Plz Discuss this assignment
Assignment No. 05
CS501: Advanced Computer Architecture
Total Marks: 20
Due Date: 03-07-13
Please read the following instructions carefully before assignment submission.
It should be clear that your assignment will not get any credit if:
The objective of this assignment is:
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.
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)
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
easy way thx fr telng B sis
dear sister Q no. 1 last column
(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.
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]+......
here common ratio =1.5
as performance improves so it wud decrease that is
90/1.5= 60 and so on so forth....
Elapsed CPU time
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.
The following code returns the CPU time used to run surf(peaks(40)).
t = cputime; surf(peaks(40)); e = cputime-t
DMA: A mechanism that provides a device controller with the ability to transfer data directly to or from the memory without involving the processor.
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
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