Solution for Assignment 4 


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1. Compare and contrast the LLC and MAC sublayers.

LLC sublayer performs framing, flow control, and error control.  The MAC sublayer specifies how nodes share a single communication channel.

2. Give an example of when autonomous error correction is appropriate.

It is appropriate in simplex communication, e.g. TV and Radio.

3. Explain the various fields of an IEEE 802.3 frame.

Preamble, used for synchronization and uses the bit string 101010101010.  Start Frame Delimiter, used for specifying the start of the frame and uses bit string 10101011.  Refer to class notes for other fields.

4. Assume the CRC generator polynomial of x5 + x4 + x2 + 1 is used to provide error checking.  Use CRC algorithm to compute the bit stream that will be transmitted if a message 111000111000 is sent.

The transmitted string will be 11100011100010001

5. Do you think repeater's latency is less than or more than that of a bridge or a switch? Why?

Since repeater is a dumb device, its latency is less.

6. What distinguishes a network component as being labeled a layer-1 or layer-2 device?

A layer 1 device provides the physical and electrical connections between a node and the network medium; a layer 2 device provides layer 2 functionality, including framing, flow control, and error control.

7. Data link protocols almost always put the CRC in a trailer, rather than in a header.  Why?

It is easier to append CRC to the data while transmitting data, otherwise, data has to be buffered.

8. A 100 km long cable runs at the T1 data rate.  The propagation speed in the cable is 2/3 speed of light.  How many bits fit in the cable?

The propagation speed in the cable is 2/3 * 3 x 108 m/s which is 200,000 km/s or 200 km/ms, so a 100 km cable will be filled in 500 micro sec.  This corresponds to four 193-bit frames, or 772 bits on the cable.

9. A group of N stations share 56-kbps pure ALOHA channel.  Each station outputs a 1000-bit frame on an average of once every 100 sec, even if the previous one has not yet been sent (e.g., the stations are buffered.)  What is the maximum value of N?

With pure ALOHA, the usable bandwidth is .184 x 56 = 10.3 kbps.  Each station requires 10 bps, so N = 10300/10 = 1030 stations.

10. Consider the delay of pure ALOHA vs. slotted ALOHA at low load, which one is less? Explain your answer.

 With pure ALOHA transmission can start immediately. With slotted ALOHA, transmission has to wait to the next slot.

11. What is the baud rate of the standard 10-Mbps 802.3 LAN?

The Ethernet uses Manchester encoding, which means it has two signal periods per bit sent.  The data rate of the standard Ethernet is 10 Mbps, so the baud rate is twice that, or 20 megabaud.

12. Consider building a CSMA/CD network running at 1 Gbps over a 1-km cable with no repeaters.  The signal speed in the cable is 200,000 km/sec.  What is the minimum frame size?

For a 1 km cable, the on-way propagation time is 5 micro sec.  To make CSMA/CD work, it must be possible to transmit entire frame in this interval.  At 1 Gbps, minimum frame should be transmitted in 10 micro sec, or must have 10,000 bits or 1250 bytes.

13. Ethernet frames must be at least 64 bytes long to ensure that the transmitter is still going in the event of a collision at the far end of the cable.  Fast (100 Mbps) Ethernet has the same 64 64 byte minimum frame size, but can get the bits out ten times faster.  How is it possible to maintain the same minimum frame size?

The maximum wire length in Fast Ethernet is 1/10 of regular Ethernet.

14.  Imagine two LAN bridges.  The first bridge is faced with 1000 512-byte frames per second that must be forwarded.  The second is faced with 200 4096-byte frames per second.  Which bridge do you think will need the faster CPU?

The 1000 frame/sec bridge would need a faster CPU.  Although the other one has a higher throughput, the 1000 frame/sec bridge has more interrupts, more process switches, more frames passed, etc.

15.  In a token ring the sender removes the frame.  What modifications to the system would be needed to have the receiver remove the frame instead, and what would the consequences be?

First the ACK can not be be piggybacked to the sender. It has to be sent separately.  Also each station must have buffers to absorb all the bits up to the destination address before it decides to absorb it all or forward it.

16. Explain why in PPP the address field is set to HDLC broadcast address of 1111111.

The address is meaningless in ppp since it is only point to point, so it got set to all 1's.

17. Why is there a need for IPCP protocol for PPP?

PPP is designed to run different packets not only IP, but IPx, NetBios, etc.

18. Explain why WLANs can not implement CSMA/CD?  Explain the protocol for CSMA/CA?

See the notes and the book.