Engineering Optimization: Theory and Practice, Fourth Edition

9.6 Example Illustrating the Tabular Method of Solution 565

quantities (i.e.,f 2 ∗andx 2 ∗)correspondingto the various discrete values ofs 3 can be
summarized as follows:

Specified value of

s 3 (kgf)

Type of columns

corresponding to

minimum cost of

stages 2 and 1,(x∗ 2 )

Minimum cost of

stages 2 and 1,f 2 ∗

($)

Value of the

corresponding

state variable,s 2

(kgf)

150,000 (a) 9,000 220,000
130,000 (a) 7,000 180,000
110,000 (b) 5,500 140,000
100,000 (b) 3,875 115,000

Suboptimization of Stages 3, 2, and 1 (Components 3, 2, and 1)

For the suboptimization of stages 3, 2, and 1, we consider all three compo-
nents together as shown in Fig. 9.14c and minimize the cost (R 3 +R 2 +R 1 ) forany
specified value ofs 4 to obtainf 3 ∗(s 4 ) However, since there is only one value of. s 4
(initial value problem) to be considered, we obtain the following results by using the
information given in Table 9.1:

f 3 ∗(s 4 ) = min

x 3 ,x 2 ,x 1

[R 3 (x 3 , s 4 )+R 2 (x 2 , s 3 )+R 1 (x 1 , s 2 )]

=min

x 3

[R 3 (x 3 , s 4 )+f 2 ∗(s 3 )]

Specific

value of

s 4 (kgf)

Type of

tank (x 3 )

Cost of

tank R 3

($)

Corresponding

output state,s 3

(kgf)

x∗ 2 (type of

columns for

minimum

cost) f 2 ∗($) R 3 +f 2 ∗($)

100,000 (a) 5,000 145,000 (a) 8 , 500 ∗∗ 13,500

(b) 8,000 130,000 (a) 7,000 15,000

(c) 6,000 125,000 (a) 6 , 625 ∗∗ 12 , 625

(d) 9,000 115,000 (b) 5 , 875 ∗∗ 14,875

(e) 15,000 105,000 (b) 4 , 68712 ∗∗ 19,687^12

(f) 12,000 110,000 (b) 5,500 17,500

(g) 10,000 115,000 (b) 5 , 875 ∗∗ 15,875

Here also the double-starred quantities indicate the interpolated values and the boxed
quantity the minimum cost solution. From the results above, the minimum cost solution
is given by

s 4 = 001 ,000 kgf

x 3 ∗= ype (c) tankt

f 3 ∗(s 4 = 001 , 000 )=$12, 625

s 3 = 251 ,000 kgf