1. Milestone 4: Courier Company

2. Problem Definition C B D C A D A B
Given N deliveries (pick up, drop off, weight)
N = 4 here, all intersections
and Given M courier depots, each with a truck of capacity C
M = 3 here, all intersections
Return: low travel time path Starting and ending at some depot Making all N deliveries by traversing intersections while Picking up each package before delivering it, and never overloading truck

3. No  dropping off packages before they’re picked up!
Possible Solution? C B D C A D A B
No  dropping off packages before they’re picked up!

4. Legal Solution C B D C A D A B Output: vector of courier paths
Courier path: start intersection, end intersection,
deliveries picked up at start, vector of street segments to follow

5. Simple Solution: Re-use m3 path-finder
// Go from first depot to first package pick up path = find_path (depot[0], delivery[0].pickUp); // Complete deliveries, in order for (i = 0; i < N-1; i++) { path += find_path (delivery[i].pickUp, delivery[i].dropOff); path += find_path (delivery[i].dropOff, delivery[i+1].pickUp); } // Drop off last package path += find_path (delivery[N-1].pickUp, delivery[N-1].dropOff); // Go back to the first depot to drop off the truck path += find_path (delivery[N-1].dropOff, depot[0]);

6. Possible Solution C B D C A D 2 A B 1
Lots of wasted travel!

7. Need to optimize delivery order!
More Logical Solution C B D C A D A B
Need to optimize delivery order!

8. Exhaustive Algorithm? Try all possible delivery orders
Pick the one with lowest travel time
How many combinations?
M truck depots
N deliveries  2N pick-up + drop-off intersections
Pick one of M starting locations
Then pick one of 2N pick-up/drop-off intersections
Then one of 2N-1 for the second intersection
… (repeat until last delivery)
Then drop off at closest depot
M * 2N * (2N-1) * (2N-2) * … * 1
= M * (2N)!
Some of these are illegal orders  say algorithm checks legality after generating the solution

9. Exhaustive Algorithm? Say M = 10, N = 100 10 * (2N)!  10377
Invoke find_path () 2N+1 times to get path between each intersection
Say find_path takes 0.1 s (very good!)
10377 * 201 * 0.1 s = 1.6 x s
5 x years!
Lifetime of universe: ~14 x 109 years!

10. Traveling Salesman Problem
We are solving a variation of the traveling salesman problem
Computationally hard problem
For N deliveries, no guaranteed optimal (lowest travel time solution) in polynomial time
i.e. > O(Nk), for any k
Means at least O(2N)
Need to use heuristics to solve
Most research problems are computationally hard
Integrated circuit design, drug design, transportation network design, …

11. Stephen Cook: Pioneer of Complexity Theory
U of T professor in Computer Science
NP-complete problems
No polynomial time solution known on conventional computers
Proved:
If a method to solve any of these problems in polynomial time found,
Then all these problems can be solved in polynomial time
P vs. NP: most famous open problem in computer science
1970: denied tenure at UC Berkeley
1971: published most famous paper
1982: won Turing award