5.5 Dichotomous Search 259whereδ is a small quantity, say 0.001, andnis the number of experiments. If the
middle point of the final interval is taken as the optimum point, the requirement can
be stated as
1
2
Ln
L 0≤
1
10
i.e.,
1
2 n/^2
+
δ
L 0(
1 −
1
2 n/^2)
≤
1
5
Sinceδ= 0 .001 andL 0 = 1. 0 , we have
1
2 n/^2+
1
1000
(
1 −
1
2 n/^2)
≤
1
5
i.e.,
999
1000
1
2 n/^2≤
995
5000
or 2n/^2 ≥999
199
≃ 5. 0
Sincenhas to be even, this inequality gives the minimum admissible value ofnas 6.
The search is made as follows. The first two experiments are made at
x 1 =L 0
2
−
δ
2= 0. 5 − 0. 0005 = 0. 4995
x 2 =L 0
2
+
δ
2= 0. 5 + 0. 0005 = 0. 5005
with the function values given by
f 1 =f(x 1 )= 0. 4995 (− 1. 0005 )≃ − 0. 49975
f 2 =f(x 2 )= 0. 5005 (− 0. 9995 )≃ − 0. 50025Sincef 2 < f 1 , the new interval of uncertainty will be (0.4995, 1.0). The second pair
of experiments is conducted at
x 3 =(
0. 4995 +
1. 0 − 0. 4995
2
)
− 0. 0005 = 0. 74925
x 4 =(
0. 4995 +
1. 0 − 0. 4995
2
)
+ 0. 0005 = 0. 75025
which give the function values as
f 3 =f(x 3 )= 0. 74925 (− 0. 75075 )= − 0. 5624994375f 4 =f(x 4 )= 0. 75025 (− 0. 74975 )= − 0. 5624999375Sincef 3 >f 4 , we delete (0.4995,x 3 ) nd obtain the new interval of uncertainty asa
(x 3 , 1. 0 )=( 0. 74925 , 1. 0 )