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MODEL:7 U X. g% b) W! Y6 l. |
* X# O6 y& z$ l) `% j8 t! The Vehicle Routing Problem (VRP); + [! a8 y# P/ ^, w5 l
1 V; [; m; x) f$ L( z' T6 i8 o; E
!************************************;
* b0 w5 G& d) T- G, k! WARNING: Runtimes for this model ;$ u! V0 r+ }; d
! increase dramatically as the number;
+ A: j" _3 N' S! of cities increase. Formulations ;2 |: V4 c2 m! N0 ]& c6 a
! with more than a dozen cities ;
% X9 E5 G5 I" g% `- u4 s! WILL NOT SOLVE in a reasonable ;
8 O6 N: Q, w3 k! amount of time! ;
& i$ K7 G g* u+ x) \6 s: z6 E* F: v!************************************; D8 F/ ^/ G! z
1 A8 W$ E# e) V& W4 n. D
SETS:2 O6 |3 v% w, J5 H! T) |
! Q(I) is the amount required at city I,
* w$ v) P+ B/ o u/ S+ z' @- {/ {0 I2 p U(I) is the accumulated delivers at city I ;0 s2 ]2 N8 P0 I, ]
CITY/1..8/: Q, U;
$ t) C: A0 n" D5 A: D' M
- d4 \2 w0 c6 W4 i8 ~ ! DIST(I,J) is the distance from city I to city J
' A; \8 f' ?# z X(I,J) is 0-1 variable: It is 1 if some vehicle4 T/ z; W/ P" a& w3 {) ~
travels from city I to J, 0 if none;+ H0 |" Y. I7 V; j2 e: L
CXC( CITY, CITY): DIST, X;
' ]& j0 t2 g% B& w( A" {1 Q ENDSETS
" k( Y% d( w- [0 K- j2 K3 z' l, g" ]# z
DATA:
4 w* r1 ?3 w. Y" b0 @# B% Y. D! y4 Z/ k ! city 1 represent the common depo;6 P2 a( E; H! ^2 ^4 z
Q = 0 6 3 7 7 18 4 5;6 V- Y: h# ^3 \3 g/ m% g
3 ]; v( o( I/ f1 s5 H- g/ o ! distance from city I to city J is same from city
) T% Q% H6 V: ]1 W J to city I distance from city I to the depot is4 X+ G- {. W: b b/ {' k7 H* B/ h; E
0, since the vehicle has to return to the depot;
/ s) U; g N* q2 r
* q& \/ J) W. ~! V! I; O. g9 T DIST = ! To City;
8 ~* ^" E! d5 x* I/ a1 s; V3 V, w1 X ! Chi Den Frsn Hous KC LA Oakl Anah From;7 l T5 V% v* u {$ I
0 996 2162 1067 499 2054 2134 2050!Chicago;1 j& x' w! ]. L' D3 A) f
0 0 1167 1019 596 1059 1227 1055!Denver;
% ?: S4 c2 t$ u4 A R m 0 1167 0 1747 1723 214 168 250!Fresno;
$ t/ j: |8 @9 U I1 M7 }$ Q- C 0 1019 1747 0 710 1538 1904 1528!Houston;
9 L1 r/ A0 a% ~. S2 Y( m& x5 X( u 0 596 1723 710 0 1589 1827 1579!K. City;6 w ~( i( k3 D( ~$ D3 S# P' l
0 1059 214 1538 1589 0 371 36!L. A.;
+ W; X) z6 o+ u3 G 0 1227 168 1904 1827 371 0 407!Oakland;
4 ]9 t- L" v/ z9 |7 m3 g$ C 0 1055 250 1528 1579 36 407 0;!Anaheim;# |# |% U2 q1 z( r1 [ e' e$ Q
% k" N7 P" J1 _$ E" V
! VCAP is the capacity of a vehicle ;
6 a, |! E5 \ ` VCAP = 18;" b& w( D/ l7 I8 W+ n4 U k7 y" H
ENDDATA% j% N2 p/ l. x, K
8 t4 L/ _; X" I; E ! Minimize total travel distance;
9 [ _* X+ m! Z* L3 @ MIN = @SUM( CXC: DIST * X);
/ o# B$ M" c: [" v0 p) S. S' w( \: W: r o
! For each city, except depot....;
0 Y7 M" P4 p' n& S+ f% m0 ^7 U @FOR( CITY( K)| K #GT# 1:
2 \) t% ^- ~1 `/ u, B; a4 \
8 Y! T. `3 r0 D; }# `5 `6 E9 g' f ! a vehicle does not travel inside itself,...;
v7 @6 X9 E3 i P7 P0 Z7 s X( K, K) = 0;3 H; m% D9 `3 A1 B
' Y$ T# g9 y. t: z k% U ! a vehicle must enter it,... ;. n6 g3 o7 v4 u4 n
@SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#
5 a$ r0 M# d- L) @ Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;
- a( {& o" e1 _' E$ ]7 R/ T1 T% ^4 q* C& P: c, b2 A5 ]
! a vehicle must leave it after service ;
8 m9 q$ h& O0 F, Q5 D" b! e @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#! C3 {" E& ?! s4 U# \+ k
Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;
1 c; G5 q, P$ r$ o q; Z7 M7 I1 E
' `' ^* x2 J$ o$ H' w( f$ [ ! U( K) is at least amount needed at K but can't 8 @1 R: K1 Y! T6 P
exceed capacity;! t% S. g- p3 p9 O# Q; m7 c
@BND( Q( K), U( K), VCAP);
" |, d7 |! t# X/ Y
, ~$ m) ?$ X: B/ b ! If K follows I, then can bound U( K) - U( I);) @3 G/ z' k% x7 Q3 t# p4 |' E" O
@FOR( CITY( I)| I #NE# K #AND# I #NE# 1:
% [6 _# [) e3 y2 M; ]4 L8 y U( K) >= U( I) + Q( K) - VCAP + VCAP *
* N* o9 {) M7 Z! z ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
$ I# \% ~" m6 R( l+ S * X( K, I);0 j- f! P* Q2 M9 i
);. E+ b' x2 b, |0 V9 E" I# h" `
0 \2 B& X. K$ u8 Y- e5 {
! If K is 1st stop, then U( K) = Q( K);( ?1 b3 m( @4 R8 O) G. `+ W6 L+ J
U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);
# M7 P7 I. L1 N7 j+ p- }! o" j5 |) p u: e+ u* p. V5 h
! If K is not 1st stop...;
1 D! l7 Q/ F8 Q# M. U2 M, W# ~6 u U( K)>= Q( K)+ @SUM( CITY( I)| $ G5 {3 z' ~8 [
I #GT# 1: Q( I) * X( I, K));
. N( t5 `* P; H2 l );8 e. Z2 P/ x/ `
& f' X+ L C1 ~" ^
! Make the X's binary;, Q5 w9 R! X) V* ^& w
@FOR( CXC: @BIN( X));: e( C, j9 x% n& F6 e
; H( S6 {' [4 b
! Minimum no. vehicles required, fractional
' ]; W1 m7 n6 r% z and rounded;
+ \, c1 u+ E) v" Z# J VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;
: E2 ?% K9 y' T! H3 h% j/ J' u VEHCLR = VEHCLF + 1.999 -
' h G' D' ]+ G4 g+ Y% n @WRAP( VEHCLF - .001, 1);8 G5 ^- c& V4 S: E
) L) [. J2 f$ }( J. _
! Must send enough vehicles out of depot;# A! M' Q' h7 [5 ?" I- j( M/ }/ f
@SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;0 [) V' | \5 H6 f; x/ N& f
END
# r5 Z' f \ T, t6 a; J+ t: \" f, _ 请问大家里面U(I)的公式如何理解啊 U(I)是城市I 的累积交付量么?谢谢 |
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