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1-
Single precision :
1 10000100 00101010000000000000000
Double precision :
1 1000000010 0010101000000000000000000000000000000000000000000000
2- Peak performances of P1 is only class F and peak performance of P2 is only C class because they executes fastest set of instructions.
Peak of P1= (1* 10^9 cycles /1 second) * (1 instructions/1 cycle) = 1*10^9 instruction/second
Peak of P2=(3*10^9 cycles / 1 second) * (1 instructions/ 3 cycle) = 1*10^9 instruction/second
3- Let’s look at execution time for each computer
Computer A has totally 760 seconds.
Computer B has totally 890 seconds.
Computer C has totally 550 seconds.
Computer D has totally 480 seconds.
Therefore fastest computer is Computer D.
Computer D is,
760 /480 = 1.58 times as fast as A
890 /480 = 1.85 times as fast as B
550 /480 = 1.14 times as fast as C
4-
a- Time = 10^9 instruction * 1 * 10^(-12) seconds = 1000 microseconds
b- It is 20 times faster than non-pipelined processor. Moreover, one instruction indicates every 5 picoseconds.
c- The answer is both because instructions will take more than than 1 picoseconds to reach the end of the pipeline. Moreover it also be more than 5 picoseconds between following one after another instructions completing.
5- a- The second instruction is dependent upon the first ($3) . The third instruction is dependent upon the second ($4). The fifth instruction is dependent upon the fourth ($6). The sixth instruction is dependent upon the fifth ($7). The seventh instruction is dependent upon the sixth ($6).
b- Dependencies between the first instruction ($3) and second instruction, dependencies between the fourth instruction ($6) and fifth instruction, dependencies between the fifth instruction ($7) and sixth instruction can be resolved via forwarding.
c-Dependencies between the second instruction ($4) and third instruction, dependencies between the sixth instruction ($6) and seventh instruction cannot be resolved via forwarding, so it will cause a stall.
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