1. A simple network connecting two machines
Machine A
Machine B

2. Message format 1 bit header Pay load (32bit) Address 1 Data 0= Request
Address 1
Data
0= Request
1= Reply

3. Software steps for sending a message
Application data
Operating system buffer
Check sum
Interface hardware
Network

4. Message Reception Send acknowledge to sender
and message to application layer
yes No
Check sum
Correct ?
Delete the message
Operating system buffer
Interface hardware
Network

5. Sender Reactions on Acknowledge
From network
Acknowledge
Receive ? No
yes
Send new message
Send copy of
previous message
network

6. Header with 2-bits 2-bit header 4-bit trailer Data 00 = Request
01 = Reply
10 = Ack. Request
11 = Ack. Reply

7. The protocol must handle many more issues than reliability.
If two machines are from different manufactures, they might order byte in differently within a word.

8. Performance issues Bandwidth Time of flight Transmission time
Transport latency
Sender overhead
Receiver overhead

9. Example#1
Assume a network with a bandwidth of 1000Mbits/sec has a sending overhead of 80μsec and receiving overhead of 100μsec. Assume two machines. One wants to send 10,000 byte message to the other (including header), and the message format allows 15,000 bytes in a single message. Let’s compare SAN, LAN, and WAN by changing the distance between machines. Calculating the total latency to send the message from one machine to another in a SAN assuming they are 10m apart. Next, perform the same calculation but assume the machines are 500m apart. Finally, assume they are 1000Km apart.

10. Solution
Signals propagate at about 63% to 66% of the speed light in a conductor.
Assumptions
We will round the speed of light to 300,000Km/sec.
We can achieve 2/3 of that in a conductor.

11. By using these two assumptions we get:
Distance between two machines in Km
Time of flight =
2/3 x 300,000Km/sec
Total Latency = Sender overhead + Time of flight
+ Message size/bandwidth
+ Receiver overhead

12. For SAN Total latency = 80μsec + (0.010Km/(2/3 x 300,000Km/sec))
+ 10,000bytes/ 1000Mbits/sec
+ 100μsec
= 80μsec μsec + 80μsec
= 260μsec

13. For LAN Total latency = 80μsec + (0.5Km/(2/3 x 300,000Km/sec))
+ 10,000bytes/ 1000Mbits/sec
+ 100μsec
= 80μsec + 2.5μsec + 80μsec
= 262μsec

14. For WAN Total latency = 80μsec + (1000Km/(2/3 x 300,000Km/sec))
+ 10,000bytes/ 1000Mbits/sec
+ 100μsec
= 80μsec μsec + 80μsec
= 5260μsec

15. Example#2
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