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Assume that we are using the classic MIPS five-stage(IF, ID, EX, MEM and WB) integer pipeling.

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Answer is 12.

There are 2 load hazards which stalls the pipeline for 1 cycle each. There are 6 instructions hence total number of clock cycles to finish = 6 + 5 – 1 = 10.

Load hazards are

instruction pair (2, 3) and (4, 5)

Adding 2 stalls cycles we get 10 + 2 = 12.

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