Pick a single link state advertisement packet captured by ethereal, and describe how to interpret the information contained in the link state advertisement.

What will be an ideal response?

```
Frame 33 (94 on wire, 94 captured)
Arrival Time: Aug 6, 2003 15:41:15.275958
Time delta from previous packet: 0.000000 seconds
Time relative to first packet: 117.870000 seconds
Frame Number: 33
Packet Length: 94 bytes
Capture Length: 94 bytes
Ethernet II
Destination: 01:00:5e:00:00:05 (01:00:5e:00:00:05)
Source: 00:00:c0:58:28:00 (00:00:c0:58:28:00)
Type: IP (0x0800)
Internet Protocol, Src Addr: 10.0.1.2 (10.0.1.2), Dst Addr: 224.0.0.5 (224.0.0.5
)
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: 80
Identification: 0x2917
Flags: 0x00
.0.. = Don't fragment: Not set
..0. = More fragments: Not set
Fragment offset: 0
Time to live: 1
Protocol: OSPF (0x59)
Header checksum: 0xa537 (correct)
Source: 10.0.1.2 (10.0.1.2)
Destination: 224.0.0.5 (224.0.0.5)
Open Shortest Path First
OSPF Header
OSPF Version: 2
OSPF Packet Type: 4 (LS Update)
Packet Length: 60
Source OSPF Router ID: 10.0.1.2
Area ID: 0.0.0.1
Packet Checksum: 0x4667 (correct)
Auth Type: Null
Auth Data (none)
LS Update Packet
Number of LSAs: 1
Network LSA (Type: 2)
LS Age: 1 seconds
Options: 0x2 (E)
LSA Type: 2 (Network LSA)
Link State ID: 10.0.1.2
Advertising Router: 10.0.1.2
LS Sequence Number: 0x80000001
```