1. Example#1
Suppose you have a 25 magnetic tapes, each containing 40GB. Assuming that you have enough tape readers to keep any network busy. How long it will take to transmit the data over a distance of 1Km. Assume the choices are category 5 twisted-pair wires at 100Mbits/sec, multimode fiber at 1000Mbits/sec and single-mode fiber at 2500Mbits/sec. How do they compare to deliver the tapes by car?

2. Solution The total amount of data
= total no. of mag. Tapes x capacity of each tape
= 25 x 40GB= 1000GB
The time for medium:
Twisted pair = 1000GB/100Mbits/sec
= sec = 22.8 hr

3. Multimode Fiber = 1000GB/1000Mbits/sec
= 8192sec = 2.3 hr
Single mode Fiber = 1000GB/2500Mbits/sec
= 3277sec
= 0.9hr
Car = time to load car + transport time + time to unload car
= 300sec + 1Km/30Kph sec
= 720 sec = 0.3hr
A car filled with tapes is a higher bandwidth medium!

4. Node
Node
Node
Node
Ethernet
Node
Node
Switch
Node
Node
Switched Media

5. Example#2
Compare 16 nodes connected by three ways: a single 100Mbits/sec shared medium; a switch connected via cat5, each segment running at 100Mbits/sec; and a switch connected via optical fiber, each running at 1000Mbits/sec. The shared medium is 500m long, and the average length of each segment to a switch is 50m. Both switch can support full bandwidth. Assume each adds 5μsec to the latency. Assume the average message size is 125bytes, and ignore the overhead of sending or receiving a message and contention for the network.

6. Solution First we will calculate the aggregate bandwidth:
For shared medium
Aggregate bandwidth = 100Mbits/sec

7. For switched twisted pair
Aggregate bandwidth = 16 x 100Mbits/sec
= 1600Mbits/sec
For switched optical fiber
Aggregate bandwidth = 16 x 1000Mbit/sec
= 16,000Mbits/sec
Transport time = Time of flight + (message size/BW)

8. (500/1000)Km Transport time shared = ---------------------- x 106μsec
(2/3 x 300,000)Km
+ (125 x 8bits / 100Mbits/sec)
= 2.5μsec + 10μsec = 12.5μsec
For the switches, the distance is twice the average segment.
We must also add latency for the switch.

9. (50/1000)Km Transport time switch = 2x ---------------------- x 106μs
(2/3 x 300,000)Km
+ 5μsec
+ (125 x 8bits / 100Mbits/sec)
= 0.55μsec + 5μsec +10μsec
= 15.5μsec

10. (50/1000)Km
Transport time fiber = 2x x 106μs
(2/3 x 300,000)Km
+ 5μsec
+ (125 x 8bits / 1000Mbits/sec)
= 0.55μsec + 5μsec +1μsec
= 6.5μsec
Although the bandwidth of the switch is many times the shared media, the latency for unloaded network is comparable.

11. Connection Oriented Communication
Source
Destination
Messages
Connectionless Communication
Source
Destination

12. Three Network Topologies
Bus
Star
Ring

13. 7-Layer OSI Model Internet A p p l i c a t i o n A p p l i c a t i o nw o r k N e t w o r k D a t a l i n k D a t a l i n k P h y s i c a l P h y s i c a l
Internet

14. Routing
Source-Based Routing
Virtual Circuit
Destination-based Routing

15. Source-Based Routing Message specifies the destination path.
Network really follows the directions.
Simpler method.

16. Virtual Circuit
A circuit is established between source and destination.
Message simply names the circuit to follow.
ATM is the example of virtual circuit.

17. Destination-based Routing
Message contains the destination address.
Switch decides a path to deliver the message.
Example of destinations based routing is IP.

18. Destination-based Routing
It may be deterministic.
Always follows the same path.
It may be adaptive.
Closely related to the adaptive routing is randomized routing.

19. TCP/IP vs. OSI Model A p l i c a t o n O S I m d e P r s T N w k h y Df t p T e l n e t D N S T r a n s m i s s i o n c o n t r o l p r o t o c o l ( T C P ) I n t e r n e t p r o t o c o l ( I P ) T o k e n E t h e r n e t L o c a l T a l k r i n g