Question

Q-1Suppose that a certain program takes 300 seconds of elapsed time toexecute. Out of these 300

seconds, 280seconds is the CPU time andthe rest is I/O time. What will be the elapsed time?[5]

Q-2Consider an I/O bus that can transfer 6 bytes of data in one buscycle. Suppose that a designer is

planning to attach the following twocomponents to this bus.

Harddrive with a transfer rate of 100 M bytes/sec

Videocard with a transfer rate of 128 M bytes/sec

Whatwill be the maximum bandwidth from these two components? (Themaximum frequency of the bus

is 30MHz) [5]

Q-3What percentage of time will a 20 MIPS processor spend in the busywait loop of an 85-character line

printerwhen it takes 2 msec to print a character and a total of 765instructions need to be executed to

printan 85 character line? Assume that 4 instructions are executed inthe polling loop? [5]

Answer 1

We can use the following formula to compute theexecution time of this loop.

         ET = CPI × IC × T
         ET = (CPI × TC) / f ……..(1)

Where,
         CPI = Clocks per Instruction
         IC = Instruction Count
         T = Time period of the clock
         f = Frequency of the clock

Putting the values in (1),we get
         ET = ((5+6) × 65535) / (2 ×10⁶)
              = (11 ×65535) / (2 × 10⁶)
              = 720885 / (2 ×10⁶)
         ET = 0.36 sec