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MODEL:& B) {# x! |& |/ G3 f- F' O
0 L) [! r; M& d- s) I! The Vehicle Routing Problem (VRP); & E; }6 J! s* m/ \& ~) r
2 O6 `4 P+ F, c- z3 Q. G7 |; m!************************************;
4 Y+ k, [9 t z0 u! WARNING: Runtimes for this model ;# j, m" T. S: H$ K/ s0 f
! increase dramatically as the number;% X# i, I1 U1 l
! of cities increase. Formulations ;
* i3 v$ l! i1 Z* W; Y1 U! with more than a dozen cities ;3 h+ u2 Q G/ `
! WILL NOT SOLVE in a reasonable ;: S' m$ K; o" x2 {( w
! amount of time! ;; j4 |: S5 H G! g4 i
!************************************;
; J) k9 L; @6 t4 G9 B, N" U7 b& {, B+ c. K
SETS:
4 q, ?* [% v" @% E) ^6 v5 u ! Q(I) is the amount required at city I,
3 j+ Q* G/ B8 R) R9 ^6 J( ^; {" C U(I) is the accumulated delivers at city I ;
7 t' d( F5 ]' b% m* R* v1 q CITY/1..8/: Q, U;
+ |- p. _: M! D' k5 h" d; q% e, L9 I( e
! DIST(I,J) is the distance from city I to city J
6 Q; F6 |( l& j, \5 g X(I,J) is 0-1 variable: It is 1 if some vehicle
# t9 @5 t" ^8 ^; W; W2 x* j# a& ? travels from city I to J, 0 if none;7 m5 D- ^0 t$ B9 {
CXC( CITY, CITY): DIST, X;; M) b8 C1 k& T9 Y
ENDSETS* c; n5 [1 y2 t" C- K/ h
m1 E+ k H5 u! _8 } DATA:* v T' D" q) }' G5 a5 ?( B" \
! city 1 represent the common depo;, o2 Q5 h' ^4 O2 r& |% t6 j! q7 a
Q = 0 6 3 7 7 18 4 5;
- D. D! X- R6 _: y C) C
( w9 e# w, \! j+ x1 Y7 E ! distance from city I to city J is same from city) F* i- [7 L' o- e7 @) Q5 M
J to city I distance from city I to the depot is; N/ c" L# w& z/ f" h9 U6 N
0, since the vehicle has to return to the depot;# w8 H* Q7 l3 O: z
; w# W3 M/ h; I! b DIST = ! To City;4 ?! y) P$ K& @7 h
! Chi Den Frsn Hous KC LA Oakl Anah From;4 x0 i0 P, v/ x5 B8 [! X* m" c
0 996 2162 1067 499 2054 2134 2050!Chicago;2 T. V( E! s' c2 C Q6 z1 O( h2 t
0 0 1167 1019 596 1059 1227 1055!Denver;6 X5 B7 T6 o9 ^
0 1167 0 1747 1723 214 168 250!Fresno;
. k6 B: n7 O/ b; g3 u" A, ~ 0 1019 1747 0 710 1538 1904 1528!Houston;
* |) P9 ]# I3 a$ j& t! T 0 596 1723 710 0 1589 1827 1579!K. City;( E. I& S; J4 P4 H7 c, B" d4 A4 }
0 1059 214 1538 1589 0 371 36!L. A.;
9 ^' D* C4 Y5 H 0 1227 168 1904 1827 371 0 407!Oakland;
# c' R5 r- F* P2 Y 0 1055 250 1528 1579 36 407 0;!Anaheim;
+ A7 \# p: L: g* I) {
/ g+ R* U J- d' j ! VCAP is the capacity of a vehicle ;
, `( @2 M3 k# X. G, J% Z6 X. | VCAP = 18;
# @: h! A; p6 R4 \) D% W) H ENDDATA- R X p |6 j% K8 g
7 y' x2 y8 H8 A
! Minimize total travel distance;8 {! `; p3 u7 F& B, l
MIN = @SUM( CXC: DIST * X);
" T1 P) y" z0 S, I% T+ a! o9 C! B7 V0 B/ p* U
! For each city, except depot....;
3 m6 h; v1 l% y% y, ?. I @FOR( CITY( K)| K #GT# 1:( M& p0 N0 K* s' \7 @
0 o, ~4 y: H8 \6 m9 T3 F' S1 Z ! a vehicle does not travel inside itself,...;5 i( g3 q* T. P
X( K, K) = 0;# D0 q! y" _* c( p/ s
& }2 C" }. t' o5 w+ a( } ! a vehicle must enter it,... ;- a2 T5 o. V7 Y6 ]) a) R6 q6 Q( p: t
@SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#' e4 u3 j$ H( N" N% ^: e3 M0 K8 p
Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;2 Z9 C! {, x8 Z9 F: F- l, D j
( y! `9 T8 p l, F3 b' g/ J
! a vehicle must leave it after service ;
4 j8 D! m! U2 l( ] @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#7 f9 z5 _# b. j) S# c( J
Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;' `% O- J# Q2 Y# m2 a! a1 H, x7 {
; R7 u! K) J- P& S6 F ! U( K) is at least amount needed at K but can't
3 L f1 N" _" m! ^ exceed capacity;
% H4 f$ i$ w* p: \6 }0 B. V! { @BND( Q( K), U( K), VCAP);
+ X* D: s. D+ h" S, J+ ]9 ?8 m& A7 h8 j: d
! If K follows I, then can bound U( K) - U( I);
3 B% l% s$ D3 |4 U- V% J0 X @FOR( CITY( I)| I #NE# K #AND# I #NE# 1: 1 h! M# j @' f( G2 {" T
U( K) >= U( I) + Q( K) - VCAP + VCAP * % }. F' I% c# z) z5 V. {* u
( X( K, I) + X( I, K)) - ( Q( K) + Q( I))3 @2 i l5 X/ i
* X( K, I);: M. E u; B4 o7 ]9 X
);7 D) A9 k4 _( V! x3 d5 V; G' {. w, F
1 \, Q3 E, Z+ W ! If K is 1st stop, then U( K) = Q( K);0 o" F; `6 n' K
U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);
# W; B/ s: p) l$ M4 P. W$ d6 q) T
! If K is not 1st stop...;" {4 V& p: _9 B* K& o
U( K)>= Q( K)+ @SUM( CITY( I)| 7 a+ U8 S f3 x! e7 C
I #GT# 1: Q( I) * X( I, K));5 H' R: @7 C! J( w8 P, K* ?0 }5 U
);
! J% g: L# t; h/ R7 i @& Q6 L% X; M1 g3 y( j" Q0 A
! Make the X's binary;
3 Z+ j4 Y% R9 T" D( A* y @FOR( CXC: @BIN( X));
4 z" a( T/ Y0 S: ]; ?* A- \" u. A8 X6 ^3 ~* x* M2 E
! Minimum no. vehicles required, fractional 0 x- a# l0 H W- Z5 n: k/ M+ [4 G, D: r
and rounded;3 O1 H* ~% l' k) r: ^
VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;8 u5 R' o- ~* G, Z
VEHCLR = VEHCLF + 1.999 -
$ c) }4 ^: n5 C6 X( r0 V- R$ p @WRAP( VEHCLF - .001, 1);: I% B/ j. A9 E" j! L
6 X! m8 S, Y8 d2 y# w8 g+ U ! Must send enough vehicles out of depot;1 Y( ~0 u4 I0 b. @
@SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
0 t1 ^* I; b: N% j3 U* [. l( g X; K. R" j/ Y END/ j# F3 s1 n5 h3 s4 o
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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