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Instructor manual for computer networking by keith
Typology: Exercises
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Chapter 1 Review Questions
a) Wifi (802.11) In a wireless LAN, wireless users transmit/receive packets to/from an base station (i.e., wireless access point) within a radius of few tens of meters. The base station is typically connected to the wired Internet and thus serves to connect wireless users to the wired network. b) 3G and 4G wide-area wireless access networks. In these systems, packets are transmitted over the same wireless infrastructure used for cellular telephony, with the base station thus being managed by a telecommunications provider. This provides wireless access to users within a radius of tens of kilometers of the base station.
networks. Third, if ISPs decide to charge more money to highly profitable content providers (in countries where net neutrality doesn't apply), the content providers can avoid these extra payments.
components, and process Transport layer as well.) Link layer switches process link and physical layers (layers 1 through2). Hosts process all five layers.
Chapter 1 Problems
Problem 1
There is no single right answer to this question. Many protocols would do the trick. Here's a simple answer below:
Messages from ATM machine to Server Msg name purpose
HELO
At time N(L/R) the first packet has reached the destination, the second packet is stored in the last router, the third packet is stored in the next-to-last router, etc. At time N(L/R)
a) A circuit-switched network would be well suited to the application, because the application involves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and not bursty, bandwidth can be reserved for each application session without significant waste. In addition, the overhead costs of setting up and tearing down connections are amortized over the lengthy duration of a typical application session.
b) In the worst case, all the applications simultaneously transmit over one or more network links. However, since each link has sufficient bandwidth to handle the sum of all of the applications' data rates, no congestion (very little queuing) will occur. Given such generous link capacities, the network does not need congestion control mechanisms.
a) Between the switch in the upper left and the switch in the upper right we can have 4 connections. Similarly we can have four connections between each of the 3 other pairs of adjacent switches. Thus, this network can support up to 16 connections.
b) We can 4 connections passing through the switch in the upper-right-hand corner and another 4 connections passing through the switch in the lower-left-hand corner, giving a total of 8 connections. c) Yes. For the connections between A and C, we route two connections through B and two connections through D. For the connections between B and D, we route two connections through A and two connections through C. In this manner, there are at most 4 connections passing through any link.
Tollbooths are 75 km apart, and the cars propagate at 100km/hr. A tollbooth services a car at a rate of one car every 12 seconds.
a) There are ten cars. It takes 120 seconds, or 2 minutes, for the first tollbooth to service the 10 cars. Each of these cars has a propagation delay of 45 minutes (travel 75 km) 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 the total delay is 96 minutes.
b) Delay between tollbooths is 812 seconds plus 45 minutes, i.e., 46 minutes and 36 seconds. The total delay is twice this amount plus 812 seconds, i.e., 94 minutes and 48 seconds.
a) (^) d (^) prop m / s seconds.
b) dtrans L / R seconds.
c) d (^) end to end ( m / s L / R )seconds.
d) The bit is just leaving Host A. e) The first bit is in the link and has not reached Host B. f) The first bit has reached Host B. g) Want
2. 5 10 536 56 10
(^) Rs 3
m km.
Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires
sec=7msec.
The time required to transmit the packet is
sec= 224 sec.
Propagation delay = 10 msec.
link requires L/R 2 to transmit the packet onto the second link; the packet propagates over the second link in d 2 /s 2. Similarly, we can find the delay caused by the second switch and the third link: L/R 3 , dproc , and d 3 /s 3. Adding these five delays gives dend-end = L/R 1 + L/R 2 + L/R 3 + d 1 /s 1 + d 2 /s 2 + d 3 /s 3 + dproc+ dproc
To answer the second question, we simply plug the values into the equation to get 6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec.
Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Thus, dend-end = L/R + d 1 /s 1 + d 2 /s 2 + d 3 /s 3
For the values in Problem 10, we get 6 + 20 + 16 + 4 = 46 msec.
The arriving packet must first wait for the link to transmit 4.5 *1,500 bytes = 6,750 bytes or 54,000 bits. Since these bits are transmitted at 2 Mbps, the queuing delay is 27 msec. Generally, the queuing delay is ( nL + ( L - x ))/ R.
a) 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 = L/(RN) * (1 + 2 + ..... + (N-1)) = L/(RN) * N(N-1)/ = LN(N-1)/(2RN) = (N-1)L/(2R)
Note that here we used the well-known fact:
1 + 2 + ....... + N = N(N+1)/
b) It takes LN / R seconds to transmit the N packets. Thus, the buffer is empty when a 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.
a) The transmission delay is L / R. The total delay is
b) Let x L / R.
Total delay = ax
x 1 For x=0, the total delay =0; as we increase x, total delay increases, approaching infinity as x approaches 1/a.
Total delay aL R a a
The total number of packets in the system includes those in the buffer and the packet that is being transmitted. So, N=10+1.
Because N a d , so (10+1)=a(queuing delay + transmission delay). That is, 11=a(0.01+1/100)=a*(0.01+0.01). Thus, a=550 packets/sec.
a) There are Q nodes (the source host and the Q 1 routers). Let dqproc denote the
processing delay at the q th node. Let Rq be the transmission rate of the q th link and let q q d (^) trans L / R. Let q dprop be the propagation delay across the q th link. Then
Q
q
q prop
q trans
q d (^) end toend dproc d d 1
b) Let dqueueq denote the average queuing delay at node q. Then
Q
q
q queue
q prop
q trans
q d (^) end toend dproc d d d 1
Traceroutes between San Diego Super Computer Center and www.poly.edu
a) The average (mean) of the round-trip delays at each of the three hours is 71.18 ms, 71.38 ms and 71.55 ms, respectively. The standard deviations are 0.075 ms, 0.21 ms, 0.05 ms, respectively.
b) In this example, the traceroutes have 12 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours.
c) Traceroute packets passed through four ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs.
Traceroutes from www.stella-net.net (France) to www.poly.edu (USA).
d) The average round-trip delays at each of the three hours are 87.09 ms, 86.35 ms and 86.48 ms, respectively. The standard deviations are 0.53 ms, 0.18 ms, 0.23 ms, respectively. In this example, there are 11 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours. Traceroute packets passed three ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs.
An example solution:
b) In this example of traceroutes from one city in France and from another city in Germany to the same host in United States, three links are in common including the transatlantic link.
Traceroutes to two different cities in China from same host in United States
c) Five links are common in the two traceroutes. The two traceroutes diverge before reaching China
Throughput = min{Rs, Rc, R/M}
If only use one path, the max throughput is given by:
max{min{ R 11 , R 21 ,, R^1 N },min{ R 12 , R 22 ,, RN^2 },,min{ R 1 M , R 2 M ,, RN^ M }}.
If use all paths, the max throughput is given by
M
k
k N Rk^ Rk R 1
min{ 1 , 2 ,, }.
Probability of successfully receiving a packet is: ps= (1-p)N.
The number of transmissions needed to be performed until the packet is successfully received by the client is a geometric random variable with success probability ps. Thus, the average number of transmissions needed is given by: 1/ps. Then, the average number of re-transmissions needed is given by: 1/ps -1.
Let’s call the first packet A and call the second packet B.
a) If the bottleneck link is the first link, then packet B is queued at the first link waiting for the transmission of packet A. So the packet inter-arrival time at the destination is simply L/Rs.
b) If the second link is the bottleneck link and both packets are sent back to back, it must be true that the second packet arrives at the input queue of the second link before the second link finishes the transmission of the first packet. That is,
L/Rs + L/Rs + dprop < L/Rs + dprop + L/Rc
a) ttrans + tprop = 400 msec + 80 msec = 480 msec. b) 20 * ( ttrans + 2 tprop ) = 20*(20 msec + 80 msec) = 2 sec. c) Breaking up a file takes longer to transmit because each data packet and its corresponding acknowledgement packet add their own propagation delays.
Recall geostationary satellite is 36,000 kilometers away from earth surface. a) 150 msec b) 1,500,000 bits c) 600,000,000 bits
Let’s suppose the passenger and his/her bags correspond to the data unit arriving to the top of the protocol stack. When the passenger checks in, his/her bags are checked, and a tag is attached to the bags and ticket. This is additional information added in the Baggage layer if Figure 1.20 that allows the Baggage layer to implement the service or separating the passengers and baggage on the sending side, and then reuniting them (hopefully!) on the destination side. When a passenger then passes through security and additional stamp is often added to his/her ticket, indicating that the passenger has passed through a security check. This information is used to ensure (e.g., by later checks for the security information) secure transfer of people.
a) Time to send message from source host to first packet switch = sec 4 sec 2 10
6
6
With store-and-forward switching, the total time to move message from source host to destination host = 4 sec 3 hops 12 sec
b) Time to send 1 st^ packet from source host to first packet switch =.
sec 5 sec 2 10
6
4 m
. Time at which 2nd^ packet is received at the first switch = time
at which 1st^ packet is received at the second switch = 2 5 m sec 10 m sec
c) Time at which 1 st^ packet is received at the destination host = 5 m sec 3 hops 15 m sec^. After this, every 5msec one packet will be received; thus time at which last (800th) packet is received = 15 m sec 799 * 5 m sec 4. 01 sec. It can be seen that delay in using message segmentation is significantly less (almost 1/3rd).
d) i. Without message segmentation, if bit errors are not tolerated, if there is a single bit error, the whole message has to be retransmitted (rather than a single packet). ii. Without message segmentation, huge packets (containing HD videos, for example) are sent into the network. Routers have to accommodate these huge packets. Smaller packets have to queue behind enormous packets and suffer unfair delays.
e) i. Packets have to be put in sequence at the destination. ii. Message segmentation results in many smaller packets. Since header size is usually the same for all packets regardless of their size, with message segmentation the total amount of header bytes is more.
Yes, the delays in the applet correspond to the delays in the Problem 31.The propagation delays affect the overall end-to-end delays both for packet switching and message switching equally.
There are F / S packets. Each packet is S=80 bits. Time at which the last packet is received
at the first router is S
sec. At this time, the first F/S-2 packets are at the
destination, and the F/S-1 packet is at the second router. The last packet must then be transmitted by the first router and the second router, with each transmission taking
sec. Thus delay in sending the whole file is ( 2 )
delay
To calculate the value of S which leads to the minimum delay,
delay S F dS
d 0 40
Problem 34
The circuit-switched telephone networks and the Internet are connected together at "gateways". When a Skype user (connected to the Internet) calls an ordinary telephone, a circuit is established between a gateway and the telephone user over the circuit switched network. The skype user's voice is sent in packets over the Internet to the gateway. At the gateway, the voice signal is reconstructed and then sent over the circuit. In the other direction, the voice signal is sent over the circuit switched network to the gateway. The gateway packetizes the voice signal and sends the voice packets to the Skype user.