the simplex method
Post on 03-Jan-2016
46 Views
Preview:
DESCRIPTION
TRANSCRIPT
The Simplex Method
Standard Linear Programming Problem
Standard Maximization Problem
1. All variables are nonnegative.
2. All the constraints (the conditions) can be expressed as inequalities of the form:
ax + by ≤ c, where c is a positive constant
Illustrating Example (1)
Maximize the objective function:P(x,y) = 5x + 4ySubject to:x + y ≤ 202x + y ≤ 35-3x + y ≤ 12x ≥ 0y ≥ 0
Solution
0
12
35
20
100045
010013
001012
000111
:
:
045
:
123,352,20
::,,
"",Re
45123,352,20
:
:
tsanconstpwvuyx
tablefollowingtheconstructWe
Second
pyx
followsasfunctionobjectivetheofformulatherewriteand
wyxvyxuyx
followsasfollowsaswandvu
riablesvaslacktroducinginbyesinequalitithewrite
yxpandyxyxyx
hadWe
formulaandconditionsgiventherewriteWe
First
)(2,
,2/35.2/35201/20:,
)
(
.)(
.
.).5(
:)(
:.
:),(
:
0
12
35
20
100045
010013
001012
000111
entryelementpivottheistheiswhichentryingcorrespond
thesoandsmallesttheisquotientTheandarequotientstheHere
rowpivottheislocated
isitwhererowtheandelementpivottheisquotientsmallestthetoingcorrespondentryThe
quotientahavenotdowethennegativeisentrythatIfquotientpositiveaobtainto
columnpivottheinentrypositiveingcorrespondthebyroweachinntconstathedivideWeb
columnxtheiscolumnpivottheThushereiswhch
linethetoleftrowlasttheinentrynegativemostthecontainingcolumntheiswhich
columnpivotthelocateWea
followsasentryelementpivotthelocateWe
Third
tsanconstpwvuyx
2175
2129
235
25
25
23
23
25
21
21
21
21
)5(,)3(,)(
235
21
21
2
1
1000
0100
0001
0010
0
12
20
100045
010013
0001
000111
0
12
35
20
100045
010013
001012
000111
.;'0
1
423212
2
tsanconstpwvuyx
tsanconstpwvuyx
tsanconstpwvuyx
columnunitatocolumnthatingtransformthusstocolumnpivottheinentriesotherthe
andtoelementpivotthetransformtooperationsrowelementarynecessarytheperformWe
Fourth
RRRRRR
R
)(2
1,,
2
55
425
5
129
2
5/
2
12935.
2
1/
2
35,5
2
1/
2
5:,
)
(
.)(
.
.).2
3(
:)(
:.
)3(
)4()3(
1000
0100
0001
0010
2
3:,,
.,,
:
2175
2129
235
25
25
23
23
25
21
21
21
21
entryelementpivottheisrowfirstthein
theiswhichentryingcorrespondthesoandsmallesttheisquotientThe
andarequotientstheHere
rowpivottheislocated
isitwhererowtheandelementpivottheisquotientsmallestthetoingcorrespondentryThe
quotientahavenotdowethennegativeisentrythatIfquotientpositiveaobtainto
columnpivottheinentrypositiveingcorrespondthebyroweachinntconstathedivideWeb
columnytheiscolumnpivottheThushereiswhch
linethetoleftrowlasttheinentrynegativemostthecontainingcolumntheiswhich
columnpivotthelocateWea
Step
andstepsepeatR
tsanconstpwvuyx
iswhichrowlasttheinentrynegativeahaveweHere
fourandthreesteprepeatweentriesnegativcontainsstillrowlastthethatafterIf
Fifth
95
52
15
5
101300
014500
001101
001110
1000
0100
0001
0010
.;
'01
)4(
4131211 )3(,)5(,)(,2
2175
2129
235
25
25
23
23
25
21
21
21
21
tsanconstpwvuyx
tsanconstpwvuyx
columnunitatocolumnthatingtransformthus
stocolumnpivottheinentriesothertheandtoelementpivot
thetransformtooperationsrowelementarynecessarytheperformWe
Step
RRRRRRR
520,0,5,15max
95max
9552,0,5,15
:
).(
1&
))((
:
.
,
:
wandvuyxwhenoccursvalueThis
ispfunctionobjectivetheofvaluetheTherfore
pandwvuyx
Thus
columnunitawithassiciatedonesthe
ariablesvbasicforthecontainingrowtheinlyingntstaconingcorrespondthe
columnunitaheadingwithassiciatednotonesthe
ariablesvnonbasicallforzero
aresolutinoptialthistoariablesvtheofvaluesingcorrespondThe
reachedbeenhassolutionoptimalthethatmeanswhich
positivearelinetheofleftthetorowlasttheinetriesallNow
Sixth
What about when all of the constraints (the inequalities) are of
the type “≤ positive constant”But we want to minimize the objective function instead of
maximizing.
Minimization with “≤” constraintsIllustrating Example (2)
Minimize the objective function:p(x,y) = -2x - 3ySubject to:5x + 4y ≤ 32x + 2y ≤ 10x ≥ 0y ≥ 0
SolutionLetq(x) = - p(x) = - ( -2x -3y) = 2x + 3yTo minimize p is to maximize q. Thus, we solve the
following standard maximization linear programming problem:
Maximize the objective function:q(x) = 2x + 3ySubject to:5x + 4y ≤ 32x + 2y ≤ 10x ≥ 0y ≥ 0
Rewriting the inequalities as equations, by introducing the “slack” variables u and v and the formula of the objective function as done in example (1).
5x + 4y ≤ 32 , x + 2y ≤ 10 and q = 2x +3y
Are transformed to:
5x + 4y + u = 32
x + 2y + v = 10
- 2x - 3y + q = 0
)3,4(),(17,min
:,
)3,4(),(17,max
173,4:
17
.
3
4
100
.....
010
001
:),1(
0
.
10
32
10032
.....
01021
00145
:tableaussimplextheconstructWe
0 q3y2x- 10,v2y x32,u4y 5x
,
67
61
65
61
32
31
yxatiswhichvaluetheattainsp
thenqpSince
yxatiswhichvaluetheattainsq
qandyxThus
antsconstqvuyx
atarriveweexampleinxplainedeasmethodtheApplying
antsconstqvuyx
haveWe
Standard Linear Programming Problem
Standard Minimization Problem
1. All variables are nonnegative.
2. All the constraints (the conditions) can be expressed as inequalities of the form:
ax + by ≥ c, where c is a positive constant
Solving
The Standard Minimization Problem
We use the fundamental theorem of Duality
Illustrating Example (3)
Minimize the objective function:p(x,y) = 6x + 8ySubject to:40x + 10y ≥ 240010x + 15y ≥ 21005x + 15y ≥ 1500x ≥ 0y ≥ 0
Minimize the objective function: p(x,y) = 6x + 8ySubject to:40x + 10y ≥ 2400, 10x + 15y ≥ 2100 , 5x + 15y ≥ 1500, x ≥ 0 and y ≥ 0We will refer to the above given problem by the primal (original) problem
First: We construct the following table, which we will refer to by the “primal” table:x y constant---------------------------------40 10 240010 15 21005 15 1500---------------------------------6 8
Second: We construct a dual (twin) table from interchanging the rows and columns in the primal table:
x' y' z' constant----------------------------------------------------------- 40 10 5 6 10 15 15 8---------------------------------------------------------2400 2100 1500
Third: We interpret the “dual table” as a standard maximization problem, which will refer to as the “dual problem” or “twin problem” of the “primal problem” or the “original problem”
Miaximoze the objective function: q( x ' , y ' , z ' ) = 2400x' + 2100y' + 1500z'Subject to:40x' + 10y' + 5z' ≤ 6, 10x' + 15y' + 15z' ≤ 8 , x' ≥ 0 and y' ≥ 0, z' ≥ 0
Fourth: We apply the simplex method explained in example (1) to solve this problem
Maximize the objective function: q(x,y,z) = 2400x' + 2100y' + 1500z'
Subject to:
40x' + 10y' + 5z' ≤ 6, 10x' + 15y' + 15z' ≤ 8 , x' ≥ 0 and y' ≥ 0, z' ≥ 0
4.a.Rewriting the inequalities and the formula of the objective function, with the slack variables being the same x and y (in that order) of the original (minimization) problem :
40x' + 10y' + 5z' + x = 6
10x' + 15y' + 15z' + y = 8
- 2400x' - 2100y' - 1500z‘ + q = 0
4.b. We construct the simplex table for this problem
1140
.
11203045000
......
010
001
:
..4
0
.
8
6
100150021002400
......
010151510
00151040
0 q z1500 - 2100y' - 2400x' -
8 y 15z' 15y' 10x'
6 x 5z' 10y' 40x'
2513
501
252
501
1011
501
1003
203
antconstqyxzyx
tablefollowingthetoleadwillwhichmethodsimplextheby
dnecessiateoperationsrowelementarytheallperformWec
antconstqyxzyx
Fifth: We read the solution from the table
4)2(20))2,0(
2,0:max
:
440)2(20)2,0(
:
4maxmin
20:
:""
min)(
4
.
2
3
12001
......
01011
01101
q
yxwhenisq
Checking
p
pofformulathefromanswerthisgetalsocanWe
qofvaluethepofvalueThe
yandxThus
yandxariablesvslacktheunderrowlasttheinappears
problemimizationoriginalprimalthetosolutionThe
antconstqyxyx
Illustrating Example (4)
Minimize the objective function:p(x,y) = x + 2ySubject to:-2x + y ≥ 1- x + y ≥ 2x ≥ 0y ≥ 0
Minimize the objective function: p(x,y) = x + 2ySubject to:-2x + y ≥ 1, - x + y ≥ 2 We will refer to the above given problem by the primal (original) problem
First: We construct the following table, which we will refer to by the “primal” table:x y constant----------------------------------2 1 1-1 1 2---------------------------------1 2
Second: We construct a dual (twin) table from interchanging the rows and columns in the primal table:
x' y' constant------------------------------------------- -2 -1 1 1 1 2-----------------------------------------1 2
Third: We interpret the “dual table” as a standard maximization problem, which will refer to as the “dual problem” or “twin problem” of the “primal problem” or the “original problem”
Maximize the objective function: q( x ' , y ‘ ) = x' + 2y' Subject to:-2x' - y' ≤ 1, x' + y' ≤ 2 , x' ≥ 0 and y' ≥ 0
Fourth: We apply the simplex method explained in example (1) to solve this problem
Maximize the objective function: q( x ' , y ‘ ) = x' + 2y'
Subject to:
- 2x' - y' ≤ 1, x' + y' ≤ 2 , x' ≥ 0 and y' ≥ 0
4.a.Rewriting the inequalities and the formula of the objective function, with the slack variables being the same x and y (in that order) of the original (minimization) problem :
- 2x' - y' ' + x = 1
x' + y' + y = 2
- x' - 2y' + q = 0
4.b. We construct the simplex table for this problem
4
.
2
3
12001
......
01011
01101
:
..4
0
.
2
1
10021
......
01011
00112
0 q 2y' - x'-
2 y y' x'
1 x ' y' - 2x' -
antconstqyxyx
tablefollowingthetoleadwillwhichmethodsimplextheby
dnecessiateoperationsrowelementarytheallperformWec
antconstqyxyx
Homework
1. Using the simplex method, maximize: p = x + (6/5)y subject to:2x + y ≤ 180 , x + 3y ≤ 300 , x ≥ 0 , y ≥ 0Solution: p(48,84) = 148.8
2. Minimize: p(x,y) = - 5x - 4y Subject to: x + y ≤ 20 , 2x + y ≤ 35 , -3x + y ≤ 12 , x ≥ 0y ≥ 0Solution: p(15,5) = - 95
3. Using the dual theorem, minimize: p = 3x + 2y subject to:8x + y ≥ 80 , 8x + 5y ≥ 240 , x + 5y ≥ 100, x ≥ 0 , y ≥ 0Solution: p(20,16) = 92Maximize the objective function:
top related