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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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we have to divide it. for 32bit it should be divided by 4 and for 16bit it should be divided by 2.

then we get the ans 2000K words for 32bit processor and 4000K words for 16bit processor

Na bhai na, galat bol rhy ho ap, assignment me 32 bit nhi dia hoa, 32 byte dia hoa he, means 256 bits, aur isi trha 16 byte means 128 bits

yup.. u r right...

Is assignment me jo main problem he wo ye k MIPS kia hoon gy, aur unhy calculate kis trha krna he.

kuch samjh nhi aa rhi. ajeeb hi questions hain. koi solution nhi ban rha.

ek cheez clear hoti hai tu dusre me issue ho jata hai

here were are give GHZ instead of MIPS so we have to use 1.5 GHZ CPU instead of MIPS

Nhi brother me b 1.5 GHz use kr rha hoon, lkn ye sahi solution nhi he, 1.5 GHz se MIPS calculate krny k lye CPI need he jo k assignment me given nhi he, 1.5 GHz to tb use kr skty hen agr hr instruction 1 clock cycle me complete ho jay, lkn aysa nhi hota. Ksi b processor ki avg CPI 1 clock cycle se zeyada hi hoti he.

phir kiya karen..this is very confusing.. today is last day aur kuch ban bhi nhi rha T-T 

check this example

  • The floppy disk transfers data to the processor in 16-bit units and has a data rate of 50KB/s.
  • Polling rate = (50KB/s)/(2 Bytes/polling)
    = 25K polling/sec
  • Fraction of CPU time = 25Kx400/(500x10^6) = 2%

for hard disk

this is an example i got from somewhere

  • Transfer in 4-word blocks
  • transfer rate: 4MB/s
  • Polling rate = (4MB/s)/(4x4 Bytes/polling)
    = 250K polling/sec
  • Fraction of CPU time = 250Kx400/(500x10^6) = 20%

Ye example kaha se li he aapny?

internet se li

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