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Let average clocks for non-pipelined instruction be x

Let each clock takes 1ns

Time to execute one non-pipelined instruction is (3+8+5+6+4) = 26 clocks

and it is given that each instruction takes x clocks, so Time to execute one non-pipelined instruction is 26*x

Time to execute pipelined instruction is 1 clock = 8ns

speedup = Tnon-pipelined / Tpipelined

4 = 26*x/8  => x = 1.23

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