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MODEL:* P: k# D- B) ~
# C* ^0 d, r( ]2 ?$ u7 e, A" {4 l! The Vehicle Routing Problem (VRP);
_ Q' S+ y/ B8 N! g9 g" a% |! e2 n$ A8 J R' m
!************************************;
4 u4 p1 _. u% q! l) z4 \( j! WARNING: Runtimes for this model ;
3 p& B$ ^9 U7 n4 {" j8 m! z! increase dramatically as the number;2 v$ T- u2 }2 |
! of cities increase. Formulations ;
! X! p" K2 o/ F) M1 X9 }$ ^" T7 Q7 q: i! with more than a dozen cities ;
3 D) c7 ?) E; h3 Q+ D7 U' {! WILL NOT SOLVE in a reasonable ;, U# Y2 f9 U) |- R1 X$ m; u, H ?* }+ d
! amount of time! ;5 i. E' ]3 F$ N1 L
!************************************;
" y( F6 ^) A) m0 X: c$ ?9 v* ^& K6 N/ w2 X, E
SETS:
; M" v$ q z1 ^9 ? ! Q(I) is the amount required at city I,
* D+ C) Z* X9 W4 t; h# [ y: Y U(I) is the accumulated delivers at city I ;
: a& p, m: ~. a/ ]4 k( t7 _. _ t! K CITY/1..8/: Q, U;
6 L$ } R3 ^- O1 J
( c& T3 S& m) Y. y ! DIST(I,J) is the distance from city I to city J: @9 w. T7 m+ m* h; `( a9 T8 Y4 W2 |3 \
X(I,J) is 0-1 variable: It is 1 if some vehicle
3 A2 b( p6 F3 x2 K" m# l+ q travels from city I to J, 0 if none;1 L0 I# `6 K9 g1 M" M
CXC( CITY, CITY): DIST, X;
9 A, @, L" W2 w& c, a* _( Q ENDSETS2 a0 M r, Q( Z) n# K8 o' E# m
! Q% L# w1 o$ }; n
DATA:
5 B5 s% _4 i0 n5 u4 q ! city 1 represent the common depo;% V. G, O5 w2 _: x+ l
Q = 0 6 3 7 7 18 4 5;, b$ z) S) Z5 X2 X% n4 G. u
: }; l+ \) W# E# I3 r
! distance from city I to city J is same from city% f! |) p* G5 W0 i5 g3 N# g* w
J to city I distance from city I to the depot is
9 y0 ~+ f9 v% j, a7 w 0, since the vehicle has to return to the depot;
+ s7 \; h4 y2 M6 G- g3 l9 K/ A0 D3 g* E
DIST = ! To City;: K6 R. z* U! R" b. s
! Chi Den Frsn Hous KC LA Oakl Anah From;# q" x- k$ I4 @. ?; v7 J1 W, b
0 996 2162 1067 499 2054 2134 2050!Chicago;( r) G; i5 r3 y' ?
0 0 1167 1019 596 1059 1227 1055!Denver;
9 r$ p( a, j, _7 { 0 1167 0 1747 1723 214 168 250!Fresno;: u) m6 {! z# C
0 1019 1747 0 710 1538 1904 1528!Houston;; n' G& ]! i( U+ A/ U8 E. N% g3 J
0 596 1723 710 0 1589 1827 1579!K. City;
" d0 H" q F* ]8 M/ [" ` 0 1059 214 1538 1589 0 371 36!L. A.;+ t3 I% d; Q4 C" T3 g4 J; L
0 1227 168 1904 1827 371 0 407!Oakland;
4 `0 D& e5 z" K( x 0 1055 250 1528 1579 36 407 0;!Anaheim;
9 g% X- }# \2 y5 }# v! {' C. U
" g" M0 I* C! Q: P) p8 s2 U. H% O ! VCAP is the capacity of a vehicle ;
' \' k/ k4 a% u VCAP = 18;
) _: l, q0 Y7 k. t7 \8 U ENDDATA3 n2 ^! a4 a( M0 S* O
! x' j4 p5 x; i4 j* U& h ! Minimize total travel distance;
. N6 E9 \0 q; l! o I7 K MIN = @SUM( CXC: DIST * X);
E* P( h; U5 \9 _' u, l7 |6 K
8 f3 w! p5 Z* ~- {& e& @1 e ! For each city, except depot....;
' \9 S+ y, |! I, @# U6 G { @FOR( CITY( K)| K #GT# 1:6 ?& F% o. u: u! s
% L, n: O! O- x% n
! a vehicle does not travel inside itself,...;. u. F/ L- [4 s
X( K, K) = 0;8 Z1 B" J, f3 z/ |
. j* ~- u" X. R4 @$ \ ! a vehicle must enter it,... ;
$ a2 x/ W! |# m2 F u, a: a @SUM( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR#0 C+ a' S o. D: S3 N1 ~
Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1;
, ]5 H. x; @3 o$ {& p, _( h8 S
+ p2 H6 E- t8 c* |0 K, \ ! a vehicle must leave it after service ;2 H$ r4 g0 P# h5 o2 u/ K
@SUM( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR#: O% L+ }; W3 A& q9 J& D
Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1;
/ K4 A" t8 E0 N" g+ B2 G; Y3 P4 g" e& h z
! U( K) is at least amount needed at K but can't
4 }6 Q* x& l7 M. L# @ exceed capacity;
3 h8 r+ t8 K4 c9 ?% v7 s4 ~' R6 e! z @BND( Q( K), U( K), VCAP);
4 o' x* R) ?1 N T4 V E
9 e: y. U/ f% n- B ! If K follows I, then can bound U( K) - U( I); O( H7 {0 N8 |* p/ e8 v
@FOR( CITY( I)| I #NE# K #AND# I #NE# 1: : W4 a% V& @& S+ s; w
U( K) >= U( I) + Q( K) - VCAP + VCAP *
$ P1 u, Z- g, k- U9 D- Y ( X( K, I) + X( I, K)) - ( Q( K) + Q( I))
# D% N+ n/ `# n% {+ G: Y * X( K, I);
1 a4 S9 ~8 r+ K- k );
' h7 v: W* m, h9 N0 Y; G2 p q4 d* q
! If K is 1st stop, then U( K) = Q( K);
3 h; ]/ E: k: A% Q! J& h U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K);
3 }: P7 s4 N( A/ k
) {' r% n4 P2 Y" I; Y2 S* A# R ! If K is not 1st stop...;
' G$ w7 y% N3 {1 h# q9 L, S4 S U( K)>= Q( K)+ @SUM( CITY( I)|
3 W5 M. o/ O# K I #GT# 1: Q( I) * X( I, K));1 _4 A3 e1 S: k5 C4 y0 ?5 V
);
% {+ p$ |) a) [& H3 M
4 R& Z' i+ W+ ^$ | ! Make the X's binary;9 ]9 H# _' ^1 I8 ?
@FOR( CXC: @BIN( X));
1 ~6 i7 A) Y9 ^8 i- y) q1 U0 J" a, t% f9 b1 p* m
! Minimum no. vehicles required, fractional 1 B7 o8 N: ~+ |- w2 Q+ Q7 k
and rounded;5 N6 F% I) R* q& F
VEHCLF = @SUM( CITY( I)| I #GT# 1: Q( I))/ VCAP;% G% R! J4 m( p |
VEHCLR = VEHCLF + 1.999 - ; I O0 n2 A% u* g* D! S
@WRAP( VEHCLF - .001, 1);0 X: k) e+ F# z2 Y+ w1 Y
) `+ B$ u+ ^/ W8 l ! Must send enough vehicles out of depot;7 w4 a* v1 r% l& y2 P( W6 K7 `7 T
@SUM( CITY( J)| J #GT# 1: X( 1, J)) >= VEHCLR;' ?4 `) \% z. X: c
END Y7 m/ O; Y( Y/ x+ I% 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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