treasurer's affidavit

46
Chapter 4 Simplex Method for Linear Programming Shi-Shang Jang Chemical Engineering Department National Tsing-Hua University

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Page 1: Treasurer's Affidavit

Chapter 4 Simplex Method for Linear Programming

Shi-Shang Jang

Chemical Engineering Department

National Tsing-Hua University

Page 2: Treasurer's Affidavit

Example 1: Inspector Problem

Assume that it is desired to hire some inspectors for monitoring a production line. A total amount of 1800 species of products are manufactured every day (8 working hours), while two grades of inspectors can be found. Maximum, 8 grade A inspector and 10 grade B inspector are available from the job market. Grade A inspectors can check 25 species/hour, with an accuracy of 98 percent. Grade B inspectors can check 15 species/hour, with an accuracy of 95 percent. Note that each error costs $2.00/piece. The wage of a grade A inspector is $4.00/hour, and the wage of a grade B inspector is $3.00/hour. What is the optimum policy for hiring the inspectors?

Page 3: Treasurer's Affidavit

Problem Formulation

Assume that the x1 grade A inspectors x2 grade B inspects are hired, then

total cost to be minimized 48 x1 +38 x2 +2580.022 x1 +1580.052 x2

=40 x1 +36 x2 manufacturing constraint 258 x1 +158 x2 1,800 200 x1 +120 x2 1,800 no. of inspectors available: 0 x1 8 0 x2 10

Page 4: Treasurer's Affidavit

The Graphical Solution

X2=10

X1 =8

B(8,10)

(8,5/3)

C(3,10)

15

10

5

5 10

Page 5: Treasurer's Affidavit

Theorem

Property: If there exists an optimal solution to a LP, then at least one of the corner point of the feasible region will always qualify to be an optimal solution.

Page 6: Treasurer's Affidavit

Special Cases

Alternate Solutions (non-unique solutions) Max x1+2x2

s.t. x1+2x210

x1+ x21

x24

x10, x20

X1+2X

2=10

X1+2X

2=6X1+2X

2=2X1+X

2=1

5

4

106

X2=4

X2

X1

Page 7: Treasurer's Affidavit

Special Cases - continued

Unbounded Optima : A system has a feasible region with open boundaries such that the optima may appear at the infinity.

Example: For the previous example, in case the constraint x1+2x210 is not given, then moving far away from the origin increases the objective function x1+2x2, and the maxim Z would be +

Page 8: Treasurer's Affidavit

Unbounded Optima

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

x1

x2

x1+x2=1

x2=4

x1+2x2=2

x1+2x2=6

x1+2x2=10

x1+2x2=15

Page 9: Treasurer's Affidavit

Example 2: Student Fab

The RIT student-run microelectronic fabrication facility is taking orders for four indigenously developed ASIC chips that can be used in (1) touch sensors ($6, s4hr, m1hr, v30), (2)LCD($10,s9hr, m1hr, v40), (3) pressure sensors($9, s7hr, m3hr, v20), and (4) controllers($20, s10hr, m8hr, v10).

Constraints : Student hr 600, machine ≦hr 420, space 800≦ ≦

Page 10: Treasurer's Affidavit

The LP Problem

80010204030

42083

60010794..

209106max

4321

4321

4321

4321

xxxx

xxxx

xxxxts

xxxxx

Page 11: Treasurer's Affidavit

4-2 The Basic Approach

Standard Form of Linear Programming

mnbbbxxx

bxaxaxa

bxaxaxa

bxaxaxats

xcxcxcy

mn

mnmnmm

nn

nn

nnxxx n

,0,,, ,0,,,

..

min

2121

2211

22222121

11212111

2211,,, 21

Page 12: Treasurer's Affidavit

Handling of in-equality Constraints

Case 1: Slack Variable:

x1+2x2+3 x3 +4x425 – Modified to

x1+2x2+3x3+4x4+ x5=25 x50 is a slack variable.

Case 2: Surplus Variable:

2 x1+ x2-3 x312– Modified to

2 x1+ x2-3 x3- x4=12

x40 is a surplus variable.

Page 13: Treasurer's Affidavit

Handling of Equality Constraints

If s is unrestricted, i.e., s can be positive or negative, then we set

s=s+-s-

such that s+0, s- 0.

Page 14: Treasurer's Affidavit

Example

0,

533

2

7 ..

32max

21

321

321

321

321

xx

xxx

xxx

xxxts

xxxz

Modify to

0,,,,,

5333

2

7 ..

332max

765421

5421

75421

65421

5421

xxxxxx

xxxx

xxxxx

xxxxxts

xxxxz

Page 15: Treasurer's Affidavit

Definitions

Definition: A feasible region, denoted by S is the set of all feasible solution. Mathematically, .

Definition: An optimal solution is a vector x*S, s.t. z0=cTx* is maximum or minimum in where Z is termed by the optimal value.

Definition: Alternate optimal solution is a set XS, s.t. all xX has the same objective value z0 and for all xS, and z=cTx, z z0.

Definition: If the solution set of LP contains only one element, it is termed the unique optimum.

Definition: If the optimum value z approaches to infinity, then the LP is said to have unbounded optimum.

0, xbAxxS

Page 16: Treasurer's Affidavit

4-3 The Simplex Method

sconstraint for theApproach Variable Artificial

andn EliminatioJordan Gauss

0,,,

..

min

21

2211

22222121

11212111

2211,,, 21

n

mnmnmm

nn

nn

nnxxx

xxx

bxaxaxa

bxaxaxa

bxaxaxats

xcxcxcyn

mnmnsmsmmmm

nnssmm

nnssmm

bxaxaxax

bxaxaxax

bxaxaxax

11,

22211,22

11111,11

Form Canonical

Page 17: Treasurer's Affidavit

Definitions

Definition: A pivot operation is sequence of elementary row operations that reduce the coefficients of a specified variable to unity in one of the equation and zero elsewhere.

Definition: In the above canonical form, x1,,xm are termed the basic variables or dependent variables, xm+1, ,xn are called nonbasic variables or the independent variables.

Definition: The solution obtained from a canonical form is by setting the nonbasic variable or independent variable to zero is called a basic solution.

Definition: A basic feasible solution is a basic solution in which the basic or dependent variables are non-negative.

Page 18: Treasurer's Affidavit

Property

Remark: )!(!

!solutions basic of no.

mnm

n

m

n

where nmjxnonbasicmibxbasic iii ,,1,0 ,,,1,

Definition: mBmB ccxx ,,,,, 11 cx

Property: A feasible basic solution is a simplex of the feasible region.

Note: Given a canonical form and feasible basic solution, then the objective function:

m

iiiB

TB bcz

1

bc

where xB is a basic variable.

Page 19: Treasurer's Affidavit

Approach (Simplex Method):

Start with an initial basic feasible solution in canonical form.

Improve the solution by finding another basic feasible solution if possible.

When a particular basic feasible solution is found, and cannot be improved by finding new basic feasible solution, the optimality is reached.

Definition: An adjacent basic solution differs from a basic solution is exactly one basic variable.

Question: If one wants to find an adjacent feasible basic solution from one feasible basic solution (i.e., switch to another simplex), which adjacent basic solution gives lowest objective function?

Page 20: Treasurer's Affidavit

Derivation of Inner Product Rule

msmsm

ss

ss

bxax

bxax

bxax

222

111

Supposing, one wants to replace one of the original basic variable with nonbasic variable xs, we firstly, increase xs from zero to one,then for all i=1,,m,

1

s

isii

x

abx

and, for all j=m+1, ,n, js. xj=0

Page 21: Treasurer's Affidavit

Theorem 1 (Inner Product Rule)

s

m

iisiinew cabcz

1

Relative cost,

m

iisisnew acczzz

1(inner product rule)

More: (1) In a minimization problem, a basic feasible solution is optimalif the relative costs of its all nonbasic variable are all positive or zero.(2) One should choose an adjacent basic solution from which the relative cost is the minimum.

Corollary: The alternate optima exists if z=0.

Page 22: Treasurer's Affidavit

Theorem 2: (The Minimum Ratio Rule )

Given a nonbasic variable xs is change into the basic variable set, then one of the basic variable xr should leave from the basic variable set, such that:

ia

bx

is

i

as

is

,minmax

0

The above minimum happens at i=r.

Corollary: The above rule fails if there exist unbounded optima.

Page 23: Treasurer's Affidavit

Example: The LP Problem

80010204030

42083

60010794..

209106max

4321

4321

4321

4321

xxxx

xxxx

xxxxts

xxxxx

Page 24: Treasurer's Affidavit

The Standard Form

80010204030

42083

60010794..

209106max

743213

643212

543211

4321

xxxxxg

xxxxxg

xxxxxgts

xxxxf

Page 25: Treasurer's Affidavit

Table 1(s=4,r=6,rc=2)Basic={5 6 7};Nonbasic={1 2 3 4}

6 10 9 20 0 0 0

CB Basis x1 x2 x3 x4 x5 x6 x7 constraints

0 x5 4 9 7 10 1 0 0 600 (600/10=60)

0 x6 1 1 3 8 0 1 0 420(420/8=52.5)

0 x7 30 40 20 10 0 0 1 800(800/10=80)

Z 6 10 9 20 - - - Z=0

Page 26: Treasurer's Affidavit

Table 2(s=2,r=7,rc=3)Basic={5 4 7};Nonbasic={1 2 3 6}

6 10 9 20 0 0 0

CB Basis x1 x2 x3 x4 x5 x6 x7 constraints

0 x5 2.75 7.75 3.25 0 1 -1.25 0 75(75/7.75=9.6774)

20 x4 0.125 0.125 0.375 1 0 0.125 0 52.5(52.5/0.125=420)

0 x7 28.75 38.75 16.25 0 0 -1.25 1 275(275/38.75=7.0798)

Z 3.5 7.5 1.5 - - -2.5 - Z=1050

Page 27: Treasurer's Affidavit

Table 3Basic={5 4 2};Nonbasic={1 7 3 6}

6 10 9 20 0 0 0

CB Basis x1 x2 x3 x4 x5 x6 x7 constraints

0 x5 -3 0 0 0 1 -1 -0.2 20

20 x4 0.0323 0 0.3226 1 0 0.129 -0.0032 51.6129

10 x2 0.7419 1 0.4194 0 0 -0.0323 0.0258 7.0968

Z -2.0645 - -0.1935 - - -1.6452 -2.2581 Z=1103.226

Page 28: Treasurer's Affidavit

2. The Two Phase Simplex Method-Example

0,,

12

334

112..

3min

321

31

321

321

321

xxx

xx

xxx

xxxts

xxx

Page 29: Treasurer's Affidavit

The Standard Form

0,,,,

12

324

112 ..

3min

54321

31

5321

4321

321

xxxxx

xx

xxxx

xxxxts

xxx

Page 30: Treasurer's Affidavit

Two Phase Approach-Phase I

12

324

112..

min

731

65321

4321

76

xxx

xxxxx

xxxxts

xx

Page 31: Treasurer's Affidavit

Two Phase Approach-Phase ITable 1

  0 0 0 0 0 1 1  

CB Basis x1 x2 x3 x4 x5 x6 x7 constraints

0 x4 1 -2 1 1 0 0 0 11(11/1=11)

1 x6 -4 1 2 0 -1 1 0 3(3/2=1.5)

1 x7 -2 0 1 0 0 0 1 1(1/1=1)

Z   6 -1 -3 - 1 - -  Z=4

Page 32: Treasurer's Affidavit

Two Phase Approach-Phase IFinal

  0 0 0 0 0 1 1  

CB Basis x1 x2 x3 x4 x5 x6 x7 constraints

0 x4 3 0 0 1 -2 2 -5 12

0 x2 0 1 0 0 -1 1 -2 1

0 x3 -2 0 1 0 0 0 -1 1

Z   - - - - - - -  Z=0

Page 33: Treasurer's Affidavit

Two Phase Approach-Phase IITable 1

  -3 1 1 0 0  

CB Basis x1 x2 x3 x4 x5 constraints

0 x4 3 0 0 1 -2 12(12/3=4)

1 x2 0 1 0 0 -1 1(1/0=)

1 x3 -2 0 1 0 0 1(1/-2<0)

Z   -1 - - - 5  Z=2

Page 34: Treasurer's Affidavit

Two Phase Approach-Phase IIFinal

  -3 1 1 0 0  

CB Basis x1 x2 x3 x4 x5 constraints

-3 x1 1 0 0 0.3333 -0.667 4

1 x2 0 1 0 0 -1 1

1 x3 0 0 1 0.6667 -1.333 9

Z   - - - - -  Z=-2

Page 35: Treasurer's Affidavit

Example: Multi-Products Manufacturing

A company manufactures three products: A, B, and C. Each unit of product A requires 1 hr of engineering service, 10 hr of direct labor, and 3lb of material. To produce one unit of product B requires 2hr of engineering, 4hr of direct labor, and 2lb of material. In case of product C, it requires 1hr of engineering, 5hr of direct labor, and 1lb of material. There are 100 hr of engineering, 700 hr of labor, and 400 lb of material available. Since the company offers discounts for bulk purchases, the profit figures are as shown in the next slide:

Page 36: Treasurer's Affidavit

Example- Continued

Formulate a linear program to determine the most profitable product mix.

Product A Product B Product C

Sales units Unit profit variables

Sales units Unit profit variables Sales units Unit profit variables

0-40 10 X1 0-50 6 X5 0-100 5 X8

40-100 9 X2 50-100 4 X6 Over 100 4 X9

100-150 8 X3 Over 100 3 X7

Over 150 7 X4

Page 37: Treasurer's Affidavit

Problem Formulation

Let’s denote the variables as shown in the table, then we have the following:

0

600

400

400)(1)(2)(3

700)(5)(4)(10

100)(1)(2)(1..

4534678910max

9

2

1

987654321

987654321

987654321

987654321,, 101

X

X

X

XXXXXXXXX

XXXXXXXXX

XXXXXXXXXts

XXXXXXXXXXX

Page 38: Treasurer's Affidavit

MATLAB PROGRAM

f=[-10 -9 -8 -7 -6 -4 -3 -5 -4]'; A=[1 1 1 1 2 2 2 1 1; 10 10 10 10 4 4 4 5 5;3 3 3 3 2

2 2 1 1]; b=[100;700;400]; Aeq=[];beq=[]; LB=[0 0 0 0 0 0 0 0 0]; UB=[40 60 50 Inf 50 50 Inf 100 Inf]; [X,FVAL,EXITFLAG,OUTPUT,LAMBDA]=LINPROG(

f,A,b,Aeq,beq,LB,UB)

Page 39: Treasurer's Affidavit

Solution

X’= 40.0000 22.5000 0.0000 0.0000 18.7500 0.0000 0.0000 0.0000 0.0000

FVAL =-715.0000 EXITFLAG =1 OUTPUT = iterations: 7 cgiterations: 0 algorithm: 'lipsol' LAMBDA = ineqlin: [3x1 double] eqlin: [0x1 double] upper: [9x1 double] lower: [9x1 double]

Page 40: Treasurer's Affidavit

3. Sensitivity Analysis

Shadow Prices: To evaluate net impact in the maximum profit if additional units of certain resources can be obtained.

Opportunity Costs: To measure the negative impact of producing some products that are zero at the optimum.

The range on the objective function coefficients and the range on the RHS row.

Page 41: Treasurer's Affidavit

Example

A factory manufactures three products, which require three resources – labor, materials and administration. The unit profits on these products are $10, $6 and $4 respectively. There are 100 hr of labor, 600 lb of material, and 300hr of administration available per day. In order to determine the optimal product mix, the following LP model is formulated and solve:

Page 42: Treasurer's Affidavit

Basic LP Problem

0,,

ation)(administr 300622

(material) 6005410

(labor) 100..

4610max

321

321

321

321

321

xxx

xxx

xxx

xxxts

xxxZ

Page 43: Treasurer's Affidavit

Optimal Solution and Sensitivity Analysis

x1=33.33, x2=66.67,x3=0,Z=733.33 Shadow prices for row 1=3.33, row 2=0.67,

row 3=0 Opportunity Costs for x3=2.67 Ranges on the objective function coefficients:

6 c≦ 1(10) 15, 4 c≦ ≦ 2(6) 10, ≦

-∞ c≦ 3(4) 6.67≦

Page 44: Treasurer's Affidavit

Optimal Solution and Sensitivity Analysis- Continued

60 b≦ 1(100) 150, 400 b≦ ≦ 2(600) 1000, ≦

200 b≦ 3(300) ∞≦

Page 45: Treasurer's Affidavit

100% Rules

100% rule for objective function coefficients

100% rule for RHS constants

%100

j j

j

c

c

%100

j j

j

b

b

Page 46: Treasurer's Affidavit

Examples

Unit profit on product 1 decrease by $1, but increases by $1 for products 2 and 3, will the optimum change?(δc1=-1, Δc1=-4, δc2=1, Δc2=4, δc3=1, Δc3=2.67)

Simultaneous variation of 10 hr decrease on labor 100 lb increase in material and 50hr decrease on administration

1875.067.2

1

4

1

4

1

1100

50

400

100

40

10