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Give the formula for the average access time experienced by the processor in a system with 2 levels of caches.
Ans: Formula is for the average access time experienced by the processor in a
system having two levels of caches is described follow:
tave = h1C1+(1-h1)h2C2+(1-h1)(1-h2)M
h1=> hit rate in the L1 cache
h2= > hit rate in the L2 cache
M=> it's time to access information in the main memory
C1=> it's for time to access information in the L1 cache
C2=> it's for time to access information in the L1 cache
Explain Hit and Miss?
Ans. The performance of cache memory is often measured in terms of a quantity is said hit ratio. When the CPU refers to memory and finds the word in cache, it is said hit. If the word is not found in cache, then it is in primary memory and it counts as a miss.
Simplify the given expressions using Boolean postulates Y = (A + B)(A‾ + C)(B + C) Ans. Y = (A + B)(A‾ + C)(B + C) = (A A‾ + AC + B A‾ + BC) (B + C) = (AC + B A‾ + BC) (B + C)
Reg data type as Combinational element module reg_combo_example( a, b, y); input a, b; output y; reg y; wire a, b; always @ ( a or b) begin y = a & b; e
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start 100 A Ds 3
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