Take the biggest UDP datagram and the biggest TCP segment that you observed, and compare the amount of application data that is transmitted in the UDP datagram and the TCP segment.
What will be an ideal response?
The biggest TCP segment is 1448 bytes and the biggest UDP datagram is 1024 bytes.
```
Frame 7 (1514 bytes on wire, 1514 bytes captured)
Arrival Time: Aug 23, 2003 02:56:16.672385000
Time delta from previous packet: 0.000088000 seconds
Time relative to first packet: 0.002939000 seconds
Frame Number: 7
Packet Length: 1514 bytes
Capture Length: 1514 bytes
Ethernet II, Src: 00:06:5b:4b:e0:dd, Dst: 00:06:5b:4b:e1:da
Destination: 00:06:5b:4b:e1:da (00:06:5b:4b:e1:da)
Source: 00:06:5b:4b:e0:dd (00:06:5b:4b:e0:dd)
Type: IP (0x0800)
Internet Protocol, Src Addr: 10.0.5.11 (10.0.5.11), Dst Addr: 10.0.5.22 (10.0.5.22)
Version: 4
Header length: 20 bytes
Differentiated Services Field: 0x00 (DSCP 0x00: Default; ECN: 0x00)
0000 00.. = Differentiated Services Codepoint: Default (0x00)
.... ..0. = ECN-Capable Transport (ECT): 0
.... ...0 = ECN-CE: 0
Total Length: 1500
Identification: 0xc3a0
Flags: 0x04
.1.. = Don't fragment: Set
..0. = More fragments: Not set
Fragment offset: 0
Time to live: 64
Protocol: TCP (0x06)
Header checksum: 0x535b (correct)
Source: 10.0.5.11 (10.0.5.11)
Destination: 10.0.5.22 (10.0.5.22)
Transmission Control Protocol, Src Port: 3062 (3062), Dst Port: 4444 (4444), Seq:
4012938045, Ack: 3987339891, Len: 1448
Source port: 3062 (3062)
Destination port: 4444 (4444)
Sequence number: 4012938045
Next sequence number: 4012939493
Acknowledgement number: 3987339891
Header length: 32 bytes
Flags: 0x0010 (ACK)
0... .... = Congestion Window Reduced (CWR): Not set
.0.. .... = ECN-Echo: Not set
..0. .... = Urgent: Not set
...1 .... = Acknowledgment: Set
.... 0... = Push: Not set
.... .0.. = Reset: Not set
.... ..0. = Syn: Not set
.... ...0 = Fin: Not set
Window size: 5840
Checksum: 0xae00 (correct)
Options: (12 bytes)
NOP
NOP
Time stamp: tsval 1067168, tsecr 1066965
Data (1448 bytes)
```
Captured UDP packet
```
Frame 2 (1066 bytes on wire, 1066 bytes captured)
Arrival Time: Aug 23, 2003 02:54:02.335756000
Time delta from previous packet: 0.000273000 seconds
Time relative to first packet: 0.000273000 seconds
Frame Number: 2
Packet Length: 1066 bytes
Capture Length: 1066 bytes
Ethernet II, Src: 00:06:5b:4b:e0:dd, Dst: 00:06:5b:4b:e1:da
Destination: 00:06:5b:4b:e1:da (00:06:5b:4b:e1:da)
Source: 00:06:5b:4b:e0:dd (00:06:5b:4b:e0:dd)
Type: IP (0x0800)
Internet Protocol, Src Addr: 10.0.5.11 (10.0.5.11), Dst Addr: 10.0.5.22 (10.0.5.22)
Version: 4
Header length: 20 bytes
Differentiated Services Field: 0x00 (DSCP 0x00: Default; ECN: 0x00)
0000 00.. = Differentiated Services Codepoint: Default (0x00)
.... ..0. = ECN-Capable Transport (ECT): 0
.... ...0 = ECN-CE: 0
Total Length: 1052
Identification: 0x0000
Flags: 0x04
.1.. = Don't fragment: Set
..0. = More fragments: Not set
Fragment offset: 0
Time to live: 64
Protocol: UDP (0x11)
Header checksum: 0x18b1 (correct)
Source: 10.0.5.11 (10.0.5.11)
Destination: 10.0.5.22 (10.0.5.22)
User Datagram Protocol, Src Port: 1028 (1028), Dst Port: 4444 (4444)
Source port: 1028 (1028)
Destination port: 4444 (4444)
Length: 1032
Checksum: 0xbe32 (correct)
Data (1024 bytes)
```
You might also like to view...
In SQL, you use keywords to construct statements, which are also known as ____________.
a. queries b. expressions c. determinants d. solutions
Why is it considerably more difficult to test, debug, and prove program correctness for concurrent programs than for sequential programs?
What will be an ideal response?
________ checks to see if programs that are interdependent actually work together as planned.
A) Link testing B) Desk checking C) Full systems testing D) Production testing
________ is determined according to evenness or oddness of a sum of bits.
a) ECC b) XOR c) Parity d) None of the above