FIGURE 12. First stage in finding most economical route.
C2-E, 1 + 21= 28. Thus, the minimum cost is 28. Record these minimum costs and their
corresponding towns in district C in Fig. 13.
- Determine the minimum cost of transportation from town A
to town E
Refer to Fig. 13. The cost of transportation from A to E is: for A-Bl-E, 9 + 32 = 41; for
A-B2-E, 6 + 41 = 47; for A-B3-E, 12 + 28 = 40. Thus, the minimum cost is 40, and the
corresponding town in district B is B3.
- Identify the most economical route
Refer to step 3 and Figs. 13 and 12, in that order. The most economical route is A-B3-C2-
Dl-E. From Fig. 11, the cost of transportation corresponding to this route is 12 + 7+11 +
10 = 40, which agrees with the result in step 3.
Related Calculations: The number of alternative routes from town A to town E
is 3x2x3 = 18. Therefore, the most economical route can be found by listing all 18
routes and computing their respective costs. However, the solution by dynamic program-
ming given above simplifies the work.
OPTIMAL INVENTORYLEVEL
A firm is under contract to supply 41,600 parts per year and plans to produce them in
equal lots spaced at equal intervals. The production capacity is 800 parts per day. Setup
and teardown cost for the production machines is $550 for each run. The cost of storage,
insurance, and interest on the investment is $1.40 per part for each year the part is carried
in inventory. The regular production cost, exclusive of setup and teardown, is $5 per part.
