Engineering Optimization: Theory and Practice, Fourth Edition

558 Dynamic Programming

Figure 9.12 Example 9.2 as a four-stage decision problem.

sinceδ 1 =s 2 , and

x 1 ∗=

1. 5625

s 2

(E 5 )

Lets 3 be the displacement available for allocation to the first two members,δ 2 the

displacement contribution due to the second member, andf 2 ∗(s 3 ) he minimum weightt

of the first two members. Then we have, from the recurrence relationship of Eq. (9.16),

f 2 ∗(s 3 ) =min

x 2 ≥ 0

[R 2 +f 1 ∗(s 2 )] (E 6 )

wheres 2 represents the resource available after allocation to stage2 and is given by

s 2 =s 3 −δ 2 =s 3 −

0. 6750

x 2

Hence from Eq. (E 4 ), we have

f 1 ∗(s 2 )=f 1 ∗

(

s 3 −

0. 6750

x 2

)

=

[

1. 5625

/ (

s 3 −

0. 6750

x 2

)]

(E 7 )

Thus Eq. (E 6 ) becomes

f 2 ∗(s 3 ) =min

x 2 ≥ 0

[

1. 2 x 2 +

1. 5625

s 3 − 0. 6750 /x 2

]

(E 8 )

Let

F (s 3 , x 2 )= 1. 2 x 2 +

1. 5625

s 3 − 0. 6750 /x 2

= 1. 2 x 2 +

1. 5625 x 2

s 3 x 2 − 0. 6750

For any specified value ofs 3 , the minimum ofFis given by

∂F

∂x 2

= 1. 2 −

( 1. 5625 )( 0. 6750 )

(s 3 x 2 − 0. 6750 )^2

= 0 or x∗ 2 =

1. 6124

s 3

(E 9 )

f 2 ∗(s 3 )= 1. 2 x 2 ∗+

1. 5625

s 3 − 0. 6750 /x∗ 2

=

1. 9349

s 3

+

2. 6820

s 3

=

4. 6169

s 3

(E 10 )