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- 乐观运动积极向上
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MODEL:
e8 v9 |) g2 O' U" L+ p1 Z4 z: l/ E
! The Vehicle Routing Problem (VRP); ( \# b8 X/ H3 a: r' i9 v$ P
5 X I$ V- Q- b
!************************************;
: ~$ a, C- u; F2 F+ Z, ?! WARNING: Runtimes for this model ;
+ Z/ g5 i. Y2 L# ?0 a3 v) q! increase dramatically as the number;
. M& j6 h$ N7 M1 W5 k! of cities increase. Formulations ;
7 a0 j# f5 p1 k3 C7 h! with more than a dozen cities ;
4 `2 P$ w3 F B0 S! B0 z1 S! WILL NOT SOLVE in a reasonable ;
1 V" H& P. _, `! amount of time! ;% ~3 q' U& n7 D
!************************************;
* f- E' s& p+ v2 c% {% f) J& j
' Y( C3 ?' Y9 s g SETS:
; t( Z6 N+ Z- `5 ? ! Q(I) is the amount required at city I,
9 B7 g8 ^( ^& r3 T/ T U(I) is the accumulated delivers at city I ;
& ]6 ], Q& g% S5 O) B, \! g/ { CITY/1..8/: Q, U;
+ q3 \6 v+ X( v( M
4 r5 F& F- k9 ~! {7 m" Y ! DIST(I,J) is the distance from city I to city J( n9 w+ K! I! L9 \0 K
X(I,J) is 0-1 variable: It is 1 if some vehicle
* \0 y% s6 Z6 y/ e( k! L. x travels from city I to J, 0 if none;0 u- e) S' E! R5 q* o; e( l2 F* u
CXC( CITY, CITY): DIST, X;
! p% ?. K, [: z7 o5 k6 j( l" F. p ENDSETS
# H$ x2 d: M: W ^, H* G6 {
/ s0 N) _/ y7 p- g0 J# R% M" O DATA:
! X, e8 `2 u+ u; I$ }2 W ! city 1 represent the common depo;
0 R; ]' z/ K) m% k' Q; c Q = 0 6 3 7 7 18 4 5;8 O+ i, f% e4 A. Y4 C7 n
! q6 P2 i9 J! o ! distance from city I to city J is same from city
+ @# |" H$ h. g& R4 O: |9 t9 s% q J to city I distance from city I to the depot is
! d/ z8 |. m1 d+ N 0, since the vehicle has to return to the depot;
% y0 j$ k! h5 F: n+ [$ S) `! Z. @
DIST = ! To City;* ~' T2 q: E5 n* C+ w4 F
! Chi Den Frsn Hous KC LA Oakl Anah From;
0 o9 Z9 B6 E: s2 d) M9 |/ p 0 996 2162 1067 499 2054 2134 2050!Chicago;
, h* A, y" g2 F; ~: N3 ] 0 0 1167 1019 596 1059 1227 1055!Denver;+ b1 s2 Y1 ?) o
0 1167 0 1747 1723 214 168 250!Fresno;, J' T0 [% F F) c4 e. U
0 1019 1747 0 710 1538 1904 1528!Houston;
7 _9 E8 z9 {( D c) E* z+ Q 0 596 1723 710 0 1589 1827 1579!K. City; {5 f+ d$ |2 m+ f* o9 L$ j6 T% D5 i
0 1059 214 1538 1589 0 371 36!L. A.;( t- Z4 [& {7 q$ X4 \5 m
0 1227 168 1904 1827 371 0 407!Oakland;
" n: y8 _1 s: m 0 1055 250 1528 1579 36 407 0;!Anaheim;+ G3 b# C; F$ W5 G6 M, ~1 i6 R
3 a8 Q1 Z7 }# j X4 s ^
! VCAP is the capacity of a vehicle ;
" {2 Q$ y2 z$ C VCAP = 18;
$ J( m- Y5 Q% v2 t3 z- T ENDDATA
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3 v# q7 j5 F* m ! Minimize total travel distance;
7 i! C$ f, D% P/ S! Z MIN = @SUM( CXC: DIST * X);
* q) i& d% C; H3 x/ _# }# H
$ I4 S. P- v. P Y# y ! For each city, except depot....;
' k# N- q3 m2 x* Y* O6 N @FOR( CITY( K)| K #GT# 1:
3 v7 N4 p. n) ]$ u9 P, h6 W1 ^4 m* r, F b/ o3 X- y
! a vehicle does not travel inside itself,...;/ l$ Z" j- {8 g" M
X( K, K) = 0;$ n9 H" ]6 p- m- {) [/ ?
6 L% ^! k4 ^; X1 ^: S2 z
! a vehicle must enter it,... ;
$ u4 B, K0 B( q% W4 x @SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#
. `* y% T+ d$ s5 p6 p9 ?9 w Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;; b! F5 W1 e4 `0 W& f* u C
5 M3 v5 W0 H/ s2 Y- L ! a vehicle must leave it after service ;
7 Z" g* v8 L/ ~1 b7 I @SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#
, h" d2 W Q0 W2 Y1 |: i Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;
) O+ |7 k. ?6 T% m, g
5 Y* u) j" K' k% g9 e/ S- t ! U( K) is at least amount needed at K but can't
0 _. o1 {$ C/ Y1 `( i exceed capacity;
4 r3 E: N4 |5 [) H2 } @BND( Q( K), U( K), VCAP);2 W& \# {; x o" @! V
+ H( ]2 F( I2 w' l! u4 G( l ! If K follows I, then can bound U( K) - U( I);- L E) w6 L& y: {* o: H! ]& ~- [
@FOR( CITY( I)| I #NE# K #AND# I #NE# 1: ! Y( V. x* |8 ?) E+ Z
U( K) >= U( I) + Q( K) - VCAP + VCAP * : v% H+ p% e* l4 Q
( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
$ h3 b \1 G4 K0 w * X( K, I);
8 Z; d4 |4 m; i. A+ U# R6 G );% C: }- U$ M4 [0 H, K8 w C( a
' B( h* Z! j/ n% e
! If K is 1st stop, then U( K) = Q( K);7 T7 s6 R/ w, F* [: Q
U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);+ _; H4 o0 M3 c4 }% W/ k
5 R0 u# o' |4 ~5 U8 q7 j* x( Y ! If K is not 1st stop...;
/ E' v& L0 K: z9 U& W" Q U( K)>= Q( K)+ @SUM( CITY( I)|
: t; D, z1 V, [! F/ C I #GT# 1: Q( I) * X( I, K));; X0 V, s$ k# H( T/ y
);6 Y! L7 J4 F! M$ k
- P: M* _; Y; |. r: C ! Make the X's binary;
& r0 y$ ~' j- J$ H @FOR( CXC: @BIN( X));
6 P5 ]* v8 X/ S" W2 Z9 R! N% r
4 d# a% ~+ s% y; z, ` ! Minimum no. vehicles required, fractional
* ?6 S. E1 m4 |$ g$ F and rounded;' v* v z' B4 y4 e+ d1 s
VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;
1 r- n% d' g9 | VEHCLR = VEHCLF + 1.999 - / _& Q. G+ u+ O }9 H$ F
@WRAP( VEHCLF - .001, 1);9 J6 K3 u# ~ @$ {, M- N
% y2 K6 ~/ D4 h
! Must send enough vehicles out of depot;
, Z& U( ?' W6 ]5 V: u( K7 E @SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;
. }" |$ S4 v% X* M END
8 r Y: t7 L- v3 p, z 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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