Module 7 - Linear Programming, The Simplex Method - Answers

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7 Linear Programming: The Simplex Method Teaching Suggestions Teaching Teaching Suggestion Sugges tion M7.1: Meaning of Slack Variables. Variables.

Slack variables have an important physical ph ysical interpretation and represent a valuable commodity, such as unused labor, machine time, money, space, and so forth. Teaching Teaching Suggestion Sugges tion 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 $ariables with values of "ero are called nonbasic and those with non"ero values are said to be basic. Teaching Teaching Suggestion Sugges tion 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 because they will have a clear physical ph ysical meaning. )arn )arn the students that in subse*uent tableaus the interpretation is the same but will not be as clear because we are dealing with marginal rates rates of substitution. Teaching Teaching Suggestion Sugges tion M7.4: Hand alculations in a Simplex Tableau. Tableau.

+t is almost impossible to walk through even a small simplex problem !two v ariables, two constraints# without making at least one arithmetic error. (his (his can be maddening madden ing for students who know what the correct solution should be but cant reach it. )e suggest two tips1. Encourage Encourage students students to also solve the assigned assigned problem problem by computer and to re*uest re*uest the detailed simplex output. (hey can now check che ck their work at each iteration. . Stress Stress the importance importance of interpretin interpreting g the numbers in the tableau tableau at each iteration. iteration. (he 0s and 1s in the columns of the variables in the solutions are arithmetic checks and balances at each step. Teaching Teaching Suggestion Sugges tion M7.5: Infeasibilit! Is a Ma"or Problem in Large LP Problems.

s we noted in (eaching Suggestion /., students should be aware that infeasibility commonly arises in large, realworldsi"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 is often a management issue.

M7-1

op!rig#t $%&'( Pearson) Inc.

Alternative Examples Example /.1 !see 2hapter / of Alternative Example M7.1: Simplex Solution to lternative Example Solutions 3anual for formulation and graphical solution#. 1st Iteration C j → ↓

Solution Mix

 X 1

0 0

1 1 0 6

S 1 S  * "  " 5 * " %nd Iteration C j → ↓

! X %

" S 1

4  0 7

1 0 0 0

Solution Mix

 X 1

! X %

7

X 

1

0

S 

1

* "

7

 " 5 * "

6

4  4 4

" S % #uantit$ 4 1 0

0 1 0 0 " S 1

" S %

1

1

0

#uantit$ 

0

− 1

1

4

7

7

0

84

0

− 74

4

4

0

(his is not an optimum solution since the X 1 column contains a positive value. 3ore profit remains !9

6

4 per

:1#.

rd&'inal Iteration C j → ↓

Solution Mix

 X 1

! X %

" S 1

7

X 

0

1

1

6

X 1 * "

1 6

0 7

−1



6

6

 " 5 * "

0

0

− 6





" S %

− 1 

#uantit$ 4

; 0

− 6

(his is an optimum solution since there are no positive values in the  " 5 * " row. (his says to make 4 of item : and ; of item :1 to get a profit of 90.

M7-2


initial simplex tableau, given the following following two Alternative Example M7.%: Set up an initial constraints and ob
Sub
6 X 1   X  ≥  (he constraints and ob
Solutio n Mix M +1 M + * "  " 5 * " (he second tableauC j → ↓

( X 1

 6 8 M ; 5 8 M ( X 1



Solutio n Mix X 

M

+ * "  " 5 * "

) X %

" S 1

4   M  5  M

" S %

51 0 5 M M

) X %

" S 1

M A1

0 51 5 M M

1 0 M 0

" S %

M A1

M A%

0 1 M 0 M A%

1

− 14

0

1



0

1

51

− 1

6  M



8 5  M

0

1



−

6



 1

+



5 M

6

M

−



−

0

#uantit$ 

1



M

1   M

4

1

#uantit$ ;  14 M

 M

M 6



−

1

 M

6



+

6

0



M

(he third and final tableauC j → ↓

( X 1



Solutio n Mix X 

) X %

" S 1

" S %

0

1

− 6;

1

;

X 1

1

0

1

− 1

− 14

1

* "

;



− 14

− 8

1

8

 " 5 * "

0

0

1

8

M A1

M A%

#uantit$

4

4

6

4



− 14

;

4

5 M 5

1

4

 minimal, minimal, optimum cost of 1/ can be achieved by using 1 of a type :1 and

M7-3




1/

 8



1



5 M 5 6

6



of a type :.

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

3aximi"e %rofit = 96 X 1  97 X  Sub
(he optimal solution was X 1 = ;, X  = 4. %rofit = 90. @sing software !see the printout#, we can perform a variety of sensitivity analyses on this solution.

M7-4


assembles both laptop and desktop personal Alternative Example M7.*: Aevine 3icros assembles computers. Each laptop yields 910 in profitB each desktop 900. (he firms A% primal is3aximi"e profit = 910 X 1  900 X  sub3 chips 1 X 1   X  ≤ 910 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)" X 1

+% +%"" X %

" S 1

" S %

" S 

00

X 

0

1

1

− 17

0

#uantit$ ;

10

X 1

1

0

51



0

4

0

S 6 * "

0 1 0

0  00

 40

5 16 1 6

1 0

4 9 ,40

 " 5 * "

0

0

540

−16 1 6

0

7

or X 1 = 4, X  = ;, S 6 = 94 in slack royalty fees paid %rofit = 9,40Cday ?ere is the dual formulation formulation3inimi"e * = = 0 !1  10; !  10 !6 sub
Solution Mix

%" y1

1"( y%

1% 1 %" y

" S 1

10;

!

0

1



− 7

1 0 0

0 1 0; 0

5 7 4

1 54 4

0

!1 * "  " 5 * " (his means

" S % 1

51 5; ;

labo r hour = 940 !1 = marginal value of one more labor ! = marginal value of one more >3 chip = 916.66 !6 = marginal value of one more 91 in royalty fees = 90

M7-5

7


#uantit$

16 1 6 40 9  , 4 0

Solutions to ,iscussion #uestions and Pro-lems M71. (he purpose of the simplex method is to find the optimal solution to A% problems in a systematic 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 used only when two variables are in modelB mod elB simplex can handle any dimensions. d imensions. !# Draphical method must evaluate all corner points !if the corner point method is used#B simplex checks a lesser number of corners. !6# Simplex method can be automated and computeri"ed. !4# Simplex method involves use of surplus, slack, and artificial variables but provides useful econ omic data as a by product.

Similarities- !1# oth methods find the optimal solution at a corner point. !# oth methods re*uire a feasible region and the same problem structure, that is, ob
(he graphical method is preferable when the problem has two variables and only two or three constraints !and when no computer is available#. M7. Slack ariables convert ≤ constraints into e*ualities for the simplex table. (hey represent a *uantity of unused resource and have a "ero coefficient in the ob
Surplus ariables convert ≥ constraints into e*ualities and represent a resource usage above the minimum re*uired. (hey, too, have a "ero coefficient in the ob
ph ysical meaning but are used with the constraints that are = or +rtificial ariables have no physical ≥. (hey carry a high coefficient, so they are *uickly removed from the initial solution. solution# is always e*ual to the M7*. (he number of basic variables !i.e., variables in the solution# number of constraints. So in this case there will be eight basic variables.  nonbasic variable is one that is not currently in the solution, that is, not listed in the solution mix column of the tableau. +t should be noted that while there will be eight basic variables, the valu es of some of them may be "ero. M7/. Piot column- Select the variable column with the largest positive  " 5 * " value !in a maximi"ation problem# or smallest negative  " 5 * " value !in a minimi"ation problem#.

Piot ro/- Select the row with the smallest *uantitytocolumn ratio that is a n onnegative number. Piot number- Fefined to be at the intersection of the pivot column and pivot row.

problems are *uite similar in the application of the M7). 3aximi"ation and minimi"ation problems simplex method. 3inimi"ation problems usually include constraints necessitating artificial and surplus variables. +n terms of techni*ue, the  " 5 * " row is the main difference. +n maximi"ation problems, the greatest positie  " 5 * " indicates the new pivot columnB in minimi"ation problems, its the smallest negatie  " 5 * ". (he * " entry in the &*uantity' column stands for profit contribution or cost, in maximi"ation and minimi"ation problems, respectively. M77. (he * " values indicate the opportunity cost of bringing one unit of a variable into the solution mix.

M7-6


M7(. (he  " 5 * " value is the net change in the value of the ob
iteration is important M7!. (he minimum ratio criterion used to select the pivot row at each iteration because it gives the maximum number of units of the new variable that can enter the solution. y choosing the minimum ratio, we ensure feasibility at the nex t iteration. )ithout )ithout the rule, an infeasible solution may occur. largest ob
duals ob
maximi"e profit =  1 X X 1    X X  sub
and for a dual of the natureminimi"e cost = 211 1  21  sub
M7-7


M717. a.

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

c. The shadow shadow price price = (increase (increase in profit)/ profit)/(incr (increase ease in rightright-hand hand side side value) value) = !/,00 5 ,400#C!40 5 ;0# = 4,;00C10 = 60 d. )ith the additional change, the the optimal corner point in part  is still still the optimal corner point. %rofit doesnt change. Gnce the righthand side went beyond 40, another constraint prevented any additional profit, and there is now slack for the first constraint. below. M71(. a. See the table below. Ta-le Ta-le 0or Pro-lem M71( C j → ↓

Solution Mix

+!"" X 1

+1/"" X %

+" S 1

+" S %

0 0

S 1 S  * "  " 5 * "

14 10 0 70 0

4 1 0 1 ,80 0

1 0 0 0

0 1 0 0

b.

#uantit$ 6,60 7,00 0

14 X 1  4 X  ≤ 6,60 10 X 1  1 X  ≤ 7,00 X 1, X  ≥ 0

c.

3axi 3axim mi"e i"e prof profit it = 700 700 X 1  1,800 X 

d.

asis is S 1 = 6,60, 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.

%rof %rofit it will will incr increa ease se by ! " 5 * "#!units of variable entering the solution#

M7-8


= !1,800#!;00# = 1,00,000 M71!. 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
 X 1  / X    X 6 5 X 4 5 S   + = 10 tableau in (able (able 3/17b below. below. b. See initial simplex tableau Ta-le Ta-le 0or Pro-lem M71!C j → ↓ 0 5 M 5 M

Solution Mix

S 1 +1 + * "  " 5 * "

".(

".*

1.%

X 1 1 0  5 M 0.;   M

X %  1 / 5; M 0.4  ; M

".1 X  X * 1 8 54 ;  51  M 5/ M 1. 5  M 50.1  / M

"

"

S 1 1 0 0 0 0

S % 0 0 51 M 5 M

M M A1 A% 0 0 1 0 0 1 M 5 M 5 0 0

#uantit$ 18 1 80 /0 10 5170 M

180, +1 = /0, + = 10, all other variables = 0 M7%". Hirst tableauC j → ↓

Solution Mix

90 90

S 1 S  * "  " 5 * " Second tableauC j → ↓

Solution Mix

+ X 1

+/ X %

+" S 1

+" S %

0 6 90 96

1  90 98

1 0 90 90

0 1 90 90

+ X 1

+/ X %

+" S 1

+" S %

0 X  6 S  90 * " 96  " 5 * " (hird and optimal tableau-

1 0 98 90

1 5 98 59 8

0 1 90 90

C j → ↓

98 90

98 96

Solution Mix

+ X 1

+/ X %

+" S 1

+" S %

X  X 1

0 1

1 0

1

0

− 6

96 98 * " 90 90  " 5 * " X 1 = , X  = , S 1 = 0, S  = 0, and profit = 96

96 5 96

M7-9


1

6

91 59 1

#uantit$  1; 90

#uantit$   9 60

#uantit$  

9 6

c. S 1

=

Draphical solution to %roblem 3/0-

M7%1. a.

b. C j → ↓

Solution Mix

1" X 1

0 0

4 S 1 1 S  0 * " 10  " 5 * " (his represents the corner point !0,0#.

( X %

" S 1

" S %

#uantit$

  0 ;

1 0 0 0

0 1 0 0

;0 80 0

c. (he pivot pivot colu column mn is is the 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.

M7-10


e. (he smallest ratio is 0, so 0 units of the entering variable X !X 1# will be brought into the solution. +f the largest ratio had been selected, the next tableau would represent an infeasible solution since the point !80,0# is outside the feasible region. f. (he leaving leaving variable variable is the solution solution mix variable variable in row with with the smallest smallest ratio. ratio. (hus, (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" X 1

( X %

" S 1

" S %

#uantit$

X 1 S  * "  " 5 * " (hird iteration

1 0 10 0

0.8 1.8 8 6

0.8 50.8 .8 5.8

0 1 0 0

0 60 0 0

C j → ↓

Solution Mix

1" X 1

( X %

" S 1

" S %

#uantit$

X 1 X  * "  " 5 * "

1 0 10 0

0 1 ; 0

0.6666 50.1/  5

50.6666 0 . /  5

10 0  0

10 0

10 ;

h.

(he secon second d iterat iteration ion repr represe esents nts the the corner corner point point !0, !0,0#. 0#. (he (he thir third d !and fina final# l# iter iterati ation on represents the point !10,0#. for first first tableau- +1 = ;0 M7%%. asis for + = /8 ! X X 1 = 0, X  = 0, S 1 = 0, S  = 0# Second tableau +1 = 88 X 1 = 8 ! X X  = 0, S 1 = 0, S  = 0, + = 0# Draphical solution to %roblem 3/-

M7-11


(hird tableau- X 1 = 14 X  = 66

!S 1 = 0, S  = 0, +1 = 0, + = 0# 2ost = 91 at optimal solution is infeasible. ll  " 5 * " are "ero or negative, but an artificial variable M7%. (his problem is remains in the basis. the following simplex tableau is is foundM7%*. t the second iteration, the C j → ↓

Solution Mix

) X 1

 X %

" S 1

" S %



X 1

1

51

1

0

#uantit$ 1

0

S 

0

0

1

1



 

 5 6 0  * " 0 7 56 0  " 5 * " t 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. optimal solution solution using using simplex simplex is X 1 = 6, X  = 0. >G+ = 9. (his is illustrated in M7%/. a. (he optimal the problems final simplex tableau(ableau (ableau for %roblem 3/8a C j → ↓

Solution Mix

0

S 1

% X 1 0



X 1

1

 X %

" S 1

/

6

6



1



 

" S % 1

51

#uantit$ 

0

0

6

M A1

 6 1 0 0 9 * " 0 0 51 0 5 M  " 5 * " b. (he variable X  has a  " 5 * " value of 90, indicating an alternative optimal solution exists by inserting X  into the basis. c. (he alternative optimal solution solution is found in the the tableau in the next column to be X 1 = 6

/

= 0.4, X  =

1

/

= 1./, >G+ = 9.

(ableau (ableau for %roblem 3/8c C j → ↓

Solution Mix

6

X 

% X 1 0

 X % 1



X 1

1

0

* "



6

1

 " 5 * "

0

0

− 16

−

M7-12

" S 1

" S %

1



/

1

1

6


1

− 1/

M A1

#uantit$

− /

1

6

6

/

0

0

0

5 M

/

/

9

d. (he graphi graphical cal solu solutio tion n is shown shown below below..

lternative optimums at a and b) * = = 9. ariable X  should enter the solution next. ut the ratios are M7%). (his problem is degenerate. $ariable as follows row X 6 ro X 1

S 

row

8 =8 1 1 6

=

unacceptable

−

10 

=8

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

) X 1

98

Solutio n Mix X 6

 X %

/ X 

" S 1

0

0

1

1

9

X 1

1

0

0

6

96

X 

0

1

0

1

* "  " 5 * "

 0

6 0

8 0

  

16 51 6

M7-13

" S %

− 1

" S  /



#uantit$ 0



− 1

/



− 1

8

; 5;

16 51 6

9 1/ /

6 1


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

sub
Solution Mix

/" X 1

1" X %

7/ X 

" S 1

M A1

M A%

+1 + S 1 * "  " 5 * "

1 0 1 M 5  80 M 

51  0 M 5 M   10

0  0  M 5 M   /8

0 0 1 0 0

1 0 0 M 0

0 1 0 M 0

#uantit$ 1,000 ,000 1,800 6,000 M

Second iterationC j → ↓ M /8

0

Solution Mix

/" X 1

1" X %

7/ X 

" S 1

M A1

M A%

+1 X 6

1 0

51 1

0 1

0 0

1 0

0

S 1 * "

1 M

0 5  /8 M 

0 /8

1 0

0 M

 " 5 * "

5  80 M 

5 8 M 5

0

0

0

1

#uantit$ 1,000 1,000



0

1,800 1,000 M   /8,000

6/ 1  6/ 1 

5 M 5

(hird iterationC j → ↓ 80 /8

0

Solution Mix

/" X 1

1" X %

7/ X 

" S 1

M A1

M A%

X 1 X 6

1 0

51 1

0 1

0 0

1 0

0

S 1 * "

0 80

1 8

0 /8

1 0

51 80

 " 5 * "

0

518

0

0

5 80 M 5

1

#uantit$ 1,000 1,000



0

6/

1

800 918,000



6/

1

M 5 5



Fourth and final iteration: C j → ↓ 80 /8

10

Solution Mix

/" X 1

1" X %

7/ X 

" S 1

M A1

M A%

X 1 X 6

1 0

0 0

0 1

1 51

0 1

0

X  * "

0 80

1 10

0 /8

1 518

51 8

 " 5 * "

0

0

0

18

5 8 M 5

1

M7-14




0

6/

1

800 911/,800



6/

M 5 5

= 911/,800 X 1 = 1,800, X  = 800, X 6 = 800, * =

#uantit$ 1,800 800

1



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
Solution Mix

+! X 1

+1/ X %

+" S 1

+" S %

M A1

M A%

+1 + * "  " 5 * "

1 1  M 5 M   7

 4  M 5 M   18

51 0 5 M M

0 51 5 M M

1 0 M 0

0 1 M 0

#uantit$ 60 ;0 110 M

First iteration: C j → ↓

Solution Mix

+! X 1

+1/ X %

+" S 1

+" S %

18

X 

1

1

− 1

0

M

+ * "

51



18

0 18

51 5 M

 " 5 * "

6



 5 M



 M

0

− 18  18

  M

 5

 M

M

M A1 1

M A%

1 M

0 8  0 M



5 18

0

#uantit$ 18

 5

 M 18

6 M 5 5

0



Second iterationSolution C j → Mix ↓ 18 X 

+! X 1

+1/ X %

+" S 1

+" S %

M A1

1

1

0

− 14

0

1

0

− 1

0

1

− 1

51

1

18

4

− 18 4

0

18

0

0

S 1

4

* "

18

 " 5 * "

1

4 4

18

4

(hird and final iteration X 1 = 0 kg, X  = 0 kg, cost = 9600 M7%!. X 1 = number of mattresses

X  = number of box springs

3inimi"e cost = 0 X 1  4 X  sub
M7-15


M

M A%

#uantit$ 0

4

10



9600

4 18

5 M 5

4

+nitial tableauC j → ↓ M M

+%" X 1

+%* X %

+" S 1

+" S %

M A1

M A%

1 1  M 5 M   0

1  6 M 56 M   4

51 0 5 M M

0 51 5 M M

1 0 M 0

0 1 M 0

Solution Mix

+1 + * "  " 5 * " Second tableau-

C j → ↓ M

Solution Mix

+1

1

94

X 

1

+%" X 1

* "

1

 " 5 * "

− 1

 

 M  

1

 1 M 

+%* X %

+" S 1

+" S %

M A1

M A%

0

51

1

1

− 1

1

0

− 1

0

1

4

5 M

1

0

M

− 1



 M 5 5

1

 1 M 

0

− 1

0

6

#uantit$ 60 40 /0 M

#uantit$ 10

0



 1 M 

 M 5 5

10 M   4;0

1

Hinal tableauC j → ↓

Solution Mix

+%" X 1

90 94

1 X 1 0 X  0 * " 0  " 5 * " X 1 = 0, X  = 10, cost = 940

+%* X %

+" S 1

+" S %

M A1

M A%

0 1 4 0

5 1 51  1

1 51 54 4

 51 1 5 1 M 5

51 1 4 5 4 M 5

#uantit$ 0 10 94 0

M7". 3aximi"e profit = 7 X 1  1 X 

sub
+nitial tableauC j → ↓

90 90

Solution Mix

+! X 1

+1% X %

+" S 1

+" S %

S 1 S  * "  " 5 * "

1 1 0 7

1  0 1

1 0 0 0

0 1 0 0

M7-16


#uantit$ 10 1 90

Second tableauC j → ↓

Solution Mix

+! X 1

+1% X %

+" S 1

+" S %

90

S 1

1

0

1

− 1

91

X 

1

1

0

1

 

#uantit$ 4





* "  " 5 * "

 6

1 0

0 0

 5

Solution Mix

+! X 1

+1% X %

+" S 1

+" S %

1 0 7 0

0 1 1 0

 51  5

51 1 6 56

9/


94 91

X 1 X  * "  " 5 * " X 1 = ;, X  = , profit = 97

#uantit$ ;  97

M71. 3aximi"e profit = ; X 1   X   14 X 6

sub
+nitial tableauC j → ↓

Solution Mix

+( X 1   5 M ;   M

+) X % 1  5 M    M

+1* X  6 4 54 M 14  4 M

Solution Mix

+( X 1

+1* X 

90

S 1

8

+) X % 0

9

X 

1

0 5 M

S 1 + * "  " 5 * " Second tableauC j → ↓

* "  " 5 * "

6 6

 

/

6

1



 0

4 10

M7-17

6


" S 1 1 0 0 0 " S 1 1

0 0 0

M A1

#uantit$ 1 0 4 0 540 M

0 1 5 M 0 M A1

− 1 1



1 5 M 5 5 1

#uantit$ ;0

40 9 40


Solution Mix

+( X 1

914

X 6

8

9

X  * "  " 5 * " X 1

= 0,

X

=

1 0 /

+) X % 0

+1* X  1

− 1/

1

0

4



14

0

0

/

/

51.1 ,

X 6

=

40 /

, profit = 98; 

" S 1 6

/

− / 60

/

− 60 /

M A1

− 6

M7%. a.

X 1 = number of deluxe onebedroom units converted

onebedroo m units converted X  = number of regular onebedroom X 6 = number of deluxe studios converted X 4 = number of efficiencies converted

Gb
/00 X 1

 600 X 4 ≤ 9;,000

 00 X 

 400 X 6

,000 X 1  1,00 X   1,00 X 6  700 X 4 ≤ 948,000 1,000 X 1  400 X 

 700 X 6

 00 X 4 ≤ 917,000

X 1  X 

 X 6

 X 4

≤ 80

X 1  X 

 X 6

 X 4

≥ 8

+ +

X 4 # 

+ X1 + X1

≥ 0.40! X1 + X  ≤ 0./0! X1 + X

X X

+ X6 + X6



X 4 # 

(he last two constraints can be rewritten as0. X 1  0. X  5 0.4 X 6 5 0.4 X 4 ≥ 0 0.6 X 1  0.6 X  5 0./ X 6 5 0./ X 4 ≤ 0

M7-18


40 10

14



5 5 M 5

!which is X 1 = 0, X  = 1/.14, X 6 = 64.7, profit = 98;.;#

1,100 X 1  1,000 X   00 X 6

1 14

+ /

/

sub
#uantit$ / /

98;  / /

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
 800 X 4  S 1

= 68,000

/00 X 1

 600 X 4  S 

= ;,000

,000 X 1  1,00 X   1,00 X 6  700 X 4  S 6

= 48,000

1,000 X 1  400 X 

 00 X 

 400 X 6  700 X 6

 00 X 4  S 4

= 17,000

X 1  X 

 X 6

 X 4

 S 8

= 80

X 1  X 

 X 6

 X 4

5 S   +1 = 8

0. X 1  0. X  5 0.4 X 6 5 0.4 X 4 5 S /  +

=0

0.6 X 1  0.6 X  5 0./ X 6 5 0./ X 4  S ;

=0

M7. a. (he initial formulation is

minimi"e cost = 91 X 1  1; X   10 X 6  0 X 4  / X 8  ; X  sub
5 6 X 6

= 100 ≤ 700

8 X   X 6   X 4  ; X 8  X 1  X 

 X  ≥ 80

 4 X 4

1; X 1 5 18 X  5  X 6 5 X 4  18 X 8

≥ 180

8 X  ≤ 600  X 4   X 8

≥ /0

b. $ariable $ariable X 8 will enter the basis next. !+ts  " 5 * " value indicates the most improvement, that is, / 5 1 M # # $ariable +6 will leave the basis because its ratio !180C18# is the smallest of the three positive ratios. M7*. a. )e change 910 !the  " coefficient for X 1# to 910  ∆ and note the effect on the  " 5 * " row in the table below.

Simplex table for %roblem 3/64 C j → ↓

Solution Mix

+" +" +1" 2 X 1 X % S 1 1 4  X 1 910  ∆ 90 0  5/ S  * " 10  ∆ 40  4∆ 0  ∆ 0  " 5 * " 510 5 4∆ 50 5 ∆ Hrom the X  column, we re*uire for optimality that

510 5 4∆ ≤ 0

or

∆≥

− 1 

M7-19


+" S % 0 1 0 0

#uantit$ 1 0 00 91,00  10∆

Hrom the S 1 column, we re*uire that 50 5 ∆ ≤ 0 Since the ∆ ≥ 9

/

1

− 1  

or

∆ ≥ 510

is more binding, the range of optimalit! is

≤  " !for X 1# ≤ ∞

b. T#e range of insignificance is 5 ∞ ≤  " !for X # ≤ 940 c. Gne more unit of the the first scarce resource resource is worth 90, which is is the shadow price in the S 1 column. d. nother unit of the second resource resource is worth 90 because there there are still 00 unused units !S  = 00#. e. (his change is within within the range of insignificance, so 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 within the range of optimality found in part a. (he profit would increase by 10!# = 60, 6 0, so the new maximum profit would be 1,00  60 = 1,70. g. (he righthand side could be decreased by 00 !the !the amount of the slack# and the profit would not change. constraint B and 0 for M7/. a. (he shadow prices are- 6./8 for constraint 1B .8 for constraint constraint 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. profits. b. Fividing the >?S values by the coefficients in the S 1 column, we have 6/.8C0.18 = 600 so we can reduce the righthandside by 600 unitsB and 1.8C!50.18# = 5100, so we can increase the righthandside by 100 units and the same variables will remain in the solution mix. c. (he righthandside of this this constraint could be decreased by 10 units. (he solution solution mix variable in this row is slack variable S 6. (hus, the righthandside 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 or slack time in the inspection department. c. IesJthe IesJthe shadow price of the soldering machine time is 94. 2lapper will net 91.80 for every additional hour he rents. d. KoJthe profit added for each additional hour of inspection time time made available is only 91. Since this shadow price is less than the 91./8 per hour cost, 2lapper will lower his profit by hiring the parttimer.

M7-20


M77. 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 painting department. (he second represents the value of one additional hour in the carpentry carp entry department. c. (he range of optimality for tables ! X 1# is established from (able (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 optimality for X . See (able (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 S 1

#uantit$ 60

6



3atio 0

40 5 50 (hus the first resource can be reduced by 0 hours or increased by 0 hours without affecting the solution. (he range is from ;0 to 10 hours. h ours. f. >anging for second resourceJcarpentry time. S %

#uantit$ 60

− 1

3atio 50

40 1 40 >ange is thus from 00 hours to 600 hours !or 40 5 40 to 40  0#. Ta-le Ta-le 0or Pro-lem M77c C j → ↓

Solution Mix

/" X % 0

/0  ∆

X 1

7" 2 X 1 1

80

X  * "

0 /0  ∆

1 80

 " 5 * "

0

0

" S 1 

− 1

5

1

6

8

6

58 5

M7-21

" S %



6



∆

18 5

∆

518 


1

 1

#uantit$ 60

∆ 

∆

40 94,100  60∆

Ta-le Ta-le 0or Pro-lem M77d C j → ↓

Solution Mix

/0 80  ∆

X 1

7" X 1 1

/" 2 X % 0

X  * "  " 5 * "

0 /0 0

" S 1

" S %

#uantit$ 60



− 1

1

5

1

40

80  ∆ 0

8 5 ∆ 58  ∆

18  ∆ 518 5 ∆

94,100  40∆

6

from the sensitivity analysis since they M7(. Kote that artificial variables may be omitted from have no physical meaning. a. >ange of optimality for X 1 !phosphate#C j → ↓

Solution Mix

90 98  ∆ 9

S  X 1 X  * "  " 5 * "

15∆≥0

or

+) X % 0 0 1  0

+/ 2 X 1 0 1 0 8∆ 0

+" S 1 51 1 51 51  ∆ 15∆

+" S % 1 0 0 0 0

#uantit$ 8 80 6 00 / 00 98,/00  600∆

∆ ≤ 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

0 8 ∆

S  X 1 X  * "  " 5 * "

/ X 1 0 1 0

8 0

)2 X % 0 0 1

∆ 0

" S 1 51 1 51

" S % 1 0 0

51 5 ∆ 1∆

0 0

#uantit$ 88 0 6 00 /0 0

98,/00  /00∆

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 righthandside ranging on the slack variables S 1 !which represents number of pounds of phosphate under the 600pound limit#. S % #uantit 3atio 8 80 51 58 80 6 00 1 60 0 / 00 51 5/ 00 (his indicates that the limit may be reduced by by 600 pounds !down to "ero pounds# without changing the solution. (he *uestion asks if the resources can be increased to 400 pounds pou nds without affecting the basis. (he smallest negative ratio !5880# tells us that the limit can b e raised to ;80 pounds without changing the solution mix. ?owever, the values of X 1, X , and S  would change. X 1 would now be

M7-22


400, X  would be 00, and S  would be 480. M7!. 3inimi"e cost = 41 1  ;1  sub
(he dual of the dual is the original primal. M7*". 3aximi"e profit = 801 1  41 

sub
maximi"e profit = 0.8 X 1  0.4 X  primal constraints-  X 1  1 X  ≤ 10  X 1  6 X  ≤ 40 X 1, X  ≥ 0

primal solution- X 1 = 60, X  = 0, profit = 967 M7*. 3aximi"e profit =

10 X 1 8 X   61 X 6

 ; X 4  1/ X 8  1 X 8 ≤ ;

sub
≤ 86

 X  5 X 6  8 X 4   X 8

X  X 1

 8 X 6

5 X 8

X 1, X , X 6, X 4, X 8

≤ /0 ≤ 1; ≥ 0

variable S 6, still has  hours of unused time. M7**. a. 3achine 6, as represented by slack variable b. (here is no unused time when the optimal solution is reached. ll ll three slack variables have been removed from the basis and have "ero values. c. (he shadow price of the third third machine is the value of the dual variable in column . ?ence an extra hour of time on machine 6 is worth 90.8. d. For each extra hour of time made available at no cost on machine 2, profit will increase by $0.786 (the dual price/value or shadow price). Thus 10 hours of time will be worth $7.86.

M7-23


M7*/. (he dual is

maximi"e * = = 101 1  1181   111 6 sub
;1 1  41   71 6 ≤ 6 41 1  1   41 6 ≤ 1; 1 1, 1 , 1 6 ≥ 0

1 1 = 9.0/ is the price of each test 1 1  = 91.6 is the price of each test  1 6 = 90

is the price of each test 6 @sing the dual obounded to profit = 96//8.0. b. Kot all resources are used. Shadow prices indicate that carpentry hours and painting hours are not fully used. lso, the 40table maximum is not reached. c. (he shadow prices relate to the five constraints- 90 value to making more carpentry and painting time availableB 96.6; is the value of additional inspectionCrework hoursB 91.0 is the value of each additional foot of lumber made available. d. 3ore lumber should be purchased pu rchased if it costs less than the 91.0 shadow price. 3ore carpenters are not needed at any price. e. Hlair has a slack !L4# of ;.08 hours available daily in the painting department. +t can spare this amount. f. 2arpentry hours range- 1 to infinity. infinity. %ainting hours range- 7 to infinity. +nspectionCrework hours range-

17 1 

to 41.

M7-24


g. (able profit range- 941./ to 910 2hair profit range- 91.;/ to 9;4. illustrates the model formulation. M7*(. %rintout 1 illustrates a. %rintout  provides the optimal solution of 97,;6. Gnly the the first product !18;# is not produced. b. %rintout  also lists the shadow prices. (he (he first, for example, deals with steel alloy. 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 yet consumed. d. (wo (wo tons of steel at a total cost of 9;,000 implies a cost per pound of 9.00. +t should be purchased since the shadow price is 9./1. e. %rintout 6 illustrates illustrates that profit declines declines to 9;,; with the change to 9;.;;. f. %rintout 4 shows the new constraints. %rofit drops to to 97,6;0, and none of the E products remain. %reviously, only 18; was not produced. produced.

M7-25


M7-26


M7-27


M7-28


Module 7 - Linear Programming, The Simplex Method - Answers

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