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1 solution
Possible tours and tour lengths are
A BCD A = (100+65+70+80) = 315 A DCB A = (80+70+65+100) = 315
A BDC A =(100+55+70+50) = 275 A CDB A = (50+70+55+100) = 275
A CBD A =(50+65+55+80) = 250 A DBC A = (80+55+65+50) = 250
Shortest possible tour length = 250.
Travelling Salesman problem