Question

A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Consider the case of a source host connected to the source router, which is connected to the destination router, andthen to the destination host If either of the routers discards a packet, the source host eventually times out and tries again.If both host-router and router- router lines are counted as hops, what is the mean number of(a) hops a packet makes per transmission?(b) transmissions a packet makes?(c) hops required per received packet?

Answer 1

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.)
`k^(2)  - 3k + 3 * (1)/((1 - k)^(2) )\\\\= (k^(2) - 3k + 3 )/((1-k)^(2) )`

Step-by-step 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 =
`k^(2)  - 3k + 3 * (1)/((1 - k)^(2) )\\\\= (k^(2) - 3k + 3 )/((1-k)^(2) )`