Answer:
1.the program is moved from secondary storage to memory.
Explanation:
Secondary storage media generally have greater storage capacity than RAM (random access memory), for this reason the data and program are stored there.
but ram memories are much faster, for example, a solid state disk (SSD) with SATA III connection can have read and write speeds of approximately 500 MB/s, while a DDR III RAM can have speeds between 12 and 18 GB/S, that is, more than 20 times faster.
So that the flow of data at the time of running a program is optimal, the data from the secondary storage unit is copied to the RAM and this ensures that the speed at which the programs run is at least 20 times faster, and our processes run better.
Answer:
Explanation:
The minimum depth occurs for the path that always takes the smaller portion of the
split, i.e., the nodes that takes α proportion of work from the parent node. The first
node in the path(after the root) gets α proportion of the work(the size of data
processed by this node is αn), the second one get (2)
so on. The recursion bottoms
out when the size of data becomes 1. Assume the recursion ends at level h, we have
(ℎ) = 1
h = log 1/ = lg(1/)/ lg = − lg / lg
Maximum depth m is similar with minimum depth
(1 − )() = 1
m = log1− 1/ = lg(1/)/ lg(1 − ) = − lg / lg(1 − )
Answer:
to close an open application
Answer:
Sequence Numbers
Explanation:
In the transmission control protocol, the sequence number are used to put data back in order.
It is the number pattern that follows sequence. This is the number pattern which interprets data before they are returned by the recipient as an acknowledgement with its numbers.
When a sent data is out of order, the sequence number works towards correcting the sequence before sending its acknowledgement.
