What are the advantages of wired networks?
1 answer:
You might be interested in
Answer:
a.) k² - 3k + 3
b.) 1/(1 - k)²
c.)
Explanation:
a.) A packet can make 1,2 or 3 hops
probability of 1 hop = k ...(1)
probability of 2 hops = k(1-k) ...(2)
probability of 3 hops = (1-k)²...(3)
Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)
= (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)
= k + 2k - 2k² + 3(1 + k² - 2k)
∴mean number of hops = k² - 3k + 3
b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3
if k = 0 then number of hops is 3
if k = 1 then number of hops is (1 - 3 + 3) = 1
multiple transmissions can be needed if K is between 0 and 1
The probability of successful transmissions through the entire path is (1 - k)²
for one transmission, the probility of success is (1 - k)²
for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)
for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on
∴ for transmitting a single packet, it makes:
∞ n-1
T = ∑ n(1 - k)²(1 - (1 - k)²)
n-1
= 1/(1 - k)²
c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)
from (a) above, mean number of hops when transmitting a packet = k² - 3k + 3
from (b) above, mean number of transmissions by a packet = 1/(1 - k)²
substituting: mean number of required packet =