"Data Link Layer: Framing, Error Detection, Medium Access & Ethernet Bridging" — Practice

Question 1

State the bit stuffing rule and the worst-case overhead it imposes.

Accepted Answer

Insert a 0 after every five consecutive 1s in the payload. In the worst case of an all-ones payload, one bit is added per five, giving 20 percent overhead.

Question 2

A generator polynomial has degree 5. How many CRC bits are appended and what burst length is guaranteed detected?

Accepted Answer

Five bits are appended, and every burst error of length 5 or less is detected.

Question 3

A code has minimum Hamming distance 5. How many errors can it detect and how many can it correct?

Accepted Answer

It detects up to 4 errors, since detection needs distance d plus 1, and corrects up to 2, since correction needs 2d plus 1.

Question 4

A 2 Mbps link has 20 ms one-way propagation and 2000-bit frames. Compute a and the stop-and-wait utilisation.

Accepted Answer

Transmission time is 1 ms, so a is 20. Utilisation is 1 over (1 plus 40), which is 1 over 41, about 2.4 percent.

Question 5

For the link above, what window size gives full utilisation, and how many sequence bits does selective repeat need?

Accepted Answer

The window must be at least 1 plus 2a, which is 41. Selective repeat allows 2 to the (m minus 1), so m must satisfy 2 to the (m minus 1) at least 41, giving m equal to 7.

Question 6

How many parity bits does a Hamming code need for 16 data bits, and where are they placed?

Accepted Answer

Solve 2 to the r at least 16 plus r plus 1. With r equal to 5, 32 is at least 22, and r equal to 4 gives 16 against 21, which fails. So 5 parity bits are needed, at positions 1, 2, 4, 8 and 16.

Question 7

A CSMA/CD network runs at 100 Mbps over 1 km with signal speed 2 times 10 to the 8 metres per second. Compute the minimum frame size.

Accepted Answer

One-way propagation is 5 microseconds and the round trip is 10. At 100 Mbps that carries 1000 bits, or 125 bytes, which is therefore the minimum frame size.

Question 8

Compare pure and slotted ALOHA throughput at G equal to 0.5 and G equal to 1, and explain the factor of two in their peaks.

Accepted Answer

Pure gives 0.184 at G equal to 0.5 and 0.135 at G equal to 1. Slotted gives 0.303 at 0.5 and 0.368 at 1. Pure ALOHA has a vulnerable period of two frame times because a frame is destroyed by any transmission starting within one frame time before or after it, while slotting confines collisions to a single slot, halving the window and doubling the peak.

Question 9

Explain what fails if an Ethernet station transmits a frame shorter than the minimum size.

Accepted Answer

The station finishes transmitting and stops listening before a collision originating at the far end can propagate back to it. The frame is destroyed on the wire but the sender records success, so the loss must be detected by a higher-layer timeout, which is exactly what collision detection was meant to avoid.

Question 10

Why does a switched Ethernet need the spanning tree protocol when an IP network does not need an equivalent?

Accepted Answer

An IP header carries a time-to-live field, so a packet caught in a routing loop is eventually discarded. An Ethernet frame has no such field, so a broadcast entering a physical loop is duplicated at every switch on every pass and circulates forever, producing a broadcast storm. Spanning tree elects a root and disables redundant links to make the topology loop-free.

Practice — "Data Link Layer: Framing, Error Detection, Medium Access & Ethernet Bridging"

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