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Artyom0805 [142]
3 years ago
13

A) Suppose a computer has an instruction pipeline with 4 phases. How many cycles (if there are no delays) would it take to compl

ete 80 instructions. How many cycles would it take if the processor were not pipelined?
b) Explain the problem caused by a conditional branch instruction in a pipeline.
c) Give 2 ways of dealing with the problem of conditional branching.
Computers and Technology
1 answer:
Reika [66]3 years ago
3 0

Answer:

Explanation:

(a) Given data:

Instruction pipeline with 4 phases.

80 instruction

The number of cycles if the processor were pipelined :

80-1 = (79) instruction takes only 1 extra cycle to complete.

So, cycles counts = 4 + (79) = 83 cycles.

The number of cycles if the processor were not pipelined.  

Each instruction takes 4 cycles to complete.  

So, cycle counts = 80 * 4 = 320 cycles.

(b) One of the main problems of conditional branching in pipelines is the delay it gives even after the branch target buffer.

In pipeline, conditional branches depend on the codes which are still in estimation.

(c) Two ways to deal with problem arising due to conditional branching are as follows :

1. Delayed Branching    

2. Multiple Condition Codes

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Read 2 more answers
(Solved—15 Lines) In many jurisdictions a small deposit is added to drink containers to encourage people to recycle them. In one
andrezito [222]

Answer:

JAVA program is given below.

import java.util.Scanner;

public class Main

{

   //static variables declared and initialized as required

   static float deposit1 = 0.10f;

   static float deposit2 = 0.25f;

   static float refund1;

   static float refund2;

   static int num_one_lit;

   static int num_more;

public static void main(String[] args){

    Scanner sc = new Scanner(System.in);

 System.out.print("Enter the number of containers of 1 liter or less: ");

 num_one_lit = sc.nextInt();

 System.out.print("Enter the number of containers of more than 1 liter: ");

 num_more = sc.nextInt();

 //refund computed based on user input

 refund1 = num_one_lit*deposit1;

 refund2 = num_more*deposit2;

 //displaying refund

 System.out.printf("Total refund amount for containers of 1 liter or less is $%.2f %n", refund1);

 System.out.printf("Total refund amount for containers of more than 1 liter is $%.2f %n", refund2);

}

}

OUTPUT

Enter the number of containers of 1 liter or less: 5

Enter the number of containers of more than 1 liter: 2

Total refund amount for containers of 1 liter or less is $0.50  

Total refund amount for containers of more than 1 liter is $0.50

Explanation:

1. Variables are declared at class level hence declared as static.

2. Variables holding deposit also initialized with the given values.

3. User input is taken for the number of containers in both categories.

4. The total value of refund is computed using the given values and user entered values.

5. The total refund is displayed to the user for both categories.

6. To display two decimal places always, variables to hold deposit and refund are declared as float.

7. The keywords used for decimals are %.2f %n. The %.2f keyword indicates two decimal places to be displayed in float variable, %n indicates a numeric value is to be displayed.

8. The printf() method is used to display the formatted output for refund amount.

9. The keywords used to display the formatted output is derived from C language.

10. All the code is written inside class and execution begins from the main() method.

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