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Answers: 2
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When making changes to optimize part of a processor, it is often the case that speeding up one type of instruction comes at the cost of slowing down something else. for example, if we put in a complicated fast floating-point unit, that takes space, and something might have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay to reach that unit. the basic amdahl's law equation does not take into account this trade-off. a. if the new fast floating-point unit speeds up floating-point operations by, on average, 2Δβ, and floating-point operations take 20% of the original program's execution time, what is the overall speedup (ignoring the penalty to any other instructions)? b. now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5Δβ slowdown (or 2/3 speedup). data cache accesses consume 10% of the execution time. what is the overall speedup now? c. after implementing the new floating-point operations, what percentage of execution time is spent on floating-point operations? what percentage is spent on data cache accesses?
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Computers and Technology, 24.06.2019 12:50
Write a new lc-3 trap subroutine (i.e. a subroutine that will be invoked via the trap instruction) that will receive a numeric digit entered at the keyboard (i.e. an ascii character), echo it to the screen, and return in r0 the corresponding numeric value: so if the user types the digit '7', the character '7' will appear on the screen, but the value returned in r0 will be b0000 0000 0000 0111 (#7) you may not use any trap calls in your code - you must implement the "polling" code that interrogates the keyboard status and data registers. ; getnum_tsr ; a subroutine for obtaining a numeric value ; given ascii numeric digit input to keyboard. ; the numeric digit is echoed to the console (e.g. '7' = b0000 0000 0011 0111), ; but the value returned in r0 is the actual numeric value ; corresponding to the digit (e.g. b0000 0000 0000 0111 =
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