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Example 3: Travelling Salesman problem
Given: n associated cities and distances among them
Find: tour of minimum length that visits all of city.
Solutions: How several tours are possible?
n*(n -1)...*1 = n!
Because n! > 2(n-1)
Therefore n! = ? (2n) (lower bound)
As of now, there is no algorithm that determines a tour of minimum length plus covers all of the cities in polynomial time. But, there are many very good heuristic algorithms.
The below figure illustrates the BOM (Bill of Materials) for product A. The MPS (Material requirements Planning) start row in the master production schedule for product A calls for
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Easy algorithm for beginner for quicksort with explanation
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