7. [15 marks] Imagine a computer with no cache, but with a reasonable-size regis

Question

 7. [15 marks] Imagine a computer with no cache, but with a reasonable-size register file.  The computer has a single floating-point multiplier.  The effect of these assumptions is that each floating-point multiply (operation) will, with probability 1, find one of its two operands in the register file, but will need its other operand delivered from memory, and this for _each_ floating-point multiply.  Let the floating-point multiplier have a peak performance of 16 GFs/s. At present, the achievable bandwidth from the memory to the processor is 7 GWs/s.  (Here, 'W' stands for _word_, not Watt, and, by assumption, one word can hold one floating-point value).  K = 10^3  a) [5 marks] Describe this situation as _compute bound_ or _bandwidth bound_.                                                  situation = ______________  b) [5 marks] We buy a second DRAM module and more interconnection links with the same aggregate capacities as the first.  Describe this new situation as _compute bound_ or _bandwidth bound_.                                                  situation = ______________  c) What would it take to achieve sustained performance equal to the peak performance?                                                      answer = ______________ 

Explanation / Answer

Integral Part = 18D 18/2 => quotient=9 remainder=0 9/2 => quotient=4 remainder=1 4/2 => quotient=2 remainder=0 2/2 => quotient=1 remainder=0 1/2 => quotient=0 remainder=1 (quotient=0 stop) Hence, 18D = 10010B Fractional Part = .6875D .6875*2=1.375 => whole number is 1 .375*2=0.75 => whole number is 0 .75*2=1.5 => whole number is 1 .5*2=1.0 => whole number is 1 Hence .6875D = .1011B Therefore, 18.6875D = 10010.1011B

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