From this, we obtain the following difference equation Yn Yn+1 = 1+p Yn-m (8) where p = . From now on, we handle the difference equation (8). The unique positive equilibrium point of Eq.(8) is Example 11 Consider Eq.(8) for m 5 and p 0.7. Hence we have following difference equation Yn+1 = 1+0.7-Yn (17) Let the initial conditions are y-s = 10, y-4 = 6, y-3 = 8, y-2 4, y-1 = 3 and yo = 5. Then unique positive equilibrium point of Eq.(17) is globally asymptotically stable. See Figure 1.

Computer Networking: A Top-Down Approach (7th Edition)
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Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
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From this, we obtain the following difference equation
Yn
Yn+1 = 1+ p-
(8)
.2
Yn-m
where p = . From now on, we handle the difference equation (8). The unique
positive equilibrium point of Eq.(8) is
%3|
Example 11 Consider Eq.(8) for m = 5 and p 0.7. Hence we have following
difference equation
Yn
Yn+1 = 1+0.7
(17)
Let the initial conditions are y-5 = 10, y-4 = 6, y-3 = 8,y-2 = 4,y-1 =
3 and yo = 5. Then unique positive equilibrium point of Eq.(17) is globally
asymptotically stable. See Figure 1.
2.5
2.0
1.5
1.0
100
200
300
400
500
600
Figure 1: Plot of Eq. (17).
Transcribed Image Text:From this, we obtain the following difference equation Yn Yn+1 = 1+ p- (8) .2 Yn-m where p = . From now on, we handle the difference equation (8). The unique positive equilibrium point of Eq.(8) is %3| Example 11 Consider Eq.(8) for m = 5 and p 0.7. Hence we have following difference equation Yn Yn+1 = 1+0.7 (17) Let the initial conditions are y-5 = 10, y-4 = 6, y-3 = 8,y-2 = 4,y-1 = 3 and yo = 5. Then unique positive equilibrium point of Eq.(17) is globally asymptotically stable. See Figure 1. 2.5 2.0 1.5 1.0 100 200 300 400 500 600 Figure 1: Plot of Eq. (17).
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