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Q 1. Suppose we have a 1.5-GHz CPU to which the following three I/O devices are connected:

 

  1. Flash drive that can transfer data in 32-byte chunks with a maximum transfer rate of 16 MB/sec
  2. DVD drive which can transfer data in 16-byte chunks with a maximum transfer rate of 16 MB/sec
  3. Joystick that needs to be polled 50 times per second

 

Polling requires 300 instructions for each I/O device. Students are required to compute the percentage of CPU time required to poll each device.

 

 

Q 2. Instead of polling, we want to use interrupts for handling the DVD drive. Keeping in view the DVD drive to be active only 12% of the time, you are required to compute the percentage of CPU time for handling it.

 

Note: Interrupt and polling requires the same amount of instructions.

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plz share correct sol.........

1500 KIDER LAGAna hh...1st optn m ya 2nd m???

plzz tell me..

Idea for question 1

 

Assume that three I/O devices are connected to a 32-bit, 10 MIPS CPU. The first device is a hard drive with a maximum transfer rate of 1MB/sec. It has a 32-bit bus. The second device is a floppy drive with a transfer rate of 25KB/sec over a 16-bit bus, and the third device is a keyboard that must be polled thirty times per second. Assuming that the polling operation requires 20 instructions for each I/O device, determine the percentage of CPU time required to poll each device.

 

Solution:
The hard drive can transfer 1MB/sec or 250 K 32-bit words every second. Thus, this hard drive should be polled using at least this rate.

 

Using 1K=210, the number of CPU instructions required would be

 

250 x 210 x 20 = 5120000 instructions per second.

 

Percentage of CPU time required for polling is
(5.12 x 106)/ (10 x106) = 51.2%

 

The floppy disk can transfer 25K/2= 12.5 x 210 half-words per second. It should be polled with at least this rate. The number of CPU instructions

 

required will be 12.5 x 210 x 20 = 256,000 instructions per second.

 

Therefore, the percentage of CPU time required for polling is

 

(0.256 x 106)/ (10 x 106) = 2.56%.

 

For the keyboard, the number of instructions required for polling is

 

30 x 20 = 600 instructions per second.

 

Therefore, the percentage of CPU time spent in polling is

 

600 / (10 x 106) = 0.006%

 

It is clear from this example that while it is acceptable to use polling for a keyboard or a floppy drive, it is very risky to use polling for the hard drive. In general, for devices with a high data rate, the use of polling is not adequate.

CS501 Assignment 3 solution

CS501 Assignment 3 solution.docx

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