The determining factor of packet size is the MTU (Maximum Transmission Unit) which varies with the host or router settings. If a received datagram indicates the packet's size is too great for the MTU of the receiving router, the incoming packet is fragmented in retransmission, which means not all the data of that one packet will be in one packet thereafter (and therefore probably not available at any one network node). See notes below.

A

From the MS web site:

MTU is the Maximum Transmission Unit of a network. This is the greatest amount of data or "packet" size that can be transferred in one physical frame on a network. Fragmentation of packets occurs if a packet is sent across a network that has an MTU smaller than the packet's frame length. This leads to reduced performance levels as fragments need to be reassembled.

The Internet standard for MTU is 576 bytes. When you connect to an Internet resource, your data may have to go through a router that has an MTU of 576. This results in fragmentation if your MTU is greater than 576, which can result in a Web resource being loaded slowly.

Microsoft recommends using the default setting of Automatic for the Internet Protocol (IP) packet size.

---

Some arcane considerations from some website:

Example calculation of the segmentation performed by IP Routers
"Calculate the number of fragments which are sent when an IP datagram with payload of 3000 bytes is sent from a computer on a network A via two routers to a destination computer C. The MTU if network A is 4000 B. The MTU of network B is 508 B and for network C the MTU is 1500 B. Ensure that your answer specifies the number and size the of the IP datagrams sent on each of the LANs"

On First LAN

Total size of initial PDU = 3000 + 20 B (PCI) = 3020 B.

Network A: MTU 4000 B > 3020 B - therefore one packet is sent.

(packet sent with offset=0, more=FALSE)

On Second LAN

Network B: MTU 508 B < 3020 B - therefore fragmentation is required.

Fragment payload size = 508- 20 B = 488 B.

Total number of packets sent via network B = round(3000/488) = 7 packets.

First six packets of size 508 B, with 488B of payload

The last packet has 48 B of data, and therefore of size 20 B + 72 B=92 B.

(all packets except first sent with offset>0,)

(packets 1-6 have more=TRUE, last packet has more=FALSE)

On Final LAN

Fragments are not reassembled by an IS - i.e. at router C.

Router C therefore receives 7 packets, 6 of size 508 B, 1 of size 92 B.

Network C: MTU 1500 B > 508 B - No further fragmentation is therefore needed.

i.e. There are 7 packets as in network C, 6 of 508 B and one of 92 B.

Note

The IP network protocol requires all fragments except the last to contain an integral number of 8 B (64-bit) chunks of data. This requires the payload to be exactly divisible by 8 in all packets which have the "More" bit set.

---

Hope this helps.


A


--------------------------------------------------------------------------------