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Answer:
The answer is "0.306".
Explanation:
The Primary to Secondary Memory Instruction miss cycle
[tex]= I \times 0.01 \times 10 = 0.1 I[/tex]
Data miss cycles[tex]= I \times 0.05 \times 10 \times 0.40 = 0.2 I[/tex]
The Secondary to main Memory Instruction miss cycle[tex]= I \times 0.01 \times 200 \times 0.0002 = 0.004 I[/tex]
Data miss cycles[tex]= I \times 0.05 \times 200 \times 0.0002 = 0.002 I[/tex]
Total memory stall cycle [tex]= ( 0.1 + 0.2 + 0.004 + 0.002 ) I =0.306 I[/tex]