Homework05-sol - 2/21/03 10:04 PM Homework #05 (Solution)...

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2/21/03 10:04 PM ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2003 Page 1 of 2 Homework #05 (Solution) 1. (10 points) P&H Q6.1. The cycle time would be 10 ns in the single-cycle implementation. For the pipelined implementation it would be 4ns. The speedup obtained from pipelining the single-cycle implementation would only be 10/4 = 2.5. Grading: 3 points for each correct time, 4 for correct speedup. 2. (25 points) Using a drawing similar to Figure 6.8 on page 447, show the forwarding paths needed to execute the following instructions: lw $6,120($0) ____ ____ ____ ____ ____ add $1,$2,$3 ____ ____ ____ ____ ____ sub $5,$1,$6 ____ ____ ____ ____ ____ add $5,$5,$1 ____ ____ ____ ____ ____ There are three forwarding paths needed. First one happens between the second and the third instructions, because of dependency involving $1. Second forwarding path happens between the third and the fourth instructions, because of dependency involving $5. Finally, there must be another forwarding path between the first and the
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This note was uploaded on 04/25/2010 for the course ECSE ecse-2340 taught by Professor Wozny during the Spring '09 term at Rensselaer Polytechnic Institute.

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Homework05-sol - 2/21/03 10:04 PM Homework #05 (Solution)...

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