The <span>ARPANET </span><span>project started the development of inter-network connections using TCP/IP that has evolved in to the internet today</span>
Answer:
Replace /* Your solution goes here*/ with the following
<em>void SwapArrayEnds(int sortArray [], int lent){
</em>
<em> int temp = sortArray[0];
</em>
<em> sortArray[0] = sortArray[lent-1];
</em>
<em> sortArray[lent-1] = temp;
</em>
<em>}
</em>
<em />
Explanation:
This defines the function SwapArrayEnds with two parameter (the array and the array length)
<em>void SwapArrayEnds(int sortArray [], int lent){
</em>
This declares an integer variable "temp" and initializes it with the first element of the array
<em> int temp = sortArray[0];
</em>
The next two lines swap the last element with the first
<em> sortArray[0] = sortArray[lent-1];
</em>
<em> sortArray[lent-1] = temp;
</em>
<em>}
</em>
<em />
<em>See attachment for full program</em>
Answer:
24.167 micro seconds.
Explanation:
The given rotation rate = 7200 rpm = 7200 rounds per minute
Rotational latency is the average time taken for the head to reach starting of sector
.
Rotational latency (in micro seconds) = (1 / (RPM / 60)) * 0.5 * 1000
(1/(7200/60))* 0.5 * 1000 = 4.167 ms
Thus, rotational latency = 4.167 ms.
Seek time = 20 ms
The average time taken for the access head to get to the beginning of randomly selected sector will be equal to the average time to first reach the random track plus the average time taken to reach random sector
.
= 20 ms + 4.167 ms = 24.167 micro seconds.
Thus, it would take 24.167 micro seconds to get the access head to the beginning of a randomly selected sector on a randomly selected track.
Explanation:
The maximum throughput is equal to the maximum number of data items processed per second cumulatively by all 4 processors.
Now, for a processor, the maximum data items processed per second
= (Number of instructions per second) / (Number of instructions per data item)
Thus, maximum data items processed per second for
processor 1 :
= (1,000,000) / (50)
= 20,000
processor 2 :
= (1,000,000) / (60)
= 16,666,67
processor 3 :
= (1,000,000) / (40)
= 25,000
processor 1 :
= (1,000,000) / (30)
= 33,333.33
Thus, cumulative throughput
= 20,000 + 16,666.67 + 25,000 + 33,333.33
= 95,000 data items per second
