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Transportation ProblemsPowerPoint Presentation

Transportation Problems

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Transportation Problems

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Transportation Problems

Dr. Ron Lembke

- Linear programming is good at solving problems with zillions of options, and finding the optimal solution.
- Could it work for transportation problems?
- Costs are linear, and shipment quantities are linear, so maybe so.

- Define cij as the cost to ship one unit from i to j.
- Demand at location j is dj.
- Supply at DC i is Si
- Xij is the quantity shipped from DC i to customer j.

You have 3 DCs, and need to deliver product to 4 customers.

Find cheapest way to satisfy all demand

D 2

A 10

E 4

B 10

F 12

C 10

G 11

- Trial and Error
- Linear Programming – ooh, what’s that?!
- Tell me more!

- Create a matrix of shipment costs (in grey in example).
- Create a matrix to hold the decision variables, shipment quantities (in yellow).
- Sum amount sent to each destination.
- Sum amount sent from each DC.
- Enter demands and supplies at each location.
- Compute total cost of shipments (in blue).

- If you don’t check “assume non-negative” we get the following results:
- Solver doesn’t converge to an optimal solution. Why not?

- Use <= for shipments from DCs.
- Use >= for shipments to customers.
- Do we really need to?

- What do we do if supply is greater than demand?

- If total demand is greater than total supply, what happens?
- If demand in G is 15, we get this:

- If demand at G is 15, there are no feasible solutions, much less a best one.
- We need to add a phantom source, Z, with huge capacity. Think of it as a supplier that ships empty boxes.
- Now supply can satisfy total demand.

- What cost should we use for supplier Z?
- It should be the last resort, so it should be higher than any real costs.
- The cost of a shipment from Z is really the cost of shorting the customer.
- If all customers are created equal, give them all the same shortage cost.
- If some are more important, give them higher shortage costs, and we’ll only short them as a last resort.

- Shortage is dealt with by shorting customer A, and B.
- Demand exceeds supply by 3 units. Our first choice is to short A, because they are the cheapest. We can only short them by 2, their total demand.
- Next, short B by 1 unit.