UDP does not have delays due to flow control and congestion control. Rather, it increments by the number of bytes of data sent. Problem 27 a In the second segment from Host A to B, the sequence number is , source port number is and destination port number is Still, host A sends data into the receive buffer faster than Host B can remove data from the buffer.
The receive buffer fills up at a rate of roughly 40Mbps. On average, the long-term rate at which Host A sends data to Host B as part of this connection is no more than 60Mbps. Half-open connections are not possible since a server using SYN cookies does not maintain connection variables and buffers for any connection before full connections are established. For establishing fully open connections, an attacker should know the special initial sequence number corresponding to the spoofed source IP address from the attacker.
This sequence number requires the "secret" number that each server uses. Since the attacker does not know this secret number, she cannot guess the initial sequence number. Problem 30 a If timeout values are fixed, then the senders may timeout prematurely. Thus, some packets are re-transmitted even they are not lost. But there might be one potential problem. Queuing delay might be very large, similar to what is shown in Scenario 1. Suppose the source sends packet P1, the timer for P1 expires, and the source then sends P2, a new copy of the same packet.
Finally suppose that shortly after transmitting P2 an acknowledgment for P1 arrives. The source will mistakenly take this acknowledgment as an acknowledgment for P2 and calculate an incorrect value of SampleRTT.
Problem 34 At any given time t, SendBase — 1 is the sequence number of the last byte that the sender knows has been received correctly, and in order, at the receiver. The actually last byte received correctly and in order at the receiver at time t may be greater if there are acknowledgements in the pipe. The actual last byte SendBase or if there are other acknowledgements in the pipe.
The designers of the triple duplicate ACK scheme probably felt that waiting for two packets rather than 1 was the right tradeoff between triggering a quick retransmission when needed, but not retransmitting prematurely in the face of packet reordering. Problem 37 a GoBackN: A sends 9 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re- sent segments 2, 3, 4, and 5. B sends 8 ACKs. Selective Repeat: A sends 6 segments in total. They are initially sent segments 1, 2, 3, 4, 5 and later re- sent segments 2.
B sends 5 ACKs. They are 4 ACKS with sequence number 1, 3, 4, 5. And there is one ACK with sequence number 2. TCP: A sends 6 segments in total. They are 4 ACKS with sequence number 2. There is one ACK with sequence numbers 6. This is because TCP uses fast retransmit without waiting until time out.
With increased loss, even a larger fraction of the packets leaving the queue will be retransmissions. If there was a timeout, the congestion window size would have dropped to 1. When loss is detected during transmission round 16, the congestion windows size is Hence the threshold is 21 during the 18th transmission round.
When loss is detected during transmission round 22, the congestion windows size is Hence the threshold is 14 taking lower floor of Thus packet 70 is sent in the 7th transmission round. Thus the new values of the threshold and window will be 4 and 7 respectively. So, the total number is Problem 41 Refer to Figure 5.
In Figure 5 a , the ratio of the linear decrease on loss between connection 1 and connection 2 is the same - as ratio of the linear increases: unity. In this case, the throughputs never move off of the AB line segment. In Figure 5 b , the ratio of the linear decrease on loss between connection 1 and connection 2 is That is, whenever there is a loss, connection 1 decreases its window by twice the amount of connection 2.
We see that eventually, after enough losses, and subsequent increases, that connection 1's throughput will go to 0, and the full link bandwidth will be allocated to connection 2. Figure 5: Lack of TCP convergence with linear increase, linear decrease Problem 42 If TCP were a stop-and-wait protocol, then the doubling of the time out interval would suffice as a congestion control mechanism.
However, TCP uses pipelining and is therefore not a stop-and-wait protocol , which allows the sender to have multiple outstanding unacknowledged segments. The doubling of the timeout interval does not prevent a TCP sender from sending a large number of first-time-transmitted packets into the network, even when the end-to-end path is highly congested.
Also, because there is no loss and acknowledgements are returned before timers expire, TCP congestion control does not throttle the sender. However, the process in host A will not continuously pass data to the socket because the send buffer will quickly fill up.
Problem 45 a The loss rate, L , is the ratio of the number of packets lost over the number of packets sent. In a cycle, 1 packet is lost. Recall the window size increases by one in each RTT. Problem 47 Let W denote max window size. Let S denote the buffer size. For simplicity, suppose TCP sender sends data packets in a round by round fashion, with each round corresponding to a RTT. If the window size reaches W, then a loss occurs. Let Tp denote the one-way propagation delay between the sender and the receiver.
Problem 48 a Let W denote the max window size. In order to speed up the window increase process, we can increase the window size by a much larger value, instead of increasing window size only by one in each RTT.
Problem 49 1. Thus C1 adjusts its window size after 50 msec, but C2 adjusts its window size after msec. Assume that whenever a loss event happens, C1 receives it after 50msec and C2 receives it after msec. We further have the following simplified model of TCP. After each RTT, a connection determines if it should increase window size or not. For C1, we compute the average total sending rate in the link in the previous 50 msec.
If that rate exceeds the link capacity, then we assume that C1 detects loss and reduces its window size. But for C2, we compute the average total sending rate in the link in the previous msec.
If that rate exceeds the link capacity, then we assume that C2 detects loss and reduces its window size. Note that it is possible that the average sending rate in last 50msec is higher than the link capacity, but the average sending rate in last msec is smaller than or equal to the link capacity, then in this case, we assume that C1 will experience loss event but C2 will not.
The following table describes the evolution of window sizes and sending rates based on the above assumptions. If we look at the above table, we can see a cycle every msec, e. Problem 51 a Similarly as in last problem, we can compute their window sizes over time in the following table. Both C1 and C2 have the same window size 2 after msec. Their max window size is 2.
Thus, the link is not fully utilized recall we assume this link has no buffer. One possible way to break the synchronization is to add a finite buffer to the link and randomly drop packets in the buffer before buffer overflow. This will cause different connections cut their window sizes at different times.
Problem 52 Note that W represents the maximum window size. From the TCP throughput 1. A disadvantage of using these values is that they may be no longer accurate. Problem 55 a The server will send its response to Y. Even if the attacker were to send an appropriately timed TCP ACK segment, it would not know the correct server sequence number since the server uses random initial sequence numbers.
A network-layer packet is a datagram. Datagram-based network layer: forwarding; routing. Additional function of VC-based network layer: call setup. Routing is about determining the end-to-routes between sources and destinations. Yes, both use forwarding tables. For descriptions of the tables, see Section 4. Single packet: guaranteed delivery; guaranteed delivery with guaranteed maximum delay.
Flow of packets: in-order packet delivery; guaranteed minimal bandwidth; guaranteed maximum jitter. ABR does not provide any of these services. With the shadow copy, the forwarding lookup is made locally, at each input port, without invoking the centralized routing processor. Such a decentralized approach avoids creating a lookup processing bottleneck at a single point within the router.
Switching via memory; switching via a bus; switching via an interconnection network. An interconnection network can forward packets in parallel as long as all the packets are being forwarded to different output ports. If the rate at which packets arrive to the fabric exceeds switching fabric rate, then packets will need to queue at the input ports. If this rate mismatch persists, the queues will get larger and larger and eventually overflow the input port buffers, causing packet loss.
Packet loss can be eliminated if the switching fabric speed is at least n times as fast as the input line speed, where n is the number of input ports.
Assuming input and output line speeds are the same, packet loss can still occur if the rate at which packets arrive to a single output port exceeds the line speed. If this rate mismatch persists, the queues will get larger and larger and eventually overflow the output port buffers, causing packet loss. Note that increasing switch fabric speed cannot prevent this problem from occurring. HOL blocking: Sometimes the a packet that is first in line at an input port queue must wait because there is no available buffer space at the output port to which it wants to be forwarded.
When this occurs, all the packets behind the first packet are blocked, even if their output queues have room to accommodate them. HOL blocking occurs at the input port. They have one address for each interface. Students will get different correct answers for this question. The 8-bit protocol field in the IP datagram contains information about which transport layer protocol the destination host should pass the segment to. Typically the wireless router includes a DHCP server.
IPv6 has a fixed length header, which does not include most of the options an IPv4 header can include. Even though the IPv6 header contains two bit addresses source and destination IP address the whole header has a fixed length of 40 bytes only. Several of the fields are similar in spirit. Traffic class, payload length, next header and hop limit in IPv6 are respectively similar to type of service, datagram length, upper-layer protocol and time to live in IPv4. Yes, because the entire IPv6 datagram including header fields is encapsulated in an IPv4 datagram.
Link state algorithms: Computes the least-cost path between source and destination using complete, global knowledge about the network. Distance-vector routing: The calculation of the least-cost path is carried out in an iterative, distributed manner.
A node only knows the neighbor to which it should forward a packet in order to reach given destination along the least-cost path, and the cost of that path from itself to the destination.
Routers are organized into autonomous systems ASs. Within an AS, all routers run the same intra-AS routing protocol. The problem of scale is solved since an router in an AS need only know about routers within its AS and the subnets that attach to the AS. To route across ASes, the inter-AS protocol is based on the AS graph and does not take individual routers into account.
Each AS has administrative autonomy for routing within an AS. The advertisement tells D that it can get to z in 11 hops by way of A. However, D can already get to z by way of B in 7 hops. Therefore, there is no need to modify the entry for z in the table. If, on the other hand, the advertisement said that A were only 4 hops away from z by way of C, then D would indeed modify its forwarding table.
With OSPF, a router periodically broadcasts routing information to all other routers in the AS, not just to its neighboring routers. A RIP advertisement sent by a router contains information about all the networks in the AS, although this information is only sent to its neighboring routers.
Policy: Among ASs, policy issues dominate. It may well be important that traffic originating in a given AS not be able to pass through another specific AS. Similarly, a given AS may want to control what transit traffic it carries between other ASs. Within an AS, everything is nominally under the same administrative control and thus policy issues a much less important role in choosing routes with in AS. Within an AS, scalability is less of a concern.
For one thing, if a single administrative domain becomes too large, it is always possible to divide it into two ASs and perform inter-AS routing between the two new ASs. Performance: Because inter-AS routing is so policy oriented, the quality for example, performance of the routes used is often of secondary concern that is, a longer or more costly route that satisfies certain policy criteria may well be taken over a route that is shorter but does not meet that criteria.
Indeed, we saw that among ASs, there is not even the notion of cost other than AS hop count associated with routes. Within a single AS, however, such policy concerns are of less importance, allowing routing to focus more on the level of performance realized on a route.
A subnet is a portion of a larger network; a subnet does not contain a router; its boundaries are defined by the router and host interfaces. A prefix is the network portion of a CDIRized address; it is written in the form a. Routers use the AS-PATH attribute to detect and prevent looping advertisements; they also use it in choosing among multiple paths to the same prefix.
False IGMP is a protocol run only between the host and its first-hop multicast router. IGMP allows a host to specify to the first-hop multicast router the multicast group it wants to join.
It is then up to the multicast router to work with other multicast routers i. In a group-shared tree, all senders send their multicast traffic using the same routing tree. With source-based tree, the multicast datagrams from a given source are routed over s specific routing tree constructed for that source; thus each source may have a different source-based tree and a router may have to keep track of several source- based trees for a given multicast group.
Chapter 4 Problems Problem 1 a With a connection-oriented network, every router failure will involve the routing of that connection. Moreover, all of the routers on the initial path that are downstream from the failed node must take down the failed connection, with all of the requisite signaling involved to do this. With a connectionless datagram network, no signaling is required to either set up a new downstream path or take down the old downstream path.
We have seen, however, that routing tables will need to be updated e. We have seen that with distance vector algorithms, this routing table change can sometimes be localized to the area near the failed router.
Thus, a datagram network would be preferable. Interestingly, the design criteria that the initial ARPAnet be able to function under stressful conditions was one of the reasons that datagram architecture was chosen for this Internet ancestor.
That is, the router must have per-session state in the router. This is possible in a connection-oriented network, but not with a connectionless network. Thus, a connection-oriented VC network would be preferable. This is due to the various packet headers needed to route the datagrams through the network. But in VC architecture, once all circuits are set up, they will never change. Thus, the signaling overhead is negligible over the long run. In this manner, it is not possible that there are fewer VCs in progress than without there being any common free VC number.
Thus, a VC will likely have a different VC number for each link along its path. For a datagram forwarding table, the columns are: Destination Address, Outgoing Interface. Problem 4 a Data destined to host H3 is forwarded through interface 3 Destination Address Link Interface H3 3 b No, because forwarding rule is only based on destination address.
There are four links. One example combination is 10,00,00, Problem 6 In a virtual circuit network, there is an end-to-end connection in the sense that each router along the path must maintain state for the connection; hence the terminology connection service. In a connection-oriented transport service over a connectionless network layer, such as TCP over IP, the end systems maintain connection state; however the routers have no notion of any connections; hence the terminology connection-oriented service.
Problem 7 a No, you can only transmit one packet at a time over a shared bus. Problem 8 a n-1 D b n-1 D c 0 Problem 9 The minimal number of time slots needed is 3. The scheduling is as follows. Slot 1: send X in top input queue, send Y in middle input queue. Slot 2: send X in middle input queue, send Y in bottom input queue Slot 3: send Z in bottom input queue.
Largest number of slots is still 3. Actually, based on the assumption that a non-empty input queue is never idle, we see that the first time slot always consists of sending X in the top input queue and Y in either middle or bottom input queue, and in the second time slot, we can always send two more datagram, and the last datagram can be sent in third time slot.
NOTE: Actually, if the first datagram in the bottom input queue is X, then the worst case would require 4 time slots. Also, label D, E, F for the upper-right, bottom, and upper- left interior subnets, respectively. Each fragment except the last one will be of size bytes including IP header. The last datagram will be of size bytes including IP header. The offsets of the 4 fragments will be 0, 85, , Note that here there is no fragmentation — the source host does not create datagrams larger than bytes, and these datagrams are smaller than the MTUs of the links.
Problem 21 a Home addresses: As each host generates a sequence of IP packets with sequential numbers and a distinct very likely, as they are randomly chosen from a large space initial identification number ID , we can group IP packets with consecutive IDs into a cluster. The number of clusters is the number of hosts behind the NAT. For more practical algorithms, see the following papers. Problem 23 It is not possible to devise such a technique.
In order to establish a direct TCP connection between Arnold and Bernard, either Arnold or Bob must initiate a connection to the other. At each iteration, a node exchanges distance tables with its neighbors. Thus, if you are node A, and your neighbor is B, all of B's neighbors which will all be one or two hops from you will know the shortest cost path of one or two hops to you after one iteration i. Now consider if c x,w changes. Connecting two nodes with a link is equivalent to decreasing the link weight from infinite to the finite weight.
There is no increasing in values. If no updating, then no message will be sent out. Thus, D x is non-increasing. Since those costs are finite, then eventually distance vectors will be stabilized in finite steps. The following table shows the routing converging process. Assume that at time t0, link cost change happens. At time t1, y updates its distance vector and informs neighbors w and z. If we continue the iterations shown in the above table, then we will see that, at t27, z detects that its least cost to x is 50, via its direct link with x.
At t29, w learns its least cost to x is 51 via z. At t30, y updates its least cost to x to be 52 via w. Finally, at time t31, no updating, and the routing is stabilized.
Problem 36 The chosen path is not necessarily the shortest AS-path. Recall that there are many issues to be considered in the route selection process. It is very likely that a longer loop-free path is preferred over a shorter loop-free path due to economic reason. For example, an AS might prefer to send traffic to one neighbor instead of another neighbor with shorter AS distance.
Consider a BitTorrent file sharing network in which peer 1, 2, and 3 are in stub networks W, X, and Y respectively. This is equivalent to B forwarding data that is finally destined to stub network Y.
A should advise to C only one route, A-V. In this manner, when Y has a datagram that is destined to an IP that can be reached through Z, Y will have the option of sending the datagram through Z. However, if Z advertizes routes to Y, Y can re-advertize those routes to X. Therefore, in this case, there is nothing Z can do from preventing traffic from X to transit through Z.
Problem 45 The 32 receives are shown connected to the sender in the binary tree configuration shown above. With network-layer broadcast, a copy of the message is forwarded over each link exactly once. With unicast emulation, the sender unicasts a copy to each receiver over a path with5 hops. A topology in which all receivers are in a line, with the sender at one end of the line, will have the largest disparity between the cost of network-layer broadcast and unicast emulation.
Other solutions are possible, but in these solutions, B can not route to either C or D from A. This center-based tree is different from the minimal spanning tree shown in the figure. Problem 50 The center-based tree for the topology shown in the original figure connects t to v; u to v; w to v; x to v; and y to v all directly.
And z connected to v via x. This center-based tree is different from the minimal spanning tree. After 3 steps, 12 copies are transmitted, and so on.
For example, an application may periodically multicast its identity to all other group members in an application-layer message. Problem 54 A simple application-layer protocol that will allow all members to know the identity of all other members in the group is for each instance of the application to send a multicast message containing its identity to all other members. This protocol sends message in- band, since the multicast channel is used to distribute the identification messages as well as multicast data from the application itself.
The use of the in-band signaling makes use of the existing multicast distribution mechanism, leading to a very simple design. The transportation mode, e. Although each link guarantees that an IP datagram sent over the link will be received at the other end of the link without errors, it is not guaranteed that IP datagrams will arrive at the ultimate destination in the proper order. With IP, datagrams in the same TCP connection can take different routes in the network, and therefore arrive out of order.
TCP is still needed to provide the receiving end of the application the byte stream in the correct order. Also, IP can lose packets due to routing loops or equipment failures.
There will be a collision in the sense that while a node is transmitting it will start to receive a packet from the other node. Slotted Aloha: 1, 2 and 4 slotted ALOHA is only partially decentralized, since it requires the clocks in all nodes to be synchronized. Token ring: 1, 2, 3, 4.
It waits In polling, a discussion leader allows only one participant to talk at a time, with each participant getting a chance to talk in a round-robin fashion. A participant is only allowed to talk if the participant is holding the wine glass.
When a node transmits a frame, the node has to wait for the frame to propagate around the entire ring before the node can release the token. An ARP query is sent in a broadcast frame because the querying host does not which adapter address corresponds to the IP address in question.
No it is not possible. The three Ethernet technologies have identical frame structures. In We can string the N switches together.
The first and last switch would use one port for trunking; the middle N-2 switches would use two ports. The parity of row 2 is now correct! The parity of columns 2 and 3 is wrong, but we can't detect in which rows the error occurred! It is clear that if we divide K by G, then the reminder is not zero. Thus, a sequence not necessarily contiguous of odd-number bit errors cannot be divided by 11, thus it cannot be divided by G.
The length of a polling round is The number of bits transmitted in a polling round is NQ. Forwarding table in E determines that the datagram should be routed to interface The adapter in E creates and Ethernet packet with Ethernet destination address Router 2 receives the packet and extracts the datagram. The forwarding table in this router indicates that the datagram is to be routed to Router 2 then sends the Ethernet packet with the destination address of and source address of via its interface with IP address of The process continues until the packet has reached Host B.
This ARP response packet is carried by an Ethernet frame with Ethernet destination address Problem 15 a No. Thus, E will not send the packet to the default router R1. And it learns that A resides on Subnet 1 which is connected to S1 at the interface connecting to Subnet 1. And, S1 will update its forwarding table to include an entry for Host A.
Problem 16 Lets call the switch between subnets 2 and 3 S2. That is, router R1 between subnets 2 and 3 is now replaced with switch S2. Thus, E will not send the packet to S2. This query packet will be re-broadcast by switch 1, and eventually received by Host B. Yes, router S2 also receives this ARP request message, and S2 will broadcast this query packet to all its interfaces.
Problem 17 Wait for 51, bit times. For 10 Mbps, this wait is So A incorrectly thinks that its frame was successfully transmitted without a collision. Thus A and B do not collide. Thus the factor appearing in the exponential backoff algorithm is sufficiently large. Problem 24 Each departmental hub is a single collision domain that can have a maximum throughput of Mbps. Hence, if the three collision domains and the web server and mail server send out data at their maximum possible rates of Mbps each, a maximum total aggregate throughput of Mbps can be achieved among the 11 end systems.
Problem 25 All of the 11 end systems will lie in the same collision domain. In this case, the maximum total aggregate throughput of Mbps is possible among the 11 end sytems.
The subnet mask is The IP addresses for those three computers from left to right in CS department are: The first one is for the subnet of EE department, and the second one is for the subnet of CS department. Suppose This means that each frame that comes from subnet Host A first encapsulates the IP datagram destined to Once the router receives the frame, then it passes it up to IP layer, which decides that the IP datagram should be forwarded to subnet Then the router encapsulates the IP datagram into a frame and sends it to port 1.
Note that this frame has an Once Host B receives this frame, it will remove the Problem 29 in out out label label dest interf. A 2 D 0 R2 R1 in out out label label dest interf. You computer first creates a special IP datagram destined to A DHCP server on the Ethernet also gives your computer a list of IP addresses of first- hop routers, the subnet mask of the subnet where your computer resides, and the addresses of local DNS servers if they exist.
Your computer first will get the IP address of the Web page you would like to download. Once your computer has the IP address of the Web page, then it will send out the HTTP request via the first-hop router if the Web page does not reside in a local Web server. Your computer sends the Ethernet frames destined to the first-hop router. Once the router receives the frames, it passes them up into IP layer, checks its routing table, and then sends the packets to the right interface out of all of its interfaces.
Then your IP packets will be routed through the Internet until they reach the Web server. Those IP packets follow IP routes and finally reach your first-hop router, and then the router will forward those IP packets to your computer by encapsulating them into Ethernet frames.
Similarly, there are four links between second and fourth tier-2 switches, together providing 40 Gbps for the traffic from racks to Thus the total aggregate bandwidth is 80 Gbps, and the value per flow rate is 1 Gbps.
So the host-to-host bit rate will be 0. Problem 33 a Both email and video application uses the fourth rack for 0. Let's assume that the fourth rack has all the data and software needed for both the email and video applications. With the topology of Figure 5. From part b, both are using the fourth rack for no more than. Chapter 6 Review Questions 1. In infrastructure mode of operation, each wireless host is connected to the larger network via a base station access point.
If not operating in infrastructure mode, a network operates in ad-hoc mode. In ad-hoc mode, wireless hosts have no infrastructure with which to connect.
In the absence of such infrastructure, the hosts themselves must provide for services such as routing, address assignment, DNS-like name translation, and more.
Path loss is due to the attenuation of the electromagnetic signal when it travels through matter. Multipath propagation results in blurring of the received signal at the receiver and occurs when portions of the electromagnetic wave reflect off objects and ground, taking paths of different lengths between a sender and receiver.
Interference from other sources occurs when the other source is also transmitting in the same frequency range as the wireless network. APs transmit beacon frames. The beacon frames permit nearby wireless stations to discover and identify the AP. James F. Availability This title is out of print. Preface Preface is available for download in PDF format. A balanced presentation focuses on the Internet as a specific motivating example of a network and also introduces students to protocols in a more theoretical context.
A chapter on wireless and mobility includes insight into Principles and Practice boxes throughout demonstrate real-world applications of the principles studied. Case History boxes are sprinkled in to help tell the story of the history and development of computer networking.
Material on application programming development is included, along with numerous programming assignments. A highly developed art program enhances the descriptions of concepts.
A comprehensive Companion Website , which includes additional learning material, links to relevant online resources, and lab material. New to This Edition. The Companion Web site has been significantly expanded and enriched to include VideoNotes and interactive exercises.
An important new component of the sixth edition is the significantly expanded online and interactive learning material. Since students can generate and view solutions for an unlimited number of similar problem instances, they can work until the material is truly mastered.
As in earlier editions, the Web site contains the interactive Java applets that animate many of the key networking concepts. The site also has interactive quizzes that permit students to check their basic understanding of the subject matter. Professors can integrate these interactive features into their lectures or use them as mini labs.
Additional technical material. Programming assignments. The Web site also provides a number of detailed programming assignments, which include building a multithreaded Web server, building an e-mail client with a GUI interface, programming the sender and receiver sides of a reliable data transport protocol, programming a distributed routing algorithm, and more.
Wireshark labs. The Web site provides numerous Wireshark assignments that enable students to actually observe the sequence of messages exchanged between two protocol entities. It also takes 2 minutes for the third tollbooth to service the 10 cars.
Thus the total delay is 96 minutes. Problem 7 Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated.
Problem 8 a 20 users can be supported. Let X j be independent random variables such that PX. Problem 11 Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Problem 12 The arriving packet must first wait for the link to transmit 4. Since these bits are transmitted at 2 Mbps, the queuing delay is 27 msec. 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. Problem 16 The total number of packets in the system includes those in the buffer and the packet that is being transmitted.
In either case, you will get three delay measurements. For those three measurements you can calculate the mean and standard deviation. Repeat the experiment at different times of the day and comment on any changes. Here is an example solution:. The standard deviations are 0. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs.
Traceroutes from www. In this example, there are 11 routers in the path at each of the three hours. Traceroute packets passed three ISP networks from source to. Traceroutes from two different cities in France to New York City in United States a In these traceroutes from two different cities in France to the same destination host in United States, seven 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. 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. The right hand side represents the time needed by the first packet to finish its transmission onto the second link.
Problem 29 Recall geostationary satellite is 36, kilometers away from earth surface. This is additional information added in the Baggage layer if Figure 1. This information is used to ensure e. 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. 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. Packets have to be put in sequence at the destination. 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.
Problem 32 Yes, the delays in the applet correspond to the delays in the Problem The propagation delays affect the overall end-to-end delays both for packet switching and message switching equally. Time at which the last packet is received at the S 80 F sec. The last packet must then be transmitted by the first router S 80 and the second router, with each transmission taking sec. 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.
At the gateway, the voice signal is. In the other direction, the voice signal is sent over the circuit switched network to the gateway. The gateway packetizes the voice signal and send. Millions discover their favorite reads on issuu every month.
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