Assignment #1

Csci4211 Spring 2019

Due on Feb 18, 2019

February 24, 2019

Notes: There are five questions in this assignment. Each question has 10 points.

1. (10 pt.)Describe the following access networks connected to your home: DSL, HFC, and FTTH.What are the advantages and dis-advantages of each?

Answer:–> DSL : Communication medium which uses copper wire telecommunication lines to transfer internetto homes.Benefits- Reliable and precents interruption and heady packeet loss.- Makes use of existing telephone lines hence no additional cost wiring is required.- Fast and low user subscription.Disadvantages- Limited physical distance.- Faster when receiving data than sending data.–> HFC: Communication technology that combines optical fiber and coaxial cable to bring broadbandcontent (video, data and voice) to homes. Benefits- Provides high performance (due to optical fiber) at low cost.- No need for subscribers to change the already existing coaxial cable setup since it uses a separateconnection.- HFC possesses high bandwidth capacities.Disadvantages- There are more expensive than coaxial cables.- The speed of transmission and signal quality is reduced since more users subscribe to the sameconnection.- Can be less reliable since fiber optic cables can break easily.–> FTTH : Communication technology that purely uses optical fiber cables to bring broadband(video, data and voice) to homes.Advantages- Superior bandwidth hence faster internet connections to homes.- Resistance to electromagnetic interference.- Secure transmission.- Exceptional durability. Disadvantages- Cost: Optical fiber are very expensive and house hold users might not be able to afford paying forthis type of communication technology.- FTTH is difficult to install since it can easily be broken.

Grading:3 points each for both DSL and HFC and 4 points for FTTH. -1 for no advantage or wrong advantage.-1 if no disadvantage or wrong disadvantage-1 for no description or wrong description-2 for no description of wrong description of optical fiber.

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2. (10 pt.) There are 100 computers to be connected to each other. How many connections required ifthey are connected to each other with a direct link? Given a set of 5X5 switches (5 ports and each porthas both input and output connections), what is the configuration that has any computer connected toany other computer with the minimum maximum distance? The distance is measured by the number ofhops (switches) to be traversed.

Answer:

1)(1002

)= 100x99

2x1 = 4, 9502) The minimum maximum distance to connect 100 computers will require 33 switches is 7 hobs or 6switches away. Please see diagram below.

Grading:5 points for each sub question.-5 if you forget any part of the solution.-2 or -3 if you think there are more than 5 ports on each switch;-1 if you forget a diagram.-1 if you have other misunderstanding of ”switch”

3. . (10 pt.) Compute the time required for circuit switching and packet switching with the followingconditions:• The destination is 4 hops away from the source: three intermediate switches between the source andthe destination, with 4 links.• The distance between source (and destination) computer to its connected switch is 1 Km. Thedistance between any two adjacent switches is 10 Km.• The signal propagation speed is 5x105 m per second.• The message size is 5 Mega bits (1 Mega = 106)• The maximum packet size is 200k bits (1k = 103. You can ignore the size of the header.).• The transmission speed of source (or end) computer to its connected switch is 100 Mbps (Mega bitsper second). The link speed between adjacent switches is 10 Giga bits per second• The circuit setup time is 2x10−1 second for the case of circuit switching.• The processing time for routing decision including queuing delay at each switch is 10-3 second.In this case, which switching method has a shorter completion time? If the message size can beenlarged, is there a chance the other switching method can be better? Note: Please show yourcomputation steps.

Answer: Book 1.4.1, 1.4.3Circuit Switching:

dsetup = 0.2s

dtrans =L

R=

5Mb

100Mbps= 0.05s,Consider lowest rate.

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dprop =103m + (10 × 103m) + (10 × 103m) + 103m

5 × 105m/s= 0.044s

dend−end = 0.294s

Packet Switching:

We consider the time duration from the first packet start transmission until last packet reachdestination.

NPacket =5 × 106

200 × 103= 25

dsourcetrans =L

R1=

200k

100Mbps= 0.002s

dswitch1trans =

L

R2=

200k

10Gbps= 0.00002s

dswitch2trans =

L

R3=

200k

10Gbps= 0.00002s

dswitch3trans =

L

R4=

200k

100Mbps= 0.002s

dtrans = dsourcetrans + dswitch1trans + dswitch2

trans + dswitch3trans = 0.00404s

dprop =103m + (10 × 103m) + (10 × 103m) + 103m

5 × 105m/s= 0.044s

dend−end = dtrans + dprop + 3 ∗ dprcessing+queuing = 0.05104

Packet switching has shorter completion time.Let message be X mb, here we compare the time of circuit switching and packet switching as follows:

dCircuitSwitching = 0.2 +X

100Mbps+ 0.084 = 0.284 + 0.01X

dPacketSwitching = (X

0.2− 1) × 0.002 + 0.00404 + 0.084 + 0.001 = 0.0886 + 0.002X = 0.2(0.01X) + 0.886

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Hence we can learn that packet switching is always faster regardless of message length.Grading: 4 points for each time result, 2 points for final conclusion.At least 1 points will be given for each if you realize there are several types of delay;-1 if you forget set up time in circuit switching-1 if you talk “packet” when computing circuit switching time-1 if you think circuit switching has transmission time at each hop-2 if you think the transmission time of packet switching corresponding to the message size, or, if youcompute transmission time for each packet-1 or -2 if the answer lacks computation steps-1 if there are computation errors;For the analysis:-1 if you mistake “Message size” as “Packet size”)-1 or -2 if your explanation is not reasonable (or you don’t have an explanation)-1 if you give a wrong conclusion

4. (10 pt.) Consider the queuing delay in a router buffer (preceding an outbound link). Suppose allpackets are L bits, the transmission rate is R bps, and the N packets simultaneously arrive at the bufferevery LN/R seconds. Find the average queuing delay of a packet. (Hint: The queuing delay for thefirst packet is zero; for the second packet L/R; for the third packet 2L/R. The Nth packet has alreadybeen transmitted when the second batch of packets arrives.)

Answer:

d̄queue =1

N

(0 +

L

R+

2L

R+ ... +

(N − 1)L

R

)=

(N − 1)L

2R

Grading:A large portion of points will be deducted if you answer is wrong and there is no derivation-2 if you count from 1 to M instead of from 0 to M-1-2 if there are computation errors-1 or -2 if a final formula is not given

5. (10 pt.) Describe today’s Internet? Please pay attention to a) How is the Internet organized andmanaged? b) What is the design principles of Internet? c) How are layering and protocols used? andd) How applications are supported? Your answer should be no more than two pages.

Answer:Basically, the Internet can be divided into two main components: the Internet core and the end system.The services provided by these two parts are different. The end system can have a variety ofapplications to server for different purposes, like web browsing, video streaming, file sharing. Thusthere are a lot of protocols for people to choose, like HTTP, FTP, and SMTP. It’s more flexible and canbe update quickly and easily. While the core part of the Internet focuses exclusively on the routing andswitching of the packet. This makes the core part more efficient in doing their own job. Thus there areonly a few protocols in the Network layer. IP plays like a narrow waist in the protocol stack of theInternet. The division of the functions of core part and end system also reflects the most importantdesign principles of the Internet: push the complexity of Internet to the end system and keep the corepart simple and efficient. The two famous features of TCP protocol, flow control and congestioncontrol, both try to recognize and fix the problem at the end instead of adding extra functions to therouters. This design principle can also be observed in the evolution from IPv4 to IPv6. Some functionsoriginally performed in the router at Network layer, like checksum, are moved to the end system toimprove the efficiency of the routers. Layering is another important design principle in the Internet.This brings benefits as well as drawbacks. By dividing the design into different layers, the complexityof the lower layer would be transparent to the upper layer. Moreover, no matter what implementationis used in the lower layer, the upper layer can treat them as the same as long as they provide uniformedinterfaces to the upper layer. This makes the mobile network possible in some ways. The APPs on ourphone can perform perfectly regardless of connecting LTE of Wi-Fi. Although some disadvantages

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emerge, such as function duplication in different layers (integrity checking is performed both in thetransport layer and network layer) and difficulties in upgrading the key layer: Network layer, thestrength of layer brings the prosperity of current Internet. The Internet is said to be the network ofnetwork. This can be interpreted in many ways. First, Internet is becoming increasingly complex sincethe initial ARPANET. We have LAN, WLAN, Ethernet and mobile network. Each itself is a networkand can be a part of the whole Internet. Second, the Internet is also a commercial place. BGP4 acts asa backbone of the whole Internet and different level of AS all have their own profit and customers inthe market of the Internet. At the end of this semester, hopefully we will have a better understandingof the Internet.

Grading:No points will be deducted unless your answer is too short to convey the basic ideas.

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