"Layering (OSI & TCP/IP) and Packet, Circuit & Virtual-Circuit Switching" — Practice

Question 1

Name the four components of end-to-end delay and say which varies with network load.

Accepted Answer

Processing, queueing, transmission and propagation delay. Only queueing delay varies with load.

Question 2

Which OSI layer uses port numbers, and which uses MAC addresses?

Accepted Answer

The transport layer uses port numbers to identify processes, and the data link layer uses MAC addresses to identify interfaces on a link.

Question 3

A 1 Mbit packet is sent over a 100 Mbps link. Give the transmission delay.

Accepted Answer

One megabit divided by 100 megabits per second, which is 10 milliseconds.

Question 4

A 10 Mbit message crosses 3 store-and-forward links of 10 Mbps each. Compare sending it as one packet against 10 packets of 1 Mbit.

Accepted Answer

As one packet each link takes 1 second and the hops are sequential, giving 3 seconds. As 10 packets, each transmission is 0.1 seconds and the formula gives (3 plus 10 minus 1) times 0.1, which is 1.2 seconds.

Question 5

A link runs at 1 Gbps with 10 ms one-way propagation delay. Give the bandwidth-delay product and the utilisation achieved with a 64 KB window.

Accepted Answer

The product is 10 to the 9 times 0.01, which is 10 megabits or 1.25 MB. A 64 KB window is 512 kilobits, transmitted in 0.5 ms, followed by a 20 ms round trip wait, so utilisation is about 0.5 over 20.5, roughly 2.4 percent.

Question 6

Compute the bytes on the wire for a 1000-byte payload with a 20-byte TCP header, 20-byte IP header and 18 bytes of Ethernet framing, then repeat for a 1-byte payload.

Accepted Answer

For 1000 bytes the total is 1058, an overhead of about 5.5 percent. For 1 byte the total is 59, an overhead of 98.3 percent.

Question 7

Give three ways datagram and virtual circuit switching differ in router behaviour.

Accepted Answer

A datagram router holds state per destination prefix while a virtual circuit router holds state per connection; a datagram packet carries a full destination address while a virtual circuit packet carries a short identifier; and a datagram network reroutes around a failed router while a virtual circuit network tears down every circuit through it.

Question 8

Explain why an optimal packet size exists, giving the limiting behaviour on each side.

Accepted Answer

Total time is (N minus 1 plus M over p) times (p plus h) over R. As p tends to zero the header dominates every transmission and the time diverges; as p tends to the whole message the pipelining term collapses to one and the time approaches N full transmissions. The product therefore has an interior minimum.

Question 9

Why are link-layer and transport-layer checksums both necessary?

Accepted Answer

The link checksum is verified on arrival at a router and discarded, so corruption occurring inside that router's memory or during its internal handling goes undetected. Only a checksum computed by the source and verified by the destination covers the whole path, which is the end-to-end argument.

Question 10

Distinguish synchronous time division multiplexing from statistical time division multiplexing and connect each to a switching model.

Accepted Answer

Synchronous TDM assigns each source a fixed recurring slot whether or not it has data, which is circuit switching at the link level. Statistical TDM allocates slots on demand, so a bursty source uses capacity only when transmitting, which is packet switching at the link level and is why more users can be served than the sum of their peak rates allows.

Practice — "Layering (OSI & TCP/IP) and Packet, Circuit & Virtual-Circuit Switching"

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