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Alternatively, x(t) also increases when a peer with a corrupted copy gets a
good copy. This event occurs at the following rate:
y(t)m(1−p(t))
Combining these two cases, we can express the rate of change of x(t) as follows:
dx(t)
dt
= [M−x(t)−y(t)]m(1−p(t)) + y(t)m(1−p(t))
(7.5)
Similarly, we can also express the rate of change of y(t) as follows:
dy(t)
dt
= [M−x(t)−y(t)]mp(t)−y(t)m(1−p(t))
(7.6)
We can also visualize these observations from the state transition diagram
shown in Figure 7.2. With some simple algebraic manipulations, the two dif-
ferential equations can be rewritten as:
dx(t)
dt
=
[M−x(t)]m(1−p(t))
(7.7)
dy(t)
dt
=
[M−x(t)]mp(t)−my(t)
(7.8)
m (1 - p(t))
Polluted
Copy
Good Copy
m p(t)
m (1 - p(t))
m p(t)
No Copy
FIGURE 7.2: State transitions for a peer in getting a file in the Copy Centric
Downloading model [Kumar et al., 2006].
Kumar et al. [Kumar et al., 2006] showed that there is a unique solution:
M
M+N
c
2
M (e
mt
−
c
1
M +N
)
x(t)
=
(7.9)
M
M+N
1 + c
2
(e
mt
−
c
1
M +N
)
M−c
1
e
−mt
−x(t)
y(t)
=
(7.10)
where c
1
= M−x(0)−y(0) and
−
M+N
x(0)
M−x(0)
N + x(0) + y(0)
M + N
c
2
=
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