Graphical Worked Examples 04

8/13/2019 Graphical Worked Examples 04

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Worked Examples for Chapter 4

Example for Section 4.1Consider the following linear programming model.

Maximize Z = 3 x1 + 2 x2,

subje t to

x1 ! "

x1 + 3 x2 ! 1#

2 x1 + x2 ! 1$

and x1 % $, x2 % $.

(a) Use graphical analysis to identify all the corner-point solutions for this model.

La el each as either feasi le or infeasi le.

&he graph showing all the onstraint boundar' lines and the orner(point solutions at

their interse tions is shown below.


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&he exa t )alue of *x 1, x 2 for ea h of these nine orner(point solutions * , -, ...,

an be identified b' obtaining the simultaneous solution of the orresponding two

onstraint boundar' e/uations. &he results are summarized in the following table.

Corner(pointsolutions

*x1, x 2 0easibilit'

*$, # 0easible- *$,1$ nfeasibleC *3, " 0easible

*", 11 3 nfeasible*", 2 0easible

0 *", $ 0easible4 *#, $ nfeasible5 *1#, $ nfeasible

*$, $ 0easible

( ) Calc!late the "al!e of the o #ecti"e f!nction for each of the C$% sol!tions.

Use this information to identify an optimal sol!tion.

&he obje ti)e )alue of ea h orner(point feasible solution is al ulated in the

following table6


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Corner(point

feasible solutions*x1, x 2 7bje ti)e 8alue

9*$, # 3:$+2:# = 1$

C *3, " 3:3+2:" = 1;*", 2 3:"+2:2 = 10 solution 6

-' olution Con ept 3, we hoose the origin, point = *$, $ , to be the initial C>0

solution. -' olution Con ept 0 solutions, = *$, # with 9 = 1$ and 0 = *", $ with 9 = 12, ha)e a larger

)alue of 9 *so mo)ing toward either adja ent C>0 solution gi)es a positi)e rate ofimpro)ement in 9 . -' olution Con ept #, we hoose 0 be ause the rate of

impro)ement in 9 of 0 *= 12 " = 3 is greater than that of *= 1$ #=2 .

C>0 solution 06

&he C>0 solution 0 is not optimal be ause one adja ent C>0 solution, = *", 2

with 9 = 10 solution .

C>0 solution 6&he C>0 solution is not optimal be ause one adja ent C>0 solution, C = *3,"

with 9 = 1;, has a larger )alue of 9. @e then mo)e to C>0 solution C.

C>0 solution C6

-' olution Con ept 0 solution C is optimal sin e its adja ent C>0

solutions, and , ha)e smaller )alues of 9 so mo)ing toward either of these

adja ent C>0 solutions would gi)e a negati)e rate of impro)ement in 9.


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&herefore, the se/uen e of C>0 solutions examined b' the simplex method would

be (A 0 (A (A C.

Example for Section 4.

Be onsider the following linear programming model *pre)iousl' anal'zed in the

pre eding example .


subje t to

x1 ! "

x1 + 3 x2 ! 1#2 x1 + x2 ! 1$

and x1 % $, x2 % $.

(a) *ntrod!ce slack "aria les in order to 'rite the f!nctional constraints in

a!gmented form.

@e introdu e x 3, x " , and x # as the sla ? )ariables for the respe ti)e onstraints. &he

resulting augmented form of the model is

Maximize 9 = 3 x 1 + 2 x 2,

subje t to

x1 + x 3 = "

x1 + 3 x 2 + x " = 1#

2 x 1 + x 2 + x # = 1$

and

x1 $, x 2 $, x 3 $, x " $, x # $.

( ) %or each C$% sol!tion+ identify the corresponding ,% sol!tion y calc!lating

the "al!es of the slack "aria les. %or each ,% sol!tion+ !se the "al!es of the

"aria les to identify the non asic "aria les and the asic "aria les.

C>0 solution = *$, $ 6

>lug in x 1 = x 2 = $ into the augmented form. &he )alues of the sla ? )ariables are x 3


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= ", x " = 1#, x # = 1$.

&he -0 solution is *x 1, x 2, x 3, x " , x # = *$, $, ", 1#, 1$ .

in e x 1 = x 2 = $, we ?now that x 1 and x 2 are the two nonbasi )ariables.

in e x3 A$, x

"A$, x

#A$, we ?now that x

3, x

", and x

# are basi )ariables.

C>0 solution = *$, # 6

>lug in x 1 = $ and x 2 = # into the augmented form. &he )alues of the sla ? )ariables

are x 3 = ", x " = $, x # = #.

&he -0 solution is *x 1, x 2, x 3, x " , x # = *$, #, ", $, # .

in e x 1 = x " = $, we ?now that x 1 and x " are the two nonbasi )ariables.

in e x 2 A$, x 3A$, x#A$, we ?now that x 2, x 3, and x # are basi )ariables.

C>0 solution C = *3, " 6

>lug in x 1 = 3 and x 2 = " into the augmented form. &he )alues of the sla ? )ariables

are x 3 = 1, x " = $, x # = $.

&he -0 solution is *x 1, x 2, x 3, x " , x # = *3, ", 1, $, $ .

in e x " = x # = $, we ?now that x " and x # are the two nonbasi )ariables.

in e x 1 A$, x 2A$, x3A$, we ?now that x 1, and x 2 and x 3 are basi )ariables.

C>0 solution = *", 2 6

>lug in x 1 = " and x 2 = 2 into the augmented form. &he )alues of the sla ? )ariablesare x 3 = $, x " = #, x # = $.

&he -0 solution is *x 1, x 2, x 3, x " , x # = *", 2, $, #, $ .

in e x 3 = x # = $, we ?now that x 3 and x # are the two nonbasi )ariables.

in e x 1 A$, x 2A$, x"A$, we ?now that x 1, x 2, and x " are basi )ariables.

C>0 solution 0 = *", $ 6

>lug in x 1 = " and x 2 = $ into the augmented form. &he )alues of the sla ? )ariables

are x 3 = $, x " = 11, x # = 2.&he -0 solution is *x 1, x 2, x 3, x " , x # = *", $, $, 11, 2 .

in e x 2 = x 3 = $, we ?now that x 2 and x 3 are the two nonbasi )ariables.

in e x 1 A$, x "A$, x#A$, we ?now that x 1, x " , and x # are basi )ariables.

ummar' of results6

abel C>0 solution -0 solution Donbasi )ariables -asi )ariables

*$, $ *$, $, ", 1#, 1$ x 1, x 2 x3, x " , x #


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*$, # *$, #, ", $, # x 1, x " x2, x 3, x #

C *3, " *3, ", 1, $, $ x " , x # x1, x 2, x 3

*", 2 *", 2, $, #, $ x 3, x # x1, x 2, x "

0 *", $ *", $, $, 11, 2 x2, x

3 x

1, x

", x

#

(c) %or each ,% sol!tion+ demonstrate ( y pl!gging in the sol!tion) that+ after the

non asic "aria les are set e !al to -ero+ this ,% sol!tion also is the sim!ltaneo!s

sol!tion of the system of e !ations o tained in part (a).

-0 solution = *$, $, ", 1#, 1$ 6 >lugging this solution into the e/uations 'ields6

$ + " = "

$ + 3*$ + 1# = 1#

2*$ + $ +1$ = 1$,

so the e/uations are satisfied.

-0 solution = *$, #, ", $, # 6 >lugging this solution into the e/uations 'ields

$ + " = "

$ + 3*# + $ = 1# 2*$ + # +# = 1$,


-0 olution C = *3, ", 1, $, $ 6 >lugging this solution into the e/uations 'ields

3 + 1 = "

3 + 3*" + $ = 1# 2*3 + " + $ = 1$,


-0 solution = *", 2, $, #, $ 6 >lugging this solution into the e/uations 'ields


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" + $ = "

" + 3*2 + # = 1#

2*" + 2 + $ = 1$,


-0 solution 0 = *", $, $, 11, 2 6 >lugging this solution into the e/uations 'ields

" + $ = "

" + 3*$ + 11 = 1#

2*" + $ + 2 = 1$,


Example for Section 4.

Be onsider the following linear programming model *pre)iousl' onsidered in the

pre eding two examples .


subje t to

x1 ! "

x1 + 3 x2 ! 1#

2 x1 + x2 ! 1$

and x1 % $, x2 % $.

@e introdu e x 3, x " , and x # as sla ? the )ariables for the respe ti)e onstraints. &he

resulting augmented form of the model is


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Maximize 9 = 3 x 1 + 2 x 2,

subje t to

x1 + x 3 = "

x1 + 3 x

2 + x

" = 1#

2 x 1 + x 2 + x # = 1$

and

x1 $, x 2 $, x 3 $, x " $, x # $.

(a) Work thro!gh the simplex method (in alge raic form) to sol"e this model.

*nitiali-ation/

et x 1 and x 2 be the nonbasi )ariables, so x 1 = x 2 = $. ol)ing for x 3, x " , and x # from

the e/uations for the onstraints6

*1 x 1 + x 3 = "

*2 x 1 + 3 x 2 + x " = 1#

*3 2 x 1 + x 2 + x # = 1$

we obtain the initial -0 solution *$, $, ", 1#, 1$ .

&he obje ti)e fun tion is 9 = 3 x 1 + 2 x 2. &he urrent -0 solution is not optimal sin e

we an impro)e 9 b' in reasing x 1 or x 2.

*teration 1/

9 = 3 x 1 + 2 x 2, so e/uation *$ is

*$ 9( 3 x 1 ( 2 x 2 = $.

f we in rease x 1, the rate of impro)ement in 9 = 3.f we in rease x 2, the rate of impro)ement in 9 = 2.

5en e, we hoose x 1 as the entering basi )ariable.

Dext, we need to de ide how far we an in rease x 1. in e we need )ariables x 3, x " ,

and x # to sta' nonnegati)e, from e/uations *1 , *2 , and *3 , we ha)e

*1 x 3 = " E x 1 $ x1 ". minimum

*2 x " = 1# E x 1 $ x1 1#.


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*3 x # = 1$ E 2 x 1 $ x1 #.

&hus, the entering basi )ariable x 1 an be in reased to ", at whi h point x 3 has

de reased to $. &he )ariable x3 be omes the new nonbasi )ariable. >roper form from

4aussian elimination is restored b' adding 3 times e/uation *1 to e/uation *$ ,

subtra ting e/uation *1 from e/uation *2 , and subtra ting 2 times e/uation *1 from

e/uation *3 . &his 'ields the following s'stem of e/uations6

*$ 9 (2 x 2 + 3 x 3 = 12

*1 x 1 + x 3 = "

*2 3 x 2 E x3 + x " = 11

*3 x 2 E 2x3 + x # = 2.

&hus, the new -0 solution is *", $, $, 11, 2 with 9 = 12.

*teration /

Fsing the new e/uation *$ , the obje ti)e fun tion be omes 9 = 2 x 2 E 3 x 3 + 12. &he

urrent -0 solution is nonoptimal sin e we an in rease x 2 to impro)e 9 with the rate

of impro)ement in 9 = 2. 5en e, we hoose x 2 as the entering basi )ariable.

Dext, we need to de ide how far we an in rease x 2. in e we need the )ariables x 1, x "and x # to sta' nonnegati)e, from e/uations *1 , *2 , and *3 in iteration 1, we ha)e

*1 x 1 = " $ no upper bound on x 2

*2 x " = 11 E 3 x 2 $ x2 11 3

*3 x # = 2 E x 2 $ x2 2. minimum

&hus, x 2 an be in reased to 2, at whi h point x # has de reased to $, so x # be omes thelea)ing basi )ariable. &hus, x # be omes a nonbasi )ariable. fter restoring proper

form from 4aussian elimination, we obtain the following s'stem of e/uations6

*$ 9 ( x 3 + 2 x # = 1erforming the minimum ratio test on x 1, as

shown in the last olumn of the abo)e tableau, the lea)ing basi )ariable is x 3. fter

using elementar' row operations to restore proper form from 4aussian elimination,

the new simplex tableau with basi )ariables x 1, x ", and x # be omes

,asic0aria le E

Coefficient of/ ightSide

atio

2 x 1 x x x4 x&9 *$ 1 $ (2 3 $ $ 12x1 *1 $ 1 $ 1 $ $ "x" *2 $ $ 3 (1 1 $ 11 11 3x# *3 $ $ 1 (2 $ 1 2 2 minimum

*teration .

in e the oeffi ient for x 2 in /. *$ is E3, we an impro)e 9 b' in reasing x 2. &he

nonbasi )ariable x 2 is to be hanged to a basi )ariable. >erforming the minimum

ratio test on x 2, as shown in the last olumn of the abo)e tableau, the lea)ing basi

)ariable is x #. fter restoring proper form from 4aussian elimination, the new simplex

tableau with basi )ariables x 1, x 2, and x " be omes


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,asic0aria le E

Coefficient of/ ightSide

atio

2 x 1 x x x4 x&9 *$ 1 $ $ (1 $ 2 1hase 1 problem.


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*teration ,asic0aria le E

Coefficient of/ ightSide2 x1 x2 x3 x" x 5 x 6

9 *$ (1 (" (2 (1 1 $ $ (1;*$ x 5 *1 $ 1 1 $ $ 1 $ ;

x 6 *2 $ 3 1 1 (1 $ 1 1$9 *$ (1 $ (2 3 1 3 (1 3 $ " 3 (11 3

*1 x 5 *1 $ $ 2 3 (1 3 1 3 1 (1 3 11 3x1 *2 $ 1 1 3 1 3 (1 3 $ 1 3 1$ 39 *$ (1 $ $ $ $ 1 1 $

*2 x 2 *1 $ $ 1 ($.# $.# 1.# ($.# #.#x1 *2 $ 1 $ $.# ($.# ($.# $.# 1.#

&herefore, the optimal solution for the >hase 1 problem is

*x1, x 2, x 3, x", x 5 , x 6 = *1.#, #.#, $, $, $, $ with 9 = $.

Dow using the original obje ti)e fun tion, the >hase 2 problem is

Minimize 9 = 3 x 1 + 2 x 2 + x 3, subje t to

x1 + x 2 = ;

3 x 1 + x 2 + x 3 E x" = 1$

and

x1 $, x 2 $, x 3 $, x " $,

or e/ui)alentl',

Maximize *(9 = E 3 x 1 E 2 x 2 E x3 , subje t to

x1 + x 2 = ;

3 x 1 + x 2 + x 3 E x" = 1$

and

x1 $, x 2 $, x 3 $, x " $.

Fsing the optimal solution for the >hase 1 problem *after eliminating the artifi ial

)ariables, whi h are no longer needed as the initial -0 solution for the >hase 2

problem, we obtain the following simplex tableau.


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,asic0aria le E

Coefficient of/ ight Side

2 x 1 x x x49 *$ (1 $ $ $.# ($.# (1#.#x2 *1 $ $ 1 ($.# $.# #.#x1 *2 $ 1 $ $.# ($.# 1.#

&his tableau re)eals that the urrent -0 solution is also optimal. 5en e, the optimal

solution is *x 1, x 2, x 3, x " = *1.#, #.#, $, $ with 9 = 1#.#.

(c) Compare the se !ence of ,% sol!tions o tained in parts (a) and ( ). Which of

these sol!tions are feasi le only for the artificial pro lem o tained y

introd!cing artificial "aria les and 'hich are act!ally feasi le for the realpro lem6.

&he se/uen e of -0 solutions obtained in part *a and *b are the same. ll these -0

solutions ex ept the last one are feasible onl' for the artifi ial problem obtained b'

introdu ing artifi ial )ariables. 7nl' the final -0 solution represents a feasible

solution for the real problem.

(d) Use a soft'are package ased on the simplex method to sol"e the pro lem.

Fsing the x el ol)er *whi h emplo's the simplex method to sol)e the problem

'ields the following optimal solution6

*x1, x 2, x 3 = *1.#, #.#, $ with 9 = 1#.#, as displa'ed next..


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Example for Section 4.7

Be onsider the linear programming model pre)iousl' anal'zed in the example for

e tions ".1, ".2, ".3, and ".". &his model is again shown below, where the right(hand

sides of the fun tional onstraints now are interpreted as the amounts a)ailable of the

respe ti)e resour es.

Maximize 9 = 3 x 1 + 2 x 2,

subje t to

*1 x 1 " *resour e 1

*2 x 1 + 3 x 2 1# *resour e 2

*3 2 x 1 + x 2 1$ *resour e 3

and

x1 $, x 2 $.

&he optimal solution is *x 1, x 2 = *3, " with 9 = 1;.


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(a) Use graphical analysis as in %ig. 4.8 to determine the shado' prices for the

respecti"e reso!rces.

&he following figure summarizes the anal'sis.

0rom the figure, we an see the following.

Constraint (1) (x 1 4)/ Constraint *1 is not binding at the optimal solution *3, " ,

sin e a small hange in b 1 = " will not hange the optimal )alue of 9. 5en e, y 1* = $.

Constraint ( ) (x 1 9 x 1&)6 Constraint *2 is binding at *3, " . @e in rease b 2from 1# to 1


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( ) Use graphical analysis to perform sensiti"ity analysis on this model. *n

partic!lar+ check each parameter of the model to determine 'hether it is a

sensitive parameter (a parameter 'hose "al!e cannot e changed 'itho!t

changing the optimal sol!tion) y examining the graph that identifies the optimal

sol!tion.

0rom part *a , we ?now that b 1 is not a sensiti)e parameter, while b 2 and b 3 are

sensiti)e parameters. imilarl', sin e onstraint *1 is not binding at the optimal

solution *3, " , the oeffi ients a 11 = 1 and a 12 = $ of onstraint *1 are not sensiti)e.

in e onstraint *2 and *3 are binding at the optimal solution, the oeffi ients a 21 =

1, a 22 = 3, a 31 = 2, and a 32 = 1 are sensiti)e parameters. 0rom the following figure, we

an see that at the optimal solution, the obje ti)e fun tion 9 = 3 x 1 + 2 x 2 is not

parallel to onstraint *2 or onstraint *3 . 5en e, the oeffi ients 1 = 3 and 2 = 2 are

not sensiti)e parameters.


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(c) Use graphical analysis as in %ig. 4.; to determine the allo'a le range for each

c j "al!e (coefficient of x j in the o #ecti"e f!nction) o"er 'hich the c!rrent optimal

sol!tion 'ill remain optimal.

0rom the following graph, we an see that the urrent optimal solution will remain

optimal for 2 3 1 " *with 2 fixed at 2 and 3 2 2 H *with 1 fixed at 3 ,

sin e the obje ti)e fun tion line will rotate around to oin ide with one of the

onstraint boundar' lines at ea h of the endpoints of these inter)als.

(d) Changing #!st one b i "al!e (the right5hand side of f!nctional constraint i ) 'ill

shift the corresponding constraint o!ndary. *f the c!rrent optimal C$% sol!tion

lies on this constraint o!ndary+ this C$% sol!tion also 'ill shift. Use graphical

analysis to determine the allo'a le range for each b i "al!e o"er 'hich this C$%

sol!tion 'ill remain feasi le.

0rom the following graph, we an see the following.


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0or Constraint *1 *x 1 " 6 &he allowable range for b 1 is 3 b1 sin e *3, "

remains feasible o)er this range.

0or Constraint *2 *x1 + 3 x

2 1# 6 &he allowable range for b

2 is 1$ b

2 3$.

0or b 2 I 1$, the interse tion of x 1 + 3x 2 = b 2 and 2x 1 + x 2 = 1$ )iolates the x 1 ! "

onstraint. 0or b 2 A 3$, this interse tion )iolates the x 1 % $ onstraint.

0or Constraint *3 *2x 1 + x 2 1$ 6 &he allowable range for b 3 is # b3 3# 3.

0or b 3 I #, the interse tion of x 1 + 3x 2 = 1# and 2x 1 + x 2 = b 3 )iolates the x 1 % $

onstraint. for b 3 A 3# 3, this interse tion )iolates the x 1 ! " onstraint.

(e) 0erify yo!r ans'ers in parts (a)+ (c)+ and (d) y !sing a comp!ter package

ased on the simplex method to sol"e the pro lem and then to generate

sensiti"ity analysis information.

Fsing the x el ol)er *whi h emplo's the simplex method , the sensiti)it' anal'sis

report *whi h )erifies these answers is generated, as shown after the following

spreadsheet.


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Example for Section 4.;

Use the interior5point algorithm in yo!r < Co!rse'are to sol"e the follo'ing

model (pre"io!sly analy-ed in the examples for Sections 4.1+ 4. + 4. + 4.4+ and

4.7). Choose = :.& from the ra' a graph of the feasi le region+ and

then plot the tra#ectory of the trial sol!tions thro!gh this feasi le region.

Maximize 9 = 3 x 1 + 2 x 2,

subje t to

x1 "

x1 + 3 x 2 1#

2 x 1 + x 2 1$

and x1 $, x 2 $.

@e use the 7B tutorial with = $.#, whi h generates the following output6

ol)e utomati all' b' the nterior >oint lgorithm6 * lpha = $.#


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teration x 1 x2 9$ $.1 $." 1.11 $.3$G#" 2.

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Graphical Worked Examples 04