One method of reducing the length of stalls is through the use of caching. Memory access times to the cache are much faster than accesses to main memory, so every cache hit will be a much shorter stall than an access to main memory. For instance, the Performance Analysis Guide for Intel Xeon 5500 processor approximates the cost of an L1 cache hit to be ~4 cycles, an L3 cache hit to be ~100-300 cycles, and local DRAM to be ~60 ns.
Doesn't a cycle take ~ 1 ns to perform? These numbers indicate that the time taken to access L3 cache and DRAM are about the same. Is something off, or am I doing something wrong here?
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kayvonf
@smcqueen, your L3 hit time seems a bit off.
An L3 hit in a modern Core i7 is about 30 cycles (see the diagram from lecture 10).
@aakashr: Period is 1/frequency so a clock cycle for a 1 GHz processor is 1 ns. As most modern CPUs operate in the 2-3 GHz range, the period of a cycle is less, about 300 to 500 picoseconds.
One method of reducing the length of stalls is through the use of caching. Memory access times to the cache are much faster than accesses to main memory, so every cache hit will be a much shorter stall than an access to main memory. For instance, the Performance Analysis Guide for Intel Xeon 5500 processor approximates the cost of an L1 cache hit to be ~4 cycles, an L3 cache hit to be ~100-300 cycles, and local DRAM to be ~60 ns.
Source: Intel Perf Guide
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Doesn't a cycle take ~ 1 ns to perform? These numbers indicate that the time taken to access L3 cache and DRAM are about the same. Is something off, or am I doing something wrong here?
This comment was marked helpful 0 times.
@smcqueen, your L3 hit time seems a bit off.
An L3 hit in a modern Core i7 is about 30 cycles (see the diagram from lecture 10).
@aakashr: Period is 1/frequency so a clock cycle for a 1 GHz processor is 1 ns. As most modern CPUs operate in the 2-3 GHz range, the period of a cycle is less, about 300 to 500 picoseconds.
This comment was marked helpful 0 times.