Reverse Bucklin vs. Instant Runoff

[DEMOREP1 at aol.com]

A larger tie example

15  A>B>C>D>E>F
16  B>C>D>E>F>A
17  C>D>E>F>A>B
18  D>E>F>A>B>C
19  E>F>A>B>C>D
20  F>A>B>C>D>E
105

Assume each is acceptable to a majority of the voters using a YES/NO vote.

89 A/B  16
88 B/C  17
87 C/D  18
86 D/E  19
85 E/F   20
90 F/A  15

72 A/C  33
70 B/D  35
68 C/E  37
66 D/F  39
70 E/A  35
74 F/B  31

54 A/D 51
51 B/E  54
48 C/F  57

A>B>C>D>E>F>A

If the 3 last choices are added using Reverse Bucklin, then --
A 51
B 54  loses
C 57  loses
D 54  loses
E 51
F 48
 315

70 E/A  35
90 F/A  15
85 E/F   20

E>F>A,  E wins

Should the 51 first choice voters for B, C and D have to lose their first
choices?
If no, then an obvious tie breaker is to have the 15 A first choice voters
lose their first choices (i.e. use the Instant Runoff tiebreaker)

88 B/C  17
87 C/D  18
86 D/E  19
85 E/F   20
70 B/D  35
68 C/E  37
66 D/F  39
74 F/B  31
51 B/E  54
48 C/F  57

B>C>D>E>F>B

First choice votes--
31 B
17 C  loses
18 D
19 E
20 F
105

86 D/E  19
85 E/F   20
70 B/D  35
66 D/F  39
74 F/B  31
51 B/E  54

D>E>F>B>D

First choice votes--
31 B
35 D
19 E loses
20 F
105

70 B/D  35
66 D/F  39
74 F/B  31

B>D>F>B

First choice votes--
31 B  loses
35 D
39 F
105

66 D/F  39,  D wins.
Note the relation of D and F at the top.  In a real election of course, the
votes would probably be much more mixed (i.e. only 6 of the possible 6
factorial (720) combinations are in the example). 

Thus, should instant run-off be used repeatedly as the tiebreaker in circular
ties ?
The theory is that the smallest number of voters should have to change their
choices repeatedly to break a tie.