# 15-441 Homework #1 (2006)

Consider sending a file of F KBytes in the following settings. Setting 1 consists of two computers A and B, each equipped with a modem that is capable of sending/receiving at 33.3kbps. For A to send a file to B, it must first establish a dial-up connection with B, which takes 30 seconds. It can then send the file in 128-byte packets, with a one-byte checksum attached to each packet. The propagation delay of the phone line is negligible. Assume that A and B are directly connected, i.e. there are no intermittent routers. Setting 2 consists of two computers C and D, connected by an established wireless connection that can transmit at 8kbps through a satellite, with a quarter-second (0.25 second) propagation delay. In setting 2, files are transmitted without being split into packets.

a)      If F=16, how long does it take to send the file from A to B? How long does it take to send the file from C to D?

Answer: The total number of bits that A has to send to B is: (16*1024/128)*129*8 = 132096 bits. The time spent on transmitting those bit is: 132096/33.3/1000 = 3.967seconds. Therefore the total time is 3.967+30=33.967seconds.  The total number of bits that C has to send to D is: 16*1024*8 = 131072 bits. The time spent on transmitting those bit is: 131072/8/1000 = 16.384seconds. Therefore the total time is 16.384+0.25=16.634seconds.

b)      Now do the same computation as in (a), but for the case of F=64.

Answer: Follow the same calculation as that of a) and you will get 45.867seconds and 65.756seconds, respectively.

Answer: It depends on what kind of traffic the link is going to carry.  Roughly speaking, when total amount of data sent is small, the 2nd link wins because of the smaller setup time. When total amount of data is big and initial latency is not an issue, then the first link wins.  The crossover point is around 39 KBytes.

(You may assume there are no errors during each transmission and you may ignore acknowledgements, i.e., consider only bits flowing from the sending computer to the receiving computer. Also note that 1k bytes=1024 bytes and 1kbps = 1000 bit/second and most people agree that 1 byte = 8 bits.)

## Problem 2: Ethernet

Problem 43 in the textbook, page 159:

(a) Since a host will be able to send 10 times as much data in the time interval corresponding to the roundtrip propagation delay (transmission rate increased from 10 Mbs to 100 Mbs), the new minimum packet size will be 464 * 10 + 48 = 4688 bits (586 bytes).

(b) The drawback of a large minimum packet size is that it can be inefficient when hosts send a lot of small datagrams.  If a datagram is smaller than the minimum packet size on the network, padding is added to fill the rest of the packet.  The padding is pure overhead and it becomes more significant as the minimum packet size grows.

(c) Part (c) was replaced by the following: How might the specification be modified so as to allow a smaller minimum packet size.

We can reduce the diameter of the network, i.e. the maximum propagation delay we can have in the network. In practice this is the decision that was made for the 100-megabit Ethernet standard.

## Problem 3: Network configuration

Suppose you are in charge of building a LAN in your department, where 7 machines are placed in a row, evenly spaced by 6 meters. The machine in the middle is a file server.

a)      Suppose you are to build a 10-Mbps thin-net. Given the prices below, how much is it going to cost? You may ignore the costs of the BNC connectors on the coax runs and the cost of the T connector at each station.

Answer: 7 thin-net interface cards, 6*6=36 meters of thin-net coaxial cable, total cost is \$368

b)      Now assume your boss is not satisfied with the performance of thin-net because there are too many collisions and wants you to replace it with a 10-Mbps 8-port twisted-pair repeater. If the repeater will be co-located with the server (i.e., have the same distances to other clients as the server has) and will be connected to the server via a 1-meter twisted-pair run, how much will this setup cost? You may ignore the costs of the RJ-14 connectors on the twisted-pair runs.

Answer: 7 10-Mbps twisted-pair interface cards, (6+12+18)*2+1=73 meters of twisted-pair cable, 10 Mbps 8-port repeater, total cost is \$456.5.

c)      Assuming each client communicates with only the server and not with other clients, will this upgrade noticeably increase performance? Why or why not?

Answer: This “upgrade” will not noticeably improve the throughput.  The reason is that the repeater does not eliminate any collisions and the capacity of the repeater-server link (which is the bottleneck in this configuration) has not improved.  Overall, the same traffic pattern will be seen on the bottleneck repeater-server link.

d)      If we choose to upgrade our original thin-net installation using an Ethernet switch instead, how much would this cost? Assume that the Ethernet switch has 8 10-Mbps ports (1 for each of the 6 clients and 2 empty) and one 100-Mbps "backbone" port (to which we will attach the server).

Answer: 6 10-Mbps twisterd-pair interface cards, (6+12+18)*2+1=7e meters of twisted-pair cable, one 100 Mbps twisted-pair interface card, one 10 Mbps 8-port switch with 100 Mbps backbone port, total cost is \$496.5

e)      Approximately what is the best throughput increase we could expect to see with the switch configuration compared to the thin-net configuration? Briefly justify your answer.

Answer: Approximately a 6x improvement in throughput is expected. The reason is that the traffic on the client-switch links will not affect each other since they now form separate collision domains, and the server-switch link now has enough capacity to accommodate all six clients' traffic at once.  If you consider the fact that in a switched environment, the links can be full duplex, the potential improvement in performance can 12-fold, 6 x 10 Mbps into the server and 6 x 10 Mbps out of the server.

 Item Price 10 Mbps thin-net interface card \$50 thin-net coaxial cable \$.50/meter 10 Mbps twisted-pair interface card \$50 100 Mbps twisted-pair interface card \$60 Category-5 twisted pair \$.50/meter 10 Mbps 8-port repeater \$70 10 Mbps 8-port switch with 100 Mbps backbone port \$100

## Problem 4: Frequency Division Multiplexing

You need to multiplex 3 connections onto a single link: The first is a music channel which plays music in the range of 15Hz - 20kHz (for simplicity assume 0 Hz - 20kHz). The second is a telephone line which transmits voice in the range 400Hz - 3400Hz. The third is a data line which transmits 10Kbps of data.

How might you proportion the frequencies to multiplex all 3 connections at their full desired rate, minimizing the total frequency used? Assume that channels are separated by 2kHz guard bands.

In working out this problem, assume no noise and assume the Nyquist Limit (from lecture) applies to translating between bps and the width of the frequency band.

Answer: Given the above applications, we have to carry three channels with the following amount of frequency spectrum:

·        Music channel: 20KHz

·        Telephone line: 3KHz

·        Data channel: 5 KHz; this will support a bit rate of 2 x 5 Kbps, which is what we need.

With frequency division multiplexing, each gets its own slice of the frequency spectrum, to avoid interference, we have to separate the three channels using guard bands of 2 KHz.

A possible allocation is as follows:

·        0-20KHz: music channel

·        20-22KHz: guard band

·        22-25KHz: telephone line

·        25-27KHz: guard band

·        27-32KHz: data channel

Clearly other allocations are possible, though there is no "trick" which will reduce the total bandwidth necessary given the problem's assumptions.

Consider two computers A and B, connected by 3000 km of optical fiber in which light propagates at 2.5 x 108 m/sec. Computer A sends a 1 MB file to computer B. Assume packet-header overhead and processing time are zero. How long will it take to transfer the file when the link speed has different values.

The file transmission time consists of two part:

• The transmission time: the time period between the first bit of the file is put on the link for transmission until the last bit of the file is put on the link. This time is determined by the link speed.
• The propagation delay: the time for a bit to travel from one end of the link to the other end of the link. This time is determined the light propagation rate.

In summary, for a file with size L bytes, on a link with physical length D meters, and link speed B bit/second (bps), its transmission time is T seconds, where

T = (L * 8) / B + D / v

Where v is the light propagation rate (2.5 * 108 meter/second).

Substituting the numbers from the problem in the above equation gives the answers shown below.  Remember that 1MB (the file size) corresponds to 220.  The results if you (incorrectly) interpret 1MB as 106 is given in parenthesis.

How long will it take to transfer the file if the link speed is:

(a)

a T1 (1.5 Mbits/sec)

T = 5.6044 sec (5.3453)

(b)

a OC-3 (155 Mbits/sec)

T = 0.06612 sec (0.06361)

(c)

a OC-48 (2.5 Gbit/sec)

T = 0.015355 sec (0.0152)

(d)

a OC-192 (10 Gbit/sec - common today)

T = 0.012839 sec (0.0128)

(e)

If increasing the link speed does not decrease the latency of a file transfer, then why are carriers spending billions of dollars to replace their links with faster ones?

Comparing the results in (c) and (d), we can see that increasing the link speed from 2.5 Gbps to 10 Gbps does not significantly reduce the file transmission time. The reason that we want faster links is that the links in the network, especially those in the core networks, are share by many users (flows). One user's flow can only get a share of the link bandwidth (link speed). For example, if there are 1000 users transmitting files at the same time, and the link bandwidth is uniformly shared among them. For a link with speed 2.5Gbps as in problem (c), a single user can only get 2.5Mbps bandwidth. If we increase the link speed to 10Gbps, the same user can get 10Mbps. That will significantly reduce the file's transmission delay (to get a feeling about the decrease, compare the results between (a) and (b)).