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GATE | GATE-CS-2007 | Question 80

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  • Difficulty Level : Hard
  • Last Updated : 19 Nov, 2018
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Consider a machine with a byte addressable main memory of 216 bytes. Assume that a direct mapped data cache consisting of 32 lines of 64 bytes each is used in the system. A 50 × 50 two-dimensional array of bytes is stored in the main memory starting from memory location 1100H. Assume that the data cache is initially empty. The complete array is accessed twice. Assume that the contents of the data cache do not change in between the two accesses.
How many data cache misses will occur in total?
(A) 40
(B) 50
(C) 56
(D) 59

Answer: (C)

Explanation: Size of main memory=216 bytes

Size of cache=32*64 Bytes

=2 11 Bytes

Size of array=2500 Bytes

Array is stored in main memory but cache will be empty

Size of cache=2048 Bytes

So number of page faults=2500-2048=452

Complete array will be access twice

So for second access no. of total page faults=452*2=904

So total page faults=452+904=1356

So data cache misses will be 56

So (C) is correct option

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