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3 years ago

Answer all of the questions,

Computers and Technology

1 answer:

Phantasy [73]3 years ago

5 0

Answer:

sry but I don't know the answer

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It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.

Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

For optimal 4-way merging, we need one dummy run with size 0.

Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows $L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300$
Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

$L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000$

The resulting run has length 6000 records.

Part b

For step 1

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{I/O per block} is 2 sec

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec$

So the input/output time is 46 seconds for step 01.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 2300 for step 01
Size_block is 100 as given
Time_{CPU per block} is 1 sec

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec$

So the CPU time is 23 seconds for step 01.

Total time in step 01

$T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\$

Total time in step 01 is 69 seconds.

For step 2

Input Output Time

Input Output Time is given as

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{I/O per block} is 2 sec

$T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\, block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec$

So the input/output time is 120 seconds for step 02.

CPU Time

CPU Time is given as

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}$

Here

L_run is 6000 for step 02
Size_block is 100 as given
Time_{CPU per block} is 1 sec

$T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\, block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec$

So the CPU time is 60 seconds for step 02.

Total time in step 02

$T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\$

Total time in step 02 is 180 seconds

Merging Time (Total)

Now the total time for merging is given as

$T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\$

Total time in merging is 249 seconds seconds

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3 years ago

The following method public String removeFromString(String old, String frag) removes all occurences of the string frag from the

Mashutka [201]

private static String removeFromString(String old, String frag)

{

int i = old.indexOf(frag);

while (i> -1) {

String rest = old.substring(i + frag.length());

System.out.println("rest = " + rest);

old = old.substring(0, i);

System.out.println("rest = " + old);

old = old + rest;

System.out.println("rest = " + old);

i = old.indexOf(frag);

System.out.println("i = "+ i);

}

return old;

}

Here the index of first occurrence is obtained outside the “while loop” and if this loop runs until index value is >-1. It extracts the rest of the characters other than “frag” from the index and all the characters for which the given set of characters are removed.

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3 years ago

how could environmental and energy problem Kenya faces as far computer installations are concerned be avoided

ipn [44]

The environmental and energy problem Kenya faces as far computer installations are concerned can be avoided by

Knowing and finding out the key aspect of ICT strength.
Create an holistic judgements in the assessment of ICT capability. Promote the use of solar power.

<h3>How can we overcome ICT challenges?</h3>

There are a lot of Energy Challenges in terms of ICT in Kenyan and it is good the government and the people work towards energy in terms of sustainable ICT.

Therefore, The environmental and energy problem Kenya faces as far computer installations are concerned can be avoided by

Knowing and finding out the key aspect of ICT strength.
Create an holistic judgements in the assessment of ICT capability. Promote the use of solar power.

Learn more about computer installations from

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consider the following code segment, which appears in a method in a class other than password. the code segment does not compile

Romashka [77]

The code segment does not compile because the method 'reset' does not return a value that can be displayed.

A method refers to a block of that which only runs when it is called. It is permissible to pass data, referred to as parameters, into a method. Methods are used to accomplish certain actions, and they are also called functions. As per the given code segment that appears as a method in a class, it does not compile because the method named 'reset' is not returning a value that can be printed on the screen as an output.

Complete question:

Consider the following code segment, which appears in a method in a class other than password.

Password p = new Password("password");

System.out.println("The new password is " + p.reset("password"));

The code segment does not compile because _____________ .

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An electronic cover letter should be

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