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MODEL:9 y8 y/ a y8 u( r2 @
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! The Vehicle Routing Problem (VRP);
/ z! N% n. b8 f7 [- G0 V
( }2 M, T9 C3 l# r, r9 @!************************************;" j7 d* r$ `+ Q% M
! WARNING: Runtimes for this model ;
- \" H3 r! E; [' {5 y! increase dramatically as the number;( I$ i: j2 k$ q4 K4 L5 G) ~" I
! of cities increase. Formulations ;7 o7 J- D8 K1 g1 L( c) U) n
! with more than a dozen cities ;5 u1 j2 O; r) N3 R; Z
! WILL NOT SOLVE in a reasonable ;
: Q, K+ J) _9 O1 E/ {) D! amount of time! ;) x/ V& N" P& _; ]
!************************************;9 Q0 ]6 _# I ?* b/ `
$ t) c9 K" C: d( w' P9 [; E1 O
SETS: x" D" J! [6 r& V2 O1 A6 S
! Q(I) is the amount required at city I,
1 h, Y+ P9 @4 m1 B2 j4 m U(I) is the accumulated delivers at city I ; m ]4 I- d1 Y/ e8 B& I% [
CITY/1..8/: Q, U;
- m2 Y+ y3 w' {& N
' l. J1 R$ m8 u3 ~ ! DIST(I,J) is the distance from city I to city J
/ |3 P9 w: `& h. r) n X(I,J) is 0-1 variable: It is 1 if some vehicle
, h8 h. B5 X* I travels from city I to J, 0 if none;" O! ]' C" P9 x# B: Y
CXC( CITY, CITY): DIST, X;; A0 c* q0 p" J2 X) I) d/ g! Y) Z2 H
ENDSETS) r* }8 p# L; u; n) T# M, U5 q! X
# X& @4 {8 u ^6 P/ o& C DATA:
- e9 |# {; C) }+ _3 C" n ! city 1 represent the common depo;
( R+ `, A/ v1 I; D ^/ _ Q = 0 6 3 7 7 18 4 5;
' [# c3 O8 _# C" u( t8 u. P
- Q; U) Y6 C7 p/ s: y0 e+ w ! distance from city I to city J is same from city
3 k: p" u/ G& x! X% S% u$ H J to city I distance from city I to the depot is$ `7 E4 ^7 p" D
0, since the vehicle has to return to the depot;0 ^( C/ Z! Y$ r: s
; p) M1 Y+ h# w# Z$ o. b7 g. I8 i DIST = ! To City;% _# t& _$ t2 n$ S
! Chi Den Frsn Hous KC LA Oakl Anah From;. v7 Z C; @0 ]! g/ k' j; U
0 996 2162 1067 499 2054 2134 2050!Chicago;, j: i- ?4 J! ?0 z, @
0 0 1167 1019 596 1059 1227 1055!Denver;
# D: ]' x1 v$ o8 n2 A 0 1167 0 1747 1723 214 168 250!Fresno;& L7 Q+ F5 |5 N v
0 1019 1747 0 710 1538 1904 1528!Houston;
8 U) E: X' a- P! w2 H 0 596 1723 710 0 1589 1827 1579!K. City;
( \. N( o+ p3 |4 r9 Z4 }2 B 0 1059 214 1538 1589 0 371 36!L. A.;
+ f9 j {. _7 F1 j 0 1227 168 1904 1827 371 0 407!Oakland;
% V' W4 l4 x. R0 `' T* m; a' h 0 1055 250 1528 1579 36 407 0;!Anaheim;
1 M7 B+ }" O- k9 q; ^) U. `$ e( T4 \, z; k
! VCAP is the capacity of a vehicle ;
% x8 O4 ]( Z, w1 R VCAP = 18;! M) y' N0 u9 \2 m
ENDDATA
+ v* M7 s. V! `) U) w% S
( a: ]8 I: y) { U ! Minimize total travel distance;+ Q$ E) t9 ~# ^: k4 g
MIN = @SUM( CXC: DIST * X);: J+ w; u" j6 w5 M% D+ x- X3 O
6 q c4 ?8 h5 S: f9 P6 i
! For each city, except depot....;* d0 c* ?; i3 V
@FOR( CITY( K)| K #GT# 1:
( G- h ?! e2 G/ W( y: R: c9 U' {
( G5 s! t' X( S5 N! Q A+ Y" i ! a vehicle does not travel inside itself,...;: ]/ l; f( ~, y% ~5 J% x: u4 t
X( K, K) = 0;6 A! {6 e1 F" F; j, m
* Y o# S5 A" P) r$ z5 E# ? P: l
! a vehicle must enter it,... ;6 m s6 p3 x4 m6 X P! h6 K
@SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#
9 w0 U: \2 R# Q- |$ Q. \, M Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;
3 M% H0 {* n& b- }
. D9 {3 U6 {1 W) t) b0 g+ v/ E9 g" v2 e ! a vehicle must leave it after service ;* ^, I, B$ q1 q$ ^: a* r- r
@SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#% @0 `; C5 h7 ]5 \" ]& n
Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;) G( C: f! F5 p# y' Q( Y- M
% y' \, t3 N4 Q
! U( K) is at least amount needed at K but can't : G% |" ~4 s7 x! B) h8 Z
exceed capacity;
* n) j- `7 ^4 O1 Y* Z- D @BND( Q( K), U( K), VCAP);# z; K/ C& f& w# R, O+ ?0 [8 _. E% m
# w$ }( a$ Y7 e
! If K follows I, then can bound U( K) - U( I);
8 v; g5 g* {. |; B0 a N @FOR( CITY( I)| I #NE# K #AND# I #NE# 1: $ L8 K" `: C9 X0 I/ b- r# r
U( K) >= U( I) + Q( K) - VCAP + VCAP *
, m( e1 y; a( }# X5 C ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))' o/ U4 i! _$ H! }. T; b8 m7 x
* X( K, I);4 c- d+ D! b$ D0 K' B
);
4 Y4 A; n9 [ B) i" D ?# d# t
# d$ X G; ]0 K$ b ! If K is 1st stop, then U( K) = Q( K);
) W C. p/ i* A U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);
& g v; Z6 E0 S- h6 M6 M$ ^! O T. A: G5 C
! If K is not 1st stop...;8 r/ W7 b. D+ i8 Y" e% k. p. }( ?
U( K)>= Q( K)+ @SUM( CITY( I)| & X; Q0 C: v+ w/ l' x, e8 A5 V. Q
I #GT# 1: Q( I) * X( I, K));* a1 P, F+ K8 P! a$ w3 ?
);
6 a" v1 w& P2 }) C! [
' c: v! L' A. @& R9 T: i ! Make the X's binary;( V2 F* [: B; O* X
@FOR( CXC: @BIN( X));" c, |) t& i* i6 O) [5 }" K
1 Y6 K$ m$ a- e; }8 g5 D
! Minimum no. vehicles required, fractional
: B' x* L1 R% `: N: t and rounded; y7 x5 [5 H" [5 `
VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP; c' \( k: L" q# U
VEHCLR = VEHCLF + 1.999 -
$ j/ N, {5 u9 t+ u9 j/ a @WRAP( VEHCLF - .001, 1);
z5 Y) W& D+ w* l7 d! s0 L4 I( `1 M" |. { z- ?2 t
! Must send enough vehicles out of depot;0 w% ^0 t( M% E) i& F( ~. V6 D
@SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
7 {$ a9 H' [' E' o9 i END
) x8 ]! J, u y# _1 H, W 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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