex method
Starting at a feasible corner point solution.
Repeatedly move to a better adjacent corner point feasible solution until an optimum point is found.

15. Finding Corner Points Algebraically
How to find corner points algebraically?
How to locate adjacent corner points?
How to decide which adjacent point to move to next?
Simplex method offers a neat solution (involves the use of slack variables) to address these questions. 

16. First, constraint equations are reformulated as equalities by introducing slack variables – variables that measures how much of a constrained resource is available.
Slack variables
Si > 0 means the corresponding resource is not fully consumed.
Si < 0 means the corresponding resource is over-consumed.

17. Augmented System
Adding slack variables results in an augmented system of linear equations which is under specified (has more unknowns than equations).
6 unknowns (= 2 original variables + 4 slack variables)
4 equations
In general, if a LP problem has n variables and m constraints, the resulting augmented system will have a total of (n original variables + m slack variables) and m equations.
Continue …

18. To solve an augmented system with
6 unknowns (= 2 original variables + 4 slack variables)
4 equations
We need to set 2 of the variables constants before we can solve the system of equations.
e.g., Setting x1 = 0, x2 = 0 yields
Continue …

19. Augmented System
Solving the resulting system yields x1=0, x2=0, S1=77, S2=80, S3=9, S4=6, which represent one of the augmented solutions – values of all (original + slack) variables are given.
For this example, the values tell us that if we don't produce any regular (x1) or premium (x2) gas, we would have the following amount of unconsumed resources:
77 m3 of raw gas,
80 production hours
9 tonne of regular gas storage
6 tonne of premium gas storage

20. So far, you know how to find an augmented solution.
Augmented System
An augmented solution which corresponds to a corner point is also known as a basic solution.
The augmented solution x1=0, x2=0, S1=77, S2=80, S3=9, S4=6 is a basic solution. It corresponds to corner point A.
In addition, it is a basic feasible solution because all variables ≥ 0.
So far, you know how to find an augmented solution.
How should we set the variables so that the augmented solution is a basic solution?
Continue …

21. Characteristics of Slack Variables
The equations (or constraints) have a one-to-one relationship with the slack variables.
For example, S1 relates to the 1st equation, S4 relates to the 4th equation.
If a point is on the line 7x1 + 11x2 = 77 (line labeled "1"), what's the value of S1?
If a point is on the line 10x1 + 8x2 = 80 (line labeled "2"), which variable has the value zero?
At the corner point D, which variables have the value zero?
Show Graph & Equations

22. Characteristics of Slack Variables
Setting two variables to zero has the effect of "selecting two lines", and solving the resulting system means finding the intersecting point (a corner point) of the selected lines.
For example, let S1=0, S2=0. Then the system becomes
The value of x1 and x2 of this system is the coordinate of the corner point C (the point where line "1" and line "2" intersect.)

23. Finding Corner Points Algebraically
We now know how to find the corner points.
Next
How to find an adjacent corner point?
How to decide which adjacent corner point to move to?

24. Basic and Non-Basic Variables
Corner Point
Non-Basic Variables
Basic Variables
Some of the Adjacent Corner Points A
x1, x2
S1, S2, S3, S4
B, E B
x2, S2
x1, S1, S3, S4
A, C C
S1, S2
x1, x2, S3, S4
B, D D
S1, S4
x1, x2, S2, S3
C, E E
x1, S4
x2, S1, S2, S3
A, D F
x1, S2
x2, S1, S3, S4
Do you see a pattern between any pair of adjacent corner points?
Variables in which their values are set to zero are called non-basic variables.
Remember: Non-basic, variable set to zero, corresponding constraint is active

25. Basic and Non-Basic Variables
If two points are adjacent corner points, then
Their basic sets are identical except for one member.
Their non-basic sets are identical except for one member.
The reverse of the above statement is not always true. (e.g., F and A)
Questions:
To find an adjacent corner point, can we just switch one variable between the basic and non-basic sets?
How should we decide which corner point to move to next?

26. Basic and Non-Basic Variables
How do we make sure the following conditions are satisfied?
The corner points are adjacent.
The corner points are feasible.
The new corner point gives better value for the objective function than the current corner point.

27. Which variables to switch?
All three conditions from the previous slide can be satisfied if the following steps are taken:
Selecting an entering basic variable – a non-basic variable that will increase the objective function value the most if allowed to take on a positive value. Move this variable from the non-basic set to the basic set.
Finding the leaving basic variable – a basic variable that is forced to a value zero first after allowing the entering basic variable to increase. Move the variable that is forced to zero from the basic set to the non-basic set.

28. Which variables to switch?
Suppose we are currently at point A.
Basic set = { S1, S2, S3, S4 }, non-basic set = { x1, x2 }
Which variable should be the entering basic variable?
Which variable should be the leaving basic variable?

29. Which variables to switch?
x2 has a larger coefficient in the objective function, so we should make x2 the entering basic variable. (x1 remains a non-basic variable and has the value zero)
If we keep increasing x2, other basic variables will decrease. Which basic variable will become zero first?

30. Which variables to switch?
Show Graph & Equations
Which variables to switch?
Minimum ratio test
If we take the ratio of the RHS value to the coefficient of the entering basic variable (x2), the smallest non-negative ratio can tell us which basic variable will reach zero first if we keep increasing the value of x2.
So S4 should be made the leaving basic variable.

31. Which variables to switch?
So setting x1=0, S4=0 and solve the system of equations would yield an augmented solution correspond to point E.
Solving the system of equations at every corner point is not very efficient. Simplex method offers an efficient way to compute the value of each variable by updating the equations incrementally.

32. F 4 E D 1 C 5 3 2 A 6 B