Q 2-31

Min 3A + 4B

s.t.

1A + 3B ≧ 6

1A + 1B ≧ 4

A, B ≧ 0

B = - 1/3A + 2

B = - A + 4

Feasible Region

A

B

B = - 1/3A + 2

B = - A + 4

Objective Function = 3A + 4BOptimal Solution A = 3, B = 1

Objective Function Value = 3(3) + 4(1) = 13

Q 2-38 a.

Let

S = yards of the standard grade material / frame

P = yards of the professional grade material / frameMin 7.50S + 9.00Ps.t.

0.10S + 0.30P ≧ 6 carbon fiber (at least 20% of 30 yards)

0.06S + 0.12P ≦ 3 kevlar (no more than 10% of 30 yards)

S + P = 30 totalS, P ≧ 0

Total

Carbon fiberKevlar

Feasible region is the line segment

Extreme PointS = 10, P = 20

S

P

Extreme PointS = 15, P = 15

Q 2-38 c.

Extreme Point Cost

(15, 15) 7.50(15) + 9.00(15) = 247.50

(10, 20) 7.50(10) + 9.00(20) = 255.00

255.00 > 247.50

Therefore, the optimal solution is

S = 15, P = 15

Q 2-38 d.

Min 7.50S + 8.00Ps.t.

0.10S + 0.30P ≧ 6 carbon fiber (at least 20% of 30 yards)

0.06S + 0.12P ≦ 3 kevlar (no more than 10% of 30 yards)

S + P = 30 totalS, P ≧ 0

Changing is only this value(From 9.00 to 8.00)

Therefore, optimal solution does not change: S = 15, and P =15New optimal solution value = 7.50(15) + 8(15) = $232.50

Q 2-38 e.

At $7.40 per yard, the optimal solution is S = 10,

P = 20.

So, the optimal solution is changed.

The value of the optimal solution is reduced to

7.50 (10) + 7.40 (20) = 223.00.

A lower price for the professional grade will

change the optimal solution.

Q 3-7 a.

From figure 3.14,

Optimal solution: U = 800, H = 1200

Estimated Annual Return

3(800) + 5(1200) = $8400

Q 3-7 b.

Constraints 1 and 2 because of 0 at

Slack/Surplus column. All funds available

are being fully utilized and the maximum

risk is being incurred.

Q 3-7 c.

Constraint Dual Prices

Funds Avail 0.09

Risk Max 1.33

U.S. Oil Max 0

A unit increase in the RHS of Funds Avail makes 0.09

improvement in the value of the objective function. Risk Max

also makes 1.33 improvement in the value of the objective

function. A unit increase in the RHS of U.S. Oil Max makes

no improvement in the objective function value.

Q 3-7 d.

NO

A unit increase in the RHS of U.S. Oil Max

makes no improvement in the objective

function value.

Q 3-10 a.

From figure 3.16,

Optimal solution: S = 4,000, M = 10,000

Estimated total risk =

8(4,000) + 3(10,000) = 62,000

Q 3-10 b.

Variable Range of Optimality

S 3.75 to No Upper Limit

M No Lower Limit to 6.4

Q 3-10 c, d, e, f

(c) 5(4,000) + 4(10,000) = $60,000

(d) 60,000/1,200,000 = 0.05

(e) 0.057

(f) 0.057(100) = 5.7

Theory of Simplex Method

For any LP problem with n decision variables, each CPF (Corner Point Feasible) solution lies at the intersection of n constraint boundaries; i.e., the simultaneous solution of a system of n constraint boundary equations.

,53 21 xxZ

1x

0,0

1823

21

21

xx

xx122 2 x4

and

Max

s.t.

A two-variable linear programming problem

0

x

bAx

cx

,

0

0

0

0,,,,,, 2

1

2

1

21

nn

n

b

b

b

b

x

x

x

xcccc

Max

s.t.

mnmm

n

n

aaa

aaa

aaa

A

21

22221

11211

Original Form Augmented Form

Max s.t. bAX

0x

Maxs.t.

cX Z

0,0

0

001

S

S

S

XX

bIXAXZ

XcXZ

Matrix Form

b

X

X

Z

IA

c

S

00

0

1

(1)

(2)

Matrix Form (2) is

Max

s.t.

bIXAX

XcXZ

Z

S

S

00

(3)

orMax

s.t.

bXA

XcZ

Z

ˆˆ

0ˆˆ(4)

where

IAAX

XXcc

S

,ˆˆ0,ˆ

matrix nonbasic a :N

matrix basic a :B

)X (to tscoefficien

objective ingcorrespond theseof vector a :c

)X (to tscoefficien

objective ingcorrespond theseof vector a :c

variablesbasicnon of vector a :X

variablesbasic of vector a :X

N

N

B

B

N

B

Max

s.t.

bNXBX

XcXcZ

Z

NB

NNBB

0

Then, we have

(5)

(6)

where

NBAcccX

XX NB

N

B ,ˆ,ˆˆ

Eq. (6) becomes

bBNXBX NB11

Putting Eq. (7) into (5), we have

0)( 11 NNNB XcNXBbBcZ

(7)

(8)So,

bBcXcNBcXZ BNNBB11 )(0 (9)

become (9) Eq. and (7) Eq. ,0 Currently, NX

bBcZbBX BB11 , (10)

bB

bBc

bB

Bc

X

Z BB

B1

1

1

1 0

0

1

Eq. (10) can be expressed by

(11)

From Eq. (2),

bB

bBc

X

X

Z

BAB

BccABc

bB

bBc

X

X

Z

IA

c

B

Bc

X

Z

B

S

BB

B

S

B

B

1

1

11

11

1

1

1

1

0

1

0

01

0

1

(12)

Thus, initial and later simplex tableau are

IterationBVZOriginalVariables

SlackVariables

RHS

Z1 -c 0 00

BX0 A I b

IterationBVZOriginalVariables

SlackVariables

RHS

Z 1 cABcB 1 1BcB bBcB1

AnyBX 0 AB1 1B bB1

1. Initialization:

Same as for the original simplex method.

2. Iteration:

Step 1

Determine the entering basic variable:

Same as for the Simplex method.

The Overall Procedure

Step 2

Determine the leaving basic variable:

Same as for the original simplex method,

except calculate only the numbers required to

do this [the coefficients of the entering basic

variable in every equation but Eq. (0), and

then, for each strictly positive coefficient, the

right-hand side of that equation].

Step 3

Determine the new BF solution:

Derive and set

3. Optimality test:

Same as for the original simplex method, except

calculate only the numbers required to do this test,

i.e., the coefficients of the nonbasic variables in

Eq. (0).

1B .1bBxB

Fundamental Insight

Z

Z

RHS

Row0

Row1~N

1BcB

1B

bBcB1

bB 1

X

BX

SX

1

0 AB 1

cABcB 1

0

5x4x3x

RightSideBVIteration 3x 4x 5x2x1x

18

12

4

)(,

100

010

001

)(,

2

2

0

3

0

111 bBbBIA

-3 -5 0 0 0 0 1 0 1 0 0 4 0 2 0 1 0 12 3 2 0 0 1 18

Coefficient of:

,0,0,0,5,3,

5

4

3

SB cc

x

x

x

x

BVIteration

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 6

,

110

00

001

,0,5,0

21 ,

6

6

4

18

12

4

110

00

001

1B bB 1

Bc

21

21

RightSide3x 4x 5x2x1x

Coefficient of:

0 0 0

BVIteration

25

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 62

1

RightSide3x 4x 5x2x1x

Coefficient of:

-3 0 0 0

0

0

1

0

110

00

001

0,5,0

33

3

0

1

110

00

001

0,5,0

2

1

21

11 cZ 111 caBcB

44 cZ 441 caBcB 2

5

BVIteration

25

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 62

1

RightSide3x 4x 5x2x1x

Coefficient of:

-3 0 0 0

0

1

0

3

0

1

2

2

0

3

0

1

110

00

001

AB 12

1

BVIteration

25

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 62

1

RightSide3x 4x 5x2x1x

Coefficient of:

-3 0 0 0 30

30

6

6

4

0,5,01

bBcB

so

BVIteration

25

1

5x2x3x 1 0 1 0 0 4

0 1 0 0 6

3 0 0 -1 1 62

1

RightSide3x 4x 5x2x1x

Coefficient of:

-3 0 0 0 30

minimum

The most negative coefficient

23

6

41

4

4

6

BVIteration

2

1x2x3x 0 0 1 2

0 1 0 0 6

1 0 0 2

21

RightSide3x 4x 5x2x1x

Coefficient of:

0 0 0

,

0

00

1

,3,5,0Bc

21 ,

2

6

2

18

12

4

0

00

1

2

13

13

1

31

31

31

31

31

31

1B bB 1

31

31

31

31

BVIteration

2

1x2x3x 0 0 1 2

0 1 0 0 6

1 0 0 2

21

RightSide3x 4x 5x2x1x

Coefficient of:

0 0 0 1

31

31

31

31

10

1

0

0

0

0

0

1

3,5,0

230

0

1

0

0

0

0

1

0,5,0

44

144 caBccZ B

551

55 caBccZ B

23

21

31

31

31

31

21

31

31

31

31

BVIteration

2

1x2x3x 0 0 1 2

0 1 0 0 6

1 0 0 2

21

RightSide3x 4x 5x2x1x

Coefficient of:

0 0 0 1 36

31

31

31

31

36

2

6

2

3,5,01

bBcB

so

23

Duality Theory

One of the most important discoveries in the early development of linear programming was the concept of duality.

Every linear programming problem is associated with another linear programming problem called the dual.

The relationships between the dual problem and the original problem (called the primal) prove to be extremely useful in a variety of ways.

The dual problem uses exactly the same parameters as the primal problem, but in different location.

Primal and Dual Problems

Primal Problem Dual Problem

Max

s.t.

Min

s.t.

n

jjj xcZ

1

,

m

iii ybW

1

,

n

1jijij ,bxa

m

ijiij cya

1,

for for.,,2,1 mi .,,2,1 nj

for .,,2,1 mi for .,,2,1 nj ,0jx ,0iy

In matrix notation

Primal Problem Dual Problem

Maximize

subject to

.0x .0y

Minimize

subject tobAx cyA

,cxZ ,ybW

Where and are row vectors but and are column vectors.

c myyyy ,,, 21 b x

Example

Maxs.t.

Min

s.t.

Primal Problem Dual Problem

,53 21 xxZ ,18124 321 yyyW

1823 21 xx

122 2 x41x

0x,0x 21

522 32 yy

33 3 y1y

0y,0y,0y 321

Max

s.t.

Primal Problem in Matrix Form

Dual Problem in Matrix Form

Min

s.t.

,5,32

1

x

xZ

18

12

4

,

2

2

0

3

0

1

2

1

x

x

.0

0

2

1

x

x .0,0,0,, 321 yyy

5,3

2

2

0

3

0

1

,, 321

yyy

18

12

4

,, 321 yyyW

Primal-dual table for linear programmingPrimal Problem

Coefficient of: RightSide

Rig

ht

Sid

eDu

al P

rob

lem

Co

effi

cien

to

f:

my

y

y

2

1

21

11

a

a

22

12

a

a

n

n

a

a

2

1

1x 2x nx

1c 2c ncVI VI VI

Coefficients forObjective Function

(Maximize)

1b

mna2ma1ma

2b

mb

Coe

ffic

ient

s fo

r O

bjec

tive

Fun

ctio

n(M

inim

ize)

One Problem Other Problem

Constraint Variable

Objective function Right sides

i i

Relationships between Primal and Dual Problems

Minimization Maximization

Variables

Variables

Constraints

Constraints

0

0

0

0

Unrestricted

Unrestricted

The feasible solutions for a dual problem are

those that satisfy the condition of optimality for

its primal problem.

A maximum value of Z in a primal problem

equals the minimum value of W in the dual

problem.

Rationale: Primal to Dual Reformulation

Max cxs.t. Ax b x 0

L(X,Y) = cx - y(Ax - b) = yb + (c - yA) x

Min yb

s.t. yA c

y 0

Lagrangian Function )],([ YXL

X

YXL

)],([

= c-yA

The following relation is always maintained

yAx yb (from Primal: Ax b)

yAx cx (from Dual : yA c)

From (1) and (2), we have

cx yAx yb

At optimality

cx* = y*Ax* = y*b

is always maintained.

(1)

(2)

(3)

(4)

“Complementary slackness Conditions” are

obtained from (4)

( c - y*A ) x* = 0

y*( b - Ax* ) = 0

xj* > 0 y*aj = cj , y*aj > cj xj* = 0

yi* > 0 aix* = bi , ai x* < bi yi* = 0

(5)

(6)

Any pair of primal and dual problems can be

converted to each other.

The dual of a dual problem always is the primal

problem.

Min W = yb,

s.t. yA c

y 0.

Dual ProblemMax (-W) = -yb,

s.t. -yA -c

y 0.

Converted to Standard Form

Min (-Z) = -cx,

s.t. -Ax -b

x 0.

Its Dual Problem

Max Z = cx,

s.t. Ax b

x 0.

Converted toStandard Form

Mins.t.

64.06.0

65.05.0

7.21.03.0

21

21

21

xx

xx

xx

0,0 21 xx

21 5.04.0 xx

Mins.t.

][y 64.06.0

][y 65.05.0

][y 65.05.0

][y 7.21.03.0

321

-221

221

121

xx

xx

xx

xx

0,0 21 xx

21 5.04.0 xx

Maxs.t.

.0,0,0,0

5.04.0)(5.01.0

4.06.0)(5.03.0

6)(67.2

3221

3221

3221

3221

yyyy

yyyy

yyyy

yyyy

Maxs.t.

.0, URS:,0

5.04.05.01.0

4.06.05.03.0

667.2

321

321

321

321

yyy

yyy

yyy

yyy

Home Work

• Ch 4: Problem 1

• Ch 4: Problem 10

• Additional Problems: A-1 and A-2

• (See the proceeding PPS)

• Due Date: September 16

Theory of Simplex Method: Consider the following

problem.

0x&0x,0x,0x

4xx2xx3

5xxx2x4

,x2xx3x4Z

4321

4321

4321

4321

to subject

Maximize

Let x5 and x6 denote slack variables for the two constraints.

After you apply the simplex method, a portion of the final

simplex tableau is as follows:

A – 1

A – 1 cont’d

Basic Variable

Coefficient of:Right SideEq. Z x1 x2 x3 x4 x5 x6

Z (0) 1 1 1

x2 (1) 0 1 -1

x4 (2) 0 -1 2

• Solve the problem.

• What is B-1 ? How about B-1b and CBB-1b ?

• If the right hand side is changed from (5, 4) to (5, 5),

how is and optimal solution changed? How about an

optimal objective value?

Linear Programming: Slim-Down Manufacturing makes

a line of nutritionally complete, weight-reduction

beverages. One of their products is a strawberry shake

which is designed to be a complete meal. The

strawberry shake which is designed to be a complete

meal. The strawberry shake consists of several

ingredients. Some information about each of these

ingredients is given below.

A – 2

A – 2 cont’d

IngredientCalories from Fat (per tbsp)

Total Calories (per tbsp)

Vitamin Content (mg/tbsp)

Thickeners (mg/tbsp)

Cost (¢/tbsp)

Strawberry flavoring

1 50 20 3 10

Cream 75 100 0 8 8

Vitamin supplement

0 0 50 1 25

Artificial sweetener

0 120 0 2 15

Thickening agent

30 80 2 25 6

The nutritional requirements are as follows. The beverage must

total between 380 and 420 calories (inclusive). No more than

20 percent of the total calories should come from fat. There

must be at least 50 milligrams (mg) of vitamin content. For

taste reasons, there must be at least 2 tablespoons (tbsp) of

strawberry flavoring for each tablespoon of artificial sweetener.

Finally, to maintain proper thickness, there must be exactly 15

mg of thickeners in the beverage.

Management would like to select the quantity of each

ingredient for the beverage which would minimize cost while

meeting the above requirements.

A – 2 cont’d

• Formulate a linear programming model for this problem.

• Show its dual formulation.

• Solve this model by your computer and confirm

complementary slackness condition.

A – 2 cont’d