Answer:
The answer is a stage of system development that involves...
Explanation:
___a stage of system development_ involves those activities that ensure the orderly dissolution of a system, including disposing of all equipment in an environmentally friendly manner, closing out contracts, and safely migrating information from the system to another system or archiving it in accordance with applicable records management policies.
That is the definition of system disposal
A LAN (local area network) is a group of computers and network devices connected together, usually within the same building. By definition, the connections must be high speed and relatively inexpensive (e.g., token ringor Ethernet). Most Indiana University Bloomington departments are on LANs.
A LAN connection is a high-speed connection to a LAN. On the IUB campus, most connections are either Ethernet (10 Mbps) or Fast Ethernet (100 Mbps), and a few locations have Gigabit Ethernet (1000 Mbps) connections.
A MAN (metropolitan area network) is a larger network that usually spans several buildings in the same city or town. The IUB network is an example of a MAN.
A WAN (wide area network), in comparison to a MAN, is not restricted to a geographical location, although it might be confined within the bounds of a state or country. A WAN connects several LANs, and may be limited to an enterprise (a corporation or an organization) or accessible to the public. The technology is high speed and relatively expensive. The Internet is an example of a worldwide public WAN.
Answer:
1 new contact 2 save and close
Answer:
Two systems are connected by a router. Both systems and the router have transmission rates of 1,000bps. Each link has a propagation delay of 10ms. Also, it takes router 2ms in order to process the packet (e.g. decide where to forward it). Suppose the first system wants to send a 10,000 bit packet to the second system. How long will it take before receiver system receives the entire packet.
Transmission time for first Router = 10,000 bits / 1000 bps = 10 seconds
Receiving time for seond route r= 10,000 bits / 1000 bps = 10 seconds
Propagation delay = 10ms = .01 seconds x 2 for two delays = .02 seconds
First router 2ms to process = .002 seconds
Add all the times together and we get 20.022 seconds which is the same as or 20 seconds and 22 ms