Module 7

Réussis tes devoirs et examens dès maintenant avec Quizwiz!

Answer this question in the context of the strategy described in class and adopted by each child to manage Weekly Pocket Money Problem. This week, the child asked and received $40. Next week, the child can certainly ask for and receive ____ without being smacked for sure.

$0

Suppose that cwnd = 20 bytes. The latest segment received by a client had the advertised window set to 40 bytes. The client has in its sending buffer these equals seized segments with sequence numbers 15, 25, 35, 45, 55, 65, and 75. So far, the client sent segments with Sequence numbers 15, 25, and 35. The highest acknowledgement if received is 25. The next segments that the client could send is segment with Sequence number to ___. (Answer -1 if the client cannot send anymore new segments)

-1

Answer this question based congestion control and flow control as performed by TCP. Suppose that a TCP sender 1) has cwnd set to 8,000 bytes 2) has 4,000 bytes available in its own input buffers 3) receives a segment with an advertised window of 0 bytes. The TCP sender will set its offered window Wnd to ____ bytes.

0

Answer this question in the context of the strategy described in class and adopted by each child to manage Weekly Pocket Money Problem. A child starts numbering weeks from 0 (zero) when he get smacked. On Week 0 he got smacked when he asked 20 dollars. Assuming that the child asks for money each week, the child will ask ___ dollars on Week 1.

1

Assume a recent TCP implementation (TCP Reno and beyond). Let the congestion window be 10 segments and all 10 outstanding segments get lost. The congestion window will then be set by the sender to ___ segment(s)

1

Assume a recent TCP implementation (TCP Reno and beyond). Let the congestion window be 12 segments and all 12 outstanding segments get lost. Thee congestion window will then be set by the sender to ___ segment(s)

1

Assume a recent TCP implementation (TCP Reno and beyond). Let the congestion window be 16 segments and all 16 outstanding segments get lost. The congestion window will then be set by the sender to ___ segment(s)

1

A client sends seven equal sized segments with sequence numbers 15, 25, 35, 45, 55, 65, and 75. The size of each segment was ___ bytes

10

A TCP client receives a segment that contains a segment with 1000 bytes of data with Sequence number 0. This TCP client will acknowledge this segment with an ack number equal to ___

1000

A TCP client receives a segment that contains a segment with 100 bytes of data with Sequence number 1000. This TCP client will acknowledge this segment with an ack number equal to ___

1100

Assume a recent TCP implementation (TCP Reno and beyond). Suppose that there are 20 outstanding (sent, but not yet acknowledged) segments. The sender will receive at most ___ duplicate acknowledgments if 5 segments get lost.

15

Answer this question in the context of the strategy described in class and adopted by each child to manage Weekly Pocket Money Problem. A child starts numbering weeks from 0 (zero) when he get smacked. On Week 0 he got smacked when he asked 20 dollars. Assuming that the child asks for money each week, the child will ask ___ dollars on Week 2.

2

A client sends seven equal sized segments with sequence numbers 15, 25, 35, 45, 55, 65, and 75. Segments with sequence numbers 25 and 65 get lost. The server will send acknowledgment number ___ after receiving segment with sequence number 75.

25

Suppose that cwnd = 40 bytes. The latest segment received by a client had the advertised window set to 80 bytes. The client has in its sending buffer these equals seized segments with sequence numbers 15, 25, 35, 45, 55, 65, and 75. So far, the client sent segments with Sequence numbers 15, 25, and 35. The highest acknowledgement if received is 35. This client could send up to ____ new segments. (Answer -1 if the client cannot send anymore new segments)

3

Let the advertised window Wa (by the received) be 40 segments and the current congestion window cwnd be 32 segments. Supposed that there are 2 outstanding segments (segments sent and not yet acknowledged). The sender can send at most ___ segments

30

Suppose that cwnd = 1,000 bytes. The latest segment received by a client had the advertised window set to 40 bytes. The client has in its sending buffer these equals seized segments with sequence numbers 15, 25, 35, 45, 55, 65, and 75. So far, the client sent segments with Sequence numbers 15, 25, and 35. The highest acknowledgement if received is 35. This client could send up to ___ new bytes (Answer -1 if the client cannot send anymore new segments).

30

A client sends five equal sized segments with sequence numbers 15, 35, 55, 75, 95. Segments with sequence numbers 55 and 75 get lost. The server will send acknowledgment number ___ after receiving segment with sequence number 15.

35

A client sends seven equal sized segments with sequence numbers 15, 25, 35, 45, 55, 65, and 75. Segments with sequence numbers 35 and 65 get lost. The server will send acknowledgment number ___ after receiving segment with sequence number 25.

35

A client sends seven equal sized segments with sequence numbers 25, 35, 45, 55, 65, and 75 Segments with sequence numbers 35 and 65 get lost. The server will send acknowledgment number ___ after receiving segment with sequence number 75.

35

Answer this question in the context of the strategy described in class and adopted by each child to manage Weekly Pocket Money Problem. A child starts numbering weeks from 0 (zero) when he get smacked. On Week 0 he got smacked when he asked 32 dollars. Assuming that the child asks for money each week, the child will ask ___ dollars on Week 3.

4

The UDP header consists of ___ field (members) of information

4

A client sends seven equal sized segments with sequence numbers 15, 25, 35, 45, 55, 65, and 75. Segments with sequence numbers 45 and 65 get lost. The server will send acknowledgment number ___ after receiving segment with sequence number 25.

45

A server sends a 1000 bytes segment with sequence number 3500. If everything goes well, it will send the next segment with sequence number

4500

Assume a recent TCP implementation (TCP Reno and beyond). Let the congestion window be 10 segments and one segment is lost. The congestion window will then be set by the sender to ___ segment(s)

5

A client sends five equal sized segments with sequence numbers 15, 35, 55, 75, 95. Segments with sequence numbers 55 and 75 get lost. The server will send acknowledgment number ___ after receiving segment with sequence number 35.

55

A client sends five equal sized segments with sequence numbers 15, 35, 55, 75, 95. Segments with sequence numbers 55 and 75 get lost. The server will send acknowledgment number ___ after receiving segment with sequence number 95.

55

Assume a recent TCP implementation (TCP Reno and beyond). Let the congestion window be 12 segments and all 12 outstanding segments get lost. The slow start threshold will then be set by the sender to ___ segment(s)

6

Suppose that cwnd = 40 bytes. The latest segment received by a client had the advertised window set to 60 bytes. The client has in its sending buffer these equals seized segments with sequence numbers 15, 25, 35, 45, 55, 65, and 75. So far, the client sent segments with Sequence numbers 15, 25, and 35. The highest acknowledgement if received is 35. This client could send new segments with the Sequence number up to ___. (Answer -1 if the client cannot send anymore new segments)

65

Answer this question in the context of the strategy described in class and adopted by each child to manage Weekly Pocket Money Problem. A child starts numbering weeks from 0 (zero) when he get smacked. On Week 0 he got smacked when he asked 20 dollars. Assuming that the child asks for money each week, the child will ask ___ dollars on Week 4.

8

Assume a recent TCP implementation (TCP Reno and beyond). Let the advertised window Wa (by the received) be 8 segments and the current congestion window cwnd be 16 segments. A timeout occurs. The sender can send up to ___ segments after 5 successful bursts (since the timeout).

8

The size of a UDP header has ___ bytes

8

Answer this question based congestion control and flow control as performed by TCP. Suppose that a TCP sender 1) has cwnd set to 8,000 bytes 2) has 4,000 bytes available in its own input buffers 3) receives a segment with an advertised window of 12,000 bytes. The TCP sender will set its offered window Wnd to ____ bytes.

8,000

UDP offers these services

Addressing Error Detection

In order to detect corrupted messages sent by the sender, these mechanisms may help at the receiver

Checksum Error detection

For reliability, a time is necessary at the receiver if the channel only corrupt packet payloads

False

For reliability, a timer is necessary at the receiver if the channel can loose packets

False

In the TCP/IP model, an application message can be the packet's payload

False

Pair/Matching related expression/variables

Flow control = Wa: advertised value by Congestion control = cwnd: a variable maintaining Min(Wa, cwnd)

During its "lifetime", a TCP connection will go through these phases

Full duplex data exchange Three way hand-shake Closing

Answer this question in the context of the strategy described in class and adopted by each child to manage Weekly Pocket Money Problem. After size months using this strategy, $40 dollars is the highest sum that a child ever requested and received without being smacked. IN the future, the highest amount that a child can ask without being smacked for sure is ___.

None of these answers

In order to detect corrupted messages sent by the sender, these mechanisms may help at the sender

None of these answers

These fields on the TCP header are used to provide network routing

None of these answers

A server send a 10 bytes segment with sequence number 0. The client receives it correctly. The client will respond with Sequence number ___

None of these choices

On an Ethernet frame, the ___ field will consist of an IP packet, an ARP request/response, or some other network protocols data units

Payload

In the TCP/IP reference model, transport layers use ___ for addressing

Port numbers

The checksum on the TCP header covers ___

Protocol number (from IP header) The TCP header The IP payload The TCP payload The IP addresses (in the IP header)

These fields on the TCP header are used for addressing

Source Port Number Destination Port Number

In order to detect lost messages sent by the sender, these mechanisms may help at the sender

Timer

In order to detect and correct lost messages sent by the sender, these mechanisms may help at the sender

Timer Retransmission

In the TCP/IP reference model, a UDP datagram can be the packet's payload

True

In the TCP/IP reference model, a packet can be the frame's payload

True

These protocols are transport layer protocols used over the Internet

UDP TCP

The following frame was "sniffed" off an Ethernet network. Below, we removed the preamble and the frame header. We use only hexadecimal numbers without using "0x" or "h" symbols. Below are the IP and TCP headers. here is the IP packet in the network byte order: 45 6F 06 00 01 24 00 00 40 06 56 7A 83 CC 0E 65 83 CC 34 43 23 00 4E 21 5E 32 78 1C 12 67 89 E2 50 10 00 34 67 F1 B3 This IP packet carries ___. The source port number is ___. The destination port number is ___.

a TCP segment 2300 4E21

The following frame "sniffed" off an Ethernet network. Below, we removed the preamble and the frame header. We use only hexadecimal numbers without using "0x" or "h" symbols. Below are the IP and TCP headers. Here is the IP packet in network byte order: 45 6F 06 00 01 24 00 00 40 06 56 7A 83 CC 0E 65 83 CC 34 43 4E 21 00 78 1C 5E 12 67 89 E2 50 10 00 34 67 F1 B3 This IP Packet carries ___. The source port number is ___. The destination port number is ___.

a TCP segment 4E21 2300

The following frame "sniffed" off an Ethernet network. Below, we removed the preamble and the frame header. We use only hexadecimal numbers without using "0x" or "h" symbols. Below are the IP and TCP headers. Here is the IP packet in network byte order: 46 6F 06 00 01 24 00 00 40 06 56 7A 83 CC 0E 65 83 CC 34 43 23 00 4E 21 5E 32 78 1C 12 67 89 E2 50 10 00 34 67 F1 B3 This IP packet carries ___. The source port number is ___. The destination port number is ___

a TCP segment 5E32 781C

The following frame was "sniffed" off an Ethernet network. Below, we removed the preamble and the frame header. We use only hexadecimal numbers without using "0x" or "h" symbols. Below are the IP and TCP headers. here is the IP packet in the network byte order: 46 6F 06 00 01 24 00 00 40 06 56 7A 83 CC 0E 65 83 CC 34 43 23 00 4E 21 5E 32 78 1C 12 67 89 E2 50 10 00 34 67 F1 B3 This IP packet carries ___. This source port number is ___. The destination port number is ___.

a TCP segment 5E32 781C

The following frame was "sniffed" off an Ethernet network. Below, we removed the preamble and the frame header. We use only hexadecimal numbers without using "0x" or "h" symbols. Below are the IP and TCP headers. here is the IP packet in the network byte order: 46 6F 06 00 01 24 00 00 40 06 56 7A 83 CC 0E 65 83 CC 34 43 23 00 4E 21 78 1C 5E 32 12 67 89 E2 50 10 00 34 67 F1 B3 This IP packet carries ___. The source port number is ___. The destination port number is ___>

a TCP segment 781C 5E32

The following frame "sniffed" off an Ethernet network. Below, we removed the preamble and the frame header. We use only hexadecimal numbers without using "0x" or "h" symbols. Below are the IP and TCP headers. Here is the IP packet in network byte order: 45 6F 06 00 01 24 00 00 40 06 56 7A 83 CC 0E 65 CC 34 43 4E 4E 21 23 00 78 1C 5E 32 12 67 89 E2 50 10 00 34 F1 B3 This IP packet carries ___. The source port number is ___. The destination port number is ___

a TCP segment None of these answers None of these answers

The following frame was "sniffed" off an Ethernet network. Below, we removed the preamble and the frame header. We use only hexadecimal numbers without using "0x" or "h" symbols. Below are the IP and TCP headers. here is the IP packet in the network byte order: 45 6F 06 00 01 24 00 00 40 11 56 7A 83 CC 0E 65 83 CC 34 43 4E 4E 21 23 00 78 1C 5E 32 12 67 89 E2 50 10 00 34 F1 B3 This IP packet carries ___. The source port number is ___. The destination port number is ___.

a UDP datagram None of these answers None of these answers

The following frame was "sniffed" off an Ethernet network. Below, we removed the preamble and the frame header. We use only hexadecimal numbers without using "0x" or "h" symbols. Below are the IP and TCP headers. here is the IP packet in the network byte order: 46 6F 06 00 01 24 00 00 40 11 56 7A 83 CC 0E 65 83 CC 34 43 23 00 4E 21 5E 32 78 1C 12 67 89 E2 50 10 00 34 67 F1 B3 This IP packet carries ___. The source port number is ___. The destination port number is ___

a UDP datagram 5E32 781C

The objective(s) of congestion control is (are) to:

allocate a fair share of the available bandwidth maximize throughput

____ is necessary when ___ at the receiver is lower (than at the sender)

flow control : processing speed

In the TCP/IP mode, a ___ is for a ___ what a packet is for a segment

frame:packet

Assuming that flow control is used, congestions control is necessary to protect the input buffers of these devices from overflowing

intermediary nodes

Assuming that congestion control is not used, flow control protects the input buffers of these devices from overflowing

receiver

The acknowledgement (ack) number in the TCP header is ___

the number of the next expected byte (to be received)

The sequence number in the TCP header is

the sequence number of the first byte in a segment


Ensembles d'études connexes

Text A: What is friendship all about?

View Set

A+ Guide to Hardware 5th Ed Chapter 1

View Set

Unit 1 heavily tested items- JBL tests

View Set

What happens to matter in a Ecosystem

View Set