What is the transmission efficiency of a 5,000 byte file is sent in response to a Web request HTTP, TCP
TCP/IP (IPv4) and Ethernet. Assume the file uses a 100 byte HTTP in addition to the 5,000-bype file. Assume the maximum packet size is 1,200. Hint: Remember form Chapter 4 that efficiency = user data/total transmission size..
Necessary Steps:
1. The efficiency formula
2. Is packetizing necessary (i.e., will the message fit in a single Ethernet frame?) If so, how many frames are necessary?
3. The overhead associated with each layer (adjusted for multiple frames if packetizing was necessary).
Step #1
Number of information bits
Efficiency (%) = -------------------------------------------------------------------------------------- * 100%
Number of information bits + Number of overhead bits
Step #2
First, recall that each of the layers operate independently. Therefore, our transport (TCP) layer will treat everything from the application layer as data. The application does not care what is below it and therefore will only transmit the HTTP header once. This gives us 5,000 + 100 or 5,100 that must be considered in the packetizing decision. Because 5,100 is greater than 1,200, two or more frames will be necessary (see below).
In reality, for the Ethernet/IP/TCP combination, each frame can accommodate:
1,492 – TCP header – IP header – (Ethernet header + Ethernet trailer)
= 1,492 – 24 – 24 = 1,444 (See next page for formats) but use 1,200 as the problem states
The number of frames is therefore 5,100/1,200 rounded up or 5 frames to accommodate the original message.

1
2
3
4
5
6
Source ID
Destination ID
Sequence number
ACK number
Header length
Unused
16 bits
16 bits
32 bits
32 bits
4 bits
6 bits
7
8
9
10
11
Flags
Flow control
CRC 16
Urgent pointer
Options
Total
6 bits
16 bits
16 bits
16 bits
32 bits
192 bits (24 bytes )
1
2
3
4
5
6
7
Version number
Header length
Type of Service
Total length
Identifiers
Flags
Packet offset
4 bits
4 bits
8 bits
16 bits
16 bits
3 bits
13 bits
8
9
10
11
12
13
Hop limit
Options
Total
8 bits
Protocol
CRC 16
8 bits
16 bits
Source address 32 bits
Destination Address 32 bits
32 bits
192 bits (24 bytes )
Step #3
Overhead
33 + 24 + 24 + 100 = 181
33 + 24 + 24 = 81
33 + 24 + 24 = 81
33 + 24 + 24 = 81
33 + 24 + 24 = 81
Number of information bits
Efficiency (%) = -------------------------------------------------------------------------------------- * 100%
Number of information bits + Number of overhead bits
5,000
Efficiency (%) = ----------------------------------------------- *100% = 90.8%
5,000+ (181 + 81 + 81 + 81 + 81)