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Chapter: Embedded Systems Design : Memory and performance trade-offs

Single cycle processor with large external memory

In this example, there is a single cycle processor that has to process a large external table by reading a value, processing it and writing it back. While the processing level is small — essentially a data fetch, three processing instructions and a data store — the resulting execution time can be almost three times longer than expected. The reason is stalling due to data latency. The first figure shows how the problem can arise.

Scenario 1 — Single cycle processor with large external memory

 

In this example, there is a single cycle processor that has to process a large external table by reading a value, processing it and writing it back. While the processing level is small — essentially a data fetch, three processing instructions and a data store — the resulting execution time can be almost three times longer than expected. The reason is stalling due to data latency. The first figure shows how the problem can arise.


The instruction sequence consists of a data fetch followed by three instructions that process the information before storing the result. The data access goes to external memory where there are access delays and therefore the first processing instruction must wait until the data is available. In this case, the instruction execution is stalled and thus the fetch instruction takes the equiva lent of five cycles instead of the expected single cycle. Once the data has been received, it can be processed and the processing instructions will continue, one per clock. The final instruction stores the end result and again this is where further delays can be experienced. The store instruction experiences the same external memory delays as the fetch instruction. However, its data is not required by the next set of instructions and therefore the rest of the instruction stream should be able to continue. This is not the case. The next instruction is a fetch and has to compete with the external bus interface which is still in use by the preceding store. As a result, it must also wait until the transaction is completed.

 

The next processing instruction now cannot start until the second fetch instruction is completed. These delays mean that the total time taken at 100 MHz for the five instructions (1 fetch + 3 processing + 1 store) is not 50 ns but 130 ns — an increase of 2.6 times.

 

The solution to this involves reordering the code so that the delays are minimised by overlapping operations. This assumes that the processor can do this, i.e. the instructions are stored in a separate memory space that can be accessed simultaneously with data. If not, then this conflict can create further delays and proces-sor stalls. The basic technique involves moving the processing segment of the code away from the data access so that the delays do not cause processing stalls because the data dependencies have been removed. In other words, the data is already available before the processing instructions need it.

 

Moving dependencies can be achieved by restructuring the code so that the data fetch preceding the processing fetches the data for the next processing sequence and not the one that imme-diately follows it.


The diagram above shows the general approach. The fetch instruction is followed by the processing and storage instruction for the preceding fetch. This involves using an extra register or other local storage to hold the sample until it is needed but it removes the data dependency. The processing instructions P0 onward that follow the fetch instruction F1 do not have any data dependency and thus can carry on processing. The storage in-struction S0 has to wait one cycle until F0 has completed and similarly the fetch instruction F2 must wait until S0 has finished. These delays are still there because of the common resource that the store and fetch instructions use, i.e. the external memory interface. By reordering in this way, the five instruction sequence is completed twice in every 20 clocks giving a 100 ns timing which is a significant improvement.

 

This example also shows that the task in this case is I/O bound in that the main delays are caused by waiting for data to be fetched or stored. The processing load could almost be doubled and further interleaved with the store operations without chang-ing or delaying the data throughput of the system. What would happen, however, is an increase in the processing load that the system could handle.

 

The delays that have been seen are frequently exploited by optimising routines within many modern compilers. These com-pilers know from information about the target processor when these types of delays can occur and how to reschedule instructions to make use of them and regain some of the lost performance.


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