solutns oflp
TRANSCRIPT
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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
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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$otal operation.
X 1 , X
2, X
3, S
1, S
2≥0
Z =8 X 2+3 X
3
X 1+ 23 X 2+1
3 X 3=2
7 X 2+2 X
3+S
1=108
5 X 2−4 X
3+S
2=35
X 1 , X
2, X
3, S
1, S
2≥0
X 1
, S1∧S
2
X 2, X
3
= Γn
Γ (n−m )× Γ (m ) =
Γ (5 ) Γ (5−3 ) Γ (3 )
=10
6 X 1−2 X
2+3 X
3=11
4 X 1+7 X
2+ X
3=21
5 X 1+8 X
2+9 X
3=48
a11
X 1−1
3 X
2+1
2 X
3=11
6−−−−−−−¿ I
1=
I o
6
0+253
X 2− X
3=413
−−−−−−−−¿ II 1= II
o−4 I
1
0+29
3 X
2+13
2 X
3=233
6−−−−−−¿ III
1= III
o−5 I
1
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Maximize
Su!ect to
Solution:!
'et us consi"er two gi$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
¿¿¿¿¿
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-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
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(b) ,a-imize
Su!ect to
S"andard )anoni*al +orm
Maximize
Su!ect to
Table "
#asis $ Solution %atio
$ 0 ,6 ,7 1 1 1
1 0 6 0 1 = 6.= (mall("
1 0 0 1 0 7 7
#asis $ Solution %atio
$ 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
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Table "
#asis $ Solution %atio
$ 0 4 6 1 1 1 (mall("
1 0 ,0 0 1 0 0
1 4 ,6 1 0 8 6
Table ""
#asis $ Solution %atio
$ 0 1 = ,4 1 ,4
1 0 ,0 0 1 0 ,0
1 1 0 ,4 0 4 4 (mall("
Table """
#asis $ Solution %atio
$ 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
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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
#asis Solution %atio
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
#asis Solution %atio
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$e. (ence the solution is optimum an" feasile.
L+
e*
e*
L+
=−7 X 1−4 X
2+S
1+S
2+9
3 X 1+ X
2−S
1+ R
1=3
4 X 1+3 X
2−S
2+ R
2=6
X 1+ X 2+S3=3
X 1, X
2, S
1,S
2,R
1, R
2≥0
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$ 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
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#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
#asis $ Solution %atio
$ 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
#asis $ Solution %atio
$ 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*
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Su!ecte" to
#asis $ Solution %atio
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
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#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
#asis $ Solution %atio
$ 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$e. (ence
the solution is optimum an" feasile.
gain 70 Start pase "" of simple- metod
&'
&2
&3
&
S'
S2
S3
&2
S'
S3
L+
Z =6 X 1+7 X
2+3 X
3+5 X
4
=6 X 1+7 (10+5 X 3+6 X 4+S2 )+3 X 3+5 X 4
Z =6 X 1+38 X
3+47 X
4+7S
2+70
−2 X 1−26 X
3−31 X
4−5S
2+S
3=42
−5 X 1−27 X
3−40 X
4+S
1−6 S
2=48
X 2−5 X
3−6 X
4−S
2=10
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he "ual form of the gi$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
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1 0/6 0 1 1 0/; 06=/6 06=
#asis $ Solution %atio
$ 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
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Table "
#asis $ Solution %atio
: 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 ""
#asis $ Solution %atio
$ 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 """
#asis $ Solution %atio
$ 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 "+
#asis $ Solution %atio
$ 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
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O!ectio$e function :, Maximize
Maximize
Su!ecte" to
Standard Canonial for is
#asis $ Solution %atio
: 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
#asis $ Solution %atio
: 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
#asis $ Solution %atio
&'
&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
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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
#asis $ Solution %atio
$ 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
#asis $ Solution %atio
$ 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
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Maximize
Su!ecte" to
#asis $ Solution %atio
$ 0 ,66 ,0; 1 1 1
1 0 0 0 1 61 61/0C61
1 481 671 1 0 =A81
#asis $ Solution %atio
$ 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
#asis $ Solution %atio
$ 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
-
8/17/2019 Solutns ofLP
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
-
8/17/2019 Solutns ofLP
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 =
-
8/17/2019 Solutns ofLP
20/28
Minimize
Su!ecte" to
#asis Solution %atio
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
#asis Solution %atio
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
#asis Solution %atio
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
#asis Solution %atio
&'
&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
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8/17/2019 Solutns ofLP
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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
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8/17/2019 Solutns ofLP
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#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
-
8/17/2019 Solutns ofLP
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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
-
8/17/2019 Solutns ofLP
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$er =7 tons of oul"er stone an" 8= tons of san" in a week. Determine the numer of"ays each 5uarry shoul" e operate" "uring a week if the contractor has to fulfill his commitment at minimum total cost. 2ssume thateach 5uarry can e operate" for fraction of "ay also.
'et G0 e the nos of "ays 5uary I is operate"/week
G6 e the nos of "ays 5uary II is operate"/week
G4 e the nos of "ays 5uary III is operate"/week
Z =20 X 1+22 X
2+18 X
3
4 X 1+6 X
2+ X
3≥54
4 X 1+4 X
2+6 X
3≥65
X 1≤7
X 2≤7 X
3≤7
X 1
, X 2
, X 3≥0
Z =20 X 1+22 X
2+18 X
3
4 X 1+6 X
2+ X
3−S
1+ R
1=54
4 X 1+4 X
2+6 X
3−S
2+ R
2=65
X 1+S
3=7
X 2+S
4=7
X 3+S
5=7
X 1
, X 2
, X 3
, S1
, S2
, S3
, S4
, S5
, R1
, R2≥0
-
8/17/2019 Solutns ofLP
25/28
R1=R2
52 =>4 -S1$ = 65-#1 = 2 = 6>4 - S2$
11? - >1-10>2 - 7>4 = S1 = S2
+ow# the taular form for II Phase simplex metho"
#asis $ Solution ?atio
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
#asis $ Solution ?atio
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
-
8/17/2019 Solutns ofLP
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1 1 1 0 1 1 1 1 1 1 0 A
#asis $ Solution ?atio
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
#asis $ Solution ?atio
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
#asis $ Solution ?atio
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*
-
8/17/2019 Solutns ofLP
27/28
#asis $ Solution ?atio
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
#asis $ Solution ?atio
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
#asis $ Solution ?atio
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
#asis $ Solution ?atio
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*
-
8/17/2019 Solutns ofLP
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