CSC460 Homework Set 2 — Fall 2001
Due: Tuesday, October 2
- Suppose
that each station in a group of 5 broadcast stations broadcasts packets of
0.1 seconds duration. Stations choose at random whether or not to
broadcast in any 0.1 second period, and all stations which choose to
broadcast are synchronized to begin broadcasting at the start of the same
0.1 second intervals. If each station chooses independently to broadcast
with probability 0.2 in each slot, what proportiuon of intervals is wasted
due to multiple colliding broadcasts in those slots?
We need to calculate either:
1 - ( P(0) + P(1) ) or,
sum up all other cases.
P(0) = 0.85 = 10-5*215
= 0.32
P(1) = 5*0.2*0.84 = 0.41 Note: 5 is the
number of ways only one station can be broadcasting.
So the result we want is: 1 - 0.73 = 0.27
- A message is split into 10
packets, each of which has 80% chance of arriving undamaged. Assuming no
error control, how many attempts to send the message are required for the
entire message to arrive intact?
P(all correct) = 0.810 ~ 0.1
Therefore the average number of attempts is 1/P ~ 10
- A router is a device which
connects together a group of Local Area Networks (LAN). In a router which
connects a 10 Mbps LAN to the internet over a T1 (1.544 Mbps) link, how
much memory is needed so that on average, only 5% of frames are lost?
Assume that both LAN and T1 frames are always 1500 bytes and, that on
average, they arrive 10 msec apart.
lambda = 1/arrival time = 1/10ms = 100 frames/s
u = 1.544 * 106 bps / (1500 B * 8b / B) = 128.7/s
a = lambda/u = .78
p = (1-a)aN / (1-a(N+1))
.05 = (1-.78) * .78N / (1 - .78(N+1)
)
if N = 7 then .05 = (.22 ) * .18 / ( 1 - .14)
Therefore we need to hold 7 packets which is 7 *
1500 = 10500 bytes.
Reference: The
Finite Buffer Case from Vastola's "Probability and Queuing"
(RPI).
- In the diagram below, the
numbers represent line reliability expressed as the probability of each line
remaining intact during bad weather. To preserve (direct or indirect)
communication among all nodes, no more than 1 line may break. If the
probability of communication among all nodes must be no less than 72%
during bad weather, what is the minimum reliability required of line QT?
P = P(none bad) + (P one bad)
.72 = 0.5*0.8*0.9*X + 0.5*0.8*0.9*(1-X) +
0.5*0.8*(1-0.9)*X + 0.5*(1-0.8)*0.9*X + (1-0.5)*0.8*0.9*X
.72 = 0.36X + .36 - 0.36X + 0.04X + 0.09X + 0.36X
0.72 = 0.36+0.49X
X = 0.36/0.49 = 0.735
Therefore, the reliability of X
must be better than 0.735
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