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Assignment # 03

Advance Computer Architecture (CS 501)

Fall 2020 

Deadline Date Feb 8, 2021

Total marks = 20 

Please carefully read the following instructions before attempting the assignment.



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 the file is corrupt.
  • Strict action will be taken if the submitted solution is copied from any other student or 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 Scanned Images, PDF, Zip, Rar, Ppt and Bmp, etc will not be accepted.



The objective of this assignment is to increase the learning capabilities of the students about


·         CPU Polling

·         Direct Memory Access



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 assignments in the last hour of the deadline. It is recommended to upload the solution file at least two days before its closing date.


If you find any mistake or confusion in the 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:


Question No 01                                                                                     12 marks

Let us assume a 32-bit processor having an 800 MHz frequency to which the following I/O devices are connected:

  • A DVD writer that has a transfer rate of 24 MB/s.
  • A keyboard that can be polled 500 times per second.
  • A SATA hard drive which can transfer at 40 MB/s.

If CPU polling is used for selecting a device where each polling operation requires 10 instructions, what will be the CPU time required to poll each device?


Question No 02                                                                                    08 marks

Consider a hard drive that has a transfer rate of 20 MB/s. The hard disk is connected to a CPU having a frequency of 1 GHz through a 32-bit data bus. The CPU uses DMA interface for I/O transfers which has the following specifications:

  • CPU spends around 300 clock cycles to set up a DMA controller for every DMA request.
  • At the end of each DMA transfer, an interrupt is generated which takes additional 100 clock cycles.

If the data is transferred in a block and each block size is 16 KB, then calculate the percentage of time consumed for DMA transfers using the hard drive.


Note: Provide each step of calculation in both questions otherwise marks will be deducted.

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CS501 Assignment 3 Solution Fall 2020 - 2021

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CS501 Assignment 3 Solution Fall 2020 - 2021 || Advance Computer Architecture||

CS501 Assignment No 3 2021 Solution File| Advanced Computer Architecture

CS501 Assignment 3 Solution Fall 2020-2021


Question NO 1 Solution

The DVD writer can transfer 24 MB/s or 6000 K 32-bit words every second so this DVD should be polled using this rate.

Using 1K = 210

No of CPU instruction required = 6000 * 210 * 10

= 61440000 = 6.14 * 10' instructions per second CPU time required for polling is = (6.14 * 10)/(800 * 106)

= 0.07675

In Percentage = 7.675 %

  • For keyboard, No. of instructions required for pollingis

= 500 * 10

= 5000 instructions per second

CPU time required for polling is = 5000 / (800 * 106)

= 6.25 * 10−6


  • The Hard drive can transfer 40 MB/s or 10000 K 32-bit words everysecond

Using 1K = 210

No of CPU instruction required = 10000 * 210* 10

= 102400000

= 1.02 * 10^8 instructions per second CPU time required for polling is = (1.02 * 108/(8 * 108) = 0.1275 Percentage : 12.75 %


Question NO 2 Solution


Solution: Hard drive transfer rate 20MB/s = 20 * 1000 KB = 20000 KB Each block size = 16 KB


20000 / 16 = 1250 Block transfer per second

Every DMA transfer uses = 300 + 100 = 400 CPU Cycle This gives us: = 400 * 1250

F=500000 Cycles/s = 5 * 105

For 1 G CPU

= (5 * 105) / (1 * 10^9)

= 5 * 10−4

= 0.05 %




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