Engineering Optimization: Theory and Practice, Fourth Edition

562 Dynamic Programming

Table 9.3 Component 1 (Foundation)

s 1 =s 2 +

Type of foundation s 2 (kgf) R 1 cost ($) Self-weight (kgf) self-weight (kgf)

(a) Mat foundation 220,000 5,000 60,000 280,000

200,000 4,000 45,000 245,000

180,000 3,000 35,000 215,000

140,000 2,500 25,000 165,000

100,000 500 20,000 120,000

(b) Concrete pile foundation 220,000 3,500 55,000 275,000

200,000 3,000 40,000 240,000

180,000 2,500 30,000 210,000

140,000 1,500 20,000 160,000

100,000 1,000 15,000 115,000

(c) Steel pile foundation 220,000 3,000 10,000 230,000

200,000 2,500 9,000 209,000

180,000 2,000 8,000 188,000

140,000 2,000 6,000 146,000

100,000 1,500 5,000 105,000

Figure 9.13 Example 9.3 as a three-stage decision problem.

which yield information for the various system components as shown in Tables 9.1

to 9.3 (these values are given only for illustrative purpose).

Suboptimization of Stage 1 (Component 1)

For the suboptimization of stage 1, we isolate component 1 as shown in Fig. 9.14a

and minimize its costR 1 (x 1 , s 2 ) forany specified value of the input states 2 to obtain

f 1 ∗(s 2 ) sa

f 1 ∗(s 2 ) =min

x 1

[R 1 (x 1 , s 2 )]

Since five settings of the input state variables 2 are given in Table 9.3, we obtainf 1 ∗

for each of these values as shown below: