Bus Topology
This type of network uses a common device to connect all computers. A single cable works as backbone and acts as a shared communication medium on to which devices attach with an interface connector.
Ring Topology
It is in the form of ring where each node is connected exactly with two neighboring nodes . In this type of topology communication takes place onliy in one direction . Failure of a single node results in the failure of complete network .
Star Topology
In this topology a central connecting device called hub, switch or router is used as a medium to connect different nodes of network. In this topology whole of the communication is dependent on central hub. Failure of this central hub results in the failure of complete network.
Tree Topology
This type of topology is in the form of tree , in this various star topologies are coneected together using a bus.In this topology if the bus stops functioning , still part of network can function.
Mesh Topology
As the ame suggests Mesh topology is combined by forming various types of topologies together . It is not dependent on central hub or switch for communication , so networks can still function if a part of it stops functioning .In this a mesage can be communicated using n number of paths.
NETWORK WORKSHUBS.
In computer networking, hub is a small, simple, inexpensive device that joins multiple computers together. Many network hubs available today support the Ethernet standard. Other types including
USB hubs also exist, but Ethernet is the type traditionally used in home networking.
Working With Ethernet Hubs
To network a group of computers using an Ethernet hub, first connect an Ethernet cable into the unit, then connect other end of the cable to each computer’s network interface card (.NIC). All Ethernet hubs accept the RJ-45 connectors of standard Ethernet cables. ^
To expand a network to accommodate more devices, Ethernet hubs can also be connected to each other, to switches, or to routers.
Characteristics of Ethernet Hubs
Ethernet hubs vary in the speed (network data rate or bandwidth) they support. Earlier, Ethernet hubs offered only 10 Mbps rated speeds but now hubs offer 100 Mbps Ethernet. Some support both 10 Mbps and 100 Mbps (so-called dual-speed or 101100 hubs).
The number of ports an Ethernet hub supports also varies. Four- and five-port Ethernet hubs are most common in home networks, but eight- and 16-port hubs can be found in some home and small office environments. Older Ethernet hubs were relatively large in size and sometimes noisy as they contained built in fans for cooling the unit. Newer devices are much smaller, designed for mobility, and noiseless.
Use of an Ethernet Hub.
Ethernet hubs operate as Layer 2 devices in the QSI_.m_Q.d_eJ, the same as network switches. Although offering comparable functionality, nearly all mainstream home network equipment today utilizes network switch technology instead of hubs due to the performance benefits of switches. Hub is useful for temporarily replacing a broken network switch or when performance is not a critical factor on the network.
SWITCH.
Network switches appear nearly identical to network hubs, but a switch generally contains more intelligence (and a slightly higher price tag) than a hub. Unlike hubs, network switches are capable of inspecting data packets as they are received, determining source and destination device of each packet, and forwarding them appropriately. By delivering messages only to the connected device intended, a network switch conserves network bandwidth and offers generally better performance than a hub.
As with hubs, Ethernet implementations of network switches are the most common. Mainstream Ethernet network switches support either 10/100 Mbps Fast Ethernet or Gigabit Ethernet (10/100/1000) standards. Different models of network switches support differing numbers of connected devices. Most consumer-grade network switches provide either four or eight connections for Ethernet devices. Switches can be connected to each other, a so-called daisy chaining method to add progressively larger number of devices to a LAN.
BRIDGES.
It is a box with ports (usually two) to LAN segments. It operates in promiscuous mode at the data link layer (i.e. at the level of frames, not signals), it examines all frames and it recognizes where they came from, and where they are going to. It selectively (frame filtering) transfers frames from any port to other ports. It does not propagate noise signals and defective frames as it was the case for repeaters (at the physical layer). It adaptively recognizes which machines are reachable from a port. It reduces traffic on each port and it improves security since each port will only transmit frames directed to nodes reachable from that port (thus one does not overhear irrelevant traffic).
ROUTERS.
It is a box (usually a regular computer) with at least two ports, used to connect also dissimilar networks. It differs from bridges since it operates at the network level, It will also use different addresses, e.g. a bridge may use Ethernet addresses while a router uses IP addresses. It does all the transformations that may be required by the transfer of packets across the networks it connects.
Routing involves two basic activities:
(i) Running routing algorithms to determine routes, as expressed by routing tables
(ii) Using the routing tables to move packets across the network
Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use ?
Solution
Band width delay product
=round trip delay x bottle neck band width
= 80 x 128 x 10-3 x 1024 bits
= (80 x 128 x 10-3 x 1024)/8 byte
Optimal size window = (80 x 128 x 10-3 x 1024) / (8x32)
= 40
In a token ring network the transmission speed is 107 bps and propogation speed is 200 meters per sec. The 1 bit delay in this network is equivalent to
Solution
107 bits need 1 second
∴ 1 bit needs (1/107) seconds = 1/10 μs
Since 1 μs corresponds to 20 meter cable
∴ 1/10 μs corresponds to 20 meter cable
The distance between two stations M and N is L kilometres . All frames are K bits long. The propagation delay per kilometre is t seconds. Let Rbits/second be the channel capapcity. Assuming that processing delay is negligible ,what is the minimum number of bits for the sequence number field in a frame for maximum utilization, when sliding window protocol is used ?
Solution
Total propogation delay = Lt sec
Round trip time = 2 x Lt = 2 Lt sec
No of bits transmitted in round trip= 2LtR bits
Number of frames = 2LtR/K
Let number of bits in sequent number be b
2b={2LtR / K}
B= {log2 (2LtR / K)}
The address of class B host is to be split into subnets with a 6 bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet ?
Solution
Maximum number of subnets
=26-2=62
Maximum number of host in each subnet = 210-2=1022
What signal to noise ratio is needed to put a T carrier on a 10-Khz line ?
Solution
To send a T1 signal we need
Hlog2(1+(s/n))=1.544 x 106 with H 50,000
S / N =230-1.48 ≈93 db
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