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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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solution example of first question

Example
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.

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Q1:

a. 61.44%

b. 122.88%

c. 0.0015%

anyone have the same answers????

plz share complete formulas with explanation example is given above :) thankx

Wrong solution on youtube

a= 10%

b=20%

c=0.001

question 2

2.4 %

no. of polling/sec * (no. of cycles needed/CPU speed)

then * 100 for percent your a b c question will solve with this

polling calculated 16MB/32B

1500 * 10 ^6 = 1.5 Ghz

and question two

rate interrupt = 12/100 =0.12 *no. of polling/sec * (no. of cycles needed/CPU speed) 

Ye to sahi nhi lg rha, ye to logical nhi bnta q k percent means 100 me se, to total agr 100 ho to us me se 122.88 jawab nhi aa skta. 

point h

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