What are the two biggest strength as a student

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

So

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

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

So

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

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

So

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

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

So

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

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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1.) Business Engineering or Manufacturing :)

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Which of the following are peripheral devices?

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Explanation:

I think it's speakers,

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Write a public member function which replace that replaces one occurrence of a given item in the ArrayBag with another passed as

ira [324]

Answer:

The method is written in Java

public static boolean abc(int [] ArrayBag,int num, int replace){

Boolean done = false;

for(int i =0; i<ArrayBag.length;i++){

if(ArrayBag[i] == num){

ArrayBag[i] = replace;

done = true;

}

for(int i = 0;i<ArrayBag.length;i++){

System.out.println(ArrayBag[i]+" ");

}

return done;

}

Explanation:

The method is written in java

The arguments of the method are:

1. ArrayBag -> The array declares as integer

2. num -> The array element to check the presence

3. replace - > The replacement variable

This line defines the method as boolean.

public static boolean abc(int [] ArrayBag,int num, int replace){

This line declares a boolean variable as false

Boolean done = false;

The following iterates through the elements of the array

for(int i =0; i<ArrayBag.length;i++){

This checks if the array element exist

if(ArrayBag[i] == num){

If yes, the array element is replaced

ArrayBag[i] = replace;

The boolean variable is updated from false to true

done = true;

}

The following iteration prints the elements of the array

for(int i = 0;i<ArrayBag.length;i++){

System.out.println(ArrayBag[i]+" ");

}

This prints returns the boolean thats indicates if replacement was done or not.

return done;

}

Attached to this solution is the program source file that includes the main method of the program

Download txt

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Two electronics technicians are discussing electrical quantities. Technician A says that resistance is an opposition to electric

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Neither are correct.

Resistance is opposition to CURRENT not power.
Technition B Is wrong about the voltage thingy.

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