Green computing is the environmentally responsible and eco-friendly use of computers and their resources. In broader terms, it is also defined as the study of designing, manufacturing/engineering, using and disposing of computing devices in a way that reduces their environmental impact.
Answer:
1)PERFORMANCE
2)RANGE
Explanation:
A mesh network can be regarded as local network topology whereby infrastructure nodes connect dynamically and directly, with other different nodes ,cooperate with one another so that data can be efficiently route from/to clients. It could be a Wireless mesh network or wired one.
Wireless mesh network, utilize
only one node which is physically wired to a network connection such as DSL internet modem. Then the one wired node will now be responsible for sharing of its internet connection in wireless firm with all other nodes arround the vicinity. Then the nodes will now share the connection in wireless firm to nodes which are closest to them, and with this wireless connection wide range of area can be convered which is one advantage of wireless mesh network, other one is performance, wireless has greater performance than wired one.
Some of the benefits derived from configuring a wireless mesh network is
1)PERFORMANCE
2)RANGE
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
