Module 7 - Linear Programming, The Simplex Method - Answers

8/18/2019 Module 7 - Linear Programming, The Simplex Method - Answers

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M O D

U L E

7

Linear Programming: The Simplex Method

Teaching Suggestions

Teaching Suggestion M7.1: Meaning of Slack Variables.

Slack variables have an important physical interpretation and represent a valuable commodity,such as unused labor, machine time, money, space, and so forth.

Teaching Suggestion M7.2: Initial Solutions to LP Problems.

Explain that all initial solutions begin with X 1 = 0, X = 0 !that is, the real variables set to "ero#,and the slacks are the variables with non"ero values. $ariables with values of "ero are called

nonbasic and those with non"ero values are said to be basic.

Teaching Suggestion M7.3: Substitution Rates in a Simplex Tableau.

%erhaps the most confusing pieces of information to interpret in a simplex tableau are&substitution rates.' (hese numbers should be explained very clearly for the first tableau becausethey will have a clear physical meaning. )arn the students that in subse*uent tableaus theinterpretation is the same but will not be as clear because we are dealing with marginal rates ofsubstitution.

Teaching Suggestion M7.4: Hand alculations in a Simplex Tableau.

+t is almost impossible to walk through even a small simplex problem !two variables, two

constraints# without making at least one arithmetic error. (his can be maddening for studentswho know what the correct solution should be but cant reach it. )e suggest two tips-

1. Encourage students to also solve the assigned problem by computer and to re*uest thedetailed simplex output. (hey can now check their work at each iteration.

. Stress the importance of interpreting the numbers in the tableau at each iteration. (he 0sand 1s in the columns of the variables in the solutions are arithmetic checks and balancesat each step.

Teaching Suggestion M7.5: Infeasibilit! Is a Ma"or Problem in Large LP Problems.

s we noted in (eaching Suggestion /., students should be aware that infeasibility commonlyarises in large, realworldsi"ed problems. (his module deals with how to spot the problem !and

is very straightforward#, but the real issue is how to correct the improper formulation. (his isoften a management issue.

op!rig#t $%&'( Pearson) Inc.M7-1


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Alternative Examples

Alternative Example M7.1: Simplex Solution to lternative Example /.1 !see 2hapter / ofSolutions 3anual for formulation and graphical solution#.

1st Iteration

C j →

↓

Solution

Mix

! " "

#uantit$

X 1 X % S 1 S %

0 S 1 1 4 1 0 40 S 1 0 1 1

* " 0 0 0 0 0 " 5 * " 6 7 0 0

%nd Iteration

C j →

↓

Solution

Mix

! " "

#uantit$ X 1 X % S 1 S %

7 X 14

1 14

0

0 S 1

0 − 1

1 4

* " 74

7 74

0 84

" 5 * " 64

0 − 7

40

(his is not an optimum solution since the X 1 column contains a positive value. 3ore profit

remains !9 64 per :1#.

rd&'inal Iteration

C j →

↓

Solution

Mix

! " "


7 X 0 1 1

− 1

4

6 X 1 1 0 −1 ;

* " 6 7 6

6

0

" 5 * " 0 0 − 6

− 6

(his is an optimum solution since there are no positive values in the " 5 * " row. (his says tomake 4 of item : and ; of item :1 to get a profit of 90.



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Alternative Example M7.%: Set up an initial simplex tableau, given the following twoconstraints and ob


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Alternative Example M7.: >eferring back to ?al, in lternative Example /.1, we had aformulation of-

3aximi"e %rofit = 96 X 1 97 X

Sub


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Alternative Example M7.*: Aevine 3icros assembles both laptop and desktop personalcomputers. Each laptop yields 910 in profitB each desktop 900.

(he firms A% primal is-

3aximi"e profit = 910 X 1 900 X

sub3 chips

1 X 1 X ≤ 910 royalty fees

where X 1 = no. laptops assembled daily

X = no. desktops assembled daily

?ere is the primal optimal solution and final simplex tableau.

C j →

↓

Solution

Mix

+1)" +%"" " " "

#uantit$ X 1 X % S 1 S % S

00 X 0 1 1 − 17

0 ;

10 X 1 1 0 51 7

0 4

0 S 6 0 0 5 1 4 * " 10 00 40

16 1

60 9,40

" 5 * " 0 0 540 −16 1

60

or X 1 = 4, X = ;, S 6 = 94 in slack royalty fees paid

%rofit = 9,40Cday

?ere is the dual formulation-

3inimi"e * = 0 !1 10; ! 10 !6

sub3 chip = 916.66

!6 = marginal value of one more 91 in royalty fees = 90



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Solutions to ,iscussion #uestions and Pro-lems

M71. (he purpose of the simplex method is to find the optimal solution to A% problems in asystematic and efficient manner. (he procedures are described in detail in Section 3/.6.

M7%. ,ifferences between graphical and simplex methods- !1# Draphical method can be usedonly when two variables are in modelB simplex can handle any dimensions. !# Draphical methodmust evaluate all corner points !if the corner point method is used#B simplex checks a lessernumber of corners. !6# Simplex method can be automated and computeri"ed. !4# Simplex methodinvolves use of surplus, slack, and artificial variables but provides useful economic data as a by product.

Similarities- !1# oth methods find the optimal solution at a corner point. !# oth methodsre*uire a feasible region and the same problem structure, that is, ob


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M7(. (he " 5 * " value is the net change in the value of the ob


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M717. a.

b. (he new optimal corner point is !0,0# and the profit is /,00.

c. The shadow price = (increase in profit)/(increase in right-hand side value)

= !/,00 5 ,400#C!40 5 ;0#

= 4,;00C10

= 60

d. )ith the additional change, the optimal corner point in part is still the optimal corner point. %rofit doesnt change. Gnce the righthand side went beyond 40, anotherconstraint prevented any additional profit, and there is now slack for the first constraint.

M71(. a. See the table below.

Ta-le 0or Pro-lem M71(

C j →

↓

Solution

Mix

+!"" +1/"" +" +"

#uantit$ X 1 X % S 1 S %0 S 1 14 4 1 0 6,60

0 S 10 1 0 1 7,00

* " 0 0 0 0 0

" 5 * " 700 1,800 0 0

b. 14 X 1 4 X ≤ 6,60

10 X 1 1 X ≤ 7,00

X 1, X ≥ 0

c. 3aximi"e profit = 700 X 1 1,800 X

d. asis is S 1 = 6,60, S = 7,00.

e. X should enter basis next.

f. S will leave next.

g. ;00 units of X will be in the solution at the second tableau.

h. %rofit will increase by ! " 5 * "#!units of variable entering the solution#



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= !1,800#!;00# = 1,00,000

M71!. a. 3aximi"e earnings = 0.; X 1 0.4 X 1. X 6 5 0.1 X 4 0S 1 0S 5 M+1 5 M+

sub


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Draphical solution to %roblem 3/0-

M7%1. a.

b.

C j →

↓

Solution

Mix

1" ( " " #uantit$

X 1 X % S 1 S %

0 S 1 4 1 0 ;0

0 S 1 0 1 80

* " 0 0 0 0 0

" 5 * " 10 ; 0 0

(his represents the corner point !0,0#.

c. (he pivot column is the X 1 column. (he entering variable is X 1.

d. >atios- >ow 1- ;0C4 = 0

>ow - 80C1 = 80

(hese represent the points !0,0# and !80,0# on the graph.



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e. (he smallest ratio is 0, so 0 units of the entering variable ! X 1# will be brought into thesolution. +f the largest ratio had been selected, the next tableau would represent aninfeasible solution since the point !80,0# is outside the feasible region.

f. (he leaving variable is the solution mix variable in row with the smallest ratio. (hus, S 1 is the leaving variable. (he value of this will be 0 in the next tableau.

g.Second iteration

C j →

↓

Solution

Mix

1" ( " " #uantit$

X 1 X % S 1 S %

10 X 1 1 0.8 0.8 0 0

0 S 0 1.8 50.8 1 60

* " 10 8 .8 0 00

" 5 * " 0 6 5.8 0

(hird iteration

C j →

↓

Solution

Mix

1" ( " " #uantit$

X 1 X % S 1 S %

10 X 1 1 0 0.6666 50.6666 10

; X 0 1 50.1/ 0./ 0

* " 10 ; 0

" 5 * " 0 0 5 5

h. (he second iteration represents the corner point !0,0#. (he third !and final# iterationrepresents the point !10,0#.

M7%%. asis for first tableau- +1 = ;0

+ = /8

! X 1 = 0, X = 0, S 1 = 0, S = 0#

Second tableau- +1 = 88

X 1 = 8

! X = 0, S 1 = 0, S = 0, + = 0#

Draphical solution to %roblem 3/-



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(hird tableau- X 1 = 14

X = 66

!S 1 = 0, S = 0, +1 = 0, + = 0#

2ost = 91 at optimal solution

M7%. (his problem is infeasible. ll " 5 * " are "ero or negative, but an artificial variableremains in the basis.

M7%*. t the second iteration, the following simplex tableau is found-

C j →

↓

Solution

Mix

) " "


X 1 1 51 1

0 1

0 S 0 0 1

1

* " 5 6 0

" 5 * " 0 7 56 0t this point, X should enter the basis next. ut the two ratios are 1C51 = negative and C0 =undefined. Since there is no nonnegative ratio, the problem is unbounded.

M7%/. a. (he optimal solution using simplex is X 1 = 6, X = 0. >G+ = 9. (his is illustrated inthe problems final simplex tableau-

(ableau for %roblem 3/8a

C j →

↓

Solution

Mix

% " " M

#uantit$ X 1 X % S 1 S % A1

0 S 1 0 /

6

1 51

X 1 1 6 1 0 0 6

* " 6 1 0 0 9

" 5 * " 0 0 51 0 5 M

b. (he variable X has a " 5 * " value of 90, indicating an alternative optimal solutionexists by inserting X into the basis.

c. (he alternative optimal solution is found in the tableau in the next column to be X 1 =

6/ = 0.4, X =

1/ = 1./, >G+ = 9.

(ableau for %roblem 3/8c

C j →

↓

Solution

Mix

% " "

M #uantit$ X 1 X % S 1 S % A1

6 X 0 1 1/

1

− /

1/

X 1 1 0 − 1

1 − 1

/ 6/

6/

* " 6 16

0 0 9

" 5 * " 0 0 − 1

60 5 M



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d. (he graphical solution is shown below.

lternative optimums at a and b) * = 9.

M7%). (his problem is degenerate. $ariable X should enter the solution next. ut the ratios areas follows-

6

8 row 8

1 X =

1

1 row unacceptable

6 X =

−

108

S =

Since X 6 and S are tied) we can select one at random, in this case S . (he optimal solution isshown below. +t is X 1 = /, X = 8, X 6 = 0, profit = 91//.

C j →

↓

Solutio

n

Mix

) / " " "

#uantit$ X 1 X % X S 1 S % S

98 X 6 0 0 1 1

− 1

/

0

9 X 1 1 0 0 6

6 −

1 /

96 X 0 1 0 1

1

− 1

8

* " 6 8 16 ; 16 91//

" 5 * " 0 0 0 516 5; 516



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M7%7. 3inimum cost = 80 X 1 10 X /8 X 6 0S 1 M+1 M+

sub


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M7%(. X 1 = number of kilograms of brand added to each batch

X = number of kilograms of brand added to each batch

3inimi"e costs = 7 X 1 18 X 0S 1 0S M+1 M+

sub


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+nitial tableau-

C j →

↓

Solution

Mix

+%" +%* +" +" M M

#uantit$ X 1 X % S 1 S % A1 A%

M +1 1 1 51 0 1 0 60

M + 1 0 51 0 1 40

* " M 6 M 5 M 5 M M M /0 M

" 5 * " 5 M 0 56 M 4 M M 0 0

Second tableau-

C j →

↓

Solution

Mix

+%" +%* +" +" M M


M +1 1

0 51 1

1 − 1

10

94 X 1

1 0 − 1

0

1

0

* " 1 M 1

4 5 M 1 M 5 1

0 − 1

M 110 M 4;0

" 5 * " −

1 M 1

0 M −

1 M 1

0 6 M 5 1

Hinal tableau-

C j →

↓

Solution

Mix

+%" +%* +" +" M M


90 X 1 1 0 5 1 51 0

94 X 0 1 1 51 51 1 10

* " 0 4 51 54 1 4 940

" 5 * " 0 0 1 4 M 5 1 M 5 4

X 1 = 0, X = 10, cost = 940

M7". 3aximi"e profit = 7 X 1 1 X

sub


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Second tableau-

C j →

↓

Solution

Mix

+! +1% +" +"


90 S 1 1

0 1 − 1

4

91 X 1 1 0

1

* " 1 0 9/

" 5 * " 6 0 0 5

Hinal tableau-

C j →

↓

Solution

Mix

+! +1% +" +"


94 X 1 1 0 51 ;

91 X 0 1 51 1

* " 7 1 6 97

" 5 * " 0 0 5 56

X 1 = ;, X = , profit = 97

M71. 3aximi"e profit = ; X 1 X 14 X 6

sub


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Hinal tableau-

C j →

↓

Solution

Mix

+( +) +1* " M

#uantit$ X 1 X % X S 1 A1

914 X 6 8/

0 1 6/

− 114

40/

9 X −1

/ 1 0 −

/ 6

14 10

/

* " 4

/ 14

60

/ +

/ 98;

/

" 5 * " 51.1 0 0 − 60

/ 5 M 5 /

1 6

10 40 0, , , profit 98;// /

X X X = = = =

!which is X 1 = 0, X = 1/.14, X 6 = 64.7, profit = 98;.;#

M7%. a.

X 1 = number of deluxe onebedroom units converted

X = number of regular onebedroom units converted

X 6 = number of deluxe studios converted

X 4 = number of efficiencies converted

Gb


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b. 3aximi"e profit = ;,000 X 1 ,000 X 8,000 X 6 6,800 X 4 0S 1 0S 0S 6 0S 4 0S 8 0S 0S / 0S ; 5 M+1 5 M+

sub


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Hrom the S 1 column, we re*uire that

50 5 ∆ ≤ 0 or ∆ ≥ 510

Since the ∆ ≥ − 1

is more binding, the range of optimalit! is

9 /1 ≤ " !for X 1# ≤∞

b. T#e range of insignificance is

5 ∞ ≤ " !for X #≤ 940

c. Gne more unit of the first scarce resource is worth 90, which is the shadow price inthe S 1 column.

d. nother unit of the second resource is worth 90 because there are still 00 unusedunits !S = 00#.

e. (his change is within the range of insignificance, so the optimal solution would not

change. +f the 60 in the " row were changed to 68, the " 5 * " would still be positive,and the current solution would still be optimal.

f. (he solution mix variables and their values would not change, because 91 is withinthe range of optimality found in part a. (he profit would increase by 10!# = 60, sothe new maximum profit would be 1,00 60 = 1,70.

g. (he righthand side could be decreased by 00 !the amount of the slack# and the profit would not change.

M7/. a. (he shadow prices are- 6./8 for constraint 1B .8 for constraint B and 0 forconstraint 6. (he shadow price is 0 for constraint 6 because there is slack for this

constraint. (his means there are units of this resource that are available but are not being utili"ed. (herefore, additional units of this could not increase profits.

b. Fividing the >?S values by the coefficients in the S 1 column, we have 6/.8C0.18 =600 so we can reduce the righthandside by 600 unitsB and 1.8C!50.18# = 5100, so wecan increase the righthandside by 100 units and the same variables will remain in thesolution mix.

c. (he righthandside of this constraint could be decreased by 10 units. (he solutionmix variable in this row is slack variable S 6. (hus, the righthandside can be decreased by this amount without changing the solution mix.

M7). a. %roduce 1; of model 10 and 4 of model ?6.

b. S 1 represents unused or slack time on the soldering machineB S represents unused orslack time in the inspection department.

c. IesJthe shadow price of the soldering machine time is 94. 2lapper will net 91.80for every additional hour he rents.

d. KoJthe profit added for each additional hour of inspection time made available isonly 91. Since this shadow price is less than the 91./8 per hour cost, 2lapper will lower his profit by hiring the parttimer.



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M77. a. (he first shadow price !in the S 1 column# is 98.00. (he second shadow price !in the S column# is 918.00.

b. (he first shadow price represents the value of one more hour in the paintingdepartment. (he second represents the value of one additional hour in the carpentrydepartment.

c. (he range of optimality for tables ! X 1# is established from (able 3/6/c.

58 5 6C∆ ≤ 0 or ∆ ≥ 56.666 from S 1 column

518 1C∆ ≤ 0 or ∆ ≤ 60 from S column

?ence the " for X 1 must decrease by at least 96.66 to change the optimal solution. +t must

increase by 960 to alter the basis. (he range of optimality is 9./ ≤ " ≤ 9100.00 for X 1.

d. (he range of optimality for X . See (able 3/6/d.

58 ∆ ≤ 0 or ∆ ≤ .8 from S 1 column

518 5∆

≤

0 or∆

≥

58 from S column(he range of optimality for profit coefficient on chairs is from 968 != 80 5 18# to 98.80 != 80 .8#.

e. >anging for first resourceJpainting department

#uantit$ S 1 3atio

60 6

0

40 5 50

(hus the first resource can be reduced by 0 hours or increased by 0 hours without affecting thesolution. (he range is from ;0 to 10 hours.

f. >anging for second resourceJcarpentry time.

#uantit$ S % 3atio

60 − 1

50

40 1 40

>ange is thus from 00 hours to 600 hours !or 40 5 40 to 40 0#.

Ta-le 0or Pro-lem M77c

C j →

↓

Solution

Mix7" 2 /" " "


/0 ∆ X 1 1 0

6

− 1

60

80 X 0 1 5 1 40

* " /0 ∆ 808

6 ∆ 18 5

1 ∆

94,100 60∆

" 5 * " 0 0 58 5

6 ∆ 518

1 ∆



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Ta-le 0or Pro-lem M77d

C j →

↓

Solution

Mix

7" /" 2 " "


/0 X 1 1 0 6

− 1

60

80 ∆ X 0 1 5 1 40 * " /0 80 ∆ 8 5 ∆ 18 ∆ 94,100 40∆

" 5 * " 0 0 58 ∆ 518 5 ∆

M7(. Kote that artificial variables may be omitted from the sensitivity analysis since theyhave no physical meaning.

a. >ange of optimality for X 1 !phosphate#-

C j →

↓

Solution

Mix+/ 2 +) +" +"


90 S 0 0 51 1 880

98 ∆ X 1 1 0 1 0 600

9 X 0 1 51 0 /00

* " 8 ∆ 51 ∆ 0 98,/00 600∆

" 5 * " 0 0 1 5 ∆ 0

1 5 ∆ ≥ 0 or ∆ ≤ 1

+f the " value for X 1 increases by 91, the basis will change. ?ence 5 ∞ ≤ " !for X 1# ≤ 9.

>ange of optimality for X !potassium#-

C j →

↓

Solution

Mix

/ ) 2 " "


0 S 0 0 51 1 880

8 X 1 1 0 1 0 600

∆ X 0 1 51 0 /00

* " 8 ∆ 51 5 ∆ 0 98,/00 /00∆

" 5 * " 0 0 1 ∆ 0

1 ∆ ≥ 0 or ∆ ≥ 51

+f the " value for X decreases by 91, the basis will change. (he range is thus 98 ≤ " !for X # ≤

∞.

b. (his involves righthandside ranging on the slack variables S 1 !which represents number of pounds of phosphate under the 600pound limit#.

#uantit S % 3atio880 51 5880600 1 600/00 51 5/00

(his indicates that the limit may be reduced by 600 pounds !down to "ero pounds# withoutchanging the solution.

(he *uestion asks if the resources can be increased to 400 pounds without affecting the basis.(he smallest negative ratio !5880# tells us that the limit can be raised to ;80 pounds withoutchanging the solution mix. ?owever, the values of X 1, X , and S would change. X 1 would now be



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400, X would be 00, and S would be 480.M7!. 3inimi"e cost = 41 1 ;1

sub


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M7*/. (he dual is

maximi"e * = 101 1 1181 111 6

sub


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g. (able profit range- 941./ to 910

2hair profit range- 91.;/ to 9;4.

M7*(. %rintout 1 illustrates the model formulation.

a. %rintout provides the optimal solution of 97,;6. Gnly the first product !18;# isnot produced.

b. %rintout also lists the shadow prices. (he first, for example, deals with steel alloy.(he value of one more pound is 9./1.

c. (here is no value to adding more workers, since all 1,000 hours are not yetconsumed.

d. (wo tons of steel at a total cost of 9;,000 implies a cost per pound of 9.00. +tshould be purchased since the shadow price is 9./1.

e. %rintout 6 illustrates that profit declines to 9;,; with the change to 9;.;;.

f. %rintout 4 shows the new constraints. %rofit drops to 97,6;0, and none of the E products remain. %reviously, only 18; was not produced.



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Module 7 - Linear Programming, The Simplex Method - Answers

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