Answer:
The comparison is based on memory organization schemes of contiguous memory allocation and paging with respect to External fragmentation, Internal fragmentation and Ability to share code across processes.
Explanation:
Memory organization schemes of contiguous memory allocation:
Contiguous memory allocation schemes suffers from external fragmentation. The reason is that address space is distributed contiguously and the holes and gaps keep growing when the old processes die and new processes are introduced. The variable size partition suffers from external fragmentation however the fixed size partitions do not suffer from external fragmentation. Contiguous memory allocation with variable size partitions does not encounter an internal fragmentation but with fixed size partitions suffers from internal fragmentation. Contiguous memory allocation does not support sharing code across processes. This is because the virtual memory segment of a process is not fragmented into non-contiguous fine grained blocks.
Paging:
Paging does not encounter external fragmentation as pages are of the fixed or equal size. So this reduces external fragmentation. However paging suffers from internal fragmentation. This is because a process can request more space or it can request for a less space. When page is allocated to the such a process that page is no longer utilized.This results in internal fragmentation because of the wastage of space even when the page has internal space but cannot be fully utilized. Paging allows to share code across processes.
Answer:
When the transmission exceeds 667 packets
Explanation:
In computer networking, a packet is a chunk of data transmitted across the network. The packet size of an Ethernet network is 1.5kilobytes, while the packet size of an IP packet payload is 64 kilobytes.
A switch is a physical network device that connects nodes or workstations while communicating the packets (or frames). The I/O bus size bandwidth is 1Gbps which allows approximately 667 packets. Once this packet size is crossed, the bus becomes a limiting factor or bottle neck.
Answer:
(a)Applications Time Stamp Events
(b)S=0.5(W-
)+, where ≤W≤+8.
Explanation:
Some applications assume that clocks always advance, so they could timestamp events under this assumption.
In our case we have the wrong timed clock, say W and the hardware clock H which is supposed to advance at a perfect rate.
We proceed to construct a software clock such that after 8 seconds we can replace the wrong timed clock with the software clock in good conditions.
Let us denote the software clock with S.
Then, S=c(W-)+ where:
=The current Time(10:27:54) and;
c is to be found.
We already know that S=+4 when W= +8,
So:
S=c(W-)+
+4=c(+8-)+
4=8c
c=0.5
We obtain the formula
S=0.5(W-)+, where ≤W≤+8.
Answer:
public class Main
{
public static void main(String[] args) {
int userNum = 40;
while(userNum > 1){
userNum /= 2;
System.out.print(userNum + " ");
}
}
}
Explanation:
*The code is in Java.
Initialize the userNum
Create a while loop that iterates while userNum is greater than 1. Inside the loop, divide the userNum by 2 and set it as userNum (same as typing userNum = userNum / 2;). Print the userNum
Basically, this loop will iterate until userNum becomes 1. It will keep dividing the userNum by 2 and print this value.
For the values that are smaller than 1 or even for 1, the program outputs nothing (Since the value is not greater than 1, the loop will not be executed).
Answer:
a). SLE =$37.5
b). ARO =75
c). ALE = $2,812.5
Explanation:
a).Single loss Expectancy (SLE) is starting point in determining the single loss of an asset that will occur and calculated this;
SLE = asset value * exposure factor.
Asset value =$500,
Exposure factor is simply the percentage of asset lost.
In this case out of 1000 phones, 75 were damaged or loss.
In percentage;
75 ÷ 1000 =0.075, 0.075×100=7.5%(exposure factor).
Therefore,
SLE = $500×7.5%= $37.5.
b). ARO - Annual Rate of Occurrence is the number of times a threat on a single asset is expected to occur in one year.
In the case the damage or loss occured in 75 devices in one year.
c). ALE - Annualized loss Expectancy is the product of SLE and ARO.
Therefore;
ALE = $37.5 × 75 = $2,812.5.
