Consider a 1GHz processor, and a hard disk that can transfer data at 16 MB/sec.

Question

Consider a 1GHz processor, and a hard disk that can transfer data at 16 MB/sec. Take 1K = 1000, 1M = 10^6, and 1G = 10^9. Compute the fraction of CPU time consumed by disk interrupts for the following cases: (a) Case 1: Data is transferred in 8-byte chunks using interrupt-driven I/O. Assume the overhead of each interrupt is 500 clock cycles. (b) Case 2: Data is DMA transferred in 4 KB blocks. Assume a DMA setup and completion consumes 1000 cycles. (ii) Consider a 1GHz processor where an instruction that accesses memory and hits the cache takes 1 clock cycle to execute. If an instruction misses the cache, the miss penalty (including detecting the cache miss) is 100 clock cycles. Assuming the miss ratio is 3%, what is the average memory access time?

Explanation / Answer

5.

     (i) Given 1 GHz processor hence

          Clock period is = 1 / 109 = 10-9 s = 1 ns = Cycle Time

           (a)

                 We are transfering data in 8 byte chunks and given data transfer rate is 16MB /s

                 hence Data transfer time is = 8 B / 16 MBps

                                                                   = 1 / 2 * 106

                                                                   = 0.5 * 10-6

                                                                   = 500 ns

                      Hence data transfer time is = 500 ns

             Given overhead of each interrupt is = 500 cycles = 500 * 1ns = 500 ns

                      Hence interrupt overhead time is = 500 ns

            Fraction of CPU time consumed is

                                                 = interrupt overhead time / (interrupt overhead time +data transfer time)

                                                 = 500 / (500+500)

                                                 = 500 / 1000

                                                 = 1/2 (50 %)

          Hence 50 % of CPU time will be consumed.

      

       (b)

      We are transfering data in 4 KB blocks using DMA and given data transfer rate is 16MB /s

                 hence Data transfer time is = 4 KB / 16 MBps

                                                                   = 1 / 4 * 103

                                                                   = 0.25 * 10-3

                                                                   = 250 microsecs

                      Hence data transfer time is = 250 microsecs

             Given overhead of DMA is = 1000 cycles = 1000 * 1ns = 1 microsec

                      Hence DMA overhead time is = 1 microsec

            Fraction of CPU time consumed is

                                                 = DMA overhead time / (DMA overhead time +data transfer time)

                                                 = 1 / (1+250)

                                                 = 1 / 251

                                                 = 0.39

          Hence 0.39 % of CPU time will be consumed.

      

  

(ii)

        Given 1 GHz processor and hence the clock period is = 1 ns

         Hence tge clock cycle time is = 1 ns

         Given Miss ratio is = 3 % = 0.03 implies Hit ratio is = 97 % = 0.97

        Average Memory access time is = Hit ratio * 1 clock cycle + Miss ratio * 100 clock cycles

                                                                   = 0.97 * 1 clock cycle + 0.03 * 100 clock cycles

                                                                   = 0.97 * 1 ns + 0.03 * 100 ns

   = 0.97 + 3

   = 3.97 ns

   Hence Average Memory access time is = 3.97 ns

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