ÐÏࡱá>þÿ  þÿÿÿ ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ) = 0; LINK( CITY, CITY): DIST, ! The distance matrix; X; ! X( I, J) = 1 if we use link I, J; ENDSETS DATA: ! Distance matrix need not be symmetric; ! However, city 1 is base of the tree; !to: Atl Chi Cin Hou LA Mon ; DIST = 0 702 454 842 2396 1196 !from Atl; 702 0 324 1093 2136 764 !from Chi; 454 324 0 1137 2180 798 !from Cin; 842 1093 1137 0 1616 1857 !from Hou; 2396 2136 2180 1616 0 2900 !from LA; 1196 764 798 1857 2900 0; !from Mon; ENDDATA ! The model size: Warning, may be slow for N >= 8; N = @SIZE( CITY); ! Minimize total distance of the links; MIN = @SUM( LINK: DIST * X); ! For city K, except the base, ... ; @FOR( CITY( K)| K #GT# 1: ! It must be entered; @SUM( CITY( I)| I #NE# K: X( I, K)) = 1; ! If there are 2 disjoint tours from 1 city to another, we can remove a link without breaking connections. Note: These are not very powerful for large problems; @FOR( CITY( J)| J #GT# 1 #AND# J #NE# K: U( J) >= U( K) + X ( K, J) - ( N - 2) * ( 1 - X( K, J)) + ( N - 3) * X( J, K); ); ); ! There must be an arc out of city 1; @SUM( CITY( J)| J #GT# 1: X( 1, J)) >= 1; ! Make the X's 0/1; @FOR( LINK: @BIN( X); ); ! The level of a city except the base is at least 1 but no more than N-1, and is 1 if it links to the base; @FOR( CITY( K)| K #GT# 1: @BND( 1, U( K), 999999); U( K) <= N - 1 - ( N - 2) * X( 1, K); ); END MODEL: !Given the number of nodes and the distance between them, finding the shortest total distance of links on the network to connect all the nodes is the classic problem called minimal spanning tree (MST). This model finds the (MST) connecting Atlanta, Chicago, Cincinnati, Houston, LA, and Montreal so that messages can be sent from Atlanta (base) to other cities through the network at minimum cost; SETS: CITY / 1.. 6/: U; ! U( I) = level of city I; ! 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Make the X's 0/1;\cf2 \par \cf1 @FOR\cf2 ( LINK: \cf1 @BIN\cf2 ( X); ); \par \par \cf3 ! The level of a city except the base is at least \par 1 but no more than N-1, and is 1 if it links to \par the base;\cf2 \par \cf1 @FOR\cf2 ( CITY( K)| K #GT# 1: \par \cf1 @BND\cf2 ( 1, U( K), 999999); \par U( K) <= N - 1 - ( N - 2) * X( 1, K); ); \par \cf1 END\cf2 \par \par } ì‹{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fnil\fcharset0 Courier New;}} {\colortbl ;\red0\green0\blue255;\red0\green0\blue0;\red0\green175\blue0;} \viewkind4\uc1\pard\cf1\f0\fs20 MODEL\cf2 : \par \cf3 !Given the number of nodes and the distance between \par them, finding the shortest total distance of links \par on the network to connect all the nodes is the \par classic problem called minimal spanning tree (MST). \par \par This model finds the (MST) connecting Atlanta, \par Chicago, Cincinnati, Houston, LA, and Montreal so \par that messages can be sent from Atlanta (base) to \par other cities through the network at minimum cost;\cf2 \par \par \cf1 SETS\cf2 : \par CITY / 1.. 6/: U; \cf3 ! U( I) = level of city I;\cf2 \par \cf3 ! U( 1) = 0;\cf2 \par LINK( CITY, CITY): \par DIST, \cf3 ! The distance matrix;\cf2 \par X; \cf3 ! X( I, J) = 1 if we use link I, J;\cf2 \par \cf1 ENDSETS\cf2 \par \par \cf1 DATA\cf2 : \cf3 ! Distance matrix need not be symmetric;\cf2 \par \cf3 ! However, city 1 is base of the tree;\cf2 \par \cf3 !to: Atl Chi Cin Hou LA Mon ;\cf2 \par DIST = 0 702 454 842 2396 1196 \cf3 !from Atl;\cf2 \par 702 0 324 1093 2136 764 \cf3 !from Chi;\cf2 \par 454 324 0 1137 2180 798 \cf3 !from Cin;\cf2 \par 842 1093 1137 0 1616 1857 \cf3 !from Hou;\cf2 \par 2396 2136 2180 1616 0 2900 \cf3 !from LA;\cf2 \par 1196 764 798 1857 2900 0; \cf3 !from Mon;\cf2 \par \cf1 ENDDATA\cf2 \par \par \cf3 ! The model size: Warning, may be slow for N >= 8;\cf2 \par N = \cf1 @SIZE\cf2 ( CITY); \par \par \cf3 ! Minimize total distance of the links;\cf2 \par \cf1 MIN\cf2 = \cf1 @SUM\cf2 ( LINK: DIST * X); \par \par \cf3 ! For city K, except the base, ... ;\cf2 \par \cf1 @FOR\cf2 ( CITY( K)| K #GT# 1: \par \par \cf3 ! It must be entered;\cf2 \par \cf1 @SUM\cf2 ( CITY( I)| I #NE# K: X( I, K)) = 1; \par \par \cf3 ! If there are 2 disjoint tours from 1 city to \par another, we can remove a link without breaking \par connections. Note: These are not very powerful \par for large problems;\cf2 \par \cf1 @FOR\cf2 ( CITY( J)| J #GT# 1 #AND# J #NE# K: \par U( J) >= U( K) + X ( K, J) - \par ( N - 2) * ( 1 - X( K, J)) + \par ( N - 3) * X( J, K); ); \par ); \par \par \cf3 ! There must be an a