Identify three types of intranet communication that can be accomplished with an EHR?

Given six memory partitions of 100 MB, 170 MB, 40 MB, 205 MB, 300 MB, and 185 MB (in order), how would the first-fit, best-fit,

nlexa [21]

Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
P5 will not be allocated to any of the available space because none can contain it.
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

8 0

3 years ago

Please NEED HELP ASAP WILL MARK BRAINLIEST ONLY #8

ValentinkaMS [17]

Answer:

I think its A

Explanation:

6 0

2 years ago

jelaskan tiga kemungkinan sebab pengasah pensil itu tidak dapat berfungsi secara tiba-tiba (translate: explain three possible re

boyakko [2]

Explanation:

Internal gear wear must be significant (visually obvious) to be the cause of off-center sharpening. Cutter carrier is rotating but the pencil is not sharpening (doesn't feel like the cutter is engaging the pencil) This is usually caused by a foreign object (e.g., an eraser or broken pencil lead) inside the pencil bore.

3 0

3 years ago

Wap in java to complete the following (16*1) +(14*2) +(12*3) +(10*4) +(8*5) +(6*6) +(4*7)

vladimir1956 [14]

Answer:

It goes like:

public class Program

{

public static void main(String[] args)

{

int j=18;

int sum=0;

for (int i =1; i<7; i++)

{

sum=sum+(i*(j-2));

j=j-2;

}

System.out.println(sum);

}

Explanation:

<u>Variables used: </u>

j : controls the first number in product and decreases by 2 each time the loop runs.

sum: saves the values of addition as the loop runs.

3 0

2 years ago

In the SDLC's third phase, the way in which a proposed information system will deliver the general abilities described in the pr

kifflom [539]

Answer: Detailed

Explanation:

SDLC consist of 7 phases. They are:

1. Planning.

2. System analysis and requirement

3. Design:

4. Coding:

5. Testing:

6. Installation

7. Maintenance

Here we are talking of the third phase. In this phase it is the design phase which consist of high and low level design. Here the preliminary design is included in high level design where it describes the required hardware, software, network capabilities and the modelling of the interface. However the detailed design in low level design will implement the coding and will finding of any errors in the implemented design as described by the preliminary design.

3 0

3 years ago