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 .
Maximize Z = 3 x1 + 2 x2,
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$,
so the e/uations are satisfied.
-0 olution C = *3, ", 1, $, $ 6 >lugging this solution into the e/uations 'ields
3 + 1 = "
3 + 3*" + $ = 1# 2*3 + " + $ = 1$,
so the e/uations are satisfied.
-0 solution = *", 2, $, #, $ 6 >lugging this solution into the e/uations 'ields
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" + $ = "
" + 3*2 + # = 1#
2*" + 2 + $ = 1$,
so the e/uations are satisfied.
-0 solution 0 = *", $, $, 11, 2 6 >lugging this solution into the e/uations 'ields
" + $ = "
" + 3*$ + 11 = 1#
2*" + $ + 2 = 1$,
so the e/uations are satisfied.
Example for Section 4.
Be onsider the following linear programming model *pre)iousl' onsidered in the
pre eding two examples .
Maximize Z = 3 x1 + 2 x2,
subje t to
x1 ! "
x1 + 3 x2 ! 1#
2 x1 + x2 ! 1$
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.