Networking HW #2

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Consider the problem in P25 but now with a link of R = 1 Gbps. C) What is the width (in meters) of a bit in the link?

.25 meters

Consider the discussion in Section 1.3 of packet switching versus circuit switching in which an example is provided with a 1 Mbps link. Users are generating data at a rate of 100 kbps when busy, but are busy generating data only with probability p = 0.1. Suppose that the 1Mbps link is replaced by a 1Gbps link. A) What is N, the maximum number of users that can be supported simultaneously under circuit switching?

1,000,000 / 100 = 10,000 users

Suppose there is a 10 Mbps microwave link between a geostationary satellite and its base station on Earth. Every minute the satellite takes a digital photo and sends it to the base station. Assume a propagation speed of 2.4 * 10^8 m/s. The geostationary satellite is 36,000 km away from Earth B) What is the bandwidth-delay product, R*d(prop)?

1,500,000 bits

A periodic signal completes one cycle in 0.001 seconds. What is its frequency?

1000Hz

150 ms

Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 10^8 meters/second B) Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time?

160,000 bits

How many KHz are in one Hz, one MHz, and one GHz?

1Hz = .001 KHz 1 MHz = 1,000 KHz 1 GHz = 1,000,000 KHz

Refer again to problem P25 B) Suppose now the file is broken up into 20 packets with each packet containing 40,000 bits. Suppose that each packet is acknowledged by the receiver and the transmission time of an acknowledgment packet is negligible. Finally, assume that the sender cannot send a packet until the preceding one is acknowledged. How long does it take to send the file?

20 * (t(trans) + 2t(prop)) = 20*(20ms + 80ms) = 2 seconds

20,000 * (2.5*10^8) = 160,000 bits

Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 percent of the time. A) When circuit switching is used, how many users can be supported?

3,000 / 150 = 20 users

Wireshark reports the Header Length of the example packet as 20 bytes. What value is actually stored in the header length field? Why?

5, Header length counts the number of 32-bit words

If the bandwidth of a signal is 5KHz and the lowest frequency is 52KHz, what is the highest frequency of the signal?

57Khz

600,000,000 bits

Consider a voice signal with frequencies from 300 to 3400Hz. What's the minimum sampling frequency that would allow me to create a perfect replica of the original signal? Assume my measurements are infinitely accurate.

6200 samples per second

Consider the problem in P25 but now with a link of R = 1 Gbps. A) Calculate the bandwidth-delay product, R*d(prop)

80,000,000 bits

Consider the problem in P25 but now with a link of R = 1 Gbps. B) Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one big message. What is the maximum number of bits that will be in the link at any given time?

800,000 bits

Can a host have more than one IP address? If so, how?

A host may have multiple IP addresses. In fact an interface may be assigned multiple IP addresses.

What does a multicast address identify? How is an Ethernet address identified as multicast?

A multicast address identifies a group of receivers. If the most significant bit is a 1 and at least one other bit is a 0, it is a multicast address

What does an IP address identify?

A network interface

A broadcast address identifies all interfaces. What is the Ethernet broadcast address? Exactly which interfaces will receive an Ethernet frame sent to the broadcast address?

An Ethernet address with all bits set to 1 is the broadcast address. All interfaces on the same Ethernet segment receives an Ethernet broadcast; however such messages will not be forwarded across routers.

Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 percent of the time. D) Find the probability that there are 21 or more users transmitting simultaneously

An even more ridiculous statistics formula. Like seriously, its disgusting.

C) Compare the results from A) and B)

Breaking up a file takes longer to transmit because each data packet and its corresponding acknowledgement packet add their own propagation delays.

Consider parsing an IP packet B) How do you know where (octet offset) the data starts?

Data starts at byte headerlength

Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 100 km/hour. B)Repeat A), now assuming that there are eight cars in the caravan instead of ten

Delay between the tollbooths is 8*12 seconds plus 45 minutes; i.e. 46 minutes and 46 seconds. Total delay is twice this amount plus 8*12 seconds, i.e. 94 minutes and 48 seconds.

Describe a mechanism to keep Ethernet addresses universally unique, given that there are multiple manufacturers producing a large number of cards.

Each manufacturer is given a dedicated range of Ethernet addresses. The first three bytes, called the Organizationally Unique Identifier (OUI), identify the organization. All NICs from the manufacturer are assigned an address from that range.

If the FCS check fails, how does Ethernet determine precisely which bit is damaged?

Ethernet cannot determine which bit was damaged using the FCS. It only knows some bits were damaged.

Why does Ethernet have a minimum frame size?

Ethernet has a minimum frame size to insure transmission lasts long enough to detect collision.

If I ask Ethernet to send 5 bytes of data, how much padding will it add? How many bytes will Ethernet deliver to the layer above?

If I send 5 bytes of data, Ethernet will add 41 bytes of padding.

Consider parsing an IP packet A) How do you know if there are options?

If the header length is greater than 20 bytes

A unicast address identifies a particular interface. How is an Ethernet address identified as unicast?

If the most significant bit is a 0, it is a unicast address

Now suppose that N such packets arrive to the link every LN/R seconds. What is the average queuing delay of a packet?

It takes LN/R seconds to transmit the N packets. Thus, the buffer is empty when each batch of N packets arrive. Thus, the average delay of a packet across all batches is the average delay within one batch, i.e. (N-1)L/2R

Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 percent of the time. C) Suppose there are 120 users. Find the probability that at any given time, exactly n users are transmitting simultaneously.

Some binomial distribution garbage

Consider the discussion in Section 1.3 of packet switching versus circuit switching in which an example is provided with a 1 Mbps link. Users are generating data at a rate of 100 kbps when busy, but are busy generating data only with probability p = 0.1. Suppose that the 1Mbps link is replaced by a 1Gbps link. B) Now consider packet switching and a user population of M users. Give a formula (in terms of p, M, and N) for the probability that more than N users are sending data

Some stupid probability function

What is the advantage of having the IP checksum cover only the header and not the data?

The advantage of having the IP checksum cover only the header and not the data is the quicker speed with which it can be calculated.

The bandwidth-delay product of a link is the maximum number of bits that can be in the link at a given time

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. D) Suppose Host A begins to transmit the packet at time t=0. At the time t = d(trans), where is the last bit of the packet?

The bit is just leaving Host A

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. F) Suppose d(prop) is less than d(trans). At the time t = d(trans), where is the first bit of the packet?

The first bit has reached Host B

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. E) Suppose d(prop) is greater than d(trans). At time t = d(trans), there is the first bit of the packet?

The first bit is in the link and has not reached B

What is the minimum size of an Ether net frame in Octets? Does this minimum size match the size of the Frame in ethernet.pcap? If not, explain why

The minimum Ethernet frame size(not including the preamble) is 64 bytes. The size of the frame in wireshark is 60 bytes so it doesn't match; however, Wireshark doesn't show the 4 bytes FCS.

Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queued. Each packet is of length L and the link has transmission rate R. What is the average queuing delay for the N packets?

The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay for the N packets is (L/R + 2L/R + .. + (N-1)L/R)/N = LN(N-1)/(2RN) = (N-1)L/(2R)

Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 10^8 meters/second D) What is the width (in meters) of a bit in the link? Is it longer than a football field?

The width of a bit = length of link/bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football field.

Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 100 km/hour. A) Supposed the caravan travels 150 km, beginning in front of one tollbooth, passing through a second tollbooth, and finishing just after the third toll-booth. What is the end-to-end delay?

There are ten cars. It takes 120 seconds, or 2 minutes, for the first tollbooth to service the 10 cars. Each of the cars has a propagation delay of 45 minutes before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 47 minutes. The whole process repeats itself for traveling between the second and third tollbooths. It also takes 2 minutes for the third tollbooth to service the 10 cars. Thus total delay is 96 minutes. 45(between 1-2) + 45 (between 2-3) + 6(2 per station) = 96 minutes

Consider parsing an IP packet C) How do you know when the data ends such that a complete IP packet has been received?

Use the total length

Is it possible for a packet sniffer like Wireshark to capture and analyze network data addressed to a machine other than the one Wireshark is running on? How?

Yes, if my NIC is in promiscuous mode, it will intercept all traffic and provide it to Wireshark.

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. C) Ignoring processing and queuing delays, obtain an expression for the end-to-end delay

d(end-to-end) = (m/s + L/R) seconds

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. A) Express the propagation delay, d(prop), in terms of m and s

d(prop) = m/s seconds

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. B) Determine the transmission time of the packet, d(trans), in terms of L and R

d(trans) = L/R seconds

Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. G) Suppose s = 2.5 x 10^8, L = 120 bits, and R = 56kbps. Find the distance m so that d(prop) = d(trans)

m = (L/R)*s = (120/56x10^3)*(2.5x10^8) = 536 km

Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 percent of the time. B) For the remainder of this problem, suppose packet switching is used. Find the probability that the given user is transmitting.

p = 0.1

Refer again to problem P25 A) How long does it take to send the file, assuming it is sent continuously?

t(trans) + t(prop) = 400 + 80 = 480 ms

Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 10^8 meters/second E) Derive a general expression for the width of a bit in terms of the propagation speed, s, the transmission rate R, and the length of the link m.

Networking HW #2

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