chap006 linear programming

8/22/2019 Chap006 linear programming

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Supplement to Chapter 06 - Linear Programming

SUPPLEMENT TO CHAPTER 6:LINEAR PROGRAMMING

Answers to Discussion and Reiew !uestions

1. Linear programming is well-suited to an environment of certainty.

2. he !area of feasi"ility#$ orfeasible solution spaceis the set of allcom"inations of values of the decision varia"les which satisfy the constraints. %ence#this area is determined "y the constraints.

&. 'edundant constraints do not affect the feasi"le region for a linearprogramming pro"lem. herefore# they can "e dropped from a linear programmingpro"lem without affecting the optimal solution.

(. )n iso-cost line represents the set of all possi"le com"inations of two inputsthat will result in a given cost. Li*ewise# an iso-profit line represents all of the possi"lecom"inations of two outputs which will yield a given profit.

+. Sliding an o",ective function line towards the origin represents a decrease in

its value i.e.# lower cost# profit# etc.. Sliding an o",ective function line away from theorigin represents an increasein its value.

6. a. /asic varia"le n a linear programming solution# it is avaria"le not reuired to eual 3ero.

". Shadow price t is the change in the value of the o",ective function per unitincrease in a constraint right hand side.

c. 'ange of feasi"ility he range of values over which a constraint4s right handside value may vary without changing the optimal "asic feasi"le solution.

d. 'ange of optimality he range of values over which a varia"le4s o",ectivefunction value may vary without changing the current optimal "asic feasi"le solution.

6S-1


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So"ution to Pro#"e$s

1. a.

1. he optimal values of the decision varia"les are 51 2# 52 7 and theoptimal o",ective function value 3 &+.

2. 8one of the constraints have any slac*. /oth constraints are "inding.

&. 8either of the constraints have any surplus there are no greater thanor eual to constraints.

(. 8o# there are no redundant constraints.

Simultaneous solution92 651: ( 52 (; 1251: ;52 76 9


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Solutions (continued)


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6S-(

?2

=ptimum

2(

22

20

1;

16

1(

12

10

;

6

(

2

00 2 ( 6 ; 10 12 1( 16 1; 20 22 2(

S

'

?1


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c. 1. he optimal values of the decision varia"les are ) 2(# / 20 and the optimal o",ective function value 3 20(.

2. he third constraint has a slac* of 120. n other words# there are 120hours of unutili3ed la"or hours.

&. here are no greater than or eual to constraints# therefore no surplusis possi"le.


)t the intersection of >achinery and >aterial constraints

9+ 20) : 6/ 600 >aterial

9achinery

100) : &0/ &000


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Brom the material constraint

20) : 6/ 600 su"stituting / 20

20) : 620 600 20) (;0 ) 2(

2. a. 1. he optimal values of the decision varia"les are S ; and 20. he optimal o",ective function value 3 +;.(.

2. Since all of the constraints are greater than or eual to type# none ofthe constraints have any slac*.

&. he third and fourth constraints have surpluses of 72 grams and 10grams respectively.

(. es# the third constraint is redundant. t does not affect the feasi"leregion.

". 1. he optimal values of the decision varia"les are 51 (.2 and52 1.6. he optimal value of the o",ective function 3 1&.2.

2. es# the constraint B has a slac* of (.6.

&. 8o# there is no surplus.


6S-6

D

B

E

=ptimum

?10 1 2 & ( + 6 @ ; 7 10 11 12

12

11

10

7

;

@

6

+

(

&

2

1

0

?2


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E A D E A B

(51 : 252 20 (51: 252 20

251 : 652 1; 51: 252 12

(51 : 252 20 (51: 252 20


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(. Peanuts cost F.60Il". Eelu5e revenue is F2.70Il".

'aisins cost F1.+0Il". Standard revenue is F2.++ Il".

Eelu5e mi5 has 1I& l". peanuts# 2I& l". raisins. %ence# delu5e mi5 cost is

1

F.60 :

2

F1.+0 F1.20Il".& &

he standard mi5 has J l". peanuts and J l". raisins. %ence# the standard mi5 cost is J F.60: J F1.+0 F1.0+Il".

Profits are F2.70 < F1.20 F1.@0Il". for delu5e and F2.++ < F1.0+ F1.+0Il". for the standardmi5. hus# the o",ective function is

>a5imi3e H F1.@0E : F1.+0S

Su",ect to

raisins2

E :1

S 70 l".& 2

peanuts1

E :1

S 60 l".

& 2E 110 l".

S 110 l".

=ptimum

E 70 l".

S 60 l".

Profit F2(&

+. >a5imi3e F1.+0) : F1.20K

Su",ect to

sugar 1.+) : 2.0K 1#200 cups

flour &.0) : &.0K 2#100 cupstime 6.0) : &.0K min.

=ptimum

) +00 pieces

K 200 pieces

'evenue F770

6S-;

Standard

S 110

Eelu5e

peanuts

raisins

E

6110

20 (0 60 ;0 100 120 1(0 160 1;0 200 220 2(0

220

200

1;0

160

1(0

120

100

;0

60

(0

20

0

Krape

200 (00 600 ;00 1000

)pple

sugar

time

1200

1000

;00

600

(00

200

0

flour


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nused supplies

sugar 1.++00 : 2200 1#1+0 cups used. %ence# 1#200 < 1#1+0 +0 cupsremaining.

flour &.0+00 : &.0200 2#100 cups used. 8o flour remains.

time 6.0+00 : &.0200 minutes. 8o time remains.

6. a. he optimal value of the decision varia"les are 51 (# 52 0# 5& 1;.he optimal value of the o",ective function value 3 106.

". he optimal value of the decision varia"les are 51 1+# 52 10# 5&0. he optimal value of the o",ective function value 3 210.

@. a. Bor pro"lem 6a# the range of feasi"ility for the three constraints are asfollows

Constraint 1 22 to infinity

Constraint 2 10 to (@.+

Constraint & 20 to (+

". Bor pro"lem 6a# the range of optimality for the three coefficients of the o",ective functionare

Maria"le 1 51 2.+ to 1+

Maria"le 2 52


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52 1.7+ to &.@+

5& 2 to +


10. 51 num"er of containers of orange ,uice

52 num"er of containers of grapefruit ,uice5& num"er of containers of pineapple ,uice

5( num"er of containers of )ll-in-=ne

=range Nuice Krapefruit Nuice Pineapple Nuice )ll-in-=ne

'evenue per t. F1.00 F.70 F.;0 F1.10

Cost per t. .+0 .(0 .&+ .(1@

Profit per t. F.+0 F.+0 F.(+ F.6;&

>a5imi3e .+051: .+052: .(+5&: .6;&5(

s.t.

=range ,uice 151 :.&&5( 1600 t.

Krapefr. ,uice 152 :.&&5( 1200 t.

Pineapple ,uice 15& :.&&5( ;00 t.

Krapefr. cont.


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11. 51 ty. of chopping "oards

52 ty. of *nife holders

ma5imi3e 251 : 652

s.t.Cutting 1.(51 : .;52 +6 minutes

Kluing +51 : 1&52 6+0

Binishing 1251 : &52 &60

51# 52 0

12. 51 ty. ham spread

52 ty. deli spread

ma5imi3e 251 : (52 profit or minimi3e &51: &52cost

s.t.

mayo 1.(51 : 1.052 @0 l".

mayo 1.(51 : 1.052 112 l".

ham 51 20 pans

deli 52 1; pans

51# 52 0

a. 51 &@.1(# 52 1;# Cost F16+.(2

". 51 20# 52 ;(# Profit F&@6.

6S-11

120

finishing

gluing

cutting

52

@0

+0

=pt.

0 &0 (0 1&05

1

=pt. is 51 0

52 +0

3 &00

Slac*Cutting 16 minutesKluing 0Binishing 210 minutes

mayo 112

mayo @0

52

51

20 +0 ;00

112

@0

1;


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Solutions (Continued)

1&. ) uantity of product )

/ uantity of product /

C uantity of product C

) / C

'evenue F;0 F70 F@0

Cost

>at4l O1 2 5 F + F10 1 5 F + F + 6 5 F + F&0

>at4l O2 & 5 F ( F12 + 5 F ( F20

La"or &.2 5 F10 F&2 1.+ 5 F10 F1+ 2 5 F10 F20

otal F+( F(0 F+0

Profit F26 F+0 F20

>a5imi3e 26) : +0/ : 20C profit

s.t.>at4l O1 2) : 1/ : 6C 200 l".

>at4l O2 &) : +/ &00 l".

La"or &.2) : 1.+/ :2.0C1+0 hr.

) output 2I&) < 1I&/

a5imi3e H (51: +52: 0s1: 0s2

Su",ect to

1 51 : &52 : 1s1 : 0s2 12

2 (51 : &52 : 0s1 : 1s2 2(

his forms the "asis of our initial ta"leau# which is shown in a"le +S


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a"le 1 Partial nitial a"leau

Profit per unitfor varia"les

in solutionEecisionMaria"les

C ( + 0 0

=",ective

row

Maria"lesin solution 51 52 s1 s2

Solutionuantity

0 s1 1 & 1 0 12

0 s2 ( & 0 1 2(

he completed ta"leau is shown in a"le 2.

he est for =ptimality

f all the values in the C < H row of any ta"leau are 3ero or negative# the optimal solution has "eeno"tained. n this case# the C < H row contains two positive values# ( and +# indicating thatimprovement is possi"le.

Eeveloping the Second a"leau

Malues in the C < H row reflect the profit potential for each unit of the varia"le in a given column. Bor

instance# the ( indicates that each unit of varia"le 51added to the solution will increase profits "y F(.

Similarly# the + indicates that each unit of 52will contri"ute F+ to profits. Kiven a choice "etween F(

per unit and F+ per unit# we select the larger and focus on that column# which means that 52will come

into the solution. 8ow we must determine which varia"le will leave the solution. )t each ta"leau# one

varia"le will come into the solution# and one will go out of solution# *eeping the num"er of varia"lesin the solution constant. 8ote that the num"er of varia"les in the solution must always eual the

num"er of constraints. hus# since this pro"lem has two constraints# all solutions will have two

varia"les.

o determine which varia"le will leave the solution# we use the num"ers in the "ody of the ta"le in the

column of the entering varia"le i.e.# & and &. hese are called row pivot values. Eivide each one into

the corresponding solution uantity amount# as shown in a"le &. he smaller of these two ratios

indicates the varia"le that will leave the solution. hus# varia"le s1will leave and "e replaced with 52.

n graphical terms# we have moved up the 52a5is to the ne5t corner point. /y determining the smallest

ratio# we have found which constraint is the most limiting. n Bigure 1# note that the two constraints

intersect the 52a5is at ( and ;# the two row ratios we have ,ust computed. he second ta"leau willdescri"e the corner point where 52 ( and 51 0V it will indicate the profits and uantities associated

with that corner point. t will also reveal if the corner point is an optimum# or if we must develop

another ta"leau.

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a"le 2 Completed nitial a"leau.

C ( + 0 0

Maria"les

in solution 51 52 s1 s2

Solutionuantity

0 s1 1 & 1 0 120 s2 ( & 0 1 2(

H 0 0 0 0 0

C < H ( + 0 0

C ( + 0 0


Solutionuantity

0 s1 1 & 1 0 12I& ( Smallest positive

ratio0 s2 ( & 0 1 2(I& ;

H 0 0 0 0 0

C < H ( + 0 0

Largestpositive

)t this point we can "egin to develop the second ta"leau. he row of the leaving varia"le will "e

transformed into the new pivot rowof the second ta"leau. his will serve as a foundation on which todevelop the other rows. o o"tain this new pivot row# we simply divide each element in the s1row "y

the row pivot value intersection of the entering column and leaving row# which is &. he resulting

num"ers are

51 52 s1 s2

Solutionuantity

Pivot-row value 1I& 1 1I& 0 (

hese num"ers "ecome the new 52row of the second ta"leau.

he pivot-row num"ers are used to compute the values for the other constraint rows in this instance#the only other constraint row is the s2row. he procedure is

1. Bind the value that is at the intersection of the constraint row i.e.# the s2rowand the entering varia"le column. t is &.

2. >ultiply each value in the new pivot row "y this value.

&. Su"tract the resulting values# column "y column# from the current row values.

6S-2&


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51 52 s1 s2 Ruantity

Current value ( & 0 1 2(


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he completed second ta"leau is shown in a"le +. t tells us that at this point ( units of varia"le 5 2arethe most we can ma*e see column Solution uantity# row 52 and that the profit associated with 52 (#51 0 is F20 see row H# column Solution uantity.

he fact that there is a positive value in the C < H row tells us that this is not the optimal solution.

Conseuently# we must develop another ta"leau.

Eeveloping the hird a"leau

he third ta"leau will "e developed in the same manner as the previous one.

1. Eetermine the entering varia"le Bind the column with the largest positivevalue in the C < H row @I in the 51 column.

2. Eetermine the leaving varia"le Eivide the solution uantity in each row "ythe row pivot. %ence#

( 12 12I& (

1I&

he smaller ratio indicates the leaving varia"le# s2. See a"le +S


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(. Compute values for the 52row >ultiply each new pivot-row value "y the 52rowpivot value i.e.# 1I& and su"tract the product from corresponding current values. hus#

51 52 s1 s2 Ruantity

Current value 1I& 1 1I& 0 (


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a"le @. =ptimal Solution

C ( + 0 0


Solutionuantity

+ 52 0 1 (I7


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would "e rewritten as eualities

1 &51: 252: (5&< 1s1< 0s2< 0s&=;0

2 +51: (52: 5&< 0s1< 1s2< 0s&=@0

& 251: ;52: 25&< 0s1< 0s2< 1s&=6;

)s eualities# each constraint must then "e ad,usted "y inclusion of an artificial varia"le. he finalresult loo*s li*e this

1 &51: 252: (5&< 1s1< 0s2< 0s&: 1a1: 0a2: 0a&=;0

2 +51: (52: 5&< 0s1< 1s2< 0s&: 0a1: 1a2: 0a&=@0

& 251: ;52: 25&< 0s1< 0s2< 1s&: 0a1: 0a2: 1a&=6;

f the o",ective function happened to "e

+51: 252: @5&

it would "ecome

+51: 252: @5&: 0s1: 0s2: 0s&: >a1: >a2: >a&

Summary of >a5imi3ation Procedure

he main steps in solving a ma5imi3ation pro"lem with only constraints using the simple5 algorithm

are these

1. Set up the initial ta"leau.

a. 'ewrite the constraints so that they "ecome eualitiesV add a slac* varia"le to each constraint.

". 'ewrite the o",ective function to include the slac* varia"les. Kive slac* varia"les coefficientsof 0.

c. Put the o",ective coefficients and constraint coefficients into ta"leau form.

d. Compute values for the H rowV multiply the values in each constraint row "y the row4s C value.)dd the results within each column.

e. Compute values for the C < H row.

2. Set up su"seuent ta"leaus.

a. Eetermine the entering varia"le the largest positive value in the C< H row. f a tie e5ists#choose one column ar"itrarily.

". Eetermine the leaving varia"le Eivide each constraint row4s solution uantity "y the row4spivot valueV the smallest positive ratio indicates the leaving varia"le. f a tie occurs# divide thevalues in each row "y the row pivot value# "eginning with slac* columns and then othercolumns# moving from left to right. he leaving varia"le is indicated "y the lowest ratio in thefirst column with uneual ratios.

c. Borm the new pivot row of the ne5t ta"leau Eivide each num"er in the leaving row "y the

row4s pivot value. Dnter these values in the ne5t ta"leau in the same row positions.d. Compute new values for remaining constraint rows Bor each row# multiply the values in the

new pivot row "y the constraint row4s pivot value# and su"tract the resulting values# column "ycolumn# from the original row values. Dnter these in the new ta"leau in the same positions asthe original row.

e. Compute values for H and C < H rows.

f. Chec* to see if any values in the C < H row are positiveV if they are# repeat 2a


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>inimi3ation Pro"lems

he simple5 method handles minimi3ation pro"lems in essentially the same way it handles

ma5imi3ation pro"lems. %owever# there are a few differences. =ne is the need to ad,ust for

constraints# which reuires "oth artificial varia"les and surplus varia"les. his tends to ma*e manual

solution more involved. ) second ma,or difference is the test for the optimum ) solution is optimal ifthere are no negative values in the C < H row.

D5ample

Solve the following pro"lem for the uantities of 51and 52that will minimi3e cost.

>inimi3e H 1251 : 1052

Su",ect to 51 : (52 ;

&51 : 252 6

51# 52 0

Solution to e5ample

1. 'ewrite the constraints so that they are in the proper form

51: (52; "ecomes 51: (52< 1s1< 0s2: 1a1: 0a2 ;

&51: 2526 "ecomes &51: 252< 0s1< 1s2: 0a1: 1a2 6

2. 'ewrite the o",ective function coefficients of C row

1251: 1052: 0s1: 0s2: 777a1: 777a2

&. Compute values for rows H and C < H

C 51 52 s1 s2 a1 a2 Ruantity

777 1777 (777


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6. Eivide each num"er in the leaving row "y the pivot value (# in this case to o"tainvalues for the new pivot row of the second ta"leau

1I( (I( 1


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e. Eetermine values for new 52row

0 1


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Problems for te enricment module (simple!)

1. Kiven this information

>a5imi3e H 10.+05 : 11.@+y : 10.;03

Su",ect to

Cutting +5 : 12y : ;3 1#(00 minutesStapling @5 : 7y : 73 1#2+0 minutes

Grapping (5 : &y : 63 @20 minutes

5# y# 3 0

Solve for the uantities of products 5# y# and 3 that will ma5imi3e revenue.

2. se the simple5 method to solve these pro"lems

a. >inimi3e H 2151: 1;52

Su",ect to 1 +51: 1052100

2 251: 15210

51# 520

". >inimi3e H 25 : +y : &3

Su",ect to 1 165 : 10y : 1;3 &(0

2 115 : 12y : 1&3 &00

& 25 : 6y : +3 120

5# y# 3 0

&. se the simple5 method to solve the following pro"lem.

>inimi3e H &51: (52: ;5&

Su",ect to 251: 526

52: 25&(

51# 52# 5&0

(. se the simple5 method to solve the following pro"lem.

>a5imi3e H ;51: 252

Su",ect to (51: +5220

251: 6521;

51# 5208ote 'ow operations in pro"lems & and ( are computationally easy.

6S-&2


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Solutions"#nricment $odule (S%$P')

1. C 10.+ 11.@+ 10.;0 0 0 0

Mar ? y 3 S1 S2 S& "i ratio

0 S1 + 12 ; 1 0 0 1#(00 116.6@

0 S2 @ 7 7 0 1 0 1#2+0 1&;.;7( S& ( & 6 0 0 1 @20 2(0

H 0 0 0 0 0 0 0

C


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2. a. >inimi3e H 2151 : 1;52

s.t. +51 : 1052 : )1< S1 100

251 : 152 : )2< S2 10

C 21 1; > 0 > 0

. C Mar 51 52 )1 S1 )2 S2 "i ratio

> )1 + 10 1 )2 2 1 0 0 1 11> > > 110>

C 0 > 0

. C Mar 51 52 )1 S1 )2 S2 "i ratio

1; 52 0.+ 1 0.1 )2 1.+ 0 1;0

C


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2. ".

. C 2 + & > 0 > 0 > 0

Mar ? y 3 )1 S1 )2 S2 )& S& "i

> )1 16 10 1; 1 )2 11 12 1& 0 0 1 )& 2 6 + 0 0 0 0 :1 2;> &6> > > > @60>

C 0 > 0 >

. C Mar ? y 3 )1 S1 )2 S2 )& S& "i

& H .;;;7 .+++6 1 .0++6 )2 0 >

. C Mar ? y 3 S1 )2 S2 S& "i

& 3 1.&1 0 1 )2 &.0679 0 0 .&10& 1 :;&.1T

C


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M. C Mar 5 y 3 S1 S2 S& "i

& 3 .( 1.2 1 0 0 10>

C, > 0 0

C & ( ; 0 0

Mar 51 52 5& S1 S2 "i "iIai,

& 51 1 J 0 )2 0 1 2 0 :7

C,

C & ( ; 0 0

Mar 51 52 5& S1 S2 "i "iIai,

& 51 1 J 0


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(. C, ; 2 0 0

Mar 51 52 S1 S2 "i

0 S1 ( + 1 0 20

0 S2 2 6 0 1 1;

H, 0 0 0 0 0

C,

chap006 linear programming

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