module 1 blp simplex revised simplex

Part 1: Foundation

Module 1bLinear ProgrammingLinear ProgrammingGeometric SolutionSimplex MethodSimplex Method Revised Simplex Method

Dr. Gajendra Kumar Adil, SJMSOM

Linear Programming

A Linear Programme (LP) is of the following formA Linear Programme (LP) is of the following form

1 1 2 2Minimize/Maximize ...... n nz c x c x c x= + + +

≤⎧ ⎫

1 1 2 2Subject to ..... , 1, 2,....,i i in n ia x a x a x b i m≤⎧ ⎫

⎪ ⎪+ + + = =⎨ ⎬⎪ ⎪≥⎩ ⎭

1 2, ,..., 0nx x x≥⎩ ⎭

≥

In Compact form:Minimize/Maximize ∑(j=1 to n) cj xj

Subject to


∑(j=1 to n) aij xj {<, = , >}bi, i=1 to m

xj ≥ 0, j =1 to n

Standard and Canonical Forms of LPMINIMIZATION PROBLEM MAXIMIZATION PROBLEM

STANDARD Minimize ∑(j=1 to n) cj xj Maximize ∑(j=1 to n) cj xjFORM

(j 1 to n) j j

Subject to

∑(j=1 to n) aij xj = bi, i=1 to m

xj ≥ 0, j =1 to n

(j 1 to n) j j

Subject to

∑(j=1 to n) aij xj = bi, i=1 to m

xj ≥ 0, j =1 to nj , j j , j

CANONICALFORM

Minimize ∑(j=1 to n) cj xj

Subject to Maximize ∑(j=1 to n) cj xj

Subject to

∑(j=1 to n) aij xj ≥ bi, i =1 to m

xj ≥ 0, j =1 to n

∑(j=1 to n) aij xj ≤ bi, i =1 to m

xj ≥ 0, j =1 to n

Notice that the structure of LP in the standard form: 1.Objective function row


2.Technology constraints (system of simultaneous linear equations.)

3. Non-negativity constraints.

Problem Manipulation to Convert an LP into its Standard FormLP into its Standard Form

A) Converting Inequality into Equation

Add Slack variable, s (s has coefficient of 0 in the objective function)Less than type: 5 x1 + 2 x2 ≤ 10, x1, x2 ≥ 0 Add slack variable: 5 x1 + 2 x2 + s = 10, x1, x2 ≥0, s≥0

Add Surplus Variable, s (s has coefficient of 0 in the objective function)Greater than type: 5 x1 + 2 x2 ≥ 10, x1, x2 ≥ 0Add surplus variable: 5 x1 + 2 x2 - s = 10, x1, x2 ≥ 0, s ≥ 0

B) Converting Variables with Unristricted in Sign (urs) into Nonnegate Variables

5 x1 + 2 x2 = 10, x1 ≥ 0, x2 urs

Replace urs variable with difference of two non-negative variables.5 x1 + 2 * (x3 - x4) = 10, x1 ≥ 0, x3, x4 ≥0

C) Converting Minimization Problem to Maximization Problem and Vice Versa

Maximize z1 = cx ⇔ Minimize z2 = –cx


Geometric Solution of a Linear Programmeg• Draw feasible region defined by half-spaces satisfying linear inequalities corresponding

to technological and non-negativity constraints.• Draw contours of constant objective function value in the direction where objective

function value improves until the contour has at least one point lying in feasible region.

x2

Max z = x1 + 2 x2 (0)s.t.

x1 + x2 < 5 (1)z = x1 + 2 x2 = 10

1 2 ( )x1 > 0 (2)x2 > 0 (3)

21

1

x1 31z = x1 + 2 x2 = 0

+ 2 5


z = x1 + 2 x2 = 5

Geometric Solution of a Linear ProgrammeCase 1. Unique Finite Optimal Solution

Max z = x1 + 2 x2 (0) s t Min z = x1 + 2 x2 (0)s.t.

x1 + x2 <= 5 (1)x1 >= 0 (2)x2 >= 0 (3)

x2

2 1

0

2 1

1 2 (0)s.t.

x1 + x2 >= 5 (1) x1 >= 0 (2)

x2 >= 0 (3)

x13 30Bounded regin Unbounded region

Case 2. Alternative Finite Optimal Solutions

Max z = x1 + x2 (0) s t

x2 Min z = x1 + x2 (0) s t0s.t.

x1 + x2 <= 5 (1)x1 >= 0 (2)x2 >= 0 (3)

21

1

0

21

s.t.x1 + x2 >= 5 (1)x1 >= 0 (2)x2 >= 0 (3)


x1 31

31 1 Bounded region

Unbounded region

Geometric Solution of a Linear ProgrammeCase 3. Unbounded Solution

Max z = x1 + x2 (0) s.t.

x2

x1 >= 0 (1) x2 >= 0 (2)

13

0 01

Case 4 Empty Feasible Region

x1 21

Unbounded region

Case 4.Empty Feasible Region

Max z = x1 + x2 (0) s.t.

5 (1)

x2

x1 + x2 <= 5 (1)x1 + x2 >= 6 (2) x1 >= 0 (3) x2 >= 0 (4)

21

1 31


x141An empty feasible

i

Some Observations to Make from Geometric Solution of LPGeometric Solution of LP

• If optimum solution exists, it lies on one or more hyperplanes (not in the interior of feasible region). Will this be true if- (i) Objective function is non-linear? (ii) e s b e eg o ). W s be ue ( ) Objec ve u c o s o e ( )Variables are not continuous (i.e., discrete)?

• By scanning just the corner feasible points one can find the optimum solution. If it is a unique optimum then one of the corner point feasible solutions (CPFS) isis a unique optimum then one of the corner point feasible solutions (CPFS) is optimum otherwise the edge joining two adjacent corner point feasible solutions (CPFS) is optimal. Thus corner points are very useful in LP. To prove this for bounded case simply substitute the variables in terms of corner points (known) and lambdas (convex combination) then you have just one constraint sum of lambdas

l t 1 Th l bd ith hi h t ti ffi i t (f i i i tiequal to 1. The lambda with highest negative coefficient (for minimization problem) can be set to 1 (This is left as an exercise).

• There are a finite number of corner point feasible solutions (CPFS). What is the b d b f CPFS?upper bound on number of CPFS?

• If a CPFS has no adjacent CPFS that are better, then such a CPFS is guaranteed to be an optimal solution. What happens if feasible set is non-convex?


Simplex BasicsGeometric solution is not possible beyond three dimensions for which simplex can be used. Therefore, we develop the f ll i k id d i i l l ithfollowing key ideas used in simplex algorithm:

1. Corner feasible points (CPFS). Concept of adjacent CFPS.p ( ) p j

2. Develop optimality criteria (a check if a CPFS is optimal, that is no adjacent CFPS is better)that is, no adjacent CFPS is better)

3. How to move from one CPFC to better adjacent CPFS. j

4. The process of detecting situations, un-boundedness and infeasibility


infeasibility.

Basic Feasible Solutions(Ch t i ti f CPFS)(Characterization of CPFS)

Geometrically: Consider the polyhedral set defined by following inequalities.a x + +a x < ba11x1+… +a1p xp < b1 a21x1+… +a2p xp < b2 m halfspaces (of form ai1x1 +…+aip xp < bi); and

: m associated hyper planes (of form ai1x1 +…+aip xp = bi)am1x1+…+amp xp < bm

x1 > 0x2 > 0 p halfspaces (of form x1 > 0) ; and

: p associated hyper planes (of form xi = 0)x > 0xp > 0

This can be geometrically interpreted in p dimensional space as follows. • The feasible region is formed by the intersection of p+m half spaces. Hence, p+m

associated hyperplanes or boundaries of half spaces. yp p p• At Corner points, p independent hyperplanes (of dimension p) are binding or intersect.

For example, (i) p=2, intersection of two hyperplanes (i.e. lines) defines a point; (ii) p=3, intersections of three hyperplanes (i.e planes) is a point and so on.

• Corner points where p hyperplanes are binding are called face of dimension zero.Ed i f ith di i 1 d b d fi d b i t ti f 1 h l


• Edge is a face with dimension 1 and can be defined by intersection of p-1 hyperplanes.

Basic Feasible Solutions(Characterization of CPFS)(Characterization of CPFS)

Example Problem 1: Consider the polyhedral set (P1) defined by following inequalities in 2 dimensional space (p=2).

x1 + x2 ≤ 6 (1)x2 ≤ 3 (2) P1

x1 ≥ 0, x2 ≥ 0 (3-4)

• This can be geometrically shown in 2 dimensions, i.e., in x1-x2 space.

• The feasible region is formed by the intersection of four half spaces defined by the inequalities (1) to (4)

x2 x1=0

E

defined by the inequalities (1) to (4). • Four (=p+m) hyperplanes or

boundaries of half spaces (replacing the inequality sign by equality sign in 1 to 4 gives the hyperplanes) are shown i fi

x1+x2 =6

D C x2=3

x =0in figure.• There are five corner points (A, B, C,

D and E) out of which A, B, C and D are four feasible corner points. One can see that the corner points are

x2=0

A B x1


pnothing but intersections of 2 (p = 2) hyperplanes.

Basic Feasible Solutions (Characterization of CPFS)

Al b i llAlgebraically: Consider the polyhedral set defined by the following equations and/ inequalities.

a11x1+… +a1p xp + xp+1 = b1 slack, xp+1 > 0 added m halfspaces (of form ai1x1 +…+aipxp < bi );

a21x1+… +a2p xp + xp+2 = b2 slack, xp+2 > 0 added and;

: : associated m hyper planes

am1x1+…+amp xp + xp+m = bm slack, xp+m > 0 added [of form xp+i = bi –(ai1x1 +…+aipxp)=0]

0 R t h lf ( f f 0 1 i ) dx1 > 0 Represent p halfspaces (of form xi > 0, 1 < i < p) ; and x2 > 0 p associated hyper planes (of form xi = 0) for structural variables. : and m half spaces (xp+i > 0, 1 < i < m) and m hyper planes for slack variables. xp+m > 0

Solution:• Number of variables, n = p + m• Number of equations = m • Basic Solution (all variables have unique values): ( q )

– Number of basic variables = m – Number of variables required to be nonbasic = p from p+m (these indicate the binding

hyperplane)

• Corner Point (unique values for each variable): intersection of p hyperplanes i e basic solution


Corner Point (unique values for each variable): intersection of p hyperplanes., i.e., basic solution gives a corner point.

Basic Feasible Solutions (Characterization of CPFS)Basic Feasible Solutions (Characterization of CPFS)Basic Solution • To get a unique solution for n (= p + m) variables some n linearly independent

equations are required. Out of required n equations the first m comes from q q q qtechnological constraints [A]mxn[x]n = [b]m and additional p (= n-m) come from the non-negativity constraints [x]n ≥ [0]n.

We choose a set of p variables and set each of these variables to zero to satisfy non-p ynegative constraints. These variables are called non-basic or independent variables, xN. Then [A]mxn[x]n = [b]m are solved to get unique solution for remaining variables called basic variable xB. Such a solution is called basic solution.

• Thus, effectively, partition columns of A into two sets, B and N, where B is a submatrix that consists of m linearly independent columns of A and N is another submatrix that consists of the remaining p (=m-n) columns.

[B , N ] [xB , xN]T = bB = basis or basic matrix. N = non-basic matrix.xB = basic variables; xN = non-basic or independent variables

• Given B, the solution of the form xB = B–1b and xN = 0 is called a basic solution.



Definition 1- Basic Solution: A basic solution to [A]mxn[x]n = [b]m and [x]n ≥ [0]n; is a solution with at least n - m zero Variables. The solution is a non-degenerate basic solution if exactly n-m variables are zero. The choice of the m non-zero variables is called the basis. Variables in the basis are called basic variables; the others are called non-basic or independentin the basis are called basic variables; the others are called non-basic or independent variables.

Definition 2- basic feasible Solution (bfs): If a basic solution satisfies xB > 0 then it is called a basic feasible Solution (bfs).

Example Problem 2: Consider polyhedral set of Example Problem 1 (P1) and convert them into standard form. Constraints Ax = b are two (=m) equations in four (n=m+p) unknowns.

x1 + x2 + x3 = 6 (1) x2 + x4 = 3 (2)x2 + x4 3 (2)

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0 (3-6)

A = , x = [x1, x2, x3, x4]T , b = [6, 3] T . 1 1 1 00 1 0 1

⎡ ⎤⎢ ⎥⎣ ⎦

It is possible to express any two (=m) basic variables in terms of other two (p=n-m) non-basic variables, assuming that equations 1 and 2 are independent (there are a maximum of 4C2 = 6 ways to select basis). The following illustrates the enumeration.

⎣ ⎦



Non-Basic Variables, xN

(set to be 0)

Basic Variables, xB

(solved for )

Basic Solution Is it a Basic Feasible Solution (xB ≥ 0?)

Correspondence to Corner Point

x1 x2 x3 x4(xB ≥ 0?) Point

x1, x2 x3, x4 0 0 6 3 Yes Ax2, x3 x1, x4 6 0 0 3 Yes B

x3 x4 x1 x2 3 3 0 0 Yes Cx3, x4 x1, x2 3 3 0 0 Yes C

x1,x4 x2, x3 0 3 3 0 Yes D

x1, x3 x2, x4 0 6 0 -3 No E

x2,x4 x1, x3 Does not exist, as a1=a3 Not None

Correspondence to the GeometryAs illustrated above each feasible basis corresponds to a corner point (A B C D or E in

x2,x4 x1, x3 Does not exist, as a1 a3

(linearly dependent columns)Not

applicableNone

As illustrated above, each feasible basis corresponds to a corner point (A, B, C, D or E in x1-x2 space). Further, the corner points are intersection of p (=2) hyperplanes. It is convenient to visualize the hyperplanes defining the boundary of feasible region by associating an indicating variable to each hyperplane such that the indicating variable


assumes a value of zero on that hyperplane. Each variable (including slack) is associated to a hyperplane in a unique way.


In x1-x2 space, the 4 lines (hyperplanes) defining the feasible set can be written x1= 0, x2 = 0, x3 = 0, x4 = 0 (see figure). Thus the corner point can be viewed in terms of hyperplanes intersecting at it or the indicating variables having zeroterms of hyperplanes intersecting at it or the indicating variables having zero values at it which are non-basic variables also. At each corner point represents a unique combination of non-basic variables.



How to associate variables to the hyperplanes?

1. Write the LP in canonical form. Let there be m technological constraints of inequality type carrying p structural variables (excluding slack variables) and m slack variables.(excluding slack variables) and m slack variables.

2. Associate each slack variable to the hyperplane corresponding to the technological constraint.

3 Associate each structural variable to the non-negativity3. Associate each structural variable to the non-negativity constraint.

Example Problem 3x – x + x ≤ 10 (1) [Introduce slack x ≥ 0]x1 – x2 + x3 ≤ 10 (1) [Introduce slack x4 ≥ 0]

2x1 – x2 + 2x3 ≤ 40 (2) [Introduce slack x5 ≥ 0]3x1 – 2x2 + 3x3 ≤ 50 (3) [Introduce slack x6 ≥ 0]x1 ≥ 0, x2 ≥ 0, x3 ≥ 0


1 , 2 , 3

Basic Feasible Solutions (Characterization of CPFS)Category of Constraint

Half space** Defining Hyperplane *Indicating Variable

Technological x1 – x2 + x3 ≤ 10 x1 – x2 + x3 = 10 x4 (slack variable)

2x1–x2 + 2x3 ≤ 40 2x1–x2 + 2x3 = 40 x5 (slack variable)

3x1–2x2 + 3x3 ≤ 50 3x1–2x2 + 3x3 =50 x6 (slack variable)

Non-negativity of problem variables

x1 ≥ 0 x1 = 0 x1

x2 ≥ 0 x2 = 0 x22 0 2 0 2

x3 ≥ 0 x3 = 0 x3

Non-negativity of slack variable

x4 ≥ 0x5 ≥ 0x6 ≥ 0

Already considered in technological constraints. x4 =0 ≡ x1 – x2 + x3 =10

Not applicable

6 4 1 2 3

*Indicating Variable having value 0 on the hyperplane

** for technology constraint of other type the indicating variable is chosen as follows:

≥ type (introduce x x – x + x ≥ 20 x – x + x = 20 – x + x≥ type (introduce x6surplus and x7 artificial variable)

x1 x2 + x3 ≥ 20 x1 x2 + x3 20 x6 + x7

= type (introduce x8 artificial variable)

x1 – x2 - x3 = 30 x1– x2 - x3 = 30 x8



Example Problem 4: Add another constraint, x1 + 2 x2 ≤ 9 in the Example Problem 1 (P1) considered earlier. What happens?

You can check it out. It results in the following degenerate bases.

xB = [x1, x2, x5]T=[3, 3,0] T, xN = [x3, x4] T = [0 0] TB [ 1 2 5] [ ] N [ 3 4] [ ]

&

xB = [x1, x2, x3]T=[3, 3,0] T, xN = [x4, x5] T = [0 0] T

&

xB = [x1, x2, x4]T=[3, 3,0] T, xN = [x3, x5] T = [0 0] T

Theorem Basic feasible solutions ≡ extremeTheorem. Basic feasible solutions ≡ extreme points of the feasible set or CFPS.

Every basic feasible solution corresponds to corner feasible point (extreme


corresponds to corner feasible point (extreme point). (However, an extreme point may have several basic feasible solution representations, e.g., in case of degeneracy.)

Optimum Basic Feasible Solution How can one conclude that a basic feasible solution is optimum? The bfs is optimum if it is proved that increasing value of any of the nonbasic variables from current level of 0 does not improve the objective function value. We need to recast the problem in0 does not improve the objective function value. We need to recast the problem in nonbasic variables for this purpose.

Key Idea of Simplex- Reduction of LP Problem in Terms of Nonbasic Variables:

The system of equations (constraints and objective row) representing a basic feasibleThe system of equations (constraints and objective row) representing a basic feasible solution is written in canonical form of nonbasic variables. We perform Pivotingconsidering basic variables as pivots.

Note that the pivoting is necessary to compute the values of basic variables. The resulting system can then be easily written in terms of non basic variables in canonicalresulting system can then be easily written in terms of non-basic variables in canonical form because basic variables just act as slack. Transforming matrix this way, achieves the following:

i) It helps making checks if the current bfs is optimum, ) p g pii) If current bfs is not optimum then it identifies the adjacent bfs(s) that can improve

the objective function? and iii) It also helps in doing a ratio test that identifies which basic variable should leave

h b i dj bf


the current basis to move to an adjacent bfs.

Optimum Basic Feasible SolutionExample Problem 5: Problem Reduction in term of Nonbasic Variables


Optimum Basic Feasible SolutionLet’s take a basic feasible solution, say, xB=[x1, x4, x5]T, xN=[x2, x3]T . 1) Eliminate x1 from equations (0), (2) and (3); 2) Eliminate x4 from equations (0), (1) and (3) (already in that form); and3) Eliminate x5 from equations (0), (1) and (2) (already in that form) respectively.


Optimum Basic Feasible SolutionThe resulting system becomes:z + x2 -x3 = -6 (0)2 3 ( )

x1 +x2 +x3 = 6 (1)x2 + x4 = 3 (2)x2 -x3 +x5 = 3 (3)

The above can be re-written as a problem reduced in terms of nonbasic variables (xN) as the basic variables (xB) just act as slack variables (i.e., they appear in ( N) ( B) j ( y ppjust one equation with coefficient 1).


Optimum Basic Feasible SolutionOptimality Check: The new problem is not optimal as increasing x2 will improve

z further from -6. By observing the sign of coefficient of x2 and x3 in the objective function row (or row 0), (called reduced costs) conclusions whether j ( ), ( )optimum has reached can be made.

How much can x2 be increased? As much as the constraints allow. For this problem one can graphically show in (x x ) space in which origin (0 0) is theproblem one can graphically show in (x2, x3) space in which origin (0, 0) is the cbfs. Move from origin along x3 (so that x3 stays at 0 or NBV and x2 increases) until it reaches the adjacent vertex (it will hit one hyperplane corresponding to one of the three constraints). One may also note that one of the basic variables will then assume a value of 0 at this hyperplane (verify it graphically)will then assume a value of 0 at this hyperplane. (verify it graphically).

Expressing the Elements in the Transformed Tableau in Terms of B-1 and the Original LP. g

An alternative to pivoting just described can be used using matrix operations. This alternate method computes B-1 and uses it to get the equations in the desired form


desired form.

Optimum Basic Feasible Solution

Original LP in Matrix FormConsider (Min z = [c]1 [x] 1) s.t. [A] [x] = [b] and [x] ≥ 0)Consider (Min z [c]1xn [x]nx1) s.t. [A]mxn[x]n [b]m and [x]n ≥ 0)Min z = cB xB + cN xN s.t. BxB + NxN = b xB, xN > 0 Where,

[ ][ ]

1 2

1 2 1 2 ( )

(original columns)

, (original costs)n

B B Bm N N N m nc c c c c c −

=

⎡ ⎤= = ⎣ ⎦⎡ ⎤

B N

A a a a

c c

L

L L

[ ]1 2 1 2 ( )

11 1

,

00

B B Bm N N N m n

NB xx bxx b

−⎡ ⎤= = ⎣ ⎦⎡ ⎤⎡ ⎤ ⎡⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥

B a a a N a a aL L

⎤⎢ ⎥⎢ ⎥22 2

( )

0, ,

0

NB

N m nBm m

xx b

xx b−

⎢ ⎥⎢ ⎥ ⎢ ⎥= = = =⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

B Nx x 0 ,bM MM M

⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Dr. Gajendra Kumar Adil, SJMSOMExample Problem 6 (in the next slide):

Optimum Basic Feasible SolutionExample Problem 6Example Problem 6


Optimum Basic Feasible Solutionbbb

Problem Reduction in terms of nonbasic variables

I. Expressing the constraints columns using B-1 and the original constraints columns

Multiplying the constraints by B-1, we obtain the columns of the transformed matrix.

B-1 B xB + B -1 N xN = B -1 b

= x + B -1 N x = B -1 b= xB + B 1 N xN = B 1 b

1j R

+ jx∈

=∑B j mx y b (T -T )j R

1, 1 1, 2 1, ( )1

(where R is the current set of indices of the nonbasic variable)

N N N m nB y y yx

∈

−⎡ ⎤⎡ ⎤⎢⎢ ⎥

L 1 1Nx b⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎥

( )

2, 1 2, 2 2, ( )2 + N N N m nB y y yx −⎢⎢ ⎥⎢⎢ ⎥⎢⎢ ⎥⎢⎢ ⎥

⎢ ⎥ ⎢⎣ ⎦ ⎣

L

M M M MM

2 2Nx b⎢ ⎥⎢ ⎥⎥⎢ ⎥⎢ ⎥⎥ = ⎢ ⎥⎢ ⎥⎥⎢ ⎥⎢ ⎥⎥⎢ ⎥⎥ ⎢ ⎥⎦ ⎣ ⎦

M M


, 1 , 2 , ( )m N m N m N m nBm y y yx −⎢ ⎥ ⎢⎣ ⎦ ⎣ L ( )N m n mx b− ⎢ ⎥⎥ ⎢ ⎥⎦ ⎣ ⎦ ⎣ ⎦

Optimum Basic Feasible SolutionWe are interested in computing column yj of non-basic variable xjand the right side column. These can be computed by premultiplying the columns from the original LP by B-1 as followspremultiplying the columns from the original LP by B , as follows.yj = B -1 aj and = B -1 bb

1 1 1 aB B B− − − ⎡ ⎤⎡ ⎤ ⎡ ⎤y 1 1 1111 12 1

1 1 1221 22 2

jm

jm

aB B BaB B B− − −

⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥= = • = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥

1j

2j -1j j

y

yy B a

L

L

M M M MM M1 1 1

1 2

1 1 111 12 1

1 1 1

m m mm mj

m

B B B a

B B B

− − −

− − −

⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦

⎡ ⎤⎢ ⎥

mj

1

y

b

L

L 1bb

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥1 1 1

2 21 22 2

1 11

m

m mm

B B B

B B

− − −

− −

⎢ ⎥⎢ ⎥= = • =⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

-1bb B b

b

L

M M M M M

2

12 mmm

b

bB−

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

M

L


1m mm⎢ ⎥⎣ ⎦ 2 mmm⎢ ⎥ ⎣ ⎦⎣ ⎦

Optimum Basic Feasible Solutionb

II. Determining the Row 0 in terms of Initial LP and B-1

Eliminate xB from row 0: pre-multiply the transformed constraints (T1-Tm) by cB or the original constraints by xB B-1 and add to the row 0.

z + (cB B–1

B - cB )xB + (cB B–1

N - cN ) xN = cB B–1

b

z + 0 xB + ∑j∈R (cB B–1

aj – cj ) xj = cB B–1

b

Or,

z + 0 xB + ∑j∈R (zj – cj ) xj = z0 (0)

(where, zj = cB B–1

aj and z0 = cB B–1

b is a constant)

Minimize z = z0 - ∑j∈R (zj – cj ) xj

Thus, Reduced Problem (RP) in Non-basic Variable Space is:


0 ∑j∈R ( j j ) jSubject to ∑j∈R yj xj ≤ , xj > 0 j∈Rb

Optimum Basic Feasible SolutionExample Problem 7: Transformation using Matrix form:

Minimize x1 + x2

S bj t t + 2 < 4 dd l kSubject to x1 + 2 x2 < 4 add slack x3x2 < 1 add slack x4

x1, x2 > 0

4⎡ ⎤

A = [a1, a2, a3, a4] = ; xB = [x1, x2]; B=[a1, a2] ; xN = [x3, x4], N=[a3, a4]1 2 1 00 1 0 1

⎡ ⎤⎢ ⎥⎣ ⎦

[ ] [ ]11 2 1 2 1 2

, 1 1 1 10 1 0 1 0 1

1 2 1 1 1 2 0 2,

0 1 0 0 0 1 1 1

− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = = = −⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦− − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= = = = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

-1 -1B

-1 -13 3 4 4

B c B

y B a y B a

[ ]

[ ]

0

3 3 3

41 1 3

1

11 1 0 1

0

z

z c c

⎡ ⎤= = − =⎢ ⎥

⎣ ⎦⎡ ⎤

− = − = − − =⎢ ⎥⎣ ⎦

-1B

-1B 3

c B b

c B a,0 1 0 0 0 1 1 1

1 2 4 20 1 1 1

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

−⎡ ⎤ ⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

3 3 4 4

-1

y y

b B b [ ]4 4 4

0

01 1 0 1

1z c c

⎣ ⎦⎡ ⎤

− = − = − − = −⎢ ⎥⎣ ⎦

-1B 4c B a


Optimum Basic Feasible SolutionExample Problem 7: Transformed problem.

System Representation in tableau form


Optimum Basic Feasible Solution

z xB1 xB2 … xBm xN1 … xj … xNp

( )

RHS (p=n-m)

z 1 0 0 … 0 zN1– cN1 … zj –cj … zNp– cNp cB B-1b = cB b

Objective Row 0

xB1

xB1

0 0

1 0

0 1

… …

0 0

y1,N1 y2,N1

… …

y1,j y2,j

… …

y1,Np y2,Np

1b

2b

C Row 1 C Row 2

: xBm

: 0

: 0

: 0

: …

: 1

: ym,N1

: …

: ym,j

: …

: ym,Np

2b

:

mb

: C Row m

Feasibility condition: The basic solution is feasible if, xB = B–1b ≥ 0 (should satisfy non-negativity constraints to be a bfs).

Optimality condition: If reduced cost (zj – cj) ≤ 0 for all j∈R theOptimality condition: If reduced cost (zj cj) ≤ 0 for all j∈R, the current basic feasible solution (cbfs) is optimal for a minimization problem.


Moving to an Adjacent CPFSWhat happens if optimal has not reached?

That is if (zj – cj) > 0 for some j = k∈R, then by bringing xk into basis objective function value improves at the rate defined by the coefficient of xk, i.e., (zk – ck). This can be observed in p y k ( k k)reduced problem.

z = z0 – (zk – ck) xkThus it is possible to find a new basis by replacing one of the existing column r with the column

of “entering variable” xkg k.Extent of Improvement in the objective function value and “leaving variable”?

1 11 kB b yxb

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ 22 2 kB

kBr rk

yx b

xx yb

⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥= − ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥

M M M

Br rkr

Bm mkm

yb

x yb

⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦

M MM


Moving to an Adjacent CPFSWhat happens if optimal has not reached? If yik < 0, then xBi increases as xk increases and so xBi continues to be non-negative. Hence

problem is unbounded if yik < 0, for all i. Otherwise, for some i, yik > 0 then xBi will decrease as x increases In order to satisfy nonnegtaivity x is increased until the first point at which aas xk increases. In order to satisfy nonnegtaivity, xk is increased until the first point at which a basic variable xBr drops to zero (equivalently hits a hyperplane for which it is the indicating variable). This gives the following condition.

bb ⎧ ⎫ In the absence of degeneracy

1 i m : 0Minimum ir

k ikrk ik

bbx y

y y≤ ≤

⎧ ⎫= ≡ >⎨ ⎬

⎩ ⎭ 0, and hence 0.rr k

rk

bb x

y> = >

In the absence of degeneracy,

A new feasible solution is obtained.

1, 2,...., ; and ik rBi i r k

rk rk

y bx b b i m x

y y= − = =

Note that at the current bfs there are p hyperplanes (including xk=0) are binding. When we set xk > 0, we have only p-1 binding hyperplanes which defines the edge along which we travel until we hit the hyperplane xBr =0, at the next corner point


and we again have p binding hyperplanes. Or if we cannot find xBr =0 the we hit unboundness and in that case the edge is an extreme direction).

Moving to an Adjacent CPFSGeometric Interpretation: In Ax < b; x > 0, each constraint, including

nonnegativity constraints can be uniquely associated with a certain variable. Thus x1 > 0 can be associated with the variables x1, and the line (hyperplane x1=0). ( yp p 1 )

Example Problem 8: Consider Example Problem 7. A constraint such as, x1+ 2 x2 < 4 can be associated with the slack variable x3, and x3 = 0 (i.e., x1 + 2 x2 = 4) is the hyperplane and so on. Intersections of the p (=n-m) h l d fi bf d th i t d d fi i i bl b ihyperplanes define bfs and the associated defining variables are nonbasic variables.


Moving to an Adjacent CPFSThe feasible region in (x1, x2) and (x3, x4) space is shown. How many such

representations are possible? As many as choices of basis (or rather non-basic variables). Introducing x3 will improve objective function value. One ) g 3 p jmoves along an edge. On edge p-1 hyperplanes are binding. How? [Hint: variables are expressed in terms of x3]

xB = B-1 b – B-1 a3 x3

13

2 11 0

xx

x⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦2 1 0x ⎣ ⎦ ⎣ ⎦⎣ ⎦

Maximum value of x3 is 2 (any larger value of x3 will force x1 to be negative). Which hyperplane it hits first? x1= 0.

New bfs [x1, x2, x3, x4] = [0,1,2,0]

(Note: old bfs =[2, 1, 0, 0]. Thus the direction of travel along the edge is: [0-2,1-1, 2-0 0-0])


2-0, 0-0])

x3 enters basis improvement in z = (z3-c3)*x3 = 2.

Moving to an Adjacent CPFSMeaning of Reduced Cost, zk-ck: It is to be noted that zk-ck is the amount of

improvement in the objective function value if value of xk is increased by 1. One unit of increase in xk reduced a basic variable xBr by yBrk,. Thus zk represents the amount by which objective function value reduces (sum total for all basic variables) and ck represents the amount by which objective function value increases due to variable xk. The net improvement in objective function for a minimization problem is (zk-The net improvement in objective function for a minimization problem is (zkck). Note that for a maximization problem the improvement will be -(zk-ck).

Simplex: Simplex method at each iteration transforms the problem into p nonbasic variable space with current solution at origin. To check the improvement in cost f i l i di i Si l i di i ifunction one can evaluate p current axis directions. Simplex in p dimensions is a convex hull of a set of p+1 non-coplanar points, that is the points not lying in the same hyperplane.

Simplex for p=1 Simplex for p

=2 Simplex for p


2 Simplex for p=3

Termination: Unique & Alternate Solutions and InfeasibilitySolutions and Infeasibility

Unique SolutionIf (z c ) < 0 [(z c ) > 0] for all non basic variables j thenIf (zj – cj) < 0 [(zj – cj) > 0] for all non-basic variables, j then unique solutions for minimization [maximization] problem.

Alt ti S l ti ( )Alternative Solution(s)If (zj – cj) = 0 for some non-basic variables, j indicates the presence of an alternate solution.

InfeasibilitySolution does not improve with artificial variable(s) still inSolution does not improve with artificial variable(s) still in the basis.


(Original) Simplex Method Step 1. Obtain the initial feasible basis.

Step 2. Check the optimality of the current basis. CBFS is optimal iff everyffi i t i 0 i iti ( ti ) f i i i ticoefficient in row 0 is nonpositive (nonnegative) for minimization

(maximization) problem. [that is, zj – cj ≤ 0 (or zj – cj > 0) for each j ∈ R for minimization (or maximization) problem]). If optimal, STOP.

St 3 If th t b i i t ti l id tif th t i i bl [th tStep 3. If the current basis is not optimal, identify the entering variable xk [that is, zk – ck > 0 (or zk – ck < 0) for minimization (or maximization) problem].

Step 4. Perform the minimum ratio test to determine the leaving variable xBr.-Select strictly positive coefficients in pivot column.-Divide coefficients into RHS of same row.-Identify row with smallest ratio, this is the pivot row.R l i bl f th t ith t i b i i bl-Replace variable of that row with entering basic variable.

Step 5. Perform a pivot operation to update the basis and go to Step 2.


Illustration of Simplex Method: Example Prob. 10


Illustration of Simplex Method


Termination: UnboundednessUnboundedness: There is no blocking hyperplane from the current solution along the

simplex or simplices. Or one moves from the current vertex along an extreme direction.

Example Problem 9: Minimize z = -x1 – 3x2Example Problem 9: Minimize z x1 3x2 Subject to

x1 – 2x2 + x3 = 4-x1 + x2 + x4 = 3

0x1, x2, x3, x4 > 0.

Coefficient Iteration BV z x1 x2 x3 x4 RHS Ratio Remarks

z 1 1 3 0 0 0 1 x3 0 1 (-2) 1 0 4 0 x2 enters x4 0 (-1) 1 0 1 3 3 x4 leaves z 1 4 0 0 (-3) (-9)( ) ( )2 x3 0 (-1) 0 1 2 10 .... x1 can enter

x2 0 (-1) 1 0 1 3 .... soln unbounded


Direction of Unboundedness

In case x1 is increased and x4 is kept at 0, the solution is unbounded. We can represent a direction along which the solution is unbounded in terms of current bfs (cbfs) as vertex as shown next.


Direction of Unboundednessx1 x1 d1

x2 x2 + λ d2

x3 = x3 d3

where, λ =xk can be increased infinitely large

x4 x4 bfs d4

xB xB dB

x x bfs d= + x

Extreme direction d = (1, 1, 1, 0).

xN xN bfs dN

xB xB bfs -yk

xj ≠ k = 0 + xk 0

= + xk For minimization problem the condition forj

xk 0 1

x1 0 1

condition for unboundedness, cd < 0.

x2 3 + x1 1 Extremex3 = 10 1 directionx4 0 0


Revised Simplex MethodRevised simplex method is a systematic procedure for implementing the steps in the

simplex method in a smaller array, thus saving storage space. Columns for nonbasic variables are not stored but generated for checking optimality and performing ratio test. First recap the simplex tableau:

As can be seen, for a given basis and hence B and B-1, we can generate the information on RHS column and column of nonbasic variables to conduct optimality check and ratio test using the following formulas:

Let simplex multipliers (shadow prices) w = cB B-1

Column of a nonbasic variable xj: (Not stored)

(Row 0 ) The reduced costs of xj = cB B-1 aj –cj = w aj –cj

(Constraint rows) Column of x in the constraints = y = B-1 a


(Constraint rows) Column of xj in the constraints = yj = B aj

Right hand side column(Row 0 ) cB B-1 b =wb

(Constraint rows) = B-1b

Revised Simplex MethodSteps in Revised Simplex Method for Minimization Problem:Step 1. Obtain the initial feasible basis, xB and B.Step 2 i) Compute B-1 the simplex multipliers w = c B-1 and current basic feasibleStep 2. i) Compute B , the simplex multipliers w = cB B and current basic feasible

solution (xB = = B-1 b, xN = 0 and z = cB B-1 b =wb= cB )ii) Calculate reduced cost zj-cj = cB B-1 aj –cj = w aj-cj for each nonbasic variable, xj. iii) Let zk-ck = maximum j ( zj - cj). If zk-ck < 0, then stop; the current solution is optimal.

Otherwise go to step 3cB

b

Otherwise go to step 3.Step 3. Calculate yk = B-1 ak. If yk < 0, then stop; the optimal solution is unbounded.

Otherwise determine the index of the variable xBr leaving the basis by ratio test.Step 4. Update the basis B by replacing aBr with ak (or more efficiently update B-1 directly)

d t t 2b

and go to step 2.The above can be carried out using a smaller array. Suppose we have a basic

feasible solution with a known B. The following array is constructed.

b

BASIS INVERSE SOLUTION Computation on xN columns

Variable Value (RHS) xj=N1 …… xj=k …… xj=Np

w = cB B-1 z = wb= cB zN1-cN1 …… zk-ck …… zNp-cNp


B-1 xB= = B-1 b Not required yk Not required

Revised Simplex Method Example Prob. 11: Illustration of Revised Simplex Method

Max z = -x1 - x2 + 4 x3Max z x1 x2 4 x3s.t. x1 + x2 + 2x3 + x4 = 9

x1 + x2 – x3 + x5 = 2 –x1 + x2 + x3 + x6 = 4

> 0 for i 1 2 3 4 5 6

Note that x4, x5 and x6 are slack variables.

xi > 0, for i = 1, 2, 3, 4, 5, 6.

j = 1 2 3 4 5 6 xj x1 x2 x3 x4 x5 x6 j cj = -1 -1 4 0 0 0

b Row No

1 1 2 1 0 0 9 R 1 1 1 2 1 0 0 9 R 1 aj = 1 1 -1 0 1 0 2 R 2 -1 1 1 0 0 1 4 R 3


Revised Simplex MethodStep 1.Assume that initial basis is xB ={x4, x5, x6}, B =[a4, a5, a6]

Iteration 1 Step 2

B = a4 a5 a6 Solution

xB = X4 X5 X6 Varble RHS cB = 0 0 0 w =cBB-1 = 0 0 0 z = 0

Basis 1 0 0 Basis Inv 1 0 0 X4 = 9

B = 0 1 0 B-1 = 0 1 0 X5 = 2 b 0 0 1 0 0 1 X6= 4

NBV, j ∈ R = 1 2 3

xj X 1 X 2 X 3 1 1 2

aj = 1 1 -1 -1 1 1

c = 1 1 4

Data from Original LP. Reproduced for sake of clarity

cj = -1 -1 4Red. Cost = w aj-cj 1 1 -4 Entering Variable, xk X 3

Step 3 i

Basis Column for Entering Variable, yk = B-1 ak

RHS Column

b Ratio

( ib /yik)

Leaving Variable

1 X 4 2 9 4.5 2 X 5 -1 2 --- 3 X 6 1 4 4 X6


X 6 1 4 4

Revised Simplex MethodStep 4 Basis updated. xB ={x4, x5, x3}, B =[a4, a5, a3]

Iteration 2 Step 2

B = a4 a5 a3 Solution VarbleValue (RHS)xB = X4 X5 X3 VarbleValue (RHS)

cB = 0 0 4 w =cBB-1 = 0 0 4 z = 16

Basis 1 0 2 Basis Inv 1 0 -2 X4 = 1

B = 0 1 -1 B-1 = 0 1 1 X5= 6 b 0 0 1 0 0 1 X3= 4

NBV, j ∈ R = 1 2 6

xj X 1 X 2 X 6 1 1 0

aj = 1 1 0 1 1 1

Data from Original LP. Reproduced for

-1 1 1cj = -1 -1 0

Reproduced for sake of clarity

Red. Cost = w aj-cj -3 5 4

Entering Variable, xk X1 Step 3 i

Basis Column for Entering Variable, yk = B-1 ak

RHS Columnb=B-1 b

Ratio ( ib /yik)

Leaving Variable

1 X 4 3 1 0.3333 X4 2 X 5 0 6 ---


2 X 5 0 6 3 X 3 -1 4 ---

Revised Simplex MethodStep 4 Basis updated. xB ={x1, x5, x3}, B =[a1, a5, a3]

Iteration 3

Step 2 B = a1 a5 a3 Solution

xB = X1 X5 X3 Vrble Value(R

HS)

cB = -1 0 4 w =cBB-1 = 1 0 2 z = 17

Basis 1 0 2 Basis Inv 0.3333 0 -0.667 X1 = 0.3333Basis 1 0 2 Basis Inv 0.3333 0 0.667 X1 0.3333B = 1 1 -1 B-1 = 0 1 1 X5= 6 b

-1 0 1 0.3333 0 0.3333 X3= 4.3333

NBV, j ∈ R = 5 2 6

xj X 4 X 2 X 6xj X 4 X 2 X 6 1 1 0

aj = 0 1 0 0 1 1

cj = 0 -1 0

Data from Original LP. Reproduced for sake of clarity

Red. Cost = w aj-cj 4 1 2 Entering Variable, xk None- Optimal reached


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