Chapter Three -Linear Programming CHAPTER THREE LINEAR PROGRAMMING PROBLEM AND MODELS (LPPM) 2.1.CHAPTER OBJECTIVES After completing this chapter, you should be able to: •
Explain what Linear Programming is;
•
Explain what is meant by the term constrained optimization; optimization;
•
List and briefly explain the components and assumptions of Linear Programming models;
•
Recogn Recognize ize proble problems ms that that can be soled soled using using Linear Linear Progra Programm mming ing !od !odels els and formulate linear programming models;
•
Explain the approaches approaches to soling Linear Programming Programming !odels; !odels;
•
"ole problems using #raphical !ethod for soling Linear Programming problems;
•
"ole problems using "implex !ethod for soling Linear Programming Problems;
•
$se sensitiity analysis to ealuate a change in the R%" of a constraint and changes in the alue of an ob&ectie function coefficient'
2.2.CHAPTER INTRODUCTION Linear Linear Programmi Programming ng is one of the most ersatil ersatile, e, powerful powerful and useful useful techni(ue techni(uess for ma)ing manageria manageriall decisions decisions'' Linear Linear programm programming ing techni(ue techni(ue may be used for soling soling broad range of problems arising in business, goernment, goernment, industry, industry, hospitals, libraries, libraries, etc'
Linear programming *LP+ model enables users to find optimal solutions to certain problems in which the solution must satisfy a gien set of re(uirements or constraints' t is a model used for optimum allocation of scarce or limited resources to competing products or actiities under such assumptions as certainty, linearity, fixed technology, and constant profit per unit' As a decision ma)ing tool, it has demonstrated its alue in arious fields such as production, finance, mar)eting, research and deelopment and personnel management' -etermination of optimal product mix *a combination of products, which gies maximum profit+, transportation schedules, Assignment problem and many more are the techni(ues managers use to ma)e optimal decisions'
1
OPERATIONS OPERATIONS RESEARCH
TEACHING MATERIAL MATERIAL
Chapter Three -Linear Programming n this chapter, you will learn about linear programming models where emphasis will be gien to such issues as familiarization of the model, problem recognition and formulation using the model, soling linear programming problem for feasibility and optimality, optimality, and post optimality analysis'
2.3. LINEAR PROGRAMMING PROGRAMMING PROBLEM AND MODELS 2.3.1. Meaning an De!ini"i#n !anagers encounter decision ma)ing situations in which the set of acceptable solutions is some how restricted' .he restrictions may be imposed internally or externally' An internal restriction might be the amount of raw materials, aailability of labor time, machine time, technical re(uirements and budgets that a department has to produce its products' An external restriction might be labour regulations *e'g' safety e(uipment, training re(uirements oertime+ that limit the options open to decision ma)er' .he restrictions are referred to as constraints' .he goal in linear programming is to find the best solution gien the constraints imposed by the problem; hence it is constrained constrained optimization' optimization' Linear Linear Programmi Programming ng is one of the most ersatil ersatile, e, powerful powerful and useful useful techni(ue techni(uess for ma)ing manageria manageriall decisions decisions'' Linear Linear programm programming ing techni(ue techni(ue may be used for soling soling broad range of problems arising in business, goernment, industry, industry, hospitals, libraries, etc wheneer we want to allocate the aailable limited resources for arious competing actiities for achieing our desired ob&ect ob&ectie ie'' As a decisi decision on ma)ing ma)ing tool, tool, it has demons demonstra trated ted its alue alue in ario arious us fields fields such such as production, finance, finance, mar)eting, research and deelopment deelopment and personnel management' management' A model, which is used for optimum allocation of scarce or limited resources to competing products or actiities under such assumptions as certainty, linearity, fixed technology, and constant profit per unit, is linear programming' t is the mathematical representation of LP problems that has been deeloped to help management in decision ma)ing, inoling the efficient allocation of scarce resources to achiee an optimization ob&ectie'
2
OPERATIONS OPERATIONS RESEARCH
TEACHING MATERIAL MATERIAL
Chapter Three -Linear Programming n this chapter, you will learn about linear programming models where emphasis will be gien to such issues as familiarization of the model, problem recognition and formulation using the model, soling linear programming problem for feasibility and optimality, optimality, and post optimality analysis'
2.3. LINEAR PROGRAMMING PROGRAMMING PROBLEM AND MODELS 2.3.1. Meaning an De!ini"i#n !anagers encounter decision ma)ing situations in which the set of acceptable solutions is some how restricted' .he restrictions may be imposed internally or externally' An internal restriction might be the amount of raw materials, aailability of labor time, machine time, technical re(uirements and budgets that a department has to produce its products' An external restriction might be labour regulations *e'g' safety e(uipment, training re(uirements oertime+ that limit the options open to decision ma)er' .he restrictions are referred to as constraints' .he goal in linear programming is to find the best solution gien the constraints imposed by the problem; hence it is constrained constrained optimization' optimization' Linear Linear Programmi Programming ng is one of the most ersatil ersatile, e, powerful powerful and useful useful techni(ue techni(uess for ma)ing manageria manageriall decisions decisions'' Linear Linear programm programming ing techni(ue techni(ue may be used for soling soling broad range of problems arising in business, goernment, industry, industry, hospitals, libraries, etc wheneer we want to allocate the aailable limited resources for arious competing actiities for achieing our desired ob&ect ob&ectie ie'' As a decisi decision on ma)ing ma)ing tool, tool, it has demons demonstra trated ted its alue alue in ario arious us fields fields such such as production, finance, finance, mar)eting, research and deelopment deelopment and personnel management' management' A model, which is used for optimum allocation of scarce or limited resources to competing products or actiities under such assumptions as certainty, linearity, fixed technology, and constant profit per unit, is linear programming' t is the mathematical representation of LP problems that has been deeloped to help management in decision ma)ing, inoling the efficient allocation of scarce resources to achiee an optimization ob&ectie'
2
OPERATIONS OPERATIONS RESEARCH
TEACHING MATERIAL MATERIAL
Chapter Three -Linear Programming
$ig%&e 2.1.T'e inea& g&a**ing +e*,*#e
2.3.2. C#*#nen"- #! LP *#eA.O+e/"i0e $%n/"i#n •
is the goal or ob&ectie of a management, stated as intent to maximize or to minimize some important (uantity such as profits or costs'
B' De/i-i#n Va&ia+eVa&ia+e•
represent the un)nown alues to be soled by the decision ma)ers'
•
are the ariables whose alues are un)nown and are searched'
C.Pa&a*e"e&- : •
are fixed alues that specify the impact of one unit on the ob&ectie function and the constraint'
D. C#n-"&ain"•
are limitations or restrictions imposed by the problems' t includes: Re-#%&/e /#n-"&ain"- are restrictions that should be clearly identifiable and measurable in (uantitatie terms, which arise from limitation of aailable resources' Examples of limited resources are:
•
Plant capacity
•
Raw materials availability
•
Labor power
•
Market demand, etc
constrain aints ts which which are sub&e sub&ect ct to indii indiidua duall decisi decision on Ini0 Ini0i% i%a a /#n-"& /#n-"&ain ain"- "- are constr ariables'
N#nnega"i0i" /#n-"&ain"- are constraints that re(uire the decision ariables not to ta)e on negatie alues'
#enerally spea)ing, a constraint consists of four elements ' .hese are:
I. A &ig'" 'an -ie 0a%e (RHS ): represents (uantity that specifies the limit for that constraint' t must be a constant, not a ariable'
II. An age+&ai/ -ign : represents whether the limit is:
An upper bound * < )that cannot be exceeded
3
OPERATIONS OPERATIONS RESEARCH
TEACHING MATERIAL MATERIAL
Chapter Three -Linear Programming
A lower bound *> )that is the lowest acceptable limit
An e(uality * 4+ that must be met exactly'
III. T'e e/i-i#n 0a&ia+e : to which the constraint applies' IV. Pa&a*e"e&-: .he impact that one unit of each decision ariable will hae on the right hand side (uantity of the constraint'
2.3.3. A--%*"i#n- #! LPP *#e.he following are some important assumptions made in formulating a linear programming model:
A. Linea&i" •
.he /b&ectie 0unction and the constraints must be linear in nature in order to deal with a Linear Programming Problems *LPP+' %ere the term linearity implies proportionality and additiely'
B' Ce&"ain" •
t is assumed that the decision ma)er here is completely certain *i'e', deterministic conditions+ regarding all aspects of the situation, i'e', aailability of resources, profit contribution of the Products, technology, courses of action and their conse(uences etc'
C' -i0i-i+ii" •
t is assumed that the decision ariables are continuous' t means that companies manufacture products in fractional units' 0or example, a company manufactures 1'2 ehicles, 3'1 barrels of oil etc'
D. N#n Nega"i0i" •
ndicate all ariables are restricted to non-negative values *i.e', their numerical alue will be 4 5+'i'e' negatie alues of ariables are unrealistic or meaningless'
2.5. $ORMULATION O$ LPPM 0ormulating linear programming problem models inoles the following steps
1. Identify the decision variables and represent them in terms of X 1 , X 2….. 2. Determine the obective f!nction :
0irst decide whether the problem is maximization or minimization problem'
"econd identify the coefficients of each decision ariable'
f the problem is a maximization problem, the profit per unit for each ariable must be determined'
f the problem is a minimization problem the cost per unit must be determined'
4
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming
!a)e sure the units of measurements of all coefficients in the ob&ectie functions must be the same'
". Identify the constraints
0irst, express each constraint in words'
"econd identify the coefficients of the decision ariables in the constraints; and the R%" alues of the constraints'
-etermine the limits for the constraints i'e' see whether the constraint is of the form *6+, *7 ) or (=+'
8rite the e(uation'
#. $sin% the above information &step 1 to "), b!ild the model.
Note •
The coecients of the variables in the Objective Function are called the prot or cost coecients. The e!press the rate at which the value of the Objective Function increases or decreases b including in the solution one unit of each of the decision variables.
•
The coecients of the constraints" variables are called the input# output coecients that indicate the rate at which the given resources are
E6a*e 1 •
A firm that assembles computers and computer e(uipment is about to start production two new microcomputers' Each type of microcomputers will re(uire assembly time, inspection time and stora%e space ' .he amount of each of these resources that can be deoted to the
production of these microcomputers is limited . 'he mana%er of the firm (o!ld lie to determine the *!antity of each microcomp!ter to prod!ce in order to ma+imie the profit %enerated by sales of these microcomp!ters.
Ai"i#na in!#&*a"i#n •
In order to develop a suitable model of the problem, the manager has met with design and manufacturing personnel. As a result of these
meetings
the
manager
has
obtained
the
following
information. T7PE ONE
.9PE .8/
Profit per unit
Bi&& 89
Bi&& :9
Assembly time per unit
hours
5 hours
5
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming nspection time per unit
1 hours
hour
"torage space per unit
3 cubic feet
3 cubic feet
•
.he manager has also ac(uired information on the aailability of company resources' .hese wee)ly resources are:
RE"/$R
A!/$=." A>ALA?LE
Assembly time
55 hours
nspection time
11 hours
"torage space
3@ cubic feet
•
.he manager also met with the firms mar)eting manager and learned that demand for the microcomputers was such that what eer combination of these two types of microcomputers is produced, all of the outputs can be sold'
Re;%i&e: •
0ormulate the LPP! of the problem'
S#%"i#n S"e 1: identify the decision ariable ⇒
the *!antity- amo!nt- !nits
of each microcomp!ter& microcomp!ter type 1 and
microcomp!ter type 2) to be prod!ced
Let614represent (uantity of microcomputer type to be produced
624represent (uantity of microcomputer type 1 to be produced S"e 2: dentify the ob&ectie function •
.he problem is maximization problem, as indicated in the problem *2 th line of the problem+
•
.o write the e(uation both the ob&ectie function and the constraints summarize the gien information in tabular form accordingly
RESOURCE
T7PE ONE
T7PE T
AVAILABILIT7
Assembly time per unit
hours
5 hours
55 hours
nspection time per unit
1 hours
hour
11 hour
"torage space per unit
3 cubic feet
3 cubic feet
3@ cubic feet
Profit per unit
?irr B5
?irr 25
%ence, the ob&ectie function is
!ax C D B5x 25 x1
S"e 3: identify each constraints and write the e(uation 6
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming
Assembly time x 5x1655
nspection time 1x x16 11
"torage space 3x 3x163@
=on negatiity constraint x F x17 5
S"e 5: write the summarized form of the model
In
summar,
the
mathematical
model
of
the
microcomputer problem is% &a! '( )*!+ -*! /ubject to 0!+ +*!1 +** !+ !1 2!+ 2!1 23 !+4 !5 *
E6a*e 2 A firm is engaged in breeding pigs' .he pigs are feed on arious products grown on the farm' n iew of the need to ensure certain nutrient constituents *call them itamins, minerals and proteins+ i" i- ne/e--a& "# +% "=# ai"i#na %/"- -a A an B ' /ne unit of product A contains 3B units of itamins, 3 units of minerals and 15 units of proteins' /ne units of product ? contains B units of itamins, 1 units of minerals and 5 units of proteins' .he minimum re(uirement of itamins, minerals and proteins is 5Gunits, 3Bunits and 55 units respectiely' Product A costs birr 15 per unit and product ? costs birr 5 per unit Re;%i&e
0ormulate the LPP! of the problem
S#%"i#n S"e 1: identify the decision ariables •
Product A and ? to be purchased to prepare the food for pigs Let x represent product A to be purchased H1 represent products ? to be purchased
S"e 2 identify the ob&ectie function ⇒
.he problem is a minimization problem, so the e(uation; 7
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming !in C D 15x 5 x1 ⇒
we can put the information in a tabular form as follows
NUTRIENT CONSTITUENTS
PRODUCT A
PRODUCT B
RE>UIREMENT
>itamins
3B units
B units
5G units
!inerals
3 units
1 units
3B units
Proteins
15 units
5 units
55 units
?irr 15
?irr 5
S"e 3: identify each constraints and write the e(uation
>itamins 3Bx Bx17 5G
!inerals 3x 1 x17 3B
Proteins 15x 5x1755
=on negatiity constraint x F x17 5
S"e 5: write the summarized form of the model
In summar, the mathematical model of the problem is% &in '(*!+ 0*! /ubject to 2)!+ )!5 +*6 2!+ + !5 2) *!+ +*!5+** !+4 !5 *
2.:. APPROACHES TO SOLVE LPPM .here are two methods used to sole LP problems ' #raphical methods 1' "implex methods
2.:.1.G&a'i/a *e"'#8
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming ⇒
s used to find solutions for LPP when the decision ariables of the problem are not greater than two'
Note: Graphical LP is a two-dimensional model P/e%&e .o apply this method, we should follow the following steps'
1. Dra( a %raph incl!din% all the constraints 2. Identify the feasible sol!tion re%ion 3. Identify the corner points and their respective coordinates 4. val!ate the obective f!nction at each corner point and obtain a point on the feasible re%ion that optimies the obective f!nction/optimal sol!tion
5. Interpret the res!lts
E6a*e 1
.o draw the graph first change the ine(uality to e(uality i'e' replace the 6 and 7 sign into D sign' !ax CDB5x 25x1 "ub&ect to 3Bx Bx17 55
3Bx Bx1D 55
3x 1 x17 11
3x 1 x1D 11
15x 5x173@
15x 5x1 D3@
xF x17 5 ⇒
x F x1 D 5
.hen find the x and y intercepts *in our case x and x1 respectiely+ I points where each constraint intersects the axis' .o do so, set x D 5 to find alues for x 1 and set x 1D 5 to find alues for x '
0or the first constraint: x 5x1 D 55
x 5x1 D 55
*5+ 5 x 1 D 55
x 5*5+ D 55
5x1D 55
xD 55 9
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming H1D 5
x D 12
.he x and x1 intercepts are *5, 5+ *12, 5+' "imilarly calculate for the second and third constraints' 0or the second constraint the intercepts are *5, 11+ *, 5+ 0or the third constraint the intercepts are *5, 3+ *3, 5+ #raph the constraints using the intercepts calculated aboe' .he graph is:
⇒
As indicated in the graph, the corner points of the feasible region *the corner points of the shaded region+ are A, ?, <, - and E' .he next
tas) is to find the coordinates of
these corner points, some are determined by obseration and some are through simultaneous e(uation' .hen we test each corner points to find the points that results the optimal solution' .hese actiities are indicated in the following table' E0a%a"ing "'e /#&ne& #in"Point
%ow determined
>alue of the ob&ectie function !ax CDB5x 25x1
H
H1
A
5
5
nspection
B5*5+ 25*5+ D 9
?
5
nspection
B5*+ 25*5+ D889
<
@
"imultaneous e(uation
B5*@+ 25*+ D?59
-
2
G
"imultaneous e(uation
B5*2+ 25*G+ D?99
E
5
5
nspection
B5*5+ 25*5+D:99
"ince the maximum alue *because the ob&ectie function of the problem is maximization+ is J5,the solution is: @14 @24 5 an T'e Ma6i*%* P!i" i- Bi&& ?59.
10
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming 8hen we interpret the result, we state the company should produce %ni"- #! *i//#*%"e& "e #ne and 5 %ni"- #! *i//#*%"e& "e "=# to get a maximum profit of ?irr J5'
E6a*e 2 Ma#' " = 25 ;+G5 1 !t : ;+1 1≤ 31 ;+1 1≤ G1 ; , 1 ≥ 5
N#"e: .ry to sole the problem by yourself, in separate piece of paper before you chec) the answer' E6a*e 3
sets; odel A and B, are produced by a company to maximize profit' .he profit realized is K355 from a .> set of model A, and K125 from that of set ?' .he limitations are: A. aailability of only 5hrs of labor each day in the production department B. a daily aailability of only 2 hrs on machine time C' ability to sale 1 set of model A
Re;%i&e %ow many sets of each model will be produced each day so that the total profit will be as large as possible Resources used per unit onstraints
odel 3
odel 4
&X 1 )
a+im!m 3vailable hrs.
&X 2 )
5abor hr.
1
5
achine hr.
3
2
aretin% hr.
5
1
6rofit
399
2:9
N#"e: .ry to sole the problem by yourself, in separate piece of paper before you chec) the answer'
S#%"i#n
11
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming ' 0ormulation of mathematical modeling of LPP !ax CD355 $125 % "t: 1 $ %6 5 $3 %6 2
566 odel
$ 6 1 $ , % & 5
1'
D 1
3' -raw the graph by intercepts 1 $ % D 5 DD7 *5, 5+ and *15, 5+ $3 %D 2DD7 *5, 2+ and *2, 5+ $
D 1DD7 *1, 5+
$ , % & 5
' dentify the feasible area of the solution which satisfies all constrains' 2' dentify the corner points in the feasible region A *5, 5+, ? *5, 2+, < *1, + and - *1, 5+
6. dentify the optimal solution point
12
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming
!ax CD355 X 1 125 X 2
A
*5, 5+
K5
B
*5, 2+
K3J25
C
(12 11)
D
*1, 5+
K83:9 K3B55
In"e&&e"a"i#n $% units o' product and $$ units o' product s*ould be produced so t*at t*e total pro'it will be +/
E6a*e 5 A manufacturer of light weight mountain tents ma)es two types of tents, 70$537 tent and 8$67 tent' Each 70$537 tent re(uires laborIhour from the cutting department and 3 laborIhours from the assembly department' Each 8$67 tent re(uires 1 laborIhours from the cutting department and laborIhours from the assembly department '.he maximum labor hours aailable per wee) in the cutting department, and the assembly department are 31 and G respectiely' !oreoer, the distributor, because of demand, will not take more t*an 1 8$67 tents per wee)' .he manufacturer sales each 70$537
tents for KB5 and costsK5 per tent to ma)e' 8here as 8$67 tent ales for K15 per tent and costs K35 per tent to ma)e'
Re;%i&e A' 0ormulate the mathematical model of the problem ?' $sing the graphic method, determine how many of each tent the company should manufacture each tent the company should manufacture each wee) so as to maximize its profit <' 8hat is this maximum profit assuming that all the tents manufactured in each wee) are sold in that wee)
S#%"i#n
99999999999999999999999999999999999999999999999999999999999999 5abor ho!rs per tent Department
70$537 &X 1 )
8$67&X 2 )
a+im!m labor/ho!rs
available per (ee
MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM
1
31
Assembly department
3
G
13
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming 8ellin% price per tent
KB5
K15
ost per tent
K5
K35
6rofit per tent
;
=
.he distributor will not ta)e more than 1 "$PER tents per wee)' .hus, the
⇒
manufacturer should not produce more than 1 "$PER tents per wee)' De/i-i#n 0a&ia+e : number of &eg%a& and -%e& tent to be produced per wee)'
Let X 1 D0*e 1 o o' 70$537 tents produced per week . X 2 D0*e 1 o o' 8$67 tents produced per week . X 1 and X 2 are called the decision ariables
Ma#' " =
25 ;+G5 1
!t : 566 odel
;+1 1≤
31
……….Cutting department constraint
;+ 1≤
G1
……….Assembly department constraint ……….Demand constraint
1 ≤ ;1 ; , 1 ≥
orners
5
……….Non-negatiity constraints
oordinates
a+ ?; X1 @=X 2
A
*5, 5+
K5
B
*5, 1+
K@B5
C
*G, 1+
K3B5
D
(29 8)
159
E
*1G, 5+
K55
In"e&&e"a"i#n 0*e manu'acturer s*ould produce and sale 2 70$537 tents and A 8$678 tents to get a ma#imum weekly profit of 1#=.
E6a*e : "uppose that a machine shop has two different types of machines; machine and machine 1, which can be used to ma)e a single product '.hese machine ary in the amount of product
! 2 produced per hr', in the amount of labor used and in the cost of operation'Assume that at ! 1 21
least a certain amount of product must be produced and that we would li)e to utilize at least
"#
the regular labor force' %ow much should we utilize each machine in order to utilize total costs and still meets the re(uirement
S#%"i#n 'he ey information inp!ts in the problem are provided as follo(s 18 B (9 12)
C(12)
OPERATIONS$ea-i+e RESEARCH Regi#n
A(99)
D (29 8)
! 2
TEACHING MATERIAL E (2 9)
"# ! 1 32
14
Chapter Three -Linear Programming 9999999999999999999999999999999999999999999999999999999 Resource used achine 1 * X 1+ achine * X 2+
!inimum re(uired hours
MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM Product producedNhr LaborNhr Bperation ost
15
2
55
1
3
2 999999
2;
"9999 MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM
'he problem is form!lated in the ne+t model Min' " =
12 ;+35 1
!t :
15 ;+;2 1≥ ;55
566 odel
1 ;+3 1≥ ;2 ; , 1 ≥
5
DD7 *5, 15N3+ and *2, 5+
DD7 *5, 2+ and *J'2, 5+
$ %& / the feasible sol!tion space and the corner points are specified on the %raph belo(
9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 orners
oordinates
in?2; X1 @ "X 2
A
*5, 15N3+
155
4
&2.;, "."")
1A2.;
<
*J'2, 5+
GJ'2
9999999999999999999999999999999999999999999999999999999999999 'he optimal sol!tion mi+ is represented by X 1 42.: X 243.33 an Min 4 1A2.;
E6a*e 8 A company owns two flour mills *A and ?+ which hae different production capacities for CI0C, DI$ and 5B grade flour' .his company has entered contract supply flour to
a firm eery wee) with 1, G, and 1 (uintals of CI0C, DI$ and 5B grade 15
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming respectiely' t costs the
Re;%i&e %ow many days per wee) should each mill be operated in order to meet the contract order most economically standardize "ole graphically'
S#%"i#n =o of days per wee) of !il A * X 1+
!inimum flour in !ill ?* X 2+
(uintals
CI0C apacity &in *!intal)
A
2
12
DI$ apacity &in *!intal)
2
2
=
#
12
2#
5B apacity &in *!intal) 7!nnin% cost-day
1
=
Ma#' " = ;55 ;+G55 1 !t :
B ;+1 1≥ ;1 1 ;+1 1≥ G ;
+ ;1 1 ≥
; , 1 ≥
1
5
O"i*a S#%"i#n 0a%e- X 1 D X 2D3 and !in CD "#
Note
$n ma%imi&ation problems' our point o( interest is loo)ing t*e (urt*est point (rom t*e origin.
3.:.2. S#0ing !#& LPP- %-ing "'e Si*e6 *e"'# .he graphical method to soling 566 s proides fundamental concepts for fully understanding the LP process' %oweer, the graphical method can handle problems inoling #n "=# e/i-i#n 0a&ia+e- (-a X 1an X 2).
n the @5s 0eor%e 4.Danti% deeloped an algebraic approach called the 8imple+ ethod, which is an efficient approach to sole applied problems containing numerous constraints; and inoling many ariables that cannot be soled by the graphical method'
16
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming .he simplex method is an I'73'IE or Ostep by step method or repetitie algebraic approach that moes automatically from one basic feasible solution to another basic feasible solution improing the situation each time until the optimal solution is reached at' "imilar to the graphical solution approach, this method can be applied to sole for LPPs of different ob&ecties with arious set of constraints'
SOLVING MA@IMIATION PROBLEMS 1. Ma6i*iFa"i#n 'a0ing a "'e /#n-"&ain"- in !#&* E6a*e 1
S#%"i#n /e%&e
.o sole the problem using the simplex approach, follow the following steps' 8tep 1: $#&*%a"e LPP M#e
n our example it is already done' 8tep 2: S"ana&iFe "'e +e*
8lac Eariables&8): is Nare added to the left hand side of a constraint to conert the
constraint into its standard form' .he alue of the slac) ariable shows %n%-e #& %ne&%"iiFe &e-#%&/e'
8!rpl!s variables&8): isNare addedto the left hand side of a constraints to conert the
constraint into its standard form' .he alue of the surplus ariable shows the excess resources used'
17
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming 3rtificial variables &3): somewhat analogous to slac) ariables in that they are added to
e(uality and a constraints in the same way that slac) ariables are added to a constraints' %oweer artificial ariables hae no physical interpretation, they merely sere as a deice to enable us to use the simplex process' "ince the aboe problem constraints hae all a 6 algebraic sign, we use slac) ariables for standardizing the problem'
Note •
A slac7 variable is alwas added for a 1 constraint to convert the constraint to a standard form.
"lac) ariables represent unused resource or idle capacity' .hus, they dont produce any product and their contribution to profit is ero'
"lac) ariables are added to the ob&ectie function with ero coefficients ' !ax CD B5x 25x15s5s15s3 "ub&ect to x 5x1sD55 1x x1s1D 11 3x 3x1 s3D3@ x, x1 ,s,s1,Fs3D 5
Note • To 8tep " standardi$e an 899, start from the constraint and nall move to the
objective function. ⇒ O+"ain "'e ini"ia -i*e6 "a+ea% :hen ouready add the slac7 variables, should bea according to the e!istence of the .o• ma)e the data for analysis, the simplex itmethod uses table called the simple+ tablea! or constraints the simple+ matri+ ' ;how it is written in the model<.
n constructing the initial simplex tableau, the search for the optimal solution begins at the origin' ndicating that nothing can be produced; .hus, based on this assumption, n# *i//#*%"e& "e#ne an *i//#*%"e& "e "=# i%/e which implies that + 1 ? and + 2?
DD7 + 15 + 2 s1 5 s2 5 s"D 55
DD7 % + 1 + 25s1 s2 5s"D 11
*5+ 5*5+ s1 5 s2 5 s"D 55
1*5+ 55s1 s2 5 s"D 11 18
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming -14 199 Q $n!sed assembly time.
-24 22 Q $n!sed inspection time.
DD7 + 1 3x15s1 5s2 s"D 3@ 3*5+ 3*5+ 5s1 5 s2 s"D 3@ -34 3 Q $n!sed stora%e space.
.herefore, a+ DB561 25 + 2 5 s1 5 s2 5 s" DB5*5+ 25*5+ 5*55+ 5*11+ 5*3@+ D
Note •
In general, whenever there are n variables and m constraints ;e!cluding the non#negativit<, where m is less than n ;m
eqa! to "ero before the solution can be solved algebraicall.
a. Ba-i/ 0a&ia+e
are ariables with nonIzero solution alues
are ariables that are in the basic solution
hae alues in the / row
+. N#n+a-i/ 0a&ia+e-
n o i s are ariables with zero solution alues i t c i e are ariables that are out of the solution ) n d s ( $ e n r l r r # o m o n e a n?; ariables *x62 s1, s2, a nd s"+ a nd m?" constraints *assembly, l i 0rom the aboe example p n $ c a r l i m $ t m e a o s l +space " c constraints+, excluding nonInegatiity' a o * inspection and $storage l o ' c r o P c
.herefore, n/mD2I3D1 ariables* + 1 and + 2+ are set e(ual to zero in the st simplex tableau' .hese are nonIbasic ariables' 3 >ariables * s1, s2, and s"+ are basic ariables *in the st simplex tableau+ because they hae nonIzero solution alues'
Pro*t per $nit row
8tep ": C#n-"&%/" "'e ini"ia -i*e6 "a+ea% Initial simple+ tablea!
OPERATIONS RESEARCH
s e l # a i r a s " n ! m $ TEACHING c MATERIAL l a l o c
n m $ l o c & n t i o i t t $ n a l o $
,onstraint e%$ation rows 19 Gross Pro*t row et Pro*t row
Chapter Three -Linear Programming
?> 8 1
5
B5 X 1
8 2
1
8 "
5
3 5 B5
3 5 25
G
25 X 2 5
5 8 1
5 8 2
5 8 "
F
5
5
55
+1
5
5
11
+2
5 5 5
5 5 5
5 5
3@ 5
+3
8tep #:C'##-e "'e incomin% #& enterin% 0a&ia+e-
Note# ⇒
The entering variable is the variable that has the most the larest
positi"e "al$e in the , - row ;indicator row. ⇒
It is the variable that has the highest contribution to prot per unit.
⇒
=+ in our case is the entering variable. ;>ecause the ma!imum number in the c#$ row is )*<
⇒
The column associated with the entering variable is called 7e or pivot
8tep ;:C'##-e "'e leavin% #& o!t%oin% 0a&ia+e an "'e i0#" ee*en" n this step, we determine the ariable that will leae the solution for X 1' .o identify the leaing ariable, we should calculate the &a"i# first and then we should select the *ini*%* n#n nega"i0e &a"i# '
Note ⇒ ⇒
The leaving variable is the variable that has the smallest replacement ratio. / in our case is the leaving variable. ;>ecause the smallest replacement ratio is
.he piot ++ element is the intersection point between the piot row and piot column; =o 1 in Theexample row associated the entering variable +eplacement +atio ++with ol$tion $antit& is called 7e or pivot row / row in the ⇒ aboe ,orrespondin "al$es in pi"ot col$mn In #%& /a-e 199 42: 5 22 4 11 i- "'e *ini*%* 2 OPERATIONS RESEARCH 3 413 3
20
TEACHING MATERIAL
Chapter Three -Linear Programming
t is interesting to note that the three ratios *12, , 3+ corresponding to the intersections of the constraints with the x*loo) the graphical solution+' =ote that the smallest of the ratios represents the extreme point of the feasible solution space; the other points lie beyond the feasible solution space' %ence by selecting the smallest ratio, the simplex procedure stays within the feasible solution space' t sometimes happens that some of the substation rates for the ariable we want to bring into solution are zero or negatie' 8e dont nee to diide the (uantity alues by a negatie or a zero substitution rates' 8tep A: ⇒
6erform ro( operations.Perform algebraic operations on the P>/.
conert the piot element into and the remaining piot column numbers into 5' /btain the new row alues through the following two operations:
!ultiply *diide+ all of the elements in a row by a constant
Add or subtract the multiple of the row to or from another row
After identifying the en"e&ingea0ing and i0#" ee*en" construct the second tableau by replacing the "1 by H in the basic solution'
2nd tablea!
B5
8E
X 1
8 1 X1
5 A
8"
5
25
5
5
5
X 2
8 1
8 2
8 "
F
77
5
G N1
5
I1 N1
5 5
2B 11
2BNGDJ NN1D11
5
3N1
5
I3N1
A
BN3N1D
+1+1 -4+2 +2+2 .2 +3+3 -3+2
21
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming B5 5
G
35 15
5 5
35 I35
5 5
BB5
Hote the follo(in% to see ho( it is calc!lated !ince t*e minimum replacement ratio (RR) is associated wit* t*e % nd row we s*ould start t*e operation on t*e second row i.e. we s*ould c*ange t*e pivot element into $. 0o do so we s*ould divide t*e w*ole elements o' t*e % nd row by t*e pivot element o' t*e 'irst tableau.
8ee the analysis belo(. 2
1 9 1 9
22 when we diide it by 1 we get
1 K 9 K 9 11 so this will be
the new set row alues of the second tableau' .hen calculate the new alues of the first row and the third row of the second tableau' 9ou can begin in ether row one or row three'
5ets start (ith the first ro( ?ring the first row alues of the initial tableau
#
1 1
1
.hen ta)e the second row alues of the 1nd tableau
1 J J 11
.hen, in order to change into 5, as) your self what algebraic operation should be done' >ery simple, multiply the whole elements of the second tableaus alue by I' 9ou get /# /2 /2 /##, then add this alue to the first row alue of initial tableau' 5oo at the s!mmary belo(. # 1 1
1
1 J J 11
# 1 1 1
#
/#& 1 J J 11)
@
1 1 1
/# /2 /2
=
1
/## /2
;A
will be the
new alues of the first row of the second tableau' .o put it in a simple formula
7 1?71 @ & /#7 2) 8imilarly calc!late for ro( three of the second tablea! " "
1 "K
1 J J 11
"
"
1 "K
/"& 1 J J 11)
" @
"
/" /"-2 /"-2
"-2
1
"K
/"" /"-2 1
A will be the
new alues of the 3rd row alue of the second tableau' .o put it in a simple formula again it will be:
7 "?7" @ & /"7 2) 22
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming Note .hen calculate theelement C row alues i'e' piot these are found multiplying the alues eachne+ column by thein Diide eac* o( t*e ro+ by by t*e piot element toinnd alues
•
corresponding t*e )ey or coefficients piot ro+.in the < column and adding them'
4E
X1
81
5*5 + D5
5 *G +D5
X1
A
B5* +DB5
A *N1+ D35
A& 5 +D5
8"
5*5 +D 5
5*3N1+ D5
5* 5 +D5
+2
A
s1
s2
5 *+D5
"
s"
5*I1 + D5
5*5+D5
B5*N1+D35
&;A)?
A *5+D5
5*I3N1+D5
F
A&11)?AA
5* +D5
&A)?
"
AA
'hen finally calc!late G ro(. It is simply abo!t follo(in% the form!la.
8tep L Repeat step IB till optimum basic feasible solution is obtained' 'e' repeat these steps till no
•
positie alue occurs in the G row'
Note • A simple% "rd simple+ tablea! solution in a ma%imi&ation problem is optimal +*en t*e C-, ro+
&erosand negatie No +*en t*ere are no positie alues consists entirely ?>
8 1
5
X 1
A 25
X 2 /
B5 X 1 5 IBN3 5
25 X 2
5 8 1 5
5 8 2
F
5 8 " B
1
5
5 5
I
IN3 1N3
@
B5
25
5
5
5N3
J5
5
5
5
I5
I 5N3
+1+1 -8+3 +2+2 -1.2+3 +3+3 .3.2
•
.he
entire G < indicatin% that no additional potential for improvement e+ists. •
/ptimal solution is reached'
•
the optimal sol!tion is (here X 1?K, X 2?#, 8 1?2#and a+ ?L#
In"e&&e"a"i#n #! "'e Re-%"-
23
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming 0*e last step is interpreting t*e result2 in order to ac*ieve t*e ma#imum weekly pro'it o' birr 34/, t*e company s*ould produce 5 units o' microcomputer type one and 4 units o' microcomputer type two. 0*is will leave no slack in eit*er inspection (s %=/) or storage space (s =/). 6ow ever, t*ere will be %4 *ours o' assembly time t*at is unused.
E6a*e 2 A uice producing
bottle of punch A re(uires 15 liters of /range uice and 2 liters of #rape uice' 8here as ?ottle of punch ? re(uires 5 liters of /range uice and 2 liters of #rape uice' '0rom each bottle of Punch A; a profit of K is made and from each bottle of Punch ?; a profit of K3 is made' "uppose that the company has 135 liters of /range uice and 15 liters of #rape uice aailable
Re;%i&e
a' 0ormulate this problem as a LPP b' %ow many bottles of Punch A and Punch ? the company should produce in order to maximize profit *$sing the simplex method+ c' 8hat is this maximum profit
S#%"i#n M!ice needed for one bottle of M!ice
6!nch 36!nch 4uice Aailable
MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM Bran%e M!ice *lt +
15
0rape M!ice *lt +
2
Profit per tent
5 2
135
15
#"
Let X 1? the H o of bottles of p!nch 3 prod!ced ' X 2? the H o of bottles of p!nch 4 prod!ced '
'he 566 odel of the problem is:
!ax CD $3 % "t: 15 $5 % 6 135
Bran%e onstraint 24
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming $2 % 6 15 $ , % & 5
0rape onstraint
Hon/ne%ativity constraint
'he8tandard formof this model (ill be a+ D61 3 + 2 5 s1 5 s2 5 s"
"t: 15 + 15 + 2 s1 5 s2 D 135
8tandard form
+ 12 + 25s1 s2 D 15 61,62 ,
s1 , s2D 5
8here,s1?$n!sed oran%e !ice s2 D$n!sed %rape !ice
nitial feasible solution 1st tablea!
4
#
"
X1
X2
81
82
81
2
1
1
82
F
77
2"
11.; 2#
;
1;
1
12
G
3
5
5
2nd simple+ tablea!
4
#
"
X1
X2
81
82
F
77
X1
#
1
J
1-2
11.;
2"
82
2;-2
/1-#
1
A2.;
;
#
2
1-;
#A
G
9
1
1,:
9
/ptimal solution mix: the 3rdsimplex tableau
<
4
3
5
X 1
X 2
8 1
5
F
8 2
H
#
1
9
3,:9
1,2:
@
H1
"
9
1
1,:9
2,2:
2
OPERATIONS RESEARCH /
5
3
9
9
9.1 9.19.9
TEACHING MATERIAL 9.9
2
25
Chapter Three -Linear Programming
O"i*a S#%"i#n
X 1? K
X 2? ;
s1?
s249
and a+ ?;1
2. Ma6i*iFa"i#n =i"' Mi6e C#n-"&ain"n the preious section, we hae learned how to sole a maximization problem with all 6constraints' %ere you will learn how to sole a maximization problem with mixed constraints *67 D+' 0or the most part the techni(ue is identical to that illustrated in the preious section' 8hat is new is the introduction of the Q! coefficients in the ob&ectie functions and the use of artificial ariables'
E6a*e1
0o standardi7e t*e model w*at we *ave learned in t*e previous e#ample is not enoug*. 6ere t*e model consists 8, &9 = sign constraints w*ic* demands t*eir uni:ue standardi7ation approac*. ;or 8 constraint you already know t*at we add a slack
26
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming variable.
;or t*e remaining two constraints, consider t*e 'ollowing two additional
variables.
S%&%- 0a&ia+e- (S) &e!e&,- "# "'e e6/e-- %"iiFa"i#n #! &e-#%&/e
A ariable inserted against a minimum constraint * &+ to create e(uality' t represents the amount of resource usage aboe the minimum re(uired leel'
"urplus ariable is subtracted from a 7 constraint in the process of conerting the constraint to standard form'
=either the slac) nor the surplus should be of negatie alue in the initial tableau' .hey must hae a positie alue' Loo) at the following illustration'
' #3 #1 " 6 2 #D / and #1D /==& s D 2 ==&s1?#;!n!sed reso!rce &all reso!rces are idle)
1' %# #1 75 #D / and #1D /(1o production)==&# 1 #1I s1D 15 ==&s1?/A& 'his is mathematically !nacceptable) •
0o avoid t*ese problem anot*er variable is inserted w*ic* is known as an rti'icial variable ()
A&"i!i/ia 0a&ia+e (A) •
tis a ariable that has no significant meaning in a physical sense but acts as a tool to create an initial feasible LP solution' t helps to ma)e the model logical and meaningful' t represents an extremely insignificant amount of resources and is included to balance the relationship between the decision ariables and R%" constraints' t is added to constraints haing D and 7 algebraic sign'
•
%as a coefficient of SI!T in the ob&ectie function for maximization type problem' t represents an artificial ery large negatie alue'
onsider the above e+ample %# #1 75 #D / and #1D /(1o production)==&# 1 #1I s1D 15 s1?/A& 'his is mathematically !nacceptable, beca!se it violates the non ne%ativity ass!mptions of 566) #1 #1I
s1D 15
#$<%#%-
s$ < $= %/
27
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming !o
w*en we set # $ ,#% and s$ = /
0*e
basic solution in t*e initial solution will be $= %/ w*ic* is
mat*ematically correct . 5et !s consider another e+ample havin% ? si%n ;+ 1 @K+ 2?" .o determine the alues of the basic ariables for the initial tableau, we set x F x1D5
;+ 1 @K+ 2?" ==& /=/
==& ;&) @K&) ?"
w*ic* is mat*ematically wrong. 0o avoid t*is problem we add an arti'icial
variable to standardi7e constraints wit* = sign. ;+ 1 @K+ 2?"
==& ;+ 1 @K+ 2@ s1?"
==& ;&) @K& @s 1 ) ?"
==&! $=/ w*ic* is logical.
Note •
For the initial basis, use arti*cial variables for constraints that have 5and ( sign. For a 1constraint use a slac! variable. ?ence, s$rpl$s variables will not
Ho(,lets standardie the problem based on the lo%ic stated above. Ma6 DB61 G62 5 -1 5 -2 5 -3I! A2I ! A3 "t: D 62 -1 D@ 61 62 A2 Bx1x1 I -3 A3 D1 All >ariables D 5 Ini"ia !ea-i+e -#%"i#n
8tandard form
1st tablea!
?
A
=
/
/
X 1
X 2
8 1
8 "
32
3"
F
77
81 A1
9
5
5
5
N5Dund'
/
1
5
5
5
@
@
3"
/
8
1
5
I
5
1
/L
/"
@
/
/
/""
L 8
"@=
G
/
5
5
28
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming "ince Jm B is the largest, x will be the entering ariable for the next tableau' "ince an artificial alue *A3+ leaes the solution it will remoed from the next table, so A 3 column will not exist in the 1nd tableau
2nd tablea!
?
81
A1 X1 /
A X 1
= X 2
8 1
8 "
/ 32
5
5
5
/ A
5 A
1N3 N3 1/"-" B1 -"
5 5
"rd table
?>
NB INB /1/1-A 1@1-A
5 / 5
F
77
2 2#/;
1JN1 1
a!
A
=
X 1
X 2 8 1
/
8 "
32
F
77
X2
=
5
5
5
und
A1
/
5
5
I1N3
NB
JN3
X1
A
5
INB
5
GN3
=eg'
#=/L-"
G
IN3
A G
A@2-"m
/1/1-A
!
5
/A/2-"m
1@1-A
5
$ina "a+ea%
?>
A
=
X 1
X 2 8 1
F
5
8 "
X2
=
5
"3
5
5
I
X1
A
5
I
5
2
A G
2
G
5/ 2
A2
8ince the val!es on / < ??>optimal sol!tion is achievedN (here, 29
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming @14:
@2 45 S149
S34 15 andMa6 4 82
E6a*e 2
!ax CD1 # #13 #3 "ub&ect to: # #1 1 #3 6 2 %# 3 #1 #3 D 1 #, #1 , #37 5
S#%"i#n
Initial feasible sol!tion: 1 st tablea!
4
<
2 X 1
5
8 1
1
"
/
X 2
X "
8 1
31
1
5
F ++
2 25
31
/
/
1
3
5
/
/2/"
I!
2@2"@1
#@"
5
1
3
/12
5
2nd simple+ tablea! <
4
2
X "
3
31
/
1
"
/
X 1
X 2
X "
8 1
31
N1
U
1
3
5
1!3N1
"-2
3N1/
3
G
1-22
/1-2
/2/"-2
F ++
5
2
1
I!
10
4
/2@1;-2
30
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming "rd simple+ tablea! <
4
2
1
"
X 1
X 2
X "
F ++
8 1
X "
3
N1
5
3N1
'2
X 2
1
5
5
I1
1
"-2
1
3
;-2
1"-2
/
1-2
/;-2
6
?nde*ned
#th simple+ tablea!
4
<
2
1
"
X 1
X 2
X "
F
8 1
X 1
1
5
1
3
3
X 2
1
5
5
I1
1
21
/
/1/#
#
=
.he optimal solution is defined in terms of X 1 ? "
X 2 ? 2
X "?
8 1 ? and ma+.6rofit &) ? =
3. Mini*iFa"i#n P+e*!anual solution procedure of minimization problems using simplex is handled in the same fashion as that of maximization problems with mixed constraints' .he two exceptions are:
.he M coefficients in the ob&ectie function are gien positie signs instead of negatie signs
.he selection of the ariable to enter the solution is based on the largest negatie alue *the negatie number far from 5+ in the
"olution is optimal when there is no negatie alue of /.
Note P; <= ,<+>/
< P? /< >(>+( =<+@
A
----------------------------------------Add a slac) ariable
-----------------------------------------Add an articial ariable
B ---------------------0ubtract a surplus ariable and add an articial OPERATIONS RESEARCH TEACHING MATERIAL
31
Chapter Three -Linear Programming
E6a*e1
!inimize CD12 # 35 #1 "ub&ect to: %/#2 #1 7 55 %# 3 #1 7 2 #F #17 5 S#%"i#n
8tep 1: S"ana&iFing "'e +e*,*#e
!inimize CD12 # 35 #1 5s5s1 !A!A1 "ub&ect to: %/#2 #1I sA D 55 %#
3 #1 Qs1A1 D 2
#, #1 , s, s1 ,A ,A1 D5 8tep 2: Ini"ia !ea-i+e -#%"i#n
.he initial basic feasible solution is obtained by setting x D #%D sD s1D5 =o production, xD #%D sD5DD7%/(/) 2*5+ I 5A D 55 DD7 A D 55 xD #%D s1D5DD7/(/)3*5+ I 5A1 D2DD7A1 D 2
1st tablea! 4
<
2;
"
X 1
X 2
8 1
8 2
31
32
F
++
31
!
15
2
I
5
5
55
32
1
3
5
I
5
2
22
1=
2; /22
"/ 1=
G
I
11;
+**@*(+-@(.-
2nd 8imple+ 'ablea! 32
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming <
4
2;
"
X 1
X 2
8 1
8 2
32
F
X 1
12
V
IN15
5
5
2
32
5
3N1
N5
I
2
G
J2N"-2
2;
I2NN5
+2+2-2 +
2NIN5
#;-#/"-2
+1+1 .20
12;@;
"rd 8imple+ 'ablea! s s*own in t*e 'ollowing tableau, all variables on t*e - "& /==&*ence, optimal solution is reac*ed w*ere, X 1?;-2X 2?1-" and in ? 1A2.;
<
4 X 1 X 2
12 "
G
F
2;
"
X 1
X 2 5
8 1 IN5
8 2 N1
5
N2
I1N3
2N1 5N3
2;
35
IN1
I2N1
1A2.;
N1
1;-2
E6a*e 2
"ole the problem proided below *!inimization with mixed constraint+ !in CD2 # 3 #1 "ub&ect to: %# #1 6 1 %# 1 #1 D 5
2x 1x175 HFH1 S#%"i#n
!in CD2 #3 #1 5s5s3 !A1!A3 "ub&ect to %# #1sD 1 %# 1 #1A1 D 5
33
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming 2x 1x1I"3 A3D5 H, H1, ", "3, A1 and A3 D 5 !in CD2 # 3 #1,
f no production
%# #1s D 1 DD7 # D #1D5DD7sD1&8ol!tion Eal!e in the initial simple+ tablea!) %#1 #1 A1 D5 DD7 # D #1D5DD7A1 D5&8ol!tion Eal!e in the initial simple+ tablea!) #1 #1 Qs3A3D5 DD7 #D1Ds3D5DD7A3D5&8ol!tion Eal!e in the initial simple+ tablea!) #, #1 , s, s1 ,A ,A1 D 5
Initial feasible sol!tion: 1 st tablea! <
4
;
"
X 1
X 2
8 1
8 2
31
32
F ++
8 1
5
1
5
5
5
1
31
1
1
5
5
5
5
32
2
1
5
I
5
5
L
#
5
; /L
"/ #
6
5
G
I
2 2
2nd simple+ tablea! 4
<
;
"
X 1
X 2
8 1
8 2
F
31
8 1
5
BN2
1N2
5
G
31
5
BN2
5
1N2
B
;
X 1
1N2
5
IN2
5
1
;
A-; @2
5
1N2 I
++
1@A 25
/
IA-;
5
I1N2
5 5
"rd simple+ tablea!
34
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming
<
4
;
"
X 1
X 2
8 1
8 2
F ++
31
X 2
3
5
2NB
NG
5
1'2
31
5
5
I3NG
N
3
X 1
;
5
5
ING
IN
5
;"
I3NG 2NB
-#/L-=
12.;@"
5
-#@L-=
/
3NG I2NB
20
12 -
#th 8imple+ tablea!
<
4
;
"
X 1
X 2
8 1
8 2
F
X 2
3
5
N1
5
8 2
5
5
I3N1
1
X 1
;
5
5
IN1
5
;"
/1
5
G
2"
'he val!es of the optimal sol!tion: X 1?#, X 2?1, 8 1?, 8 2?1 and in ? 2"
E6e&/i-eOind the optimal sol!tion for the 566s !sin% simple+ method
1 . !in CD5 # 2 #1 "ub&ect to: %# 2 #1 7 25 # #1 7 15 #, #17 5 ns2 #D25N3, #1 D15N3 and !in CDK25
2 . !in CD # 2 #1 "ub&ect to:
35
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming $ 1x1 7 G5 # x1 7 J2 #, #17 5 ns2 #D, #1 D33 and !in CDK11
"' !in CDJ # @ #1 "ub&ect to: # Bx1 7 3B ># x17 B #, #17 5 ns2 #D15N3, #1 DGN3 and !in CD11N3
Note To get an initial feasible solution &pes oC constraint Presence oC "aria#les in the initial sol$tion miD 1 A lac! es 2 B E$rpl$s o E>rti*cial es 3 >rti*cial es
2.8. SPECIAL CASES,ISSUES IN SIMPLE@ METHOD I. T=# in/#*ing 0a&ia+e- , #& Tie !#& en"e&ing 0a&ia+e-, 8here two or more ariables are represented by the same alues on the
1. ?' t*ere is a tie between two decision variables, t*en t*e selection can be made arbitrarily. %. ?' t*ere is a tie between a decision variable and a slack (or surplus) variable, t*en select t*e decision variable to enter into basis 'irst.
36
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming . ?' t*ere is a tie between slack or surplus variable, t*en selection can be made arbitrary. E6a*e
f the e(uation is max C: ?
F X 1
X 2
;2
;
8 1
8 "
/
ven if X 1has the same coefficient to " on the
entering ariable' .his is because H represents a decision ariable whereas " stands for a slac)Nsurplus ariable'
II. In!ea-i+ii" A situation with no feasible solution may exist if the problem was formulated improperly' nfeasibility comes about when there is no sol!tion that satisfies all the problems constraints' n the simplex method, an infeasible solution is indicated by loo)ing at the final tableau' n it, all G row entries will be of the proper sign to imply optimality, b!t an artificial variable &3) will
still be in the solution mix'
E6a*e
<
;
=
X 1
X 2
8 1
8 2
F
32
X 1
;
I1
3
X 2
=
5
1
5
55
32
5
5
5
I
15
2
=
/2
"1/
G
2
/"1
5
155
,G55155!
37
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming Een though all G alues are positie and 5 *i'e' the criterion for an optimal solution in a minimization case+, no feasible sol!tion is possible because an artificial variable &3 2 ) remains in the solution mix'
III. Un+#%ne S#%"i#n =o finite solution may exist in problems that are not bounded' .his means that a ariable can be infinitely large without iolating a constraint' n the simplex method, the condition of unbounded ness will be discoered prior to reaching the final tableau' 8e will note the problem when trying to decide which ariable to remoe from the solution mix' .he procedure in the simplex solution approach is to diide each (uantity column number by the corresponding piot column number to identify the leaing ariable' .he row with the smallest positive ratio is replaced' ?ut if the entire ratios turn out to be ne%ative or !ndefined , it indicates
that the problem is unbounded'
E6a*e
=ote the following LPP of maximization ob&ectie' 4
<
2
@
F
++
X 2
8 1
I
1
5
35
I
5
1=
1J5
/1=
;
8 2
Pi"ot ,ol$mn I1 5 K I@
/
5
X 1 X 2
5
1;
8 2
30.-1-30 ?naccepta#le ++s - -
.he solution in the aboe case is not optimal because not all / entries are 5 or negatie, as re(uired in a maximization problem' .he next ariable to enter the solution should be X 1'.o determine the ariable that will leae the solution, we examine the ratios of the (uantity column alues to their corresponding coefficients in the X 1or piot column' "ince both piot column alues are negatie, an unbounded solution is indicated'
IV. Degene&a/ NTie !#& ea0ing +a-i/ 0a&ia+e (e =) N f there is a tie for the smallest ratio of the (uantity column alues to their corresponding coefficients in the piot column, it is a signal that degeneracy exists' -egeneracy can occur right in the first *initial+ tableau'.his normally happens when the number of constraints is less than the number of ariables in the ob&ectie function'
38
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming E6a*e .he following tableau shows the case of degeneracy
<
4 X 2
=
;
=
2
X 1
X 2
X "
N
8 1
I1
8 2
F
8 "
5
5
5
++
8 2
5
N3
I
5
15
10.1.440
8 "
1
5
1
1N2
5
5
20.45ie Cor the
1
=
=
1A
G5
smallest ratio indicates de
/
"
/A
/1A
10.25
-egeneracy could lead to a situation )nown as cycling, in which the simplex algorithm alternaties bac) and forth between the same nonIoptimal solutions, i'e', it puts a new ariable in, then ta)es it out in the next tableau, puts it bac) in ,and so on'.his situation *cycling+ can be oercome by trial and error method ' /ne simple way of dealing with the issue is to select either row * 8 2 or 8 " in this case+ arbitrary' f we are unluc)y and cycling does occur, we simply go bac) and select the other row'
V. M%"ie O"i*a S#%"i#n!ultiple optimal solutions exist when nonIbasic ariable contains ero on its G row'
+ample: a+ ?"X 1@2X 2
4
<
"
2
F
X 1
X 2
8 1
8 2
3N1
5
B
5
N1
3
3
2
2
1
/
/2
X 2 8 2
2
X 1?, X 2?A, 8 2?" and a+ ?12 or: X 1?", X 2?"-2 and a+ ?12
.he G alue of the =onIbasic ariable * X 1+ is 5'.hus, there is alternatie optimal solution'
39
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming
E6e&/i-e1 . "ole the following LPP using the simplex algorithm !in CDB # G #1 "ub&ect to: # %#1 7 G5 # x1 7 J2 #, #17 5 Re;%i&e •
8hat are the alues of the basic ariables at each iteration
•
8hich are the nonIbasic ariables at each iteration
3ns: X 1?1#, X 2?"", and in ?221
2 . At the 3rd iteration of a particular LP maximization problem, the following tableau is established:
?
F X 1
X 2
X "
8 1
8 2
8 "
X "
;
5
I1
5
5
2
X 1
A
I3
5
5
5
1
8 2
5
1
5
I
5
B
/1"
;
;
21
1A
/;
/21
/
@J
Re;%i&e •
8hat special condition exists if you improe the profit and moe to the next iteration
•
Proceed to sole the problem for optimal solution 3ns: De%eneracyN X 1?2L, X 2?;, X "?, and a+ ?1LL
3 .
!in CD # #1 40
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming "ub&ect to: # #1 D 3 4# 3x1 7 B # 1x1 6 3 #, #17 5 3ns(er: in. D61 + 2 5 s1 5 s2 ! 31! 3"
"t: "+ 1 + 2 #+ 1 3 + 2Is1 # 1x1
D3
31
32 D B
@
s2
D3
All >ariables 7 5
5' "ole the following LPP using simplex method !axCD@ # J #1 "ub&ect to: %# x1 6 5 # 3x1 6 35 #, #17 5 3ns: X 1?1=, X 2?#, and a+?1K
:' "ole the following LPP to show that it has alteratie optimal solutions' a' !axCDB # 3 #1
3ns:
i . X 1?#, X 2?, and a+?2#
ii' X 1?;-2, X 2?", and a+?2#
"ub&ect to: %# x1 6 G # 3x1 6 G
x16 3
#, #17 5 +. !inCD1 # G #1
"ub&ect to:
3ns:
i. X 1?"2-A, X 2?1-A, and in?2# ii ' X 1?12, X 2?, and in?2#
# x1 7 5 %# 1x1 7 # x1 7
41
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming #, #17 5
8. "ole the following LPP to show that it has unbounded solution' a' !ax CDI1 # 3 #1
+' !ax CD3 # B #1
"ub&ect to:
"ub&ect to:
x6 2
# x1 7 1
% #I3x1 6 B
-%# x1 6
#, #17 5
#, #17 5
?' "ole the following LPP to show that it has no feasible solution' a' !axCDI1 # 3 #1
3ns: X 1?2, X 2?, 31?2 and a+?#/2
"ub&ect to: xI #1 7 # x1 6 B #6 1 #, #17 5 +' !axCD3 # 3 #1
3ns: X 1?, X 2?2, 32?2 and a+?#/#
"ub&ect to: %# x1 6 1 # x1 7 1 #, #175
2.?.SENSITIVIT7 ANAL7SIS (POST OPTIMALIT7 ANAL7SIS) "ensitiity Analysis is the study of how changes in the coefficients of LPP affect the optimal solution' t is initiated after the optimal solution is obtained' .he purpose is to see the effect on the alue of the optimal solution of changes in R%" of a constraint, coefficients of ob&ectie function, changes in the coefficients of the constraints, changes of the inputIoutput coefficient, additionNdeletion of constraints and the addition of a new ariable to the problem t starts from the final tableau becauseit is concerned with the study of W"ensitiity of the optimal solution of an LPP with changes in parameters of the ariables'.he degree of sensitiity of the solution due to those ariations can range from no change at all to a substantial change in the 42
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming optimal solution of the gien LPP' .hus, in sensitiity analysis, we determine the range oer which the LP model parameters can change with out affecting the current optimal solution' .he process of studying the sensitiity of the optimal solution of an LPP is called post/optimal analysis-(hat/ if analysis-.
n this moduleNleel, we will see the following two aspects post optimality analysis
A' han%es in the 7C8 F!antity of a constraint &b) ?' han%es in the coefficients of the obective f!nction *<+
2.?.1. C'ange in "'e Rig'" HanSie >%an"i" (RHS )
.he first step in determining how a change in R%" of a constraint would influence the optimal solution is to examine the shadow prices in the final simplex tableau' S'a#= &i/e
is a managers window for guessing the impact of one unit change has on the alue
o
of the ob&ectie function o
is the maximum amount that should be paid for one additional unit of resource
o
is a marginal alue, i'e' it indicates the impact a one unit change in the amount of a constraint would hae on the ob&ectie function
0or illustration purpose, lets loo) at the microcomputer *whose optimal solution has been already generated+ case' !ax CD B5x 25x1 "ub&ect to: Assembly time
x 5x1 6 55
nspection time
1x x1 6
"torage space
3x 3x16 3@
11
H F x1 7 5 ts optimal solution is represented in the final tableau as follows
B5
25
5
5
X 2
8 1
8 2
5
F
?>
X 1
8 1
5
5
5
B
IBN3
1
X 1
A
5
5
IN3
@
X 2
25
5
5
I
1N3
B5
25
5
5
5N3
5 5 OPERATIONS RESEARCH
5
/
8 "
J5
I 5 I5N3 TEACHING MATERIAL
43
Chapter Three -Linear Programming
.o deal with shadow prices you should focus on the slac) ariable column' ?ecause each slac) ariables column represents each constraint, i'e' " represents the first constraint; " 1 represents the second constraints and so on' .he alues in the slac variable col!mns of the < / row *the magnitude+ of the final tableau proide us with the shadow prices' .herefore, the shadow prices for the aboe example are:
S149
S2419 S34 59,3
8hat does this indicate 9ou )now that shadow prices signify the change in the optimal alue of the ob&ectie function for 1 !nit increases in the alue of the R%" of the constraint that represent the aailability of scarce resources' ?ased on this logic, we can interpret it as follows:
$. ?' t*e amount o' assembly time (! $ ) was increased by one *our, t*e e''ect would be no e''ect on t*e pro'it (ob@ective 'unction) %. ?' t*e amount o' inspection time (! % ) was increased by one *our, t*e e''ect would be an increase in total pro'it by $/ birr. . ?' t*e amount o' storage space (! ) was increased by one cubic 'eet, t*e e''ect would be an increase in total pro'it by 4/A birr. ?ut what the shadow prices do not tell us is the extent to which the leel of a scarce resource can be changed and still hae the same impact per unit' .herefore, we need to determine the range oer which we can change the R%" (uantities and still hae the same shadow prices' .his is called the &ange #! !ea-i+ii" or &ig'" 'an -ie &ange'
Note •
Range o$ $easi%i!it i s the range over which the B?/ value of a constraint can be changed without changing the optimal solution mi! of a problem ;still have the same shadow prices<
.he alues in the body of the final tableau in each slac) column are used to compute the range' .o compute the range of feasibility for the corresponding constraint, the en"&ie- in "'e a--#/ia"e -a/ /#%*n *%-" +e i0ie in"# "'e 0a%e- in "'e ;%an"i" /#%*n '
44
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming E6a*e 1 -etermine the range of feasibility for each of the constraints in the microcomputer problem' 8 1
8 2
8 "
F
8 1 & assembly time)
82& inspection time)
8" & stora%e space )
1
A 1 /1
/1A-" /1-" 2-"
2# K #
2#-1? 2# K-? !nd #-? !nd
2#-A? # K-1? K #-/1? /#
1NIBN3D I'2 @NIN3D I1J N1N3DB
1 /1
#-" /#-"
L#
1 ho!rs
22 ho!rs
"K c!bic feet
Hone& no !pper limit)
22 @ #?2A
"K @ #.;? #".;
199 254 ?8
22 54 1
3 84 33
$pper limit 7an%e 5o(er 5imit
.o calculate the limits, use the following general rules' *.a)e only the magnitude, dont consider the sign+
'or ma(imi"ation pro%!em )*ith a!! <+onstraints To calculate the upper range, ta7e the smallest negative ratio ;a number closest to $ero<.
'or minimi"ation pro%!em )*ith a!! +onstraints ;The rules will be reversed< To calculate the upper range, ta7e the smallest positive ratio ;a number closest to $ero.
4ased on the r!les ill!strated above •
.he upper range for assembly time *" + would be 55hours as a result of the original aailability plus the smallest negatie ratio' ?ut, since there is no negatie ratio, there is no upper limit'
•
the lower limit will be 55 hours minus the smallest positie ratio' .herefore, it will be 55 I 1D JB'
•
.he ranges of feasibility for the other constraints shall be calculated similarly'
Ho( lets see the application &rela+ation and ti%htenin% of constraints) E6a*e 2 .he manager in the microcomputer problem is contemplating one of two possible changes in the leel of the storage space constraints' •
/ne change would be an in/&ea-e #! 3 /%+i/ !ee" in its leeland
45
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming •
the other would be an in/&ea-e #! /%+i/ !ee" in its leel .
Re;%i&e •
-etermine the reised optimal solution for each possible change'
S#%"i#n •
0or purposes of simplicity first lets see the effects of a one unit change in storage space' .he analysis should focus on the alues of " 3 column because storage space is represented by "3' 4asis
8"
F
81
/1A-"
2#
X1
/1-"
K
X2
2-"
#
#-"
L#
Cence, any potential increase of stora%e space & 8 ") by 1 !nit ⇒
auses ! $ (assembly time) to decrease by $A units ( i.e. ! $ will become $>.)
⇒
auses #$(micro computer type one) to because by $A units
⇒
auses #%(micro computer type two) to increase by %A units
⇒
auses pro'it to increase by +4/A
Ho( lets come to the proposed chan%es. ase 1: a three !nit increase on stora%e space. •
#ien the analysis of the impacts of change in a unit of storage space, it is ery simple to calculate the impact of a three unit change in this constraint on the optimal solution alues'?ut we hae to first chec) if the change is within the range of feasibility'
•
8e hae already calculated the range for storage space to be between 33 and 3'2 cubic feet' n the original model of the problem the maximum aailability of this resource was specified at 3@ and the increase is 3; therefore, the changed alue is 3@3D 1; which is, within the range' "o the impact of the shadow prices is alid' .herefore, the effects of an increase of three cubic feet can be computed in thefollowing way'
4asic
8"
F!antity
!rrent
han%e
7evised sol!tion
sol!tion
81
/1A-"
2#
2#
2#@"&/1A-")
=
X1
/1-"
K
K
K@"&/1-")
=
X2
2-" #-"
# L#
# L#
#@"&2-") L#@"-")
A L= 46
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming ase 2: an increase of = c!bic feet in stora%e space •
0or an increase in storage space of G cubic feet, since the upper limit of the feasible range for storage is 3'2 *3@'2+ the amount of increase aboe '2 cubic feet will be excess or slac)' !oreoer, the new optimal solution will occur at the point where the constraint leel e(uals its upper limit'
•
.herefore, the additional amount needed to achiee the upper limit is used to compute the reised solution'
4asic
8"
F!antity
!rrent
han%e
7evised sol!tion
sol!tion
81
/1A-"
2#
2#
2#@#.; &/1A-")
X1
/1-"
K
K
K@#.; &/1-")
L.;
X2
2-"
#
#
#@#.;&2-")
L
#-"
L#
L#
L#@#.; -")
=
=ote that beyond the upper limit, " 3 would come into solution replacing " which would no longer be slac)' .he amount of slac) would be GI'2D 3'2 cubic feet'
8 "? ".; X 1? L.; X 2? L E6a*e 3 "uppose the manager in the microcomputer problem is contemplating a decrease in storage space due to an emergency situation' .here are two possibilities being considered' ⇒
A decrease of B cubic feet in its leel
⇒
A decrease of @ cubic feet in its leel
i. A e/&ea-e #! 8 /%+i/ !ee" •
4asic
"ince the change is within the range we can calculate it straight forward' 8"
F!antity
!rrent sol!tion
han%e
7evised sol!tion
47
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming 81 X1 X2
/1A-" /1-" 2-"
2# K #
2# K #
2# / A&/1A-") K / A&/1-") # / A&2-")
;A 11
#-"
L#
L#
L#/A -")
AA
ii' A e/&ea-e #! /%+i/ !ee" •
"ince the change is beyond the lower limit the substitution rates and the shadow prices do not hold'
E6a*e 5
!ax CD3 # #1 "ub&ect to: 3 #2 #1 6 2 1 # #1 6 G #161 #,
#17 5
.he optimal tableau is gien as: 4
3
5
X 1
X 2
8 1
5 8 2
5
F
8 "
X 1
3
5
I5'3
5'J
5
3'2J
8 "
5
5
I5'1GB
5'1G
'3
X 2
#
5
5'1GB
I5 I 5'1G
5
5'G2J
C
3
5'J
5'1G
/
5
5
I5'J
I5'1G
5 5
Re;%i&e
' -eterm -etermine ine the the shadow shadow price price for each each const constrai raint nt 1' -etermine -etermine the R%" ranges ranges oer oer which which the shadow prices prices are alid S#%"i#n • 3nalysis
of the 1st constraint &8 1 ) )
F 8 1 F- 8 1
3'2J
I5'3
I1'@B
48
OPERATIONS OPERATIONS RESEARCH
TEACHING MATERIAL MATERIAL
Chapter Three -Linear Programming '3
I5'1GB
I3'@@
5'G2J
5'1GB
3'55
5o(er 5imit? b1 / the smallest positive n!mber in the F- 8 1col!mn
?1;/"?12
$pper 5imit? b 1 @the smallest ne%ative n!mber in the F- 8 1 col!mn ?1; @ &".KK) ?1=.KK resource $ over w*ic* t*e s*adow price +/.3$4 per unit .herefore , 12< b1< 1=.KK (0*erange o' resource is valid). • 3nalysis
of the 2nd constraint &8 2 ) )
F 8 2F- 8 2
3'2J
5'J
2
'3
5'1G
1'BJ
5'G2J
I5'1G
I1
5o(er 5imit?b2/the smallest positive n!mber in the F- 8 2 col!mn ?=/&2.AL) ?;."" $pper 5imit?b2@ the smallest ne%ative n!mber in the F- 8 2 col!mn ?= @2 ?1
.herefore , ;.""< b1< 1 (0*erange o' resource % over w*ic* t*e s*adow price + 5'1G per unit is valid). • 3nalysis
of the "rd constraint &8 " ) )
F 8 "F- 8 "
3'2J
5
I
5o(er 5imit?b"/
'3
'3
5'G2J
5
I
the smallest positive
n!mber in the F- 8 " col!mn
?2/&1.1#")?.=;L
$pper 5imit?b" @ the smallest ne%ative n!mber in the F- 8 " col!mn ?2 @P ?P
49
OPERATIONS OPERATIONS RESEARCH
TEACHING MATERIAL MATERIAL
Chapter Three -Linear Programming resource over w*ic* t*e s*adow price + 5 per unit .herefore , .=;L< b"< P (0*erange o' resource is valid).
E6e&/i-e1' !ax CD25 #5 #1
"ub&ect to: 3 #2 #1 6 25 &3ssembly time) #1615 &6ortable display) ># 2x1 6 355 &areho!se space) #,
#17 5
'he final simple+ tablea! is: 4
25
5
X 1
X 2
5
5
8 1
8 2
5
F
8 "
X 2
5
5
GN12
5
I3N12
1
8 2
5
5
IGN12
3N12
G
X 1
;
5
I2N12
5
2N12
35
25
5
G
5
5
N2 IN2
5 5
1BN2
@G5
I1BN2
Re;%i&e
-etermine the shadow prices for the three constraints of the %igh .ech .ech
the assembly time constraint is 1#-;&i.e.1 additional assembly time over the e+istin% 1; is 2.=).
•
the portable display constraint is .
•
the areho!se space constraint is2A.; 50
OPERATIONS OPERATIONS RESEARCH
TEACHING MATERIAL MATERIAL
Chapter Three -Linear Programming .herefore, we see that obtaining more warehouse space will hae the biggest positie impact on %igh .echs profit'
2' "ole the following LPP using simplex method' A firm that manufactures both lawn mowers and snow blowers: X 1 ?the n!mber of la(n mo(ers X 2 ?the n!mber of sno( blo(ers
!ax CDK35 #KG5 #1 "ub&ect to: 1 # #1 6 555 5abor ho!rs available # 1x1 6 ,155lb of steel available #1615 sno( blo(er en%ine available #,
#17 5
Re;%i&e a' 8hat is the best product mix 8hat is the optimal profit 3ns(er: #D55, #1D155
and profit DK@,555
+' 8hat are the shadow prices 8hen the optimal solution has been reached,
which resource has the highest marginal alue 3ns(er: •
'he shado( price for 1 additional labor?1;
•
'he shado( price for 1 additional po!nd of steel?
•
'he shado( price for 1 additional sno( blo(ers en%ine made available ?2
.hus, snow blower engine hae the highest marginal alue at the optimal solution' /' /er what range in each of the R%" alues are these shadows alid 3ns(er: •
'he shado( price for labor ho!rs is valid from = ho!rs to 1,AA.AA h o!rs
•
'he shado( price for po!nds of steel is valid from 1,po!nds !p to an infinite n!mber of po!nds
•
'he shado( price for sno( blo(er en%ines ran%es from 1= en%ines !p to 2; en%ines
2.?.2. C'ange- #! "'e /#e!!i/ien"- #! "'e #+e/"i0e !%n/"i#n (/) .o calculate changes in the coefficients of ob&ectie function we should consider only De/i-i#n 0a&ia+e- .De/i-i#n 0a&ia+e- can be:
i. 4asic &variable in the sol!tion)
51
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming ii. Hon/basic &variable o!t/of the sol!tion)
Note# Instead of resolving the entire problem as a new problem with new parameters, we ma ta7e the original optimal solution table as an initial solution table for the purpose of 7nowing ranges both lower and upper within
a. Range !#& "'e /#e!!i/ien"- #! +a-i/ e/i-i#n 0a&ia+e.here are two cases to consider changes for a ariable that is not currently in the solution mix and changes for a ariable that is currently in the solution mix'
T'e Range #! O"i*ai" (R#O) s the range oer which a basic decision ariable coefficient in the ob&ectie function can change without changing the optimal solution mix' %oweer, this change will change only the optimal alue of the ob&ectie function' *.he profitNcost figure+ Note *e general rule (or calculating t*e upper and lo+er range is
>llowa#le increase t*e smallest positie ratio o( C-, alue and t*e ariable substitution rate
E6a*e1 #ien the final tableau of the microcomputers problem, determine the rangeNs of optimality of the ariables that are already in the optimal solution mix'
S#%"i#n .he final simplex tableau for the problem is repeated here for conenience
?
<
B5
25
5
5
5
X 1
X 2
8 1
8 2
8 "
F
8 1
5
5
5
B
BN3
1
X 1
A
5
5
IN3
@
X 2
25
5
5
I
1N3
B5
25
5
5N3
J5
/
5
5
5
5 I5
I 5N3
3nalysis of X 1
52
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming P/e%&e
a. .a)e the G row of the optimal solution' b. .a)e the X 1 row alues' c. G = X 1 row X 2 8 1
5
5
8 2 8 " /#"-" ?@#
IN3
.he positie ratio is 5' .herefore, the coefficients of H can be increased by 5 without changing the optimal solution' .he upper end of its range of optimality is this amount added to its current *original+ alue' .hus its upper end is B5 5 D 55'
•
the smallest negatie ratio is I5; therefore the x coefficient can be decreased by as much as 5 from its current alue, ma)ing the lower end of the range e(ual to B5I5D 25
3nalysis of X 2 P/e%&e
a' .a)e the G row alues of the optimal solution b' .a)e the X 2row alues' c. G = X 1 row
X 1
X 2
8 1
8 2
8 "
G row ?%n ?9 ?%n /1 ?19 /#"-" ?/29 X 2 row
•
5
5
I
1N3
.he smallest positie ratio is 5' .his tells us that the H 1 coefficient in the ob&ectie function could be increased by 5 to 25 5 D B5'
•
.he smallest negatie ratio is I15, which tells us that x 1 coefficient could be decreased by 15 to 25I15 D35'
•
%ence, the range of optimality for the ob&ectie function coefficient of x1 is 35 to B5'
E6a*e 2
!ax CD2 # '2 #1 #3 "ub&ect to: 53
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming 2 #2'G #1
6 25
2xB'x12x3 6 JJ 1'G #1'G #3 63B #, #1 , #3 7 5
.he optimal tableau for this solution is: 4
2
'2
X 1
X 2
X "
5 8 1
5 8 2
5
F
8 "
X 1
:
'523
5
5'5BJ
5
5
5
X "
1
5
5'BJ
I5'511
5'5BJ
5
'G
8 "
5
'@1
5
5'12G
I5'JJ3
2'1
2
2 '31
5'3
5'5BJ
5
G
5
I5'G1
5
I5'3
I5'5BJ
5
2'G
Re;%i&e
-etermine the &ange #! #"i*ai" for the coefficients of the basicIdecision ariables' S#%"i#n Ana-i- #! +a-i/ e/i-i#n 0a&ia+e •
.he analysis will be conducted on products on X 1 and X " which are in the basic solution' -iide each < / row entry for ariables not in the solution *for instance, by X 2, 8 1 and 8 2 val!es+ the remaining alues result undefined so we can s)ip it
I. 3nalysis of X 1
a' .a)e the G row of the optimal solution of the nonIbasic ariables b' .a)e the X 1 row of the nonIbasic ariables c' G row X 1 row •
$pper 5imit
.he smallest positive n!mber in the / row tells how much the profit of X 1
Hcan be increased before the solution is changed' $pper 5imit ? &for X 1+ the smallest positive val!e of / row ?;@P?P
X 1 row Hote: &for X 1 ) ?;&5oo in the BO of the 56 problem)
54
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming •
5o(er 5imit
.he smallest ne%ative ratio &the n!mber closest negatie amount closest to 5+ 5o(er 5imit ? &for X 1+ the lar%est ne%ative val!e of / row ?;@ &/.=) ? #.2
X 1 row
.herefore, the range of optimality for the coefficient of X 1 is '1< &for X 1 ) < P (0*e coe''icient o' $ in t*e ob@ective 'unction can c*ange between 4.% and B wit*out c*anging t*e optimal solution mi# $=$/, =$.> and ! =$.$%)
II. 3nalysis of X "
•
$pper 5imit ? &for X "+ 'he smallest positive val!e of / row ?1/ &1) ?
•
X 1 row
5o(er 5imit ? &for X " )@'he lar%est ne%ative val!e of / row ?1@ &1#.1") ? 1;.1"
X 1 row
.herefore, the range of optimality for the coefficient of X " is 5 < &for X " ) < 1;.1" (0*e coe''icient o' in t*e ob@ective 'unction can c*ange between / and $.$ wit*out c*anging t*e optimal solution mi# $=$/, =$.> and ! =$.$%).6owever, t*is c*ange will c*ange only t*e optimal value o' t*e ob@ective 'unction (i.e. Ma# " will c*ange)
+. T'e &ange !#& "'e n#n+a-i/ 0a&ia+e-, &ange #! in-igni!i/an/e f there is a ariable , not participating in the optimal basis , then, in order for the ariable to be included in the optimal sol!tion , its coefficient in the ob&ectie function will hae to change from the existing to a new leel .
, ;new<5 T'e Range #! In-igni!i/an/e (R#I) •
is the range oer which rates for nonIbasic ariables can ary without causing a change in the optimal solution mix *ariable+ is called the range of insignificance
E6a*e
!ax CD2 # '2 #1 #3 "ub&ect to: 55
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming 2 #2'G #1
6 25
2xB'x12x3 6 JJ 1'G #1'G #3 63B #, #1 , #3 7 5
.he optimal tableau for this solution is: 4
2
'2
X 1
X 2
X "
5
5
5
8 2
8 "
8 1
F
X 1
2
'523
5
5'5BJ
5
5
5
X "
#.;
5
5'BJ
I5'511
5'5BJ
5
'G
8 "
5
'@1
5
5'12G
I5'JJ3
2'1
2
2'31
5'3
5'5BJ
5
5
I5'G1
5
I5'3
I5'5BJ
5
G
2'G
Re;%i&e
?ecause H1 is not in the solution, its ob&ectie function coefficient would need to increase in order for it to come into solution'
•
.he amount of increase must be greater than its
•
?ecause its current alue is '2 as indicated aboe, the current solution will remain optimal as long as its ob&ectie function coefficient does not exceed '2 5'G1D 2'31'
•
"ince an increase would be needed to bring it into solution, any lesser alue of its ob&ectie function coefficients would )eep it out of solution, hence the range of insignificance for H 1 is 2'31or less
athematicallyN
Range of insignificance for H 1D the /0 coefficient of H 1 *< I C+ alue of H 1 ? *for X 2+?#.; and / * for X 2+?/.=#2 ? #.; @ &.=#2) ? ;."#2 •
f the profit contribution of X 2 is greater than 2'31, then X 2 will be included in the solution'
•
.hus, P< *ne( for X 2 )< ;."#2 is the range of insignificance for X 2.
56
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming &profit contrib!tion) forX 2canary with in the gien range without causing a change in
•
the optimal solution mix'
E6e&/i-e' !ax CD25 # 15 #1 "ub&ect to: 1 # #1 6 G5 3xx16 B5 #, #17 5 Re;%i&e: -etermine the range of optimality for the coefficient of the basic ariables for the gien problem'
4
G5
B5
5
5
X 1
X 2
8 1
8 2
F
8 2
5
2N1
5
IN
15
X 2
A
3N1
N1
5
5
@5
B5
I5
5
35
5
I35
K1,55 5
G 3ns: 'he ran%e of optimality for X 2s profit coefficient is: KBB'B < &for X 2 ) < P
CHAPTER SUMMAR7 Linear programming models are used to find optimal solutions to constrained optimization problems' n order for linear programming models to be used, the problems must inole a single ob&ectie, a linear ob&ectie function and linear constraints, and hae )nown and constant numerical alues' 8e can sole linear programming problems using either graphical solution approach or simplex approach' .he graphical method for solution is used when the problem deals with 1 decision ariables' .he ine(ualities are assumed to be e(uations *in e(ualities should be changed to e(uations+' As the problem deals with 1 ariables, it is easy to draw straight lines as the relationship between the ariables and constraints are linear' n cases where the problem deals with at least three ariables, instead of lines we hae to draw planes and it will become ery difficult to isualize the feasible area' n contrast to the graphical approach, the simplex approach is not limited to problems with only two decisions ariables' t is a general purpose method to sole problems with any number of decision ariables' 57
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming .he simplex procedure inoles deeloping a series of tableaus, each of which describes the solution at a corner point of the feasible solution space beginning with the origin after standardizing the problem' .he standard form of LP problem should hae the following characteristics: ' All the constraints should be expressed as e(uations by slac) or surplus andNor artificial ariables '
.he right hand side of each constraint should be made nonInegatie; if it is not, this
should be done by multiplying both sides of the resulting constraint by I'
+ample: 2X 13 X 2I X " X ";, we multiply both sides by negatie
' .hree types of additional ariables, namely a' 8lac Eariable* 8 + b' 8!rpl!s variable */8 +, and c' 3rtificial variables * 3+ are added in the gien LP problem to conert it into standard form for two reasons: i' to conert an ine(uality to hae a standard form of an LP model, and ii' to get an initial feasible solution represented by the columns of an identity matrix .he summery of the extra ariables needed to add in the gien LP problem to conert it into standard form is gien below: 'ypes of constraint
+tra variables to be added
oefficient of e+tra variables in the obective f!nction a+ in
<
3dd only slac variable
>
8!btract s!rpl!s variable and
3dd artificial variable 3dd artificial variable
?
6resence of variables in the initial sol!tion mi+
Qes
Ho
/
@
Qes
/
@
Qes
"ensitiity analysis *post optimality analysis+ is an analysis concerned with proiding the decision ma)er with greater insight about the sensitiity of the optimal solution to changes in arious parameters of the problem' "uch changes might inole the specified leels of constraints or coefficients of the ob&ectie function' nterest in changes may arise due to improed information relating to a problem or because of the desire to )now the potential impact of changes that are contemplated'
58
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming
1 Test o$ optima!it& i. If all C D '1*, then the basic feasible solution is optimal ;&a!imi$ation case< ii. If all C D'5*, then the basic feasible solution is optimal ;&inimi$ation case<
2 .aria%!e to enter the %asis i. A variable that has the most positive value in the C D 'row ;&a!imi$ation case< ii. A variable that has the highest negative value in the C # 'row ;&inimi$ation case<
3 .aria%!e to !ea/e the %asis The row with the non#negative and minimum replacement ratio ;For both ma!imi$ation and minimi$ation cases i.e. % ++B0
SEL$ ASSESSMENT >UESTIONS AND E@ERCISES ' A furniture company produces a ariety of products' /ne department specializes in wood tables, chairs and boo)cases' .hese are made using three resources: labour, wood and machine time' .he department has B5 hours of aailable each day, B hours of machine time and 55 board feet of wood' A consultant has deeloped a linear programming model for the department' H D (uantity of tables H1D (uantity of chairs H3D (uantity of boo) cases !aximize CD 5x 35x1 2x3SprofitT "ub&ect to Labour:
1xx11'2x3 6 B5 hours
!achine: 'Gx'Bx1'5x3 6 B hours 8ood:
35x 15x135x3 6 55feet
.ables :x
75 units
Hx1x3 7 5 Re;%i&e answer the following (uestions based on the information gien'
a' 8hat is the main purpose of the model
59
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming b' 8hat are the decision ariables c' 8hat are the system constraints *name them+ d' 8hat is the meaning of the number 1'2 in the labour constraint e' Explain what is meant by x 7 5' f'
8hat does the term 5xin the ob&ectie function represents
1' Explain the similarities and differences between graphical method of soling linear programming problem andthe simplex method' 3' ?riefly define or explain each of these terms' "hadow price ncoming ariable Range of feasibility Piot row Range of insignificance Piot column Range of optimality Xey element "lac) ariable !aximization problem "urplus ariable !inimization problem Artificial ariable !ixed constraint problem /utgoing ariable ' A farmer has 555 acres of land on which he can grow corn, wheat and soybean' Each acre of corn costs birr 55 for preparation, re(uires J manIdays wor) and yields a profit of birr 35' An acre of wheat costs birr 15 to prepare, re(uires 5 manIdays of wor) and yields a profit of birr 15' An acre of "oybeans costs birr J5 to prepare, re(uires G manIdays of wor) and yields a profit of birr 15' f the farmer has birr 55,555 for preparation and can count on G555 manIdays of wor), how many acres should be allocated to each crop to maximize profit' Re;%i&e: A' formulate the LPP! of the problem'
?' 0ind the optimal solution using the simplex algorithm
:. 0ood H contains 15 units of itamin A and 5 units of >itamin ? per gram' 0ood 9 contains 35 units each of >itamin A and ?' .he daily minimum human re(uirement of itamin A and ? are @55 units and 155 units respectiely' %ow many grams of each type of food should be consumed so as to minimize the cost, if food H costs birr 5'B5 per gram and food 9 costs birr 5'G5 per gram' Re;%i&e: A' formulate the LP! of the problem
?' 0ind the optimal solution using graphical solution
8' An air force is experimenting with three types of bombs P, Y, and R in which three )inds of Explosie iz, A, ? and < will be used' .a)ing the arious factors into consideration, it has been decided to use the maximum B55 )g of explosie A, at least G5 )g of explosie ? and Exactly 25 of explosie <' ?omb P re(uires 3,1,1 )g of A,?,< respectiely' ?omb Y re(uires , , 3 )g of A, ?, and < respectiely' ?omb R re(uires , 1, 3 )g of A, ?, < respectiely' =ow bomb P will gie the
60
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming e(uialent of 1 ton explosie, bomb Y will gie 3 ton explosie and bomb R will gie ton explosie' $nder what production schedule can the Air 0orce ma)e the biggest bang Re;%i&e: formulate the LPP! of the model
?' A furniture wor)shop ma)es des)s, chairs, and cabinet and boo)cases' .he wor) is carried out in three ma&or departmentsZdesigning, fabrication, and finishing' .ime re(uired per unit of product in hours is -epartment -es)
b' n order to maintain to maximize profits what should be the production program
' "ole the following problems using graphical method a. !ax CD2HI5H1
"t:
+' !ax CD1HH1
"t: HBH1 6 3B5
H1H1 6 5
3H5H16 G5
H H1 6 B
5H2H16 1G5
H I H16 1
H, H1 75
H I1H1 6 H, H1 75
/'
!ax
CD5H2H1 ' '!in CDH1H1 "t:
"t: 1HH1 6 1B
3HH1 7 1J
1HH16 2B
I H IH1 6 I1J
IHH16 2
H 1 H1 7 35
H, H1 75
H , H1 75
' "ole the following problems using simplex methods a. !ax CD3 # 2 #1
b' !ax CD15 # 5 #1
"ub&ect to: #16 B
"ub&ect to: # #1 6 125
#1 #1 6 G
%#2 #1 6 25 61
OPERATIONS RESEARCH
TEACHING MATERIAL
Chapter Three -Linear Programming #, #17 5
#, #17 5
62
OPERATIONS RESEARCH
TEACHING MATERIAL