9.5 Example Illustrating the Calculus Method of Solution 559Lets 4 be the displacement available for allocation to the first three members. Let
δ 3 be the displacement contribution due to the third member andf 3 ∗(s 4 ) he minimumt
weight of the first three members. Then
f 3 ∗(s 4 ) =min
x 3 ≥ 0[x 3 +f 2 ∗(s 3 )] (E 11 )wheres 3 is the resource available after allocation to stage 3 and is given by
s 3 =s 4 −δ 3 =s 4 −1. 5625
x 3From Eq. (E 10 ) we have
f 2 ∗(s 3 )=4. 6169
s 4 − 1. 5625 /x 3(E 12 )
and Eq. (E 11 ) can be written as
f 3 ∗(s 4 ) =min
x 3 ≥ 0[
x 3 +4. 6169 x 3
s 4 x 3 − 1. 5625]
(E 13 )
As before, by letting
F (s 4 , x 3 )=x 3 +4. 6169 x 3
s 4 x 3 − 1. 5625(E 14 )
the minimum ofF, for any specified value ofs 4 , can be obtained as
∂F
∂x 3= 1. 0 −
( 4. 6169 )( 1. 5625 )
(s 4 x 3 − 1. 5625 )^2= 0 or x 3 ∗=4. 2445
s 4(E 15 )
f 3 ∗(s 4 )=x 3 ∗+4. 6169 x∗ 3
s 4 x 3 ∗− 1. 5625=
4. 2445
s 4+
7. 3151
s 4=
11. 5596
s 4(E 16 )
Finally, lets 5 denote the displacement available for allocation to the first four
members. Ifδ 4 denotes the displacement contribution due to the fourth member, and
f 4 ∗(s 5 ) he minimum weight of the first four members, thent
f 4 ∗(s 5 ) =min
x 4 ≥ 0[0. 6 x 4 +f 3 ∗(s 4 )] (E 17 )where the resource available after allocation to the fourth member (s 4 ) is given by
s 4 =s 5 −δ 4 =s 5 −1. 3500
x 4(E 18 )
From Eqs. (E 16 ), (E 17 ), and (E 18 ), we obtain
f 4 ∗(s 5 ) =min
x 4 ≥ 0[
0. 6 x 4 +11. 5596
s 5 − 1. 3500 /x 4]
(E 19 )
By setting
F (s 5 , x 4 )= 0. 6 x 4 +11. 5596
s 5 − 1. 3500 /x 4