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Assignment No. 05
Semester: Spring 2013

CS501: Advanced Computer Architecture

 

 

Total Marks: 20

Due Date:  03-07-13

 

Instructions

Please read the following instructions carefully before assignment submission.

It should be clear that your assignment will not get any credit if:

  • The assignment is submitted after the due date.
  • The submitted assignment does not open or file is corrupt.
  • The assignment is found to be copied from the internet.
  • The assignment is found to be copied from other student.
  • The assignment submitted is not according to required file format (.doc).


Objective

The objective of this assignment is:

  • To assess your overall understanding of Computer Architecture and Organization
  • To assess your overall understanding of Computer processing
  • To assess your overall understanding of DMA, Polling and interrupts

Note:

  • The assignment should be in .doc format.
  • Assignment .05 covers lecture 25-31. You can also consult reference books for help.
  • Students are advised to submit their assignment as early as possible in order to avoid any sort of inconvenience like Load shedding etc.

 

 

 

 


Question No 1:
                                                                                                     (10 marks)

Suppose we have a benchmark that executes in 100 seconds of elapsed time, where 90 seconds is CPU time and the rest is I/O (Input/output) time. If CPU time improves by 50% per year for the next five years but I/O time doesn’t improve, how much faster will our program run at the end of five years?

You are required to calculate the CPU improved performance and improved elapsed time.    

After n years

CPU/ time

I/O time

Elapsed time

% I/O time

0 (Current Year)

 

 

 

 

1

 

 

 

 

2

 

 

 

 

3

 

 

 

 

4

 

 

 

 

5

 

 

 

 


NOTE: Theoretical answer will not be considered

______________________________________________________________________

Question No 2:                                                                                                      (4 marks)

Consider a 20 MIPS (Microprocessor without Interlocked Pipeline Stages) processor with several input devices attached to it, each running at 1000 characters per second. Assume that it takes 17 instructions to handle an interrupt. If the hardware interrupt response takes 1msec, what is the maximum number of devices that can be handled simultaneously?

NOTE: Theoretical answer will not be considered

______________________________________________________________________

Question No 3:                                                                                                        (6 Marks)

Scenarios discussion:

If we want the lowest latency for an I/O operation to a single I/O device; while in terms of lowest impact on processor utilization from a single I/O device then what will be the orders/arrangements of Interrupt driven, DMA(Direct Memory Access) and polling in both scenarios? Explain reasons.

NOTE: Give answer within 3-5 lines. Otherwise answer will not be considered.

______________________________________________________________________

GOOD LUCK J

 

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Replies to This Discussion

easy way thx fr telng B sis

dear sister Q no. 1 last column

I/O%???

(I/O Time X  100) / Elapsed Time

For Q.3 see this

An I/O scheme that employs interrupts to indicate to the processor that an I/O device needs attention. DMA is a mechanism that provides a device controller with the ability to transfer data directly to or from the memory without involving the processor. Polling is the process of periodically checking the status of an I/O device to determine the need to service the device. DMA Works in the background without CPU intervention while Interrupt works by asking for the use of the CPU by sending. The DMA is used for moving large files since it would take too much of CPU capacity while interrupts take up time of the CPU. Polling is best then DMA and interrupts because polling is a low level process since the peripheral device is not in need of a quick response. The major advantage is that the polling can be adjusted to the needs of the device.

Q.3 

3 to 5 lines main DMA, intrrupt i/o our polling ko compair kro .

our btao k DMA kaisy better hy interrupt i/o sy our interrupt i/o  kaisy better hy polling sy .

Q1 is the geometric series like problem n to calculate it we form a series n find the common ratio..

It is CPU time = 90 as it improves by 50% = 50/100=5/10 per year

then 90+ [90+90(5/10)]+[90+90(5/10)^2]+........

     = 90+[90(1+5/10)] + [90(1+(5/10)^2]+......

     = 90+90(1.5)+90[1.5]^2+.....

here common ratio =1.5

as performance improves so it wud decrease that is

90/1.5= 60 and so on so forth....

90/1.5

Elapsed CPU time

Syntax
cputime

Description

cputime returns the total CPU time (in seconds) used by your MATLAB® application from the time it was started. This number can overflow the internal representation and wrap around.

Examples

The following code returns the CPU time used to run surf(peaks(40)).

t = cputime; surf(peaks(40)); e = cputime-t

e =
0.4667

Interrupt Driven: An I/O scheme that employs interrupts to indicate to the processor that an I/O device needs attention.

DMA: A mechanism that provides a device controller with the ability to transfer data directly to or from the memory without involving the processor.

Polling: The process of periodically checking the status of an I/O device to determine the need to service the device.

10% 

14.29% 

20% 

27.03% 

35.71% 

45.45% 

dear fellows tell me all of u got these percentages

and Q # 2  me jo 1msec hai us ko solve kaise krna hai  plzzzzzzzzzz tell me 

yar someone plz upload complete solution there is no time left plz plz

Suppose we have a benchmark that executes in 100 seconds of elapsed time, where 90
seconds is CPU time and the rest is I/O time. If the CPU time improves by 50% per year
for the next five years but I/O time does not improve, how much faster will our program
run at the end of the five years?
Impact of I/O on System
Performance
Over five years:
After n years CPU/ time I/O time Elapsed time % I/O time
0 (Current Year) 90 Sec 10 Sec 100 Sec 10%
1 90/1.5=60 Sec 10 Sec 70 Sec 14%
2 60/1.5=40 Sec 10 Sec 50 Sec 20%
3 40/1.5=27 Sec 10 Sec 37 Sec 27%
4 27/1.5=18 Sec 10 Sec 28 Sec 36%
5 18/1.5=12 Sec 10 Sec 22 Sec 45%

CPU improvement = 90/12 = 7.5 BUT System improvement = 100/22 = 4.5

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