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(MNSPTREE);\cf2 \par \cf3 !Given a set of nodes and the distance between each pair, find \par the shortest total distance of links on the network to connect \par all the nodes. This is the classic minimal spanning tree (MST) \par problem. Typical application is designing the network of \par utilities in a community, e.g., sewers, power lines, \par phone lines, cable TV, roads, water lines, also electric \par circuit design;\cf2 \par \cf1 SETS\cf2 : \par CITY: LVL; \par \cf3 ! LVL( I) = level of city I in tree. LVL( 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 IFLINK(CITY,CITY,CITY,CITY); \cf3 ! Link 1,2 only if link 3,4;\cf2 \par ATMOST1(CITY,CITY); \cf3 ! At most one of these links;\cf2 \par \cf1 ENDSETS\cf2 \par \cf3 ! This model finds the minimum cost network connecting \par a set of cities;\cf2 \par \cf1 DATA\cf2 : \par CITY = A B C D E; \par \cf3 ! Distance matrix need not be symmetric. City 1 is base;\cf2 \par DIST = 0 20 10 15 99 \cf3 !from A;\cf2 \par 20 0 99 99 30 \cf3 !from B;\cf2 \par 10 99 0 25 5 \cf3 !from C;\cf2 \par 15 99 25 0 40 \cf3 !from D;\cf2 \par 99 30 5 40 0;\cf3 !from E;\cf2 \par \par IFLINK = A, D, D, E; \par ATMOST1 = A,D C,D, A,B; \par \cf1 ENDDATA\cf2 \par \cf3 !----------------------------------------------;\cf2 \par \cf3 ! Take care of special constraints;\cf2 \par \cf3 ! Link i,j only if link r,s;\cf2 \par \cf1 @FOR\cf2 ( IFLINK(I,J,R,S): \par X(I,J)+X(J,I) <= X(R,S)+X(S,R); \par ); \par \cf3 ! At most 1 of the links in this set;\cf2 \par \cf1 @SUM\cf2 (ATMOST1(I,J): X(I,J)+X(J,I)) <= 1; \par \cf3 ! This is a simple, small, loose formulation. \par Warning, may be slow for N > 8;\cf2 \par N = \cf1 @SIZE\cf2 ( CITY); \par \cf3 !The objective is to minimize total dist. of links;\cf2 \par \cf1 MIN\cf2 = \cf1 @SUM\cf2 ( LINK(I,J): DIST(I,J) * X(I,J)); \par \cf3 !For city K, except the base, ... ;\cf2 \par \cf1 @FOR\cf2 ( CITY( K)| K #GT# 1: \cf3 ! It must be entered;\cf2 \par \cf1 @SUM\cf2 ( CITY( I)| I #NE# K: X( I, K)) = 1; \par \cf3 !If there is a link from J-K, then LVL(K)=LVL(J)+1. \par Note:These are not very powerful for large problems;\cf2 \par \cf1 @FOR\cf2 ( CITY( J)| J #NE# K: \par LVL( K) >= LVL( J) + X( J, K) \par - ( N - 2) * ( 1 - X( J, K)) \par + ( N - 3) * X( K, J); ); ); \par LVL( 1) = 0; \cf3 ! City 1 has level 0;\cf2 \par \cf3 !There must be an arc out of city 1;\cf2 \par \cf1 @SUM\cf2 ( CITY( J)| J #GT# 1: X( 1, J)) >= 1; \par \cf3 !Make the X's 0/1;\cf2 \par \cf1 @FOR\cf2 ( LINK: \cf1 @BIN\cf2 ( X); ); \par \cf3 !The level of a city except the base is at least 1 but no more than N-1, \par and is 1 if link to the base;\cf2 \par \cf1 @FOR\cf2 ( CITY( K)| K #GT# 1: \par \cf1 @BND\cf2 ( 1, LVL( K), 999999); \par LVL( K) <= N - 1 - ( N - 2) * X( 1, K); ); \par \cf1 END\cf2 \par \par \par \par }