GATE CS 2014 Set 3 — Question 54
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Computer Organization & Architecture → Memory Hierarchy & Cache → Cache Numericals
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Question
The memory access time is 1 nanosecond for a read operation with a hit in cache, 5 nanoseconds for a read operation with a miss in cache, 2 nanoseconds for a write operation with a hit in cache and 10 nanoseconds for a write operation with a miss in cache. Execution of a sequence of instructions involves 100 instruction fetch operations, 60 memory operand read operations and 40 memory operand write operations. The cache hit-ratio is 0.9. The average memory access time (in nanoseconds) in executing the sequence of instructions is __________.
Correct answer
1.68 to 1.68
Solution
We calculate the total time taken for all memory accesses and divide by the total number of accesses.1. Instruction Fetches (Read):
- Count: 100
- Hit time: 1 ns, Miss time: 5 ns
- Hits: . Time: ns.
- Misses: . Time: ns.
- Total Fetch Time: ns.
- Count: 60
- Hit time: 1 ns, Miss time: 5 ns
- Hits: . Time: ns.
- Misses: . Time: ns.
- Total Read Time: ns.
- Count: 40
- Hit time: 2 ns, Miss time: 10 ns
- Hits: . Time: ns.
- Misses: . Time: ns.
- Total Write Time: ns.
- Total Time = ns.
- Total Accesses = .
- Average = ns.
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