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Solve a Traveling Salesman Problem as a special case \par of the Vehicle Routing problem;\cf2 \par \cf3 ! The Vehicle Routing Problem occurs in many service \par systems such as delivery, customer pick-up, repair and \par maintenance. A fleet of vehicles, each with fixed \par capacity, starts at a common depot and returns to the \par depot after visiting locations where service is demanded. \par This is a simple, loose formulation. \par Problems with more than a dozen cities can take lots of \par time with this formulation;\cf2 \par \cf1 SETS\cf2 : \par CITY: Q, U; \par \cf3 ! Q(I) = amount required at city I(given), \par must be delivered by just 1 vehicle. \par U(I) = accumulated deliveries at city I ;\cf2 \par CXC( CITY, CITY): DIST, X; \par \cf3 ! DIST(I,J) = distance from city I to city J \par X(I,J) is 0-1 variable, \par = 1 if some vehicle travels from city I to J, \par else 0 ;\cf2 \par \cf1 ENDSETS\cf2 \par \cf1 DATA\cf2 : \par CITY = 1 2 3 4 5 6 7; \par \cf3 ! city 1 represents the common depot, i.e. Q( 1) = 0;\cf2 \par Q= 0 1 1 1 1 1 1; \par \cf3 ! distance from city I to city J is same from J to I, \par distance from city I to the depot is 0, \par because vehicle need not return to the depot ;\cf2 \par DIST= \cf3 ! To City;\cf2 \par \cf3 ! 1 2 3 4 5 6 7;\cf2 \par 0 12 10 999 999 999 12 \cf3 !1;\cf2 \par 12 0 8 12 999 999 999 \cf3 !2;\cf2 \par 10 8 0 11 3 999 9 \cf3 !3;\cf2 \par 999 12 11 0 11 10 999 \cf3 !4;\cf2 \par 999 999 3 11 0 6 7 \cf3 !5;\cf2 \par 999 999 999 10 6 0 9 \cf3 !6;\cf2 \par 12 999 9 999 7 9 0;\cf3 !7;\cf2 \par \cf3 ! VCAP is the capacity of a vehicle ;\cf2 \par VCAP = 6; \par \cf3 ! A traveling salesperson problem is vehicle routing problem \par in which the vehicle capacity is large enough to carry all \par demands in one trip;\cf2 \par \cf1 ENDDATA\cf2 \par \cf3 !----------------------------------------------------------;\cf2 \par \cf3 ! The objective is to minimize total travel distance;\cf2 \par \cf1 MIN\cf2 = \cf1 @SUM\cf2 ( CXC: DIST * X); \par \cf3 ! for each city, except depot....;\cf2 \par \cf1 @FOR\cf2 ( CITY( K)| K #GT# 1: \par \cf3 ! a vehicle does not travel inside itself,...;\cf2 \par X( K, K) = 0; \par \cf3 ! a vehicle must enter it,... ;\cf2 \par \cf1 @SUM\cf2 ( CITY( I)| I #NE# K #AND# ( I #EQ# 1 #OR# \par Q( I) + Q( K) #LE# VCAP): X( I, K)) = 1; \par \cf3 ! a vehicle must leave it after service ;\cf2 \par \cf1 @SUM\cf2 ( CITY( J)| J #NE# K #AND# ( J #EQ# 1 #OR# \par Q( J) + Q( K) #LE# VCAP): X( K, J)) = 1; \par \cf3 ! U( K) = amount delivered on trip up to city K \par >= amount needed at K but <= vehicle capacity;\cf2 \par \cf1 @BND\cf2 ( Q( K), U( K), VCAP); \par \cf3 ! If K follows I, then can bound U( K) - U( I);\cf2 \par \cf1 @FOR\cf2 ( CITY( I)| I #NE# K #AND# I #NE# 1: U( K) >= \par U( I) + Q( K) - VCAP + VCAP*( X( K, I) + X( I, K)) \par - ( Q( K) + Q( I)) * X( K, I); \par ); \par \cf3 ! If K is 1st stop, then U( K) = Q( K);\cf2 \par U( K) <= VCAP - ( VCAP - Q( K)) * X( 1, K); \par \cf3 ! If K is not 1st stop...;\cf2 \par U( K) >= \par Q( K)+ \cf1 @SUM\cf2 ( CITY( I)| I #GT# 1: Q( I) * X( I, K)); \par ); \par \cf3 ! Make the X's binary;\cf2 \par \cf1 @FOR\cf2 ( CXC( I, J): \cf1 @BIN\cf2 ( X( I, J)) ;); \par \cf3 ! Must send enough vehicles out of depot;\cf2 \par \cf1 @SUM\cf2 ( CITY( J)| J #GT# 1: X( 1, J)) >= \par \cf1 @FLOOR\cf2 ((\cf1 @SUM\cf2 ( CITY( I)| I #GT# 1: Q( I))/ VCAP) + .999); \par \cf1 END\cf2 \par \par }