1. (Q1) Suppose the network layer provides the following service. The network layer in the source host accepts a segment of maximum size 1,200 bytes and a destination host address from the transport layer. The network layer then guarantees to deliver the segment to the transport layer at the destination host. Suppose many network application processes can be running at the destination host.
a. Design the simplest possible transport-layer protocol that will get application data to the desired process at the destination host. Assume the operating system in the destination host has assigned a 4-byte port number to each running application process.
b. Modify this protocol so that it provides a “return address” to the destination process.
c. In your protocols, does the transport layer “have to do anything” in the
core of the computer network.
Answer:
a. Call this protocol Simple Transport Protocol (STP). At the sender side, STP accepts from the sending process a chunk of data not exceeding 1196 bytes, a destination host address, and a destination port number. STP adds a four-byte header to each chunk and puts the port number of the destination process in this header. STP then gives the destination host address and the resulting segment to the network layer. The network layer delivers the segment to STP at the destination host. STP then examines the port number in the segment, extracts the data from the segment, and passes the data to the process identified by the port number.
b. The segment now has two header fields: a source port field and destination
port field. At the sender side, STP accepts a chunk of data not exceeding 1192 bytes, a destination host address, a source port number, and a destination port number. STP creates a segment which contains the application data, source port number, and destination port number. It then gives the segment and the destination host address to the network layer. After receiving the segment, STP at the receiving host gives the application process the application data and the source port number.
c. No, the transport layer does not have to do anything in the core; the transport layer “lives” in the end systems.
2. (Q2) Consider a planet where everyone belongs to a family of six, every family lives in its own house, each house has a unique address, and each person in a given house has a unique name. Suppose this planet has a mail service that delivers letters form source house to destination house. The mail service requires that (i) the letter be in an envelope and that (ii) the address of the destination house (and nothing more ) be clearly written on the envelope. Suppose each family has a delegate family member who collects and distributes letters for the other family members. The letters do not necessarily provide any indication of the recipients of the letters.
a. Using the solution to Problem Q1 above as inspiration, describe a protocol that the delegates can use to deliver letters from a sending family member to a receiving family member.
b. In your protocol, does the mail service ever have to open the envelope and examine the letter in order to provide its service. Answer:
a. For sending a letter, the family member is required to give the delegate the
letter itself, the address of the destination house, and the name of the
recipient. The delegate clearly writes the recipient’s name on the top of the letter. The delegate then puts the letter in an envelope and writes the address of the destination house on the envelope. The delegate then gives the letter to the planet’s mail service. At the receiving side, the delegate receives the letter from the mail service, takes the letter out of the envelope, and takes note of the recipient name written at the top of the letter. The delegate than gives the letter to the family member with this name.
b. No, the mail service does not have to open the envelope; it only examines the
address on the envelope.
3. (Q3) Describe why an application developer might choose to run an application over UDP rather than TCP.
Answer: An application developer may not want its application to use TCP’s congestion control, which can throttle the application’s sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCP’s congestion control. Also, some applications do not need the reliable data transfer provided by TCP.
4. (P1) Suppose Client A initiates a Telnet session with Server S. At about the same time, Client B also initiates a Telnet session with Server S. Provide possible source and destination port numbers for a. The segment sent from A to B. b. The segment sent from B to S. c. The segment sent from S to A. d. The segment sent from S to B.
e. If A and B are different hosts, is it possible that the source port number
in the segment from A to S is the same as that from B to S f. How about if they are the same host Answer:
a b c d A → S B → S S → A S → B source port numbers 467 513 23 23 destination port numbers 23 23 467 513 e Yes. f No.
5. (P2) Consider Figure What are the source and destination port values in the
segments flowing form the server back to the clients’ processes What are the IP addresses in the network-layer datagrams carrying the transport-layer segments Answer:
Suppose the IP addresses of the hosts A, B, and C are a, b, c, respectively. (Note that a,b,c are distinct.)
To host A: Source port =80, source IP address = b, dest port = 26145, dest IP address = a
To host C, left process: Source port =80, source IP address = b, dest port = 7532, dest IP address = c
To host C, right process: Source port =80, source IP address = b, dest port = 26145, dest IP address = c
6. (P3) UDP and TCP use 1s complement for their checksums. Suppose you have the
following three 8-bit bytes: 01101010, 01001111, 01110011. What is the 1s complement of the sum of these 8-bit bytes (Note that although UDP and TCP use 16-bit words in computing the checksum, for this problem you are being asked to consider 8-bit sums.) Show all work. Why is it that UDP takes the 1s complement of the sum; that is , why not just sue the sum With the 1s complement scheme, how does the receiver detect errors Is it possible that a 1-bit error will go undetected How about a 2-bit error Answer:
+
0 1 0 1 0 1 0 1 0 1 1 1 0 0 0 0 1 1 0 0 0 1 0 1
+
1 1 0 0 0 1 0 1
0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 1
One's complement = 1 1 1 0 1 1 1 0.
To detect errors, the receiver adds the four words (the three original words and the checksum). If the sum contains a zero, the receiver knows there has been an error. All one-bit errors will be detected, but two-bit errors can be undetected ., if the last digit of the first word is converted to a 0 and the last digit of the second word is converted to a 1).
7. (P4) Suppose that the UDP receiver computes the Internet checksum for the
received UDP segment and finds that it matches the value carried in the checksum field. Can the receiver be absolutely certain that no bit errors have occurred Explain. Answer:
No, the receiver cannot be absolutely certain that no bit errors have occurred. This is because of the manner in which the checksum for the packet is calculated.
If the corresponding bits (that would be added together) of two 16-bit words in the packet were 0 and 1 then even if these get flipped to 1 and 0 respectively, the sum still remains the same. Hence, the 1s complement the receiver calculates will also be the same. This means the checksum will verify even if there was transmission error.
8. (P5) a. Suppose you have the following 2 bytes: 01001010 and 01111001. What
is the 1s complement of sum of these 2 bytes
b. Suppose you have the following 2 bytes: and 01101110. What is the 1s complement of sum of these 2 bytes
c. For the bytes in part (a), give an example where one bit is flipped in
each of the 2 bytes and yet the 1s complement doesn’t change.
Answer:
a. Adding the two bytes gives . Taking the one’s complement gives 01100010 b. Adding the two bytes gives 00011110; the one’s complement gives . c. first byte = 00110101 ; second byte = 01101000.
9. (P6) Consider our motivation for correcting protocol . Show that the receiver,
shown in
the figure on the following page, when operating with the sender show in Figure , can lead the sender and receiver to enter into a deadlock state, where each is waiting for an event that will never occur.
Answer: Suppose the sender is in state “Wait for call 1 from above” and the receiver (the receiver shown in the homework problem) is in state “Wait for 1 from below.” The sender sends a packet with sequence number 1, and transitions to “Wait for ACK or NAK 1,” waiting for an ACK or NAK. Suppose now the receiver
receives the packet with sequence number 1 correctly, sends an ACK, and transitions to state “Wait for 0 from below,” waiting for a data packet with sequence number 0. However, the ACK is corrupted. When the sender gets the corrupted ACK, it resends the packet with sequence number 1. However, the receiver is waiting for a packet with sequence number 0 and (as shown in the home work problem) always sends a NAK when it doesn't get a packet with sequence number 0. Hence the sender will always be sending a packet with sequence number 1, and the receiver will always be NAKing that packet. Neither will progress forward from that state.
10. (P7) Draw the FSM for the receiver side of protocol
Answer: The sender side of protocol differs from the sender side of protocol in that
timeouts have been added. We have seen that the introduction of timeouts adds the possibility of duplicate packets into the sender-to-receiver data stream. However, the receiver in protocol can already handle duplicate packets. (Receiver-side duplicates in rdt would arise if the receiver sent an ACK that was lost, and the sender then retransmitted the old data). Hence the receiver in protocol will also work as the receiver in protocol rdt .
11. (P8) In protocol , the ACK packets flowing from the receiver to the sender
do not have sequence numbers (although they do have an ACK field that contains the sequence number of the packet they are acknowledging). Why is it that our ACK packets do not require sequence numbers
Answer: To best Answer this question, consider why we needed sequence numbers in the first
place. We saw that the sender needs sequence numbers so that the receiver can tell if a data packet is a duplicate of an already received data packet. In the case of ACKs, the sender does not need this info ., a sequence number on an ACK) to tell detect a duplicate ACK. A duplicate ACK is obvious to the receiver, since when it has received the original ACK it transitioned to the next state. The duplicate ACK is not the ACK that the sender needs and hence is ignored by the sender.
12. (P9) Give a trace of the operation of protocol when data packets and
acknowledgment packets are garbled. Your trace should be similar to that used in Figure
Answer: Suppose the protocol has been in operation for some time. The sender is in state “Wait
for call from above” (top left hand corner) and the receiver is in state “Wait for 0 from below”. The scenarios for corrupted data and corrupted ACK are shown in Figure 1.
13. (P10) Consider a channel that can lose packets but has a maximum delay that
is known. Modify protocol to include sender timeout and retransmit. Informally argue why your protocol can communicate correctly over this channel.
Answer: Here, we add a timer, whose value is greater than the known round-trip propagation delay. We add a timeout event to the “Wait for ACK or NAK0” and “Wait for ACK or NAK1” states. If the timeout event occurs, the most recently transmitted packet is retransmitted. Let us see why this protocol will still work with the receiver.
• Suppose the timeout is caused by a lost data packet, ., a packet on the senderto- receiver channel. In this case, the receiver never received the previous transmission and, from the receiver's viewpoint, if the timeout
retransmission is received, it look exactly the same as if the original transmission is being received.
• Suppose now that an ACK is lost. The receiver will eventually retransmit the packet on a timeout. But a retransmission is exactly the same action that is take if an ACK is garbled. Thus the sender's reaction is the same with a loss, as with a garbled ACK. The rdt receiver can already handle the case of a garbled ACK.
14. (P11) Consider the protocol. Draw a diagram showing that if the network
connection between the sender and receiver can reorder messages (that is, that two messages propagating in the medium between the sender and receiver can be reordered), then the alternating-bit protocol will not work correctly (make sure you clearly identify the sense in which it will not work correctly). Your diagram should have the sender on the left and the receiver on the right, with the time axis running down the page, showing data (D) and acknowledgement (A) message exchange. Make sure you indicate the sequence number associated with any data or acknowledgement segment.
Answer:
15. (P12) The sender side of simply ignores (that is, takes no action on) all
received packets that are either in error or have the wrong value in the ack-num field of an acknowledgement packet. Suppose that in such circumstances, were simply to retransmit the current data packet . Would the protocol still work (hint: Consider what would happen if there were only bit errors; there are no packet losses but premature timeout can occur. Consider how many times the nth packet is sent, in the limit as n approaches infinity.)
Answer: The protocol would still work, since a retransmission would be what would happen if the packet received with errors has actually been lost (and from the receiver standpoint, it never knows which of these events, if either, will occur). To get at the more subtle issue behind this question, one has to allow for premature timeouts to occur. In this case, if each extra copy of the packet is ACKed and each received extra ACK causes another extra copy of the current packet to be sent, the number of times packet n is sent will increase without bound as
n approaches infinity.
16. (P13) Consider a reliable data transfer protocol that uses only negative
acknowledgements. Suppose the sender sends data only infrequently. Would a NAK-only protocol be preferable to a protocol that uses ACKs Why Now suppose the sender has a lot of data to send and the end to end connection experiences few losses. In this second case , would a NAK-only protocol be preferable to a protocol that uses ACKs Why
Answer:
In a NAK only protocol, the loss of packet x is only detected by the receiver when packet
x+1 is received. That is, the receivers receives x-1 and then x+1, only when x+1 is received does the receiver realize that x was missed. If there is a long delay
between the transmission of x and the transmission of x+1, then it will be a long time until x can be recovered, under a NAK only protocol.
On the other hand, if data is being sent often, then recovery under a NAK-only scheme could happen quickly. Moreover, if errors are infrequent, then NAKs are only occasionally sent (when needed), and ACK are never sent – a significant reduction in feedback in the NAK-only case over the ACK-only case.
17. (P14) Consider the cross-country example shown in Figure . How big would
the window size have to be for the channel utilization to be greater than 80 percent
Answer: It takes 8 microseconds (or milliseconds) to send a packet. in order for the sender
to be busy 90 percent of the time, we must have util = = / or n approximately 3377 packets.
18. (P15) Consider a scenario in which Host A wants to simultaneously send
packets to Host B and C. A is connected to B and C via a broadcast channel—a packet sent by A is carried by the channel to both B and C. Suppose that the broadcast channel connecting A, B, and C can independently lose and corrupt packets (and so, for example, a packet sent from A might be correctly received by B, but not by C). Design a stop-and-wait-like error-control protocol for reliable transferring packets from A to B and C, such that A will not get new data from the upper layer until it knows that B and C have correctly received the current packet. Give FSM descriptions of A and C. (Hint: The FSM for B should be essentially be same as for C.) Also, give a description of the packet format(s) used.
Answer:
In our solution, the sender will wait until it receives an ACK for a pair of messages (seqnum and seqnum+1) before moving on to the next pair of messages. Data packets have a data field and carry a two-bit sequence number. That is, the valid sequence numbers are 0, 1, 2, and 3. (Note: you should think about why a 1-bit sequence number space of 0, 1 only would not work in the solution below.) ACK messages carry the sequence number of the data packet they are acknowledging.
The FSM for the sender and receiver are shown in Figure 2. Note that the sender state records whether (i) no ACKs have been received for the current pair, (ii) an ACK for seqnum (only) has been received, or an ACK for seqnum+1 (only) has been received. In this figure, we assume that the seqnum is initially 0, and that the sender has sent the first two data messages (to get things going). A timeline trace for the sender and receiver recovering from a lost packet is shown below:
Sender
Receiver
make pair (0,1) send packet 0
Packet 0 drops
send packet 1
receive packet 1 buffer packet 1 send ACK 1
receive ACK 1 (timeout) resend packet 0
receive packet 0 deliver pair (0,1) send ACK 0
receive ACK 0
19. (P16) Consider a scenario in which Host A and Host B want to send messages
to Host C. Hosts A and C are connected by a channel that can lose and corrupt (but not reorder)message. Hosts B and C are connected by another channel (independent of the channel connecting A and C) with the same properties. The transport layer at Host C should alternate in delivering messages from A and B to the layer above (that is, it should first deliver the data from a packet from A, then the data from a packet from B, and so on). Design a stop-and-wait-like error-control protocol for reliable transferring packets from A to B and C, with alternating delivery at C as described above. Give FSM descriptions of A and C. (Hint: The FSM for B should be essentially be same as for A.) Also, give a description of the packet format(s) used.
Answer:
This problem is a variation on the simple stop and wait protocol . Because the channel may lose messages and because the sender may resend a message that one of the receivers has already received (either because of a premature timeout or because the other receiver has yet to receive the data correctly), sequence numbers are needed. As in , a 0-bit sequence number will suffice here.
The sender and receiver FSM are shown in Figure 3. In this problem, the sender state indicates whether the sender has received an ACK from B (only), from C (only) or from neither C nor B. The receiver state indicates which sequence number the receiver is waiting for.
20. (P17) In the generic SR protocol that we studied in Section the sender
transmits a message as soon as it is available (if it is in the window) without waiting for an acknowledgment. Suppose now that we want an SR protocol that sends messages two at a time. That is , the sender will send a pair of messages and will send the next pair of messages only when it knows that both messages in the first pair have been receiver
correctly.
Suppose that the channel may lose messages but will not corrupt or
reorder messages. Design an error-control protocol for the unidirectional reliable transfer of messages. Give an FSM description of the sender and receiver. Describe the format of the packets sent between sender and receiver, and vice versa. If you use any procedure calls other than those in Section (for example, udt_send(), start_timer(), rdt_rcv(), and so on) ,clearly state their actions. Give an example (a timeline trace of sender and receiver) showing how your protocol recovers from a lost packet.
Answer:
21. (P18) Consider the GBN protocol with a sender window size of 3 and a sequence
number range of 1024. Suppose that at time t, the next in-order packet that the receiver is expecting has a sequence number of k. Assume that the medium does not reorder messages. Answer the following questions:
a. What are the possible sets of sequence number inside the sender’s
window at time t Justify your Answer.
b .What are all possible values of the ACK field in all possible messages
currently propagating back to the sender at time t Justify your Answer.
Answer:
a. Here we have a window size of N=3. Suppose the receiver has received packet
k-1, and has ACKed that and all other preceeding packets. If all of these ACK's have been received by sender, then sender's window is [k, k+N-1]. Suppose next that none of the ACKs have been received at the sender. In this second case, the sender's window contains k-1 and the N packets up to and including k-1. The sender's window is thus [k- N,k-1]. By these arguments, the senders window is of size 3 and begins somewhere in the range [k-N,k]. b. If the receiver is waiting for packet k, then it has received (and ACKed)
packet k-1 and the N-1 packets before that. If none of those N ACKs have been yet received by the sender, then ACK messages with values of [k-N,k-1] may still be propagating back. Because the sender has sent packets [k-N, k-1], it must be the case that the sender has already received an ACK for k-N-1. Once the receiver has sent an ACK for k-N-1 it will never send an ACK that is less that k-N-1. Thus the range of in-flight ACK values can range from k-N-1 to k-1.
22. (P19) Answer true or false to the following questions and briefly justify
your Answer.
a. With the SR protocol, it is possible for the sender to receive an
ACK for a packet that falls outside of its current window.
b. With CBN, it is possible for the sender to receiver an ACK for a
packet that falls outside of its current window.
c. The alternating-bit protocol is the same as the SR protocol with a
sender and receiver window size of 1.
d. The alternating-bit protocol is the same as the GBN protocol with a
sender and receiver window size of 1.
Answer:
a. True. Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at
t0 . At t1 (t1 > t0) the receiver ACKS 1, 2, 3. At t2 (t2 > t1) the sender
times out and resends 1, 2, 3. At t3 the receiver receives the duplicates and re-acknowledges 1, 2, 3. At t4 the sender receives the ACKs that the receiver sent at t1 and advances its window to 4, 5, 6. At t5 the sender receives the ACKs 1, 2, 3 the receiver sent at t2 . These ACKs are outside its window. b. True. By essentially the same scenario as in (a). c. True.
d. True. Note that with a window size of 1, SR, GBN, and the alternating bit
protocol are functionally equivalent. The window size of 1 precludes the possibility of out-of-order packets (within the window). A cumulative ACK is just an ordinary ACK in this situation, since it can only refer to the single packet within the window.
23. (Q4) Why is it that voice and video traffic is often sent over TCP rather
than UDP in today’s Internet. (Hint: The Answer we are looking for has nothing to do with TCP’s congestion-control mechanism. )
Answer: Since most firewalls are configured to block UDP traffic, using TCP for video and voice traffic lets the traffic though the firewalls
24. (Q5) Is it possible for an application to enjoy reliable data transfer even
when the application runs over UDP If so, how
Answer: Yes. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however.
25. (Q6) Consider a TCP connection between Host A and Host B. Suppose that the
TCP segments traveling form Host A to Host B have source port number x and destination port number y. What are the source and destination port number for the segments traveling form Host B to Host A
Answer: Source port number y and destination port number x.
26. (P20) Suppose we have two network entities, A and B. B has a supply of data
messages that will be sent to A according to the following conventions. When A gets a request from the layer above to get the next data (D) message from B, A must send a request (R) message to B on the A-to-B channel. Only when B receives an R message can it send a data (D) message back to A on the B-to-A channel. A should deliver exactly one copy of each D message to the layer above. R message can be lost (but not corrupted) in the A-to-B channel; D messages, once sent, are always delivered correctly. The delay along both channels is unknown and variable.
Design(give an FSM description of) a protocol that incorporates the
appropriate mechanisms to compensate for the loss-prone A-to-B channel and implements message passing to the layer above at entity A, as discussed above. Use only those mechanisms that are absolutely necessary.
Answer:
Because the A-to-B channel can lose request messages, A will need to timeout and retransmit its request messages (to be able to recover from loss). Because the channel delays are variable and unknown, it is possible that A will send duplicate requests ., resend a request message that has already been received by B). To be able to detect duplicate request messages, the protocol will use sequence numbers. A 1-bit sequence number will suffice for a stop-and-wait type
of request/response protocol.
A (the requestor) has 4 states:
• “Wait for Request 0 from above.” Here the requestor is waiting for a call from above to request a unit of data. When it receives a request from above, it sends a request message, R0, to B, starts a timer and makes a transition to the “Wait for D0” state. When in the “Wait for Request 0 from above” state, A ignores anything it receives from B.
• “Wait for D0”. Here the requestor is waiting for a D0 data message from B. A timer is always running in this state. If the timer expires, A sends another R0 message, restarts the timer and remains in this state. If a D0 message is received from B, A stops the time and transits to the “Wait for Request 1 from above” state. If A receives a D1 data message while in this state, it is ignored.
• “Wait for Request 1 from above.” Here the requestor is again waiting for a call from above to request a unit of data. When it receives a request from above, it sends a request message, R1, to B, starts a timer and makes a transition to the “Wait for D1” state. When in the “Wait for Request 1 from above” state, A ignores anything it receives from B.
• “Wait for D1”. Here the requestor is waiting for a D1 data message from B. A timer is always running in this state. If the timer expires, A sends another R1 message, restarts the timer and remains in this state. If a D1 message is received from B, A stops the timer and transits to the “Wait for Request 0 from above” state. If A receives a D0 data message while in this state, it is ignored.
The data supplier (B) has only two states:
• “Send D0.” In this state, B continues to respond to received R0 messages by sending D0, and then remaining in this state. If B receives a R1 message, then
it knows its D0 message has been received correctly. It thus discards this D0 data (since it has been received at the other side) and then transits to the “Send D1” state, where it will use D1 to send the next requested piece of data.
• “Send D1.” In this state, B continues to respond to received R1 messages by sending D1, and then remaining in this state. If B receives a R1 message, then it knows its D1 message has been received correctly and thus transits to the “Send D1” state.
27. (P21) Consider the GBN and SR protocols. Suppose the sequence number space
is of size k. What is the largest allowable sender window that will avoid the occurrence of problems such as that in Figure for each of these protocols
Answer:
In order to avoid the scenario of Figure , we want to avoid having the leading edge of the receiver's window ., the one with the “highest” sequence number) wrap around in the sequence number space and overlap with the trailing edge (the one with the \"lowest\" sequence number in the sender's window). That is, the sequence number space must be large enough to fit the entire receiver window and the entire sender window without this overlap condition. So - we need to determine how large a range of sequence numbers can be covered at any given time by the receiver and sender windows.
Suppose that the lowest-sequence number that the receiver is waiting for is packet m. In this case, it's window is [m,m+w-1] and it has received (and ACKed) packet m-1 and the w-1 packets before that, where w is the size of the window. If none of those w ACKs have been yet received by the sender, then ACK messages
with values of [m-w,m-1] may still be propagating back. If no ACKs with these ACK numbers have been received by the sender, then the sender's window would be [m-w,m-1].
Thus, the lower edge of the sender's window is m-w, and the leading edge of the receivers window is m+w-1. In order for the leading edge of the receiver's window to not overlap with the trailing edge of the sender's window, the sequence number space must thus be big enough to accommodate 2w sequence numbers. That is, the sequence number space must be at least twice as large as the window size,
k ≥ 2w.
28. (P22) We have said that an application may choose UDP for a transport
protocol because UDP offers finer application control (than TCP) of what data is sent in a segment and when.
a. Why does an application have more control of what data is sent in a
segment
b. Why does an application have more control on when the segment is sent
Answer:
a. Consider sending an application message over a transport protocol. With TCP,
the application writes data to the connection send buffer and TCP will grab bytes without necessarily putting a single message in the TCP segment; TCP may put more or less than a singe message in a segment. UDP, on the other hand, encapsulates in a segment whatever the application gives it; so that, if the application gives UDP an application message, this message will be the payload of the UDP segment. Thus, with UDP, an application has more control of what data is sent in a segment.
b. With TCP, due to flow control and congestion control, there may be significant
delay from the time when an application writes data to its send buffer until when the data is given to the network layer. UDP does not have delays due to flow control and congestion control.
29. (P23) Consider transferring an enormous file of L bytes from Host A to Host
B. Assume an MSS of 1460 bytes.
a. What is the maximum value of L such that TCP sequence numbers are
not exhausted Recall that the TCP sequence number field has 4 bytes.
b. For the L you obtain in (a), find how long it takes to transmit the
file. Assume that a total of 66 bytes of transport, network, and data-link header are added to each segment before the resulting packet is sent out over a 100Mbps link. Ignore flow control and congestion control so A can pump out the segments back to back and continuously.
Answer:
There are 232 = 4,294,967,296 possible sequence numbers.
a. The sequence number does not increment by one with each segment. Rather, is
increments by the number of bytes of data sent. So the size of the MSS is irrelevant – the maximum size file that can be sent from A to B is simply the number of bytes representable by 232 ≈ Gbytes .
b. The number of segments is ┌2/1460┐=2,941,758. 66 bytes of header get added
to each
segment giving a total of 194,156,028 bytes of header. The total number of bytes transmitted is 232 +194,156,028 = 3,591×107 bits.
Thus it would take 3,591 seconds = 59 minutes to transmit the file over a 10~Mbps
link.
30. (P24) Host A and B are communicating over a TCP connection, and Host B has
32
already received from A all bytes up through byte 358. Suppose Host A then sends two segments to Host B back-to-back. The first and second segments contain 50 and 80 bytes of data, respectively. In the first segment, the sequence number is 359, the source port number is 1028, and the destination port number is 80. Host B sends an acknowledgment whenever it receives a segment from Host A.
a. In the second segment sent from Host A to B, what are the sequence
number, source port number, and destination port number
b. If the first segment arrivers before the second segment, in the
acknowledgement of the first arriving segment, what is the acknowledgement number, the source port number, and the destination port number
c. If the second segment arrives before the first segment, in the
acknowledgement of the first arriving segment, what is the acknowledgement number
d. Suppose the two segments sent by A arrive in order at B. The first
acknowledgement is lost and the second acknowledgement arrives after the first timeout interval. Draw a timing diagram, showing these segments and all other segments and acknowledgements sent. (Assume there is no additional packet loss.) For each segment in your figure, provide the sequence number and the number of bytes of data; for each acknowledgement that you add, provide the acknowledgement number.
Answer:
a. In the second segment from Host A to B, the sequence number is 289, source
port number is 503 and destination port number is 80.
b. If the first segment arrives before the second, in the acknowledgement of the
first arriving segment, the acknowledgement number is 289, the source port number is 80 and the destination port number is 503.
c. If the second segment arrives before the first segment, in the acknowledgement
of the first arriving segment, the acknowledgement number is 249, indicating that it is still waiting for bytes 249 and onwards. d.
31. (P25) Host A and B are directly connected with a 200 Mbps link. There is
one TCP connection between the two hosts, and Host A is sending to Host B an enormous file over this connection. Host A can send application data into the link at 100 Mbps but Host B can read out of its TCP receive buffer at a maximum rate of 50 Mbps. Describe the effect of TCP flow control.
Answer: 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 50Mbps. When the buffer is full, Host B signals to Host A to stop sending data by setting RcvWindow = 0. Host A then stops sending until it receives a TCP segment with RcvWindow > 0. Host A will thus repeatedly stop and start sending as a function of the RcvWindow values it receives from Host B. On average, the long-term rate at which Host A sends data to Host B as part of this connection is no more than 50Mbps.
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