. 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

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.

Question 1

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 200000km/sec. What is the minimum frame size?

Answer

We must keep in mind that in CSMA/CD , for a station to get some surety of successful transmission the contention interval (time during which the station is transmitting) should have at least 2ι slot width where ι is time for signal to propagate between two farthest stations ie there must be enough time for the front of the frame to reach the end of the cable and then for an error message to be sent back to the start before the entire frame is transmitted.

As a result for a 1 km cable the one way propagation time = 1/200000

= 5 x 10-6

= 5 μsec

so for both ways it would be = 2 x 5 μsec = 10 μsec

To make CSMA/CD work, it must be impossible to transmit an entire frame in this interval. At 1 Gbps, all frames shorter than 10,000 bits can be completely transmitted in under 10 μsec, so the minimum frame is 10,000 bits or 1250 bytes.

ie 109

bps x 10 x 10-6

sec = 104

bits

104

bits / 8 = 1250 bytes

q

Ten thousand airline reservation stations are comingfor the use of a single slotted ALOHA channel. Theaverage station makes 18 request/hour. A slot is 125sec.What is theapproximatetotal channel load

A

Average requests for 10000 stations= 10^4 x 18 / (60 x 60) =50 requests/secAverage slots number = 1 / (125 x10^-6) = 8000 slots/sec.Total channel load = averagerequests / average slots number= 50 / 8000 = 0.0625Hence, the total channel load is0.0625 request/slot.

1) An upper-layer packet is split into 10 frames, each of which has an 80 percent chance of arriving undamaged. If no error control is done by the data link protocol, how many times must the message be sent on average to get the entire thing through?

Solution:Since each frame has a chance of 0.8 of getting through, the chance of the whole message getting through is (0.8)10, which is about 0.107. Call this value p. The expected number of transmissions for an entire message is then

∞ ∞E = ∑ ip(1-p)i-1 = p∑ i(1-p)i-1

i=1 i=1

To reduce this, use the well known formula for the sum of an infinite geometric series, ∞

S = ∑ ai = 1 i=1 ------

1 – aDifferentiate both sides with respect to a to get,

∞S' = ∑ iai-1 = 1

i=1 ------- (1-a)2

Now use a = 1-p to get E = 1/p. Thus, it takes an average of 1/0.107, or about 9.3 transmissions.

When the IPv6 protocol is introduced, does the ARP protocol have to be changed? If so, are the changes conceptual or technical?

There are no changes conceptually. Technically, the IP addresses requested are now bigger, so bigger fields are needed.

Many companies have a policy of having two (or more) routers connecting the company to the Internet to provide some redundancy in case one of them goes down. Is this policy still possible with NAT? Explain your answer.

Ans: It is possible if the two rouers can exchange the NAT translation table information when new entries are added. From Chris Cabuzzi, "In the case of redundancy, CISCO has a protocol called HSRP that allows two or more routers to "appear" as one virtual router. With the implementation of HSRP, more than one router could be used with NAT without issue. Otherwise, the two routers would most likely conflict IP addresses amongst each other."

Problem 2.b. With NAT, can we still claim that 1) the destination address of IP packets will not be changed end-to-end? 2) every IP address uniquely idenfites a machine worldwide? Please brief explain and give an example.

Ans: 1) We cannot, since the destionation adress was changed to a private LAN address. A web server runs in a DMZ LAN may have a 192.168.0.10 address and packets to the external 128.198.60.10 may be routed to the web server using DNAT and convert its destination address to 192.168.0.10. 2) If every IP address includes that in the private LAN addreses, then different machines at different organizations may have the same 10.0.0.10 address.