Ececs 752 spring 2008 midterm 2 page 11 a 5 pts for

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ECE/CS 752 Spring 2008 Midterm 2 -- Page 11 a. [5 pts] For steady-state execution, determine the fraction of instructions that have to be squashed and replayed under non-selective replay (Sec. 3.3 in the paper). jmp addi addi addi fmul sta bge load These ops are in flight when the load misses. All eight are squashed and replay. In steady state, this is 8/9 instructions per loop, with one miss per 16 loops. This works out to 8/9 x 1/16 = 1/18, or 5.5% b. [5 pts] For steady-state execution, determine the fraction of instructions that have to be squashed and replayed under position-based selective replay (Sec. 3.4.3) Only the fmul must be squashed, so 1/9 per loop, with one miss per 16 loops, which works out to 1/9 x 1/16 = 1/144 = 0.69%
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ECE/CS 752 Spring 2008 Midterm 2 -- Page 12 c. [5 pts] For steady-state execution, determine the fraction of instructions that have to be squashed and replayed under ID-based selective replay (Sec. 3.4.1). Only the fmul must be squashed, so 1/9 per loop, with one miss per 16 loops, which works out to 1/9 x 1/16 = 1/144 = 0.69% d. [3 pts] ID-based selective replay is difficult to scale to large instruction windows, since each in-flight load needs its own “name”. Describe how the token-based technique in Sec. 4 addresses this problem. Only loads that are likely to miss get assigned a token. Since most misses are caused by a small fraction of loads (in typical programs), we need fewer tokens in flight. e. [2 pts] Do you believe the token-based technique (Sec. 4) will be effective for the loop studied in this problem? Why or why not? No, since all misses are caused by the same static load, there will be no reduction in the number of tokens needed.
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