Solutns ofLP

8/17/2019 Solutns ofLP

1/28

M.Sc. in Power system Engineering

Department of Electrical Engineering

IOE/Pulchowk

Tutorials on Linear programming problems (Optimization)

Last date of submission : 2066/03/23

Q1

Maximize

Su!ect to

unrestricte" in sign.

Solution:

Since

is unrestricte" in sign# so it is expresse" as "ifference of two non negati$e terms of

.i.e%

&here

Stan"ar" form of the gi$en '.P. prolem is

Su!ecte" to

State the following prolem in stan"ar" form

Z =2 X 1− X

2+5 X

3

X 1−2 X 2+ X 3≤83 X

1−2 X

2≤−18

2 X 1+ X

2−2 X

3≤−4

X 1, X

2≥0, X

3

X 3 X 3

' ∧ X 3

''

X 3= X 3' − X 3

''

X 3

' X

3

''≥0∴

Z =2 X 1− X

2+5 ( X 3' − X 3'' )

Z =2 X 1− X

2+5 X

3

' −5 X

3

''

X 1−2 X 2+ X 3' − X 3

''+S1=8

−3 X 1+2 X

2−S

2=18

2X X +2 X ' 2 X '' S 4


2/28

2X X +2 X 2 X S 4

Canonial form

Maximize

Su!ecte" to

(ere# *asic $ariales are

.

+on, asic $ariales are

.

+o. of asic solutions -n# m

herefore# there are 01 asic solutions.

Solution:

Performing pi$otal operation on

4.

3in" the solution of the following system y re"ucing it to the canonical form using pi$otal operation.

X 1 , X

2, X

3, S

1, S

2≥0

Z =8 X 2+3 X

3

X 1+ 23 X 2+1

3 X 3=2

7 X 2+2 X

3+S

1=108

5 X 2−4 X

3+S

2=35

X 1 , X

2, X

3, S

1, S

2≥0

X 1

, S1∧S

2

X 2, X

3

= Γn

Γ (n−m )× Γ (m ) =

Γ (5 ) Γ (5−3 ) Γ (3 )

=10

6 X 1−2 X

2+3 X

3=11

4 X 1+7 X

2+ X

3=21

5 X 1+8 X

2+9 X

3=48

a11

X 1−1

3 X

2+1

2 X

3=11

6−−−−−−−¿ I

1=

I o

6

0+253

X 2− X

3=413

−−−−−−−−¿ II 1= II

o−4 I

1

0+29

3 X

2+13

2 X

3=233

6−−−−−−¿ III

1= III

o−5 I

1


3/28

Maximize

Su!ect to

Solution:!

'et us consi"er two gi$en e5uations for plotting the graph.he e5uations are

3or e5uation -i#

1 0 6 4

,0 1 0 6 3or e5uation -ii#

1 6 7 8

0 1 ,0 ,6

he point of intersection of these two lines is foun" y sol$ing these ao$e two e5uations.

7. 3in" the solution of the following 'P prolem graphically<

X 1=1

X 2=2

X 3=3

Z =2 X 1+6 X

2

− X 1+ X

2≤1

2 X 1+ X

2≤2

X 1 , X

2≥0

− X 1+ X

2=1. . . . . . . . . . . . . . . . . . . . . . . .( i)

2 X 1+ X

2=2. . . . . . . . . . . . . . . . . . . . . . . . . . .(ii )

− X 1+ X

2=1.

X 2

X 1

2 X 1+ X

2=2 .

X 2

X 1

−2 X 1+2 X

2=2

2 X 1

+ X 2

=2

3 X 2=4

or , X 2=

4

3=1.33

¿¿¿¿¿


4/28

-a Maximize

Su!ect to

Solution:!

Stan"ar" form of 'inear programming is

Maximize

Su!ect to

Table "

#asis $ Solution %atio

$ 0 ,= ,7 1 1 1 1 1

1 8 7 0 1 1 1 67 7 Small("

1 0 6 1 0 1 1 8 8

1 ,0 0 1 1 0 1 0 ,0

=. >se Simplex algorithm to "etermine the optimum feasile solution of the following 'P prolems<

&'

&2

S'

S2

S3

S

S'

S2

S3

e*

L+

Z =8.64

X 1=0.33

X 2=1.33

Z =5 X 1+4 X 2

6 X 1+4 X

2≤24

X 1+2 X

2≤6

− X 1+ X

2≤1

X 2≤2

X 1 , X

2≥0

Z =5 X 1+4 X

2

6 X 1+4 X

2+S

1=24

X 1+2 X

2+S

2=6

− X 1+ X

2+S

3=1

X 2+S

4=2

X 1, X

2, S

1, S

2, S

3, S

4≥0


5/28

(b) ,a-imize

Su!ect to

S"andard )anoni*al +orm

Maximize

Su!ect to

Table "


$ 0 ,6 ,7 1 1 1

1 0 6 0 1 = 6.= (mall("

1 0 0 1 0 7 7


$ 0 1 1 6 1 01

1 1.= 0 1.= 1 6.=1 1.= 1 ,1.= 0 0.=

Since all the coefficients of : row non asic $ariales are non negati$e an" all the$alues in ?(S are positi$e. he solutions are optimum an" feasile.

&'

&2

S'

S2

S'

S2

&'

&2

S'

S2

&2S

2

Since all the coefficients of : row coefficient of all non asic $ariales are non negati$e so the solution is optimum.*ut the z row coefficient non asic $ariale is zero (ence the gi$en ' P prolems ha$e infinite numer of

e*

L+

Z =2 X 1+4 X

2

X 1+2 X

2≤5

X 1+ X

2≤4

X 1

, X 2≥0

X

X 1+2 X

2+S

1=5

X 1+ X

2+S

2=4

X 1, X

2, S

1, S

2≥0

Z =2 X 1+4 X

2


6/28

Table "


$ 0 4 6 1 1 1 (mall("

1 0 ,0 0 1 0 0

1 4 ,6 1 0 8 6

Table ""


$ 0 1 = ,4 1 ,4

1 0 ,0 0 1 0 ,0

1 1 0 ,4 0 4 4 (mall("

Table """


$ 0 1 1 06 ,= ,0;

1 0 1 ,6 0 7

1 1 0 ,4 0 4

*o,,i*in" o, ro&( ar no" o(i"iv (olu"ion i( unoundd

Minimize

Su!ect to

Solution:

Stan"ar" form of 'inear programming isMinimize

Su!ect to

&'

&2

S'

S2

S'

S2

&'

&2

S'

S2

&'

S2

&'

&2

S'

S2

&'

&2

Since

6 .se to pase simple- metod to sol*e te folloing L1 problem

L+

L+

e*

Z =3 X 1+2 X

2

3 X 1+ X

2≥3

4 X 1+3 X

2≥6

X 1+2 X

2≤4

X 1, X

2≥0

Z =3 X 1+2 X

2


7/28

u ecte to

al 3

#asis Solution %atio

0 A 7 ,0 ,0 1 1 1 B

1 4 0 ,0 1 1 0 1 4 0 Small("

1 7 4 1 ,0 1 1 0 8 4/61 0 0 1 1 0 1 1 4 4


0 1 =/4 7/4 ,0 1 ,A/4 1 6

1 0 0/4 ,0/4 1 1 0/4 1 0 4

1 1 =/4 7/4 ,0 1 ,7/4 0 6 8/= Small("

1 1 6/4 0/4 1 0 ,0/4 1 6 4


0 1 1 1 1 1 ,0 ,0 1

1 0 1 ,4/= 0/= 1 4/= ,0/= 4/=

1 1 0 7/= ,4/= 1 ,7/= 4/= 8/=

1 1 1 ,0/= 6/= 0 0/= ,6/= 8/=

2 i i & 1 h II f i l h "

&'

&2

S'

S2

S3

% '

% 2

% '

% 2

S3

&'

&2

S'

S2

S3

% '

% 2

&

% 2

S3

&'

&2

S'

S2

S3

% '

% 2

&'

&2

S3

+ow# Since the &,row coefficient of non asic $ariale ? 0 ?

6are non,positi$e an" ?(S is

non negati$e. (ence the solution is optimum an" feasile.

L+

e*

e*

L+

=−7 X 1−4 X

2+S

1+S

2+9

3 X 1+ X

2−S

1+ R

1=3

4 X 1+3 X

2−S

2+ R

2=6

X 1+ X 2+S3=3

X 1, X

2, S

1,S

2,R

1, R

2≥0


8/28

$ 0 1 1 ,0/= ,4/= 1 60/=

1 0 1 ,4/= 0/= 1 4/=

1 1 0 7/= ,4/= 1 8/=

1 1 1 ,0/= 6/= 0 8/=

Since all the :,row coefficients of non,aic $ariales are non,positi$e an" ?(S column is non negati$e# the solution is optimum an" feasile.

?e5uire" solutions are<

9heck<

7

Minimize

Su!ect to

Solution:!

Stan"ar" form of the ao$e prolem is

Minimize

Su!ect to

+ow# the stan"ar" form of '.P. is

Minimise

&'

&2

&3

Sol$e the following 'P prolem y Dual simplex metho"<

X 1=3

5 X

2=6

5Z =

21

5

Z =3 X 1+2 X

2=3∗3

5+2∗6

5=21

5

Z =4 X 1+ X

2

3 X 1+ X

2=3

4 X 1+3 X

2≥6

X 1+2 X

2≤4

X 1

, X 2≥0

Z =4 X 1+ X

2

3 X 1+ X

2≤3

3 X 1+ X

2≥3

4 X 1+3 X

2≥6

X 1+2 X

2≤4

X 1

, X 2≥0


9/28

#asis $ Solution

$ 0 ,;/4 1 1 1 ,0/4 1 6

1 =/4 1 0 1 0/4 1 0

1 ,=/4 1 1 0 ,0/4 1 ,0 Minmum

1 7/4 0 1 1 ,0/4 1 6

1 ,=/4 1 1 1 6/4 0 1

?atio /5 1


$ 0 ,0 1 1 ,0 1 1 4

1 1 1 0 0 1 1 1

1 = 1 1 ,4 0 1 4

1 4 0 1 ,0 1 1 4

1 ,= 1 1 6 1 0 ,6 Minmum

?atio 1/5 1


$ 0 1 1 1 ,A/= 1 ,0/= 0A/=

1 1 1 0 0 1 1 1

1 1 1 1 ,0 0 0 0

1 1 0 1 0/= 1 4/= B/=

1 0 1 1 ,6/= 1 ,0/= 6/=

he solution is 0A/=.

&'

&2

S'

S2

S3

S

S'

S2

&2

S

&'

&2

S'

S2

S3

S

S'

S3

&2

S

&'

&2

S'

S2

S3

S

S'

S3

&2

&'

Since the coefficient of non asic $ariale in the :,row are non positi$e an" all the $alues of?(S column are non,negati$e# the solution is optimum.

L+

L+

e*


10/28

Su!ecte" to


Z 0 ,8 ,A ,4 ,= 1 1 1 1

1 ,= ,8 4 ,7 0 1 1 ,06

1 1 ,0 = 8 1 0 1 ,01

1 ,6 ,= ,0 ,0 1 1 0 ,;

-ii wo Phase Simplex metho"<

1ase ": Minimize

Su!ecte" to

'et &C auxiliary o!ecti$e function.

Minimize

Su!ecte" to

#asis ?atio

0 A 06 ,A ,0 1 1 1 ,0 ,0 ,0 41

1 = 8 ,4 7 0 1 1 ,0 1 1 06 6

&'

&2

&3

&

S'

S2

S3

S'

S2

S3

&' &2 &3 & % ' % 2 % 3 S' S2 S3 Solutio

% '

e*

Z =6 X 1+7 X

2+3 X

3+5 X

4

−5 X 1−6 X

2+3 X

3−4 X

4+S

1=−12

− X 2+5 X

3+6 X

4+S

2=−10

−2 X 1−5 X

2− X

3− X

4+S

3=−8

X 1

, X 2

, X 3

, X 4

, S1

, S2

, S3≥0

Z =6 X 1+7 X 2+3 X 3+5 X 4

5 X 1+6 X

2−3 X

3+4 X

4−S

1+ R

1=12

X 2−5 X

3−6 X

4−S

2+ R

2=10

2 X 1+5 X

2+ X

3+ X

4−S

3+ R

3=8

X 1

, X 2

, X 3

, X 4

, S1

, S2

, S3

, R1

, R2

, R3≥0

W = R1+ R

2+ R

3=−7 X

1−12 X

2+7 X

3+ X

4+S

1+S

2+S

3+30

5 X 1+6 X

2−3 X

3+4 X

4−S

1+ R

1=12

X 2−5 X

3−6 X

4−S

2+ R

2=10

2 X 1+5 X

2+ X

3+ X

4−S

3+ R

3=8


11/28

#asis ?atio

0 ,=/8 1 ,B/6 ,61/4 ,A/8 1 ,0 0/8 ,0 1 ;

1 04/8 1 ,A/6 A/4 =/8 1 ,0 ,=/8 1 0 6

1 ,=/8 1 ,B/6 ,61/4 ,0/8 0 1 0/8 ,0 1 ; 7;

1 =/8 0 ,0/6 6/4 0/8 1 1 ,0/8 1 1 6

#asis ?atio

0 1 1 1 1 ,0 ,0 ,0 1 1 1 1

1 ,6 1 ,68 ,40 1 = ,0 1 ,= 0 76

1 ,= 1 ,6A ,71 ,0 8 1 0 ,8 1 7;

1 1 0 ,= ,8 1 0 1 1 ,0 1 01

1ase ""

Minimize

Su!ecte" to


$ 0 ,8 1 ,4; ,7A 1 ,A 1 A1

1 1 0 ,= ,8 1 ,0 1 01

1 ,= 1 ,6A ,71 ,0 ,8 1 7;

1 ,6 1 ,68 ,40 1 ,= 0 76

&'

&2

&3

&

% '

% 2

% 3

S'

S2

S3

Solutio

S3

% 2

&2

&'

&2

&3

&

% '

% 2

% 3

S'

S2

S3

Solutio

S3

S'

&2 +ow the &. row coefficient of non,linear asic $ariale ?

0 # ?

6 ?

4are non positi$e an" ?(S is non negati$e. (ence

the solution is optimum an" feasile.

gain 70 Start pase "" of simple- metod

&'

&2

&3

&

S'

S2

S3

&2

S'

S3

L+

Z =6 X 1+7 X

2+3 X

3+5 X

4

=6 X 1+7 (10+5 X 3+6 X 4+S2 )+3 X 3+5 X 4

Z =6 X 1+38 X

3+47 X

4+7S

2+70

−2 X 1−26 X

3−31 X

4−5S

2+S

3=42

−5 X 1−27 X

3−40 X

4+S

1−6 S

2=48

X 2−5 X

3−6 X

4−S

2=10


12/28

he "ual form of the gi$en 'P prolem is

Minimize

Su!ecte" to

10

ime re5uire" per unit -min for

Sol"ering inspection

2 08 01 7* 01 06 ;

Solution:

O!ecti$e function

Maximize

Su!ecte" to

Stan"ar" form of the linear programming prolem is

Maximize

wo types of printe" circuit oar"s 2 an" * are to e pro"uce" in a computer manufacturing company. he component placement time# sol"ering time an" inspection time re5uire" in pro"ucing each unit of 2 an" * are gi$en elow.

9ircuit*oar"

9omponent placement

he amount of time a$ailale per "ay for component placement# sol"ering an" inspection are 0=11# 0111 an" =11minutesrespecti$ely. If each unit of 2 an" * contriutes a profit of F01 an" F0= respecti$ely# "etermine the numer of units of 2an" * to e pro"uce" per "ay to maximize the profit.

'et G0C+o.of units of type 2 printe" circuit oar".

G6C+o. of units of type * printe" circuit oar".

V =800 y1+900 y

2+1200 y

3+300 y

4

3 y1+4 y

2+5 y

3≥70

5 y1+6 y

2+8 y

3+ y

4≥110

y1

, y2

, y3

, y4≥0

Z =10 X 1+15 X

2

16 X 1+10 X

2≤1500

10 X 1+12 X

2≤1000

4 X 1

+8 X 2

≤500

X 1, X

2≥0

Z=10 X +15 X


13/28

1 0/6 0 1 1 0/; 06=/6 06=


$ 0 1 1 1 =/; 0=/08 74A=/7

1 1 1 0 ,00/7 80/; 4A=/61 0 1 1 0/7 ,4/; 06=/6

1 1 0 1 ,0/; =/08 06=/7

hus # the re5uire" solutions is

y oard

y oard

Ma8imum Pro,i"

11

?esources

&easel ?ait Dog

06A 604 4B7 0111

= A 01 =1

6 7 8 4=

6111 4111 7111

&2

&'

&2

S'

S2

S3

S'&

'

&2

Since :,row coefficient of all the non asic $ariales are non negati$e an" all ?(S column are also non negati$e# sothe column is optimal an" feasile.

2n electric company pro"uces three $ariales of 29S? con"uctors $iz. Hweasel# Hraitan" H"og. he pro"uction process re5uires two types of skille" laour for construction an" finishing. he following tale gi$es the a$ailale ofresources their uses y the three pro"ucts an" the profit per unit.

?esources re5uirement per JM length

Dailya$ailaility

?awmaterial-kg

9onstruction-manhr

3inishing-man/hr

Profit per unit-?s

X 1=125

2

X 2=125

4

Z = profit =4375

4

5 X 7 X 10 X S 500


14/28

Table "


: 0 ,6111 ,4111 ,7111 1 1 1 1

1 06A 604 4B7 0 1 1 0111Small("

1 = A 01 1 0 1 =1 =1/01C=

1 6 7 8 1 1 0 4= 4=/8C=.;

Table ""


$ 0 ,A01.88 ,;4A.=8 1 01.0= 1 1 010=6.6;

1 06A/4B7 604/4B7 0 0/4B7 1 1 0111/4B7 7.8B (mall("

1 0.A; 0.=B 1 ,1.14 0 1 67.86 0=.7=

1 1.1A 1.A8 1 ,1.16 1 0 0B.AA 68.07

Table """


$ 0 ,600.6A 1 0=7B.41 07.1; 1 1 071;7.=0

1 1.81 0 0.;= 1.11= 1 1 7.8B A.;A

1 1.;4 1 ,6.B= ,1.14 0 1 0A.07 61.A7

1 ,1.4; 1 ,0.71 ,1.16 1 0 08.66 ,76.04

Table "+


$ 0 1 4=7.44 6617.A6 0=.A= 1 1 0=A7;.14

1 ' 0.8; 4.01 1.10 1 1 A.;A71

1 1 ,0.4B ,=.=0 ,1.17 0 1 01.84

1 1 1.8= ,1.61 ,1.16 1 0 0B.6=

&'

&2

&3

S'

S2

S3

S0

0111/4B7C6.=

S6

S4

&'

&2

&3

S'

S2

S3

&3

S2

S3

&'

&2

&3

S'

S2

S3

&2

S2

S3

&'

&2

&3

S'

S2

S3

&'

S2

S3

e*

L+

e*

L+

e*

L+

5 X 1+7 X

2+10 X

3+S

2=500

2 X 1+4 X

2+6 X

3+S

3=35

X 1

, X 2

, X 3

, S1

, S2

, S3≥0


15/28

O!ectio$e function :, Maximize

Maximize

Su!ecte" to

Standard Canonial for is


: 0 ,81 ,71 1 1 1 1

1 01 ; 0 1 1 ;111

1 0 1 1 0 1 811 811/0C811

1 1 0 1 1 0 A=1


: 0 1 ,71 1 81 1 48111

1 1 ; 0 ,01 1 6111 6111/;C6=1

1 0 1 1 0 1 811

1 1 0 1 1 0 A=1 A=1/0CA=1


&'

&2

S'

S2

S3

S0

;111/01C;1

S6

S4

&'

&2

S'

S2

S3

S0

G0

S4

&'

&2

S'

S2

S3

e*

L+

L+

e*

Z =60 X 1+40 X

2

10 X 1+8 X

2≤8000

X 1≤600

X 2≤750

X 1

, X 2≥0

10 X 1+8 X

2+S

1=8000

X 1+S2=600 X

2+S

3=750

X 1

, X 2

, S1

, S2

, S3≥0


16/28

Su!ecte" to

Stan"ar" form

Maximize

Su!ecte" to

#asis $ Solution ?atio

$ 0 ,04= ,=1 1 1 1

1 6 0/6 0 1 46 46/6C08

1 ,7 0 1 0 1


$ 0 1 ,8=/7 04=/6 1 6081

1 0 0/7 0/6 1 08 87

1 1 6 6 0 87 87/6C46


$ 0 1 1 44=/7 8=/; 68;1

1 0 1 0/7 ,1.06= ;

1 1 0 0 0/6 46

&'

&2

S'

S2

S'

S2

&'

&2

S'

S2

&'

S2

&'

&2

S'

S2

&'

&2

Since all the :,row coefficient of non asic $ariales are non negati$e an" ?(S column are also non negati$e the solutionis optimum an" feasile

e*

e*

L+

L+

Z =135 X 1+50 X

2

2 X 1+1

2 X

2≤32

X 2≤4 X

1

Z =135 X 1+50 X

2

2 X 1+1

2 X

2+S

1=32

−4 X 1+ X

2+S

2=0


17/28

Maximize

Su!ecte" to


$ 0 ,66 ,0; 1 1 1

1 0 0 0 1 61 61/0C61

1 481 671 1 0 =A81


$ 0 1 ,01/4 1 00/0;1 4=6

1 1 '/3 0 ,0/481 7 7K4C06

1 0 6/4 1 0/481 08


$ 0 1 1 01 0/48 4B6

1 1 0 4 ,0/061 06

1 0 1 ,6 0/061 ;

he re5uire" solution is

: C 4B6

&'

&2

S'

S2

S'

S2

=A81/481C08

&'

&2

S'

S2

S'

&'

08K4/6C67

&'

&'

S'

S'

&2

&'

Since all the :,row coefficient of non asic $ariales are non,negati$e ?(S column are also non,negati$e the solutionis optimum an" feasile.

G0C ;

G6C06

e*

L+

e*

L+

Z =22 X 1+18 X

2

X 1+ X

2+S

1=20

360 X 1

+240 X 2

+S2

=5760


18/28

Minimize

Su!ecte" to

"n Canonial form4

Minimize

Su!ecte" to

+ow# the taular form for Dual simplex metho"

#asis $ Solution

$ 0 ,61 ,66 ,0; 1 1 1 1 1 11 ,7 ,8 ,0 0 1 1 1 1 ,=7

1 ,7 ,7 !6 1 0 1 1 1 ,8=

1 0 1 1 1 1 0 1 1 A

'et G0 e the nos of "ays 5uary I is operate"/week

G6 e the nos of "ays 5uary II is operate"/week

G4 e the nos of "ays 5uary III is operate"/week

&'

&2

&3

S'

S2

S3

S

S5

S'

S2

S

e*

L+

Z =20 X 1+22 X 2+18 X 3

4 X 1+6 X

2+ X

3≥54

4 X 1+4 X

2+6 X

3≥65

X 1≤7

X 2≤7

X 3≤7

X 1 , X 2 , X 3≥0

Z =20 X 1+22 X

2+18 X

3

−4 X 1−6 X

2− X

3+S

1=−54

−4 X 1−4 X

2−6 X

3+S

2=−65

X 1+S

3=7

X 2+

S4=7

X 3+S

5=7

X 1

, X 2

, X 3

, S1

, S2

, S3

, S4

, S5≥0


19/28

1 =/; ' 1 ,4/08 0/46 1 1 1 6=B/46

1 0/7 1 0 0/; ,0/7; 1 1 1 ;A/08

1 0 1 1 1 1 0 1 1 A

1 !5/8 1 1 4/08 ,0/46 1 0 1 ,4=/46

1 ,0/7 1 1 ,0/; 4/08 1 1 0 6=/08

%atio 07/=

#asis $ Solution

$ 0 1 1 1 ,06/= ,04/= 1 ,07/= 1 6AB

1 1 0 1 1 1 1 0 1 A

1 1 1 0 0/= ,0/41 1 6/= 1 =

1 1 1 1 4/01 ,0/61 0 ;/= 1 4=/7

1 ' 1 1 ,4/01 0/61 1 ,;/= 1 A/71 1 1 1 ,0/61 0/= 1 ,6/= 0 =4/08

he solution is

Ce

16

Solution:

'et &eight of ingre"ient P in kg.

&eight of ingre"ient L in kg

&2

&3

S3

S

S5

&'

&2

&3

S'

S2

S3

S

S5

&2

&3

S3

&'

S5

Since the :,row coefficieint of non asic $ariales are non positi$e an" all the $alues of ?(S column are nonnegati$e# the solution is optimum an" feasile.

2 firm must pro"uce 611kgs of a mixture consisting of the ingre"ients P an" L which cost ?s 4 an" ?s ; per kgrespecti$ely. +ot more than ;1 kgs of P can e use" an" at least 81 kgs of L must e use". 3in" how much of each

ingre"ient shoul" e use" in or"er to minimize the cost.

L+

Z =279

X 1=7

4

X 2=7

X 3=5

Z =20×7

4+22×7+18×5

=35+154+90

=279

x1=

x =


20/28

Minimize

Su!ecte" to


0 0 6 1 ,0 1 1 681

1 0 1 0 1 1 1 ;1

1 0 0 1 1 0 1 611 611

1 1 ' 1 ,0 1 0 81 81


0 0 1 1 0 1 ,6 071

1 ' 1 0 1 1 1 ;1 ;1

1 0 1 1 0 0 ,0 0711 1 0 1 ,0 1 0 81


0 1 1 ,0 0 1 ,6 81

1 0 1 0 1 1 1 ;1

1 1 1 ,0 ' 0 ,0 81 81

1 1 0 1 ,0 1 0 81


&'

&2

S'

S2

% '

% 2

S'

% '

% 2

&'

&2

S'

S2

% '

% 2

S'

% ' 071/0C07&

2

&'

&2

S'

S2

% '

% 2

&'

% '

&2

&'

&2

S'

S2

% '

% 2

e*

L+

L+

e*

L+

e*

W = R1+ R

2

W = R1+ R

2

=− X 1−2 X

2+S

2+260

X 1+ X

2+ R

1=200

X 1+S1=80 X

2−S

2+ R

2=60


21/28

1 1 0 ,0 1 061

1 1 1 ,0 0 81

he solution is

17

'ighting < 4B;@

9ooking < =16=@In"ustrial < 767@

3ollowing are the limitations of each of the energy sources<

-i *io,gas can not e use" for supplying in"ustrial loa".

Solution:

'et

&2

S2

Since the coefficient of non asic $ariale in o!ecti$e function row are non,positi$e an" all the $alues of ?(S columnare positi$e. he solution is optimal an" feasile.

<

here are four sources of energy generation options $iz solar# micro hy"ro# io,gas an" fuel woo" to supply in"ustrial#lighting an" cooking loa"s of a remote $illage. he cost of generation per @ for solar# micro hy"ro# io,gas an" fuelwoo" are ?s 4;1=# ?s B=0# ?s 664 an" 6.6= respecti$ely. he energy re5uirement for $arious loa"s are as follows<

-ii 3uel woo" can not e use" for supplying in"ustrial an" lighting loa"s.

-iii Micro hy"ro can e use" for all kin"s of loa"s.

-i$ Solar power can only e use" for lighting loa"s an" ;@ of energy alrea"y eing supplie" is not to e "isture".

3ormulate the 'P prolem to "etermine the amount of energy to e generate" from each source to minimize the total costof generation.

Z =1200

X 1=80 kg

X 2=120kg

Z =5S1+1200

=5×0+1200

=1200

X X +S = 398


22/28

#asis $ Solution

$ 0 ,4;1= ,B=0 ,664 ,6.6= 1 1 1 1 1 1

1 ,0 ,0 ,0 1 0 1 1 1 1 ,4B;

1 1 ,0 ,0 ,0 1 0 1 1 1 ,=16=

1 1 ,0 1 1 1 1 0 1 1 ,767

1 ,0 1 1 1 1 1 1 0 1 ,;

1 ,0 ,0 ,0 !' 1 1 1 1 0 ,=;7A

%atio 4;1= 4=0 664 6.6=

Smllest

#asis $ Solution

$ 0 1 1 1 1 1 ,6.6=

1 ,0 ,0 ,0 1 0 1 1 1 1 ,4B;

1 0 1 1 1 1 0 1 1 1 ;66

1 1 !' 1 1 1 1 0 1 1 ,767

1 ,0 1 1 1 1 1 1 0 1 ,;

1 N0 N0 N0 0 1 1 1 1 0 =;7A

%atio

&'

&2

&3

&

S'

S2

S3

S

S5

S'

S2

S3

S

S5

&'

&2

&3

&

S'

S2

S3

S

S5

S'

S2

S3

S

&

e*

L+

L+

e*

e*

− X 1− X

2+S

1=−398

− X 2− X

3− X

4+S

2=−5025

− X 2+S

3=−424

− X 1+S

4=−8

− X 1− X

2− X

3− X

4+S

5=5847

X 1

, X 2

, X 3

, S1

, S2

, S3

, S4

, S5≥0

−152114

−3795

4

−883

4

52623

4

37954


23/28

1 0 1 1 1 1 1 1 ,0 1 ;

1 1 1 0 0 1 1 0 0 ,0 =70=

%atio

he solutions are

9nd of te ssignment o '4 on d*aned ,atematis

&'

&

Since the coefficient of no asic $ariale in o!ecti$e function row are non positi$e an" all the $alues of ?(S column are

positi$e. he solution is optimal an" feasile.

Z =1783391

4

X 1=8

X 2=424

X 3=0

X 4=5415


24/28

15

Luarry -@ra"e I per "ay

I 7 7 61

II 8 7 66

III 0 8 0;

Solution:

Minimize

Su!ecte" to

"n Canonial form4

Minimize

Su!ecte" to

2 contractor operates three material 5uarries for a "am. he material from each 5uarry is separate" in to two gra"es. he "aily pro"uction capacities of the 5uarries as well as their "aily operating costs are as follows<

*oul"erstone in

tons per"ay

San" intons per

"ay

Dailyoperatin

g cost inF0111

-@ra"eII

he contractor has committe" himself to "eli$er =7 tons of oul"er stone an" 8= tons of san" in a week. Determine the numer of"ays each 5uarry shoul" e operate" "uring a week if the contractor has to fulfill his commitment at minimum total cost. 2ssume thateach 5uarry can e operate" for fraction of "ay also.

'et G0 e the nos of "ays 5uary I is operate"/week

G6 e the nos of "ays 5uary II is operate"/week

G4 e the nos of "ays 5uary III is operate"/week

Z =20 X 1+22 X

2+18 X

3

4 X 1+6 X

2+ X

3≥54

4 X 1+4 X

2+6 X

3≥65

X 1≤7

X 2≤7 X

3≤7

X 1

, X 2

, X 3≥0

Z =20 X 1+22 X

2+18 X

3

4 X 1+6 X

2+ X

3−S

1+ R

1=54

4 X 1+4 X

2+6 X

3−S

2+ R

2=65

X 1+S

3=7

X 2+S

4=7

X 3+S

5=7

X 1

, X 2

, X 3

, S1

, S2

, S3

, S4

, S5

, R1

, R2≥0


25/28

R1=R2

52 =>4 -S1$ = 65-#1 = 2 = 6>4 - S2$

11? - >1-10>2 - 7>4 = S1 = S2

+ow# the taular form for II Phase simplex metho"


0 ; 01 A 1 1 0 0 1 1 1 00B

1 7 8 0 0 1 ,0 1 1 1 1 =7 ?

1 7 7 6 0 ' 1 ,0 1 1 1 8= 65/<

1 0 1 1 1 1 1 1 0 1 1 A 1

1 1 0 1 1 1 1 1 1 0 1 A 7

1 1 1 0 1 1 1 1 1 1 0 A


0 ; 1 A 1 1 0 0 1 ,01 1 00

1 7 1 0 0 1 ,0 1 1 ,8 1 06 4

1 7 1 6 0 ' 1 ,0 1 ,7 1 4A 47/4

1 0 1 1 1 1 1 1 0 1 1 A A

1 1 0 1 1 1 1 1 1 0 1 A

Minimize

&'

&2

&3

% '

% 2

S'

S2

S3

S

S5

% '

% 2

S3

S

S5

&'

&2

&3

% '

% 2

S'

S2

S3

S

S5

% '

% 2

S3

&2

e*

L+

e*

4 X 1+6 X

2+ X

3−S

1+ R

1=54

4 X 1+4 X

2+6 X

3−S

2+ R

2=65

X 1+S

3=7

X 2+S

4=7

X 3+S

5=7

X 1

, X 2

, X 3

, S1

, S2

, S3

, S4

, S5

, R1

, R2≥0


26/28

1 1 1 0 1 1 1 1 1 1 0 A


0 1 1 = ,6 1 4 0 1 6 1 ,041 0 1 0/7 0/7 1 ,0/7 1 1 ,4/6 1 4 12

1 1 1 5 !' ' 0 ,0 1 6 1 6= 5

1 1 1 ,0/7 ,0/7 1 0/7 1 0 4/6 1 7

1 1 0 1 1 1 1 1 1 0 1 A

1 1 1 0 1 1 1 1 1 1 0 A 7


0 1 1 1 ,0 ,0 6 6 1 1 1 ,4;

1 0 1 1 4/01 ,0/61 ,4/01 0/61 1 ,;/= 1 A/7 45

1 1 1 ' !'/5 '/5 0/= ,0/= 1 6/= 1 =

1 1 1 1 ,4/01 0/61 4/01 ,0/61 0 ;/= 1 60/=

1 1 0 1 1 1 1 1 1 0 1 A

1 1 1 1 0/= ,0/= ,0/= 0/= 1 ,6/= 0 6 10


0 1 1 1 ,4 0 7 1 1 7 ,01 ,=;

1 0 1 1 0/7 1 ,0/7 1 1 ,6/4 ,0/7 =/7

1 1 1 ' 0 0 1 1 1 1 1 A

1 1 1 1 ,0/7 1 0/7 1 0 0/4 1 7A/=

1 1 0 1 1 1 1 1 1 0 1 A

1 1 1 1 0 ,0 ,0 0 1 ,6 = 01

S5

&'

&2

&3

% '

% 2

S'

S2

S3

S

S5

&'

% 2

S3

&2

S5

&'

&2

&3

% '

% 2

S'

S2

S3

S

S5

&'

&3

S3

&2

S5

&' &2 &3 % ' % 2 S' S2 S3 S S5

&'

&3

S3

&2

S2

e*

'..

e*

e*

e*

'..

'..

e*

e*


27/28


0 1 ,7 1 ,4 0 7 1 1 1 ,01 ,;8

1 0 6/4 1 0/7 1 ,0/7 1 1 1 ,0/7 A0/06

1 1 1 ' 0 0 1 1 1 1 1 A

1 1 ,1.444 1 ,0/7 1 0/7 1 0 1 1 018/0=

1 1 0 1 1 1 1 1 1 0 1 A

1 1 6 1 0 ,0 ,0 0 1 1 = 67


0 1 7/4 1 0 0 1 1 ,7 1 ,01 ,6B;8

1 0 0/4 1 1 1 1 1 0 1 ,0/7 AAB/81

1 1 1 ' 0 0 1 1 1 1 1 A

1 1 ,7/4 1 ,0 1 0 1 0 1 1 767/0=

1 1 0 1 1 1 1 1 1 0 1 A 7

1 1 6 1 0 ,0 ,0 0 1 1 = 67 12


0 1 1 1 0 0 1 1 ,7 ,7/4 ,01 ,;B;8/4

1 0 1 1 1 1 1 1 0 ,0/4 ,0/7 84B/81

1 1 1 ' 0 0 1 1 1 1 1 A1 1 1 1 ,0 1 0 1 0 7/4 1 =87/0=

1 1 0 1 1 1 1 1 1 0 1 A

1 1 1 1 0 ,0 ,0 0 1 ,6 = 01


0 1 1 1 1 6 0 ,0 ,7 ,7/4 ,01 ,;B;8/4

1 0 1 1 1 1 1 1 0 ,0/4 ,0/7 84B/81

&'

&2

&3

% '

% 2

S'

S2

S3

S

S5

&'

&3

S3

S

S2

&'

&2

&3

% '

% 2

S'

S2

S3

S

S5

&'

&3

S3

S

S2

&'

&2

&3

% '

% 2

S'

S2

S3

S

S5

&'

&3S

3

&2

S2

&'

&2

&3

% '

% 2

S'

S2

S3

S

S5

&'

e*

'..

e*

'..

e*

e*

'..

e*

e*


28/28

1 1 1 ' 0 0 1 1 1 1 1 A

1 1 1 1 1 ,0 1 0 0 ,6/4 = A07/0=

1 1 0 1 1 1 1 1 1 0 1 A

1 1 1 1 0 ,0 ,0 0 1 ,6 = 01

he solution is

Ce

&3

S3

&2

S2

Since the :,row coefficieint of non asic $ariales are non positi$e an" all the $alues of ?(S column are non negati$e# thesolution shoul" e optimum an" feasile *ut o!ecti$e function is not zero solution is not feasile.

e*

Z =279

X 1=7

4

X 2=7

X 3=5

Z =20×7

4+22×7+18×5

=35+154+90

=279

Documents

Solutns ofLP