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Dmitry [639]
3 years ago
13

Suppose that a client performs an intermixed sequence of (queue) enqueue and dequeue operations. The enqueue operations put the

integers 0 through 9 in order onto the queue; the dequeue operations print out the return value. Which of the following sequence(s) could not occur? a. 0 1 2 3 4 5 6 7 8 9 b. 4 6 8 7 5 3 2 9 0 1c. 2 5 6 7 4 8 9 3 1 0
d. 4 3 2 1 0 5 6 7 8 9
Computers and Technology
1 answer:
earnstyle [38]3 years ago
8 0

Answer:

b, c and d can't occur.

Explanation:

Order is preserved in queue. Option A is order preserving so only it will occur.

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Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress 50%, or 2 sec
kozerog [31]

Answer: bandwidth = 0.10 MB/s

Explanation:

Given

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.50 MB / Bandwidth)

Transfer Size = 0.50 MB

Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)

Total Time = 1 s + RTT + (0.50 MB / Bandwidth)

Compression Time = 1 sec

Situation B:

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.40 MB / Bandwidth)

Transfer Size = 0.40 MB

Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)

Total Time = 2 s + RTT + (0.40 MB / Bandwidth)

Compression Time = 2 sec

Setting the total times equal:

1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)

As the equation is simplified, the RTT term drops out(which will be discussed later):

1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)

Like terms are collected:

(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s

0.10 MB / Bandwidth = 1 s

Algebra is applied:

0.10 MB / 1 s = Bandwidth

Simplify:

0.10 MB/s = Bandwidth

The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.

(2) . Assume the RTT for the network connection is 200 ms.

For situtation 1:  

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB

Total Time = 1.2 sec + 5 sec

Total Time = 6.2 sec

For situation 2:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB

Total Time = 2.2 sec + 4 sec

Total Time = 6.2 sec

Thus, latency is not a factor.

5 0
3 years ago
. If you have written the following source code:
taurus [48]

Answer:

SpringBreak.java

Explanation:

Java classes are saved in files having the same name as the class name.

So if the given class structure is:

public class SpringBreak{

// lots of code here

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It needs to be saved in a file called SpringBreak.java.

The physical file should follow the package structure as provided in the class.

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True, this is a true statement. 
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Write a program that asks the user to enter a series of single digit numbers with nothing separating them. Read the input as a C
vladimir2022 [97]

Answer:

#include <stdio.h>

#include <string.h>

int main(){

   char number[100];

   printf("Number: ");

   scanf("%s", number);

   int sum = 0;

   for(int i =0;i<strlen(number);i++){

       sum+= number[i] - '0';

   }

           printf("Sum: %d",sum);

   return 0;

}

Explanation:

This declares a c string of 100 characters

   char number[100];

This prompts user for input

   printf("Number: ");

This gets user input

   scanf("%s", number);

This initializes sum to 0

   int sum = 0;

This iterates through the input string

   for(int i =0;i<strlen(number);i++){

This adds individual digits

       sum+= number[i] - '0';

   }

This prints the calculated sum

           printf("Sum: %d",sum);

   return 0;

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