Table 4.1: An example of RNABOX performance for a population with 10 strata and total sample size
n = 5110. Columns L
r
/U
r
, r = 1, . . . , 6, represent the content of sets L, U respectively, in the r-th
iteration of the RNABOX (between Step 3 and Step 4): symbols □ or ■ indicate that the stratum with label
h is in L
r
or U
r
, respectively.
h A
h
m
h
M
h
L
1
/U
1
L
2
/U
2
L
3
/U
3
L
4
/U
4
L
5
/U
5
L
6
/U
6
x
∗
1 2700 750 900 □ □ □ □ 750
2 2000 450 500 ■ □ □ □ □ 450
3 4200 250 300 ■ ■ ■ ■ ■ 261.08
4 4400 350 400 ■ ■ ■ □ 350
5 3200 150 200 ■ ■ ■ ■ ■ 198.92
6 6000 550 600 ■ ■ ■ □ □ 550
7 8400 650 700 ■ ■ ■ □ 650
8 1900 50 100 ■ ■ ■ ■ ■ ■ 100
9 5400 850 900 ■ ■ □ □ □ 850
10 2000 950 1000 □ □ □ □ □ 950
SUM 5000 5600 0/8 1/7 3/6 4/3 5/3 7/1 5110
We would like to point out that regardless of the population and other allocation parameters chosen,
the graph illustrating the operations of RNABOX algorithm will always have a shape similar to that of the
right half of a Christmas tree with the top cut off when U
r
∗
̸= ∅. This property results from that fact that
L
r
⊊ L
r+1
(see (C.1)) and U
r
⊇ U
r+1
(see (C.16)) for r = 1, . . . , r
∗
− 1, r
∗
≥ 2. For the population
given in Table 4.1, we clearly see that L
1
= ∅ ⊊ L
2
= {10} ⊊ L
3
= {10, 1, 2} ⊊ L
4
= {10, 1, 2, 9} ⊊
L
5
= {10, 1, 2, 9, 6} ⊊ L
6
= L
∗
= {10, 1, 2, 9, 6, 4, 7} and U
1
= {8, 5, 3, 7, 4, 6, 9, 2} ⊃ U
2
=
{8, 5, 3, 7, 4, 6, 9} ⊃ U
3
= {8, 5, 3, 7, 4, 6} ⊃ U
4
= {8, 5, 3} = U
5
⊃ U
6
= U
∗
= {8}. Moreover,
regardless of the population and other allocation parameters chosen, subsequent values of set function
s, which are placed at the ends of branches of the Christmas tree, above ■, form a non-increasing
sequence while moving upwards. This fact follows directly from Lemma C.3. For the example allocation
illustrated in Fig. 4.1, we have s(L
1
, U
1
) = 0.3 > s(L
2
, U
2
) = 0.204 > s(L
3
, U
3
) = 0.122 >
s(L
4
, U
4
) = 0.0803 > s(L
5
, U
5
) = 0.075 > s(L
6
, U
6
) = 0.0622 (note that L
1
= ∅). The same
property appears for values of s related to the trunk of the Christmas tree, placed above □. In this case,
this property is due to (C.2), (C.15) and (C.1). For allocation in Fig. 4.1, we have s(L
1
, ∅) = 0.127 >
11