C Minim Algorithm um ost s
Whe sim x is too difficult n ple
C putation of Table om au
putation for a LP Wewill focus on thecom proble m udocodefor a com r programto writea pute Thepse ne table give that wehavea basic fe w au, n asible solution could beas follows
e ct n st S le colum with highe valueof zj cj. This is colum k (Pk to beintroduce n d) e ct ctor in S le basis ve with m xi/xik for xik > 0. This is basis ve l (Pl to bere ove frombasis) ctor m d i l xij = xij xlj*xik/xlk (transform ations for ve ctors P0 to Pn) xkj = xlj/xlk z = c1x1 + c2x2 + ...+cmxm
Num r of C be alculations
Thefirst com parison should taken ste ps These cond com parison should takemste ps Thetransform ation take m + 1) ste s (n ps Thecalculation of z take mste s ps
Ne twork Proble m
be ps (n Thetotal num r of ste is n + m+ m + 1) + m = m + 3m+ n n twork proble , thenum r of constraints m be I n a ne is like to bee ly qual to thenum r of links, L, and be thenum r of variable m bethenum r of be s ight be flows plus constraints, pe rhaps N(N 1) + L
LargeNe twork
Thus thetotal num r of ste would beLN2 be ps LN + L2 + N2 + 4L N n twork be e ve large (e N = com s ry , g Whe thene 1000, L = 3000) this is dom inate by LN2 d e I f weassum that L is proportional to N, wehave com puting tim ~ N3 e
Num r of Ite be rations
be quire is ofte of d n Thenum r of basic solutions re theorde of N to 3N r ake ple e This m s thecom tesolution tim for such a proble ~ N4 m
Tim C ple e om xity
I n shorthand wesay com puting tim is O(N4) e r s Othe possibilitie areO(2N), O(N!) and O(NN) n , se r r ple s Whe N is large the highe orde com xitie m it im ake practicableto solvein re asonabletim e
C putation Tim om e
Orde of com xity r ple O(N2) O(N4) O(2N) O(N!) O(NN) Tim for N = 100 with 1 e GHz m achine 10-5 se conds 10-1 se conds 4*1013 ye ars -
Polynom Proble s ial m
ple e Wearelucky that thesim x m thod is "polynom in tim com xity ial" e ple m d asily Not all proble s can besolve so e se dge With our pre nt stateof knowle , weclassify proble s as follows m
C s of Proble s lasse m
m ial e Easy proble s polynom in tim m Probably difficult proble s O(2N) algorithmis known, but it is not known whe r a polynom the ial algorithme xists m re I ntractableproble s wecan provethe is no polynom algorithm ial cidableproble s cannot besolve m d Unde
Probably Difficult Proble s m
se m long to a class of proble s calle m d The proble s be "NP com te or "NP hard" ple " a re ing d This is an are whe a lot of work is be carrie out by m m athe aticians s re Weavoid such algorithm whe wecan
Lim on I te it rations
be xtre e Thenum r of e m points on a solution space for a sim x proble is thenum r of ways we ple m be can se ct mbasic ve le ctors fromn ve ctors This is
n n! = m (n - m)!m! ore ial This is m than polynom
Num r of Route be s
rage be Theave num r of links in a routewill be be e N and N twe n ach , be At e nodein a route thenum r of possible ne hops will be(L/N - 1) on theave . I n a xt rage re alistic ne twork this will bebe e about 2 twe n o be s S thenum r of possibleroute will beof the orde of 2N to 2N r
LargeNum rs of Route be s
o m s ach S a proble which include a condition for e possibleroutewill beprobably difficult re any s The areso m ways of putting theroute toge r that theproble can be e the m com astronom ically large
Avoiding theNP C ple om te
e ve ple e Wese that e n thesim x m thod could involve us in NP com tealgorithm ple s Wehavetwo ways of avoiding this situation spe cialist solutions to particular proble s and m he uristics
Minim S al panning Tre e
a cialist algorithm is "m al s inim Oneare for spe spanning tre s" e cte cte I n a conne d, undire d graph which has costs associate with e e , find thetre which d ach dge e conne all thenode so that thetotal cost of the cts s e s use is m ise dge d inim d
Minim S al panning Tre e
ost ngth in this spanning tre e C = le
Minim S al panning Tre e
re ve s The arese ral algorithm for finding the m al spanning tre inim e y pe sult alle The de nd on there that st thesm cost e incide to any nodeof G is a part of any dge nt m al spanning tre inim e
Prim Algorithm 's
tarts fromonenodeand adds ne e s to the w dge S e xisting tre until it include n 1 e s e s dge i is de d as thecost of thee be e one fine dge twe n nodewithin thee xisting tre and onenode ni, not e , ye include t d i = ni
Prim Algorithm 's
aintain value of s During thealgorithmwem and for e nodenot ye in thetre ach t e ach ration wechoosethelowe cost e st dge At e ite crossing fromthetre to there of thegraph, and e st thenodeto which it conne cts
Prim Algorithm 's
k=1 te e S p 1: Find thenodenj* not in thetre , such that j* = m j). Movenodenj* and thee in( dge conne cting it, into thetre e te S p 2: If k = n 1, stop te e in( s S p 3: S t j = m j, cjj*) for all node not in thetre . S t k = k + 1 and re e e turn to ste 1 p
Exam ple
n1 2 n2 3 1 7 n3 5 8 4 n5 6 n4
inim Find m al spanning tre for this e ne twork dge Thee costs are shown tart S with noden5
First I te ration
n1 2 n2 3 1 7 n3 5 8 4 n5 6 n4
Valueof , = 1, 2, 3, = , , 5, hoosen = 4 C = 1, 2, 3 = 6, 3, 5
4 4
S cond Ite e ration
n1 2 n2 3 1 7 n3 5 8 4 n5 6 n4
Valueof , = 1, 2, 3 = 6, 3, 5 hoosen = 2 C = 1, 3 = 2, 5
Third Ite ration
n1 2 n2 3 1 7 n3 5 8 4 n5 6 n4
Valueof , = 1, 3 = 2, 5 hoosen = 1 C =3 =1
Fourth I te ration
n1 2 n2 3 1 7 n3 5 8 4 n5 6 n4
Valueof , =3 =1 Only choiceis n = 3
UniqueS olution?
sult e ve There will bethesam , whate r nodewe start from ve e e re Howe r, if som link costs arethesam , the m bem than onem al spanning tre ay ore inim e
C ple of Algorithm om xity
te quirea num r of ope be rations S ps 1 and 3 re proportional to n at theworst re rations The aren 1 ite re ple The forethecom xity is O(n2)
C TV able
le m inim e I n te com unications, them al spanning tre is re quire to de how to providecableTV to a d cide se of custom rs t e e s ach Thesam traffic flow passe along e link, and thele ngth of therouteis not im portant
Lowe Cost Path st
n twork, we Whe weroutetraffic through a ne want to do so with thelowe possiblecost st d rtain cost and the Each link is associate with a ce proble is to find theroutewith thelowe cost m st froma sourcenodeto any (all) othe node r s
S horte Path Algorithm st s
e ral s e d, S ve algorithm havebe n propose such as the Be an-Ford, algorithmand Floyd's algorithm llm ost ly d Probably them wide use algorithmis theone propose by Dijkstra d
Dijkstra's...