iegr 516: industrial engineering principles ii iegr...

IEGR 516: Industrial Engineering Principles II

IEGR 361: Introduction to Linear Programming

Fall 2014

M. Salimian

Topic 5

Test Problem

Name: --------------

Time: --30 Minutes --

Consider the given LP below:

MIN -9X1+6X2+7X3

ST

-X1+2X2+4X3=0

2X1+X2+X3 <= 5

X1-5X2=7

X2+2X3<=1

3X1+X2+3X3=8

X1>=0, X2<=0

1. Set up the first legitimate tableau and identify the

entering and leaving variables. (40 pts)

2. Perform a partial iteration that has at least two rows

and the Z row of the new tableau. (30 pts).

3. Can the optimal solution be found in an easier fashion?

If so, show how and find the optimal solution. (30 pts)



Fall 2014

M. Salimian

KEY:

1. Set up the first legitimate tableau and identify the

entering and leaving variables. (40 pts)

Process the LP into standard form:

X2<=0, replace it with X2’ where X2’=-X2

X3 unrestricted, replace it with X3’ and X3” where

X3=X3’-X3”

Convert minimization problem into maximization by

multiplying Z by -1

–Z -9X1-6X2’+7X3’-7X3” =0

- X1-2X2’+4X3’-4X3” =0

2X1- X2’+ X3’- X3”+ S1 =5

X1+5X2’ =7

- X2’+2X3’-2X3” + S2 =1

3X1- X2’+3X3’-3X3” =8

X1>=0

X2’>=0

X3’>=0

X3”>=0

S1>=0

S2>=0

Set up the Big-M by adding artificial variables to

constraints without slack variables, penalize the objective

function row:

–Z -9X1-6X2’+7X3’-7X3” +MR1+MR2+MR3 =0 - X1-2X2’+4X3’-4X3” + R1 =0

2X1- X2’+ X3’- X3”+ S1 =5

X1+5X2’ + R2 =7

- X2’+2X3’-2X3” + S2 =1

3X1- X2’+3X3’-3X3” + R3 =8

X1, X2’, X3’, X3”, S1, S2, R1, R2, R3 >=0

M>>0

Select R1, S1, R2, S2, and R3 as basic variables for the

first tableau but to legitimize the tableau, multiply rows

with artificial variables by -1 and add them to Z-row.

-Z+ (-3M-9) X1+ (-2M-6) X2’+ (-7M+7) X3’+ (7M-7) X3” = -15M



Fall 2014

M. Salimian

-Z X1 X2’ X3’ X3” S1 S2 R1 R2 R3 RHS Min

-Z 1 -3M-9 -2M-6 -7M+7 7M-7 0 0 0 0 0 -15M Ratio

R1 0 -1 -2 4 -4 0 0 1 0 0 0 0

S1 0 2 -1 1 -1 1 0 0 0 0 5 5

R2 0 1 5 0 0 0 0 0 1 0 7 ---

S2 0 0 -1 2 -2 0 1 0 0 0 1 1/2

R3 0 3 -1 3 -3 0 0 0 0 1 8 8/3

Solution:

X1=X2’=X3’=X3”=0

S1=5, S2=1, R1=0, R2=7, R3=8, Z=-15M

Not feasible; artificial variables in the basis and >0

Entering variable: X3’; most negative in the Z-row

Leaving variable: R1; min ratio

Pivot: 4

2. Perform a partial iteration that has at least two rows

and the Z row of the new tableau. (30 pts).

Pivot row (to be multiplied by reciprocal of pivot (1/4) is

selected and R2 row that does not need any additional

calculation (since its pivot column entry is zero) and the

Z-row (multiply X3’ row by 7M-7 and add to Z-row).

-Z X1 X2' X3' X3" S1 S2 R1 R2 R3 RHS

-Z 1 (-19M-29)/4 (-11M-5)/2 0 0 0 0 (7M-7)/4 0 0 -15M

X3' 0 -1/4 -1/2 1 -1 0 0 1/4 0 0 0

S1 0

R2 0 1 5 0 0 0 0 0 1 0 7

S2 0

R3 0

3. Can the optimal solution be found in an easier fashion?

If so, show how and find the optimal solution. (30 pts)

Both Big-M and Two-phase methods involve several steps to

reach a conclusion and although two-phase involves no M

calculations, still both methods are not that easy.

However, a close observation of the original problem

reveals an interesting fact. There are three variables in



Fall 2014

M. Salimian

this problem and constraint set also has three equations.

So, if this problem had any solution it would be the result

of solving those equations. However, after solving the

three equation-three unknown set of linear equations we

need to check for feasibility with respect to other

constraints.

-X1+2X2+4X3=0

X1-5X2=7

3X1+X2+3X3=8

Solve:

X1=2X2+4X3 (from first equation)

(2X2+4X3)-5X2=7 -3X2+4X3=7

3(2X2+4X3) +X2+3X3=8 7X2+15X3=8

7(-3X2+4X3=7) -21X2+28X3=49

3(7X2+15X3=8) 21X2+45X3=24

---------------

73X3=73 X3=1

-3X2+4(1)=7 -3X2=3 X2=-1

X1=2(-1)+4(1)=2

Check the solution against other constraints:

X1>0, X2<0 and X3 unrestricted (here positive) so special

constraints are OK.

2X1+X2+X3 <= 5 2(2)+(-1)+(1)=2<5

X2+2X3<=1 (-1)+2(1)=1=1

So this solution is feasible and optimal (feasible region

is just one point).



Fall 2014

M. Salimian

Topic 5

Problem 01

Name: --MOR --

Pickup Time and Date: --M/6:00pm--

Due: in 24 hours

1. Solve the following LP using simplex method.

MIN -9X1+6X2

ST

-5X1+9X2>=7

-2X1+7X2 <= 11

X1+5X2>=2

-10X1+5X2=24

7X1+8X2<=5

X1<=0

2. Plot the feasible region (using MAPLE)

3. Identify the extreme points associated with each tableau

on the plot and show the progress path from starting point

to the optimal solution.

4. Use LINDO to solve the problem and verify your final

tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 02

Name: --ARH --


Due: in 24 hours


MIN 9X1-6X2

ST

5X1-9X2>=8

2X1-7X2 <= 11

10X1-3X2=14

-7X1-8X2<=5

-X1-2X2>=1

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 03

Name: --NGT --

Pickup Time and Date: --W/6:00pm--

Due: in 24 hours


MIN 14X1-15X2

ST

-3X2>=4

7X1-5X2 <= 34

-3X1-5X2<=6

3X1-5X2>=13

4X1+X2>=5

5X1-6X2<=30

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 04

Name: --MIF --


Due: in 24 hours


MIN 7X1-10X2

ST

-2X1-7X2 <= 9

X1-3X2>=1

-10X1-3X2=9

7X1-8X2<=10

-5X1-9X2>=8

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 05

Name: --NGT --


Due: in 24 hours


MIN -14X1+15X2

ST

-7X1-5X2 <= 34

-3X1-5X2>=13

-4X1+X2>=5

3X1-5X2<=6

-5X1-6X2=30

X1<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 06

Name: --ARH2 --

Pickup Time and Date: --R/6:00pm--

Due: in 24 hours


MIN -7X1+18X2

ST

-7X1+3X2 <= 24

-5X1+9X2=30

3X1+5X2<=3

-4X1+5X2>=11

-4X1-X2>=5

X1<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 07

Name: --MOR2 --


Due: in 24 hours


MIN -7X1-27X2

ST

X1-10X2>=5

7X1-3X2 <= 24

-3X1-6X2=5

4X1+X2>=7

5X1-9X2<=30

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 08

Name: --BBA --


Due: in 24 hours


MIN 15X1-4X2

ST

5X1-9X2>=8

3X1-7X2 <= 13

-5X1-8X2>=3

10X1-3X2=14

-7X1-9X2<=12

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 09

Name: --EDT --


Due: in 24 hours


MIN -15X1+7X2

ST

-5X1-9X2>=8

-3X1-7X2 <= 13

5X1-8X2>=3

7X1-9X2<=12

-10X1-3X2=14

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 10

Name: --CHV --


Due: in 24 hours


MIN -13X1+9X2

ST

-5X1-9X2>=14

-9X1+7X2 <= 10

X1-5X2>=2

-11X1-3X2=27

5X1-9X2<=5

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 11

Name: --ELT --


Due: in 24 hours


MIN -13X1+9X2

ST

-4X1+7X2=8

-8X1-5X2 <= 7

2X1+7X2>=2

-9X1+4X2>=10

7X1+8X2<=4

X1<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 12

Name: --OBM --


Due: in 24 hours


MIN 11X1+17X2

ST

-4X1+7X2=26

-8X1-5X2 <= 7

2X1+7X2>=5

-9X1+4X2>=10

7X1+8X2<=4

X1<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 13

Name: --NGT2 --


Due: in 24 hours


MIN -12X1+17X2

ST

-5X1-9X2>=11

-2X1-7X2 <= 11

2X1-7X2>=2

-10X1-3X2=22

7X1-9X2<=12

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 14

Name: --CHS --


Due: in 24 hours


MIN 12X1-9X2

ST

-5X1-9X2>=11

-2X1-7X2 <= 11

2X1-7X2>=2

-10X1-3X2=22

7X1-9X2<=4

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 15

Name: --MOI --


Due: in 24 hours


MIN -11X1+10X2

ST

-5X1+9X2>=11

-3X1+7X2 <= 11

2X1+7X2>=2

7X1+9X2<=3

-10X1+X2=11

X1<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 16

Name: --KYM --


Due: in 24 hours


MIN -10X1+17X2

ST

-5X1+9X2>=4

-3X1+7X2 <= 9

2X1+9X2>=10

7X1+9X2<=16

10X1-X2=7






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 17

Name: --ADC --


Due: in 24 hours


MIN 11X1-13X2

ST

3X1-7X2 <= 3

-5X1-8X2>=2

-9X1+4X2=3

-7X1-9X2<=7

5X1-9X2>=0

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 18

Name: --JUO --


Due: in 24 hours


MIN -12X1+9X2

ST

X1-9X2=8

-5X1-9X2>=1

-3X1-7X2 <= 8

5X1-8X2>=2

7X1-9X2<=7






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 19

Name: --ABS --


Due: in 24 hours


MIN 15X1-14X2

ST

-5X1+9X2=14

-7X1+3X2 <= 8

3X1+7X2<=8

-X1+10X2>=11

-8X1-X2>=1






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 20

Name: --KEM --


Due: in 24 hours


MIN -15X1+17X2

ST

7X1-3X2 <= 17

5X1-9X2=18

-3X1-6X2<=0

X1-10X2>=8

4X1+X2>=6






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 21

Name: --WER --


Due: in 24 hours


MIN -19X1+17X2

ST

-5X1+9X2>=15

-2X1+7X2 <= 11

X1+5X2>=3

-10X1+5X2=22

7X1+8X2<=0

X1<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 22

Name: --JOD--


Due: in 24 hours


MIN 15X1-11X2

ST

-5X1-9X2>=15

-2X1-7X2 <= 11

X1-5X2>=3

-12X1-7X2=27

7X1-8X2<=1

X2<=0






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 23

Name: --MIF2 --


Due: in 24 hours


MIN -15X1-11X2

ST

-6X1-7X2 <= 27

3X1-5X2=4

-3X1-5X2>=15

-5X1+6X2<=1

-4X1+X2>=6






tableau solution.



Fall 2014

M. Salimian

Topic 5

Problem 24

Name: --EDT2 --


Due: in 24 hours


MIN 15X1-11X2

ST

-5X1+9X2>=8

-3X1+7X2 <= 8

5X1+8X2>=7

-10X1+3X2=5

7X1+9X2<=11






tableau solution.

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