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Consequently system H survives in every case, while system N survives only
when M
a
(0)
−
M
b
ðÞ
\
h
b
N
b
ðÞ
.
Now, the case of mutual indifference of C
a
-and C
b
-elements is considered. In
this case, by supposing r ¼
q
¼ 0 in Eqs. (
10.17
) and (
10.18
), we obtain:
N
Ra
ð
n
Þ
¼N
Ra
ð
0
Þ
X
n
1
m
Ra
ð
i
Þ
p
1
½N
Ra
ð
i
Þ;
i¼0
ð
10
:
37
Þ
N
Rb
ð
n
Þ
¼N
Rb
ð
0
Þ
X
n
1
m
Rb
ð
i
Þ
p
2
½N
Rb
ð
i
Þ
i¼0
and
X
k
1
X
k
1
½
m
a
ð
n
Þþ
m
Ra
ð
n
Þ
¼ M
b
ð
0
Þ;
½
m
b
ð
n
Þþ
m
Rb
ð
n
Þ
¼ M
a
ð
0
Þ
ð
10
:
38
Þ
n¼0
n¼0
The solution of the game of the two systems, H and N, with the win function as
shown in Eq. (
10.16
) and under conditions as expressed in Eqs. (
10.37
) and (
10.38
),
is reduced to the problem of maximizing two functions:
Q
1
¼
X
k
1
m
a
ð
n
Þ
p
1
½N
Ra
ð
n
Þ
¼ max
Ra
ð
10
:
39
Þ
n¼0
Q
2
¼
X
k
1
m
b
ð
n
Þ
p
2
½N
Rb
ð
n
Þ
¼ max
Rb
ð
10
:
40
Þ
n¼0
From Eqs. (
10.39
) and (
10.40
) it is evident that when p
1
[N
Ra
(n)] =
p
2
[N
Rb
in = 1, the optimal strategies of both sides consist in the destruction of a-
and b-elements. If the C-elements act independently of one another, and the
probability of putting the C-elements out of operation by a single R-element is equal
to a constant value D
a
and D
b
for H and N systems, respectively, then
p
1
N
Ra
ðÞ
½
exp
d
a
N
Ra
ðÞ
½
;
p
2
N
Rb
ðÞ
½
exp
d
b
N
Rb
ðÞ
½
ð
10
:
41
Þ
− ʸ
1
ln(1
−
− ʸ
2
ln(1
−
where, d
a
=
D
a
),d
b
=
D
b
) are the effectiveness coef
cients
ʸ
1
and
ʸ
2
are the density of the location of elements
of the R
a
- and R
b
-elements; and
in the H and N system, respectively.
When k = 1, the optimal strategies of the behavior of the systems will be
m
a
ð
0
Þ
¼M
b
ð
0
Þ
and m
b
ð
0
Þ
¼M
a
ð
0
Þ
. This is natural, since in the last step it makes
no sense to destroy protective elements. When k = 2, from Eqs. (
10.37
)
(
10.40
)we
-
obtain the optimal strategies in the form:
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