CS501 Assignment 3 solution 2020 By Abid Farooq Bhutta || advance computer architecture spring 2020
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Advance Computer Architecture(CS501)
Assignment # 03
Spring 2020
Total marks = 20
Deadline Date
30th of July 2020
Please carefully read the following instructions before attempting assignment.
RULES FOR MARKING
It should be clear that your assignment would not get any credit if:
The assignment is submitted after the due date.
The submitted assignment does not open or file is corrupt.
Strict action will be taken if submitted solution is copied from any other student or from the internet.
You should consult the recommended books to clarify your concepts as handouts are not sufficient. You are supposed to submit your assignment in .doc or docx format.
Any other formats like scan images, PDF, zip, rar, ppt and bmp etc will not be accepted.
OBJECTIVE
Objective of this assignment is to increase the learning capabilities of the students about
• Direct Memory Access
• Dynamic Properties of Disk Drive
NOTE
No assignment will be accepted after the due date via email in any case (whether it is the case of load shedding or internet malfunctioning etc.). Hence refrain from uploading assignment in the last hour of deadline. It is recommended to upload solution file at least two days before its closing date.
If you find any mistake or confusion in assignment (Question statement), please consult with your instructor before the deadline. After the deadline no queries will be entertained in this regard.
For any query, feel free to email at: [email protected]
Consider a disk drive having the rotational speed of 5400 rpm. The disk has an average seek time of 4ms and the controller overhead is 0.5ms. The bandwidth of internal outer track of disk is 48MBps. How much time will be required to read a file of 2 MB from the disk? Assume that the file contents are stored in contiguous blocks and disk buffer is larger than file size.
Answer:
Average seek time = Tseek = 4 ms
Average rotational = Trot. = (0.5 * 60) / 5,400
= 0.0056 *1000 s
= 5.6 ms
Ttransfer = 2MB/48MB
= (1/24)
= 0.041666 *1000 s
= 41.7ms
Tcontroller = 0.5ms
The total time required= Tseek +Trot+ Ttsfr +Tctr
= 4 + 5.6 + 41.7 + 0.5
= 51.8 ms
Consider a 32-bit processor which has a frequency of 1 GHz. The processor has a 32-bit data bus with a speed of 30 MHz. An I/O device is connected to this bus which has a transfer at rate of 20 MB/s. If DMA technique is used, then the data is transferred in chunks of 16 KB. For each DMA request to complete, a total of 5000 clock cycles are required. Calculate the percentage of CPU time spent to handle data transfer with and without DMA.
Answer:
With DMA
Time required in handling each DMA request is 5000 cycles.
Since 1280 DMA requests are issued (20MB/16/1024=1280) the total time taken is (1280*5000) 6,400,000 cycles. As the CPU clock is 1024MHZ, the fraction of CPU time spent is 6,400,000/(1024x106) or 0.625%
Without DMA
The processor here copies the data into memory as it is sent over the bus.
I/O device sends data at a rate of 20MB/s over the 30MB/s bus,
= (20/30)*100 = 66.66 % of each second is spent transferring data. 66.66% of the CPU time is spent copying data to memory.