Question

1.When a packet arrives at a router, one other packet is halfway done being transmitted on the outbound link and three other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 1000 bytes and the link rate is 10 Mbps. What is the queuing delay for the packet in milliseconds (to one decimal place)?

2.A UDP segment has 16 bits allocated for its length. What is the maximum number of bytes of application data (that is, the payload) a UDP segment can carry?

Answer 1

Answer:-----------

1).
Given,
Packet length L = 1000 bytes
Transmission rate R = 10 Mbps = 10 x 10⁶ bps
Currently transmitted packets = x bits
x = 1000 / 10 = 100
Waiting queue = n packets
n = 4

Formula for Queuing Delay :
Queuing Delay = (nL + (L-x)] / R
So,
(nL + (L-x) =[ (4 x 1000) + ( 1000 - 100) ]
== 4000 + 900
== 4900
Packets are transmitted at 10 Mbps,
= 4900 x 10 x 4
= 196000

Queuing Delay for packet is,
= 196000 / 10 x 10⁶
= 0.0196 second
= 19.6 millisecond.

2). The maximum number of bytes of application data (that is , the payload),
a UDP segment carry is (2^16 – 1) bytes - the header bytes.
This gives, 65535 bytes – 8 bytes = 65527 bytes.