We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>
Q 1. Suppose we have a 1.5-GHz CPU to which the following three I/O devices are connected:
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.
Share This With Friends......
solution example of first question
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.
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.
Please Discuss here about this assignment.Thanks
Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions. Read More>>
Discussed & be touched with this discussion. After discussion a perfect solution will come in a result at the end.
For Important Helping Material related to this subject (Solved MCQs, Short Notes, Solved past Papers, E-Books, FAQ,Short Questions Answers & more). You must view all the featured Discussion in this subject group.
For how you can view all the Featured discussions click on the Back to Subject Name Discussions link below the title of this Discussion & then under featured Discussion corner click on the view all link.
Or visit this link
Please Click on the below link to see…
P.S: Please always try to add the discussion in proper format title like “CS101 Assignment / GDB No 01 Solution & Discussion Due Date: ___________”
anyone have the same answers????
plz share complete formulas with explanation example is given above :) thankx
Wrong solution on youtube
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.