Engineering Optimization: Theory and Practice, Fourth Edition

564 Dynamic Programming

Specific value x 1 ∗(type of foundation f 1 ∗ Corresponding value

ofs 2 (kgf) for minimum cost) ($) ofs 1 (kgf)

220,000 (c) 3,000 230,000

200,000 (c) 2,500 209,000

180,000 (c) 2,000 188,000

140,000 (b) 1,500 160,000

100,000 (a) 500 120,000

Suboptimization of Stages 2 and 1 (Components 2 and 1)

Here we combine components 2 and 1 as shown in Fig. 9.14b and minimize

the cost (R 2 +R 1 ) or any specified valuef s 3 to obtainf 2 ∗(s 3 ) sa

f 2 ∗(s 3 ) =min

x 2 ,x 1

[R 2 (x 2 , s 3 )+R 1 (x 1 , s 2 ) ]=min

x 2

[R 2 (x 2 , s 3 )+f 1 ∗(s 2 )]

Since four settings of the input state variables 3 are given in Table 9.2, we can findf 2 ∗

for each of these four values. Since this number of settings fors 3 is small, the values of

the output state variables 2 that result will not necessarily coincide with the values of

s 2 tabulated in Table 9.3. Hence we interpolate linearly the val ues ofs 2 (if it becomes

necessary) for the purpose of present computation. However, if the computations are

done on a computer, more settings, more closely spaced, can be considered without

much difficulty. The suboptimization of stages 2 and 1 gives the following results:

Value of

the

output

Specific Value of Cost of state

value of x 2 (type columns, variable x 1 ∗(Type of f 1 ∗ R 2 +f 1 ∗

s 3 (kgf) of columns) R 2 ($) s 2 (kgf) foundation) ($) ($)

150,000 (a) 6,000 220,000 (c) 3,000 9 , 000

(b) 8,000 210,000 (c) 2,750** 10,750

(c) 15,000 180,000 (c) 2,000 17,000

130,000 (a) 5,000 180,000 (c) 2,000 7 , 000

(b) 6,000 180,000 (c) 2,000 8,000

(c) 10,000 150,000 (b) 1,625** 11,625

110,000 (a) 4,000 150,000 (b) 1,625** 5,625

(b) 4,000 140,000 (b) 1,500 5 , 500

(c) 9,000 125,000 (b) 1,125** 10,125

100,000 (a) 3,000 140,000 (b) 1,500 4,500

(b) 3,000 115,000 (a) 875 ** 3 , 875

(c) 8,000 110,000 (a) 750 ** 8,750

Notice that the double-starred quantities indicate interpolated values and the boxed

quantities the minimum cost solution for the specified value ofs 3. Now the desired