Micropreemption synthesis: an enabling mechanism for multitask VLSI systems
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Citations
A Hardware Preemptive Multitasking Mechanism Based on Scan-path Register Structure for FPGA-based Reconfigurable Systems
Controller support device, simulation method of control program, support program of controller and computer-readable storage medium storing support program of controller
A Preemption Algorithm for a Multitasking Environment on Dynamically Reconfigurable Processor
References
Synthesis and optimization of digital circuits
Simultaneous multithreading: maximizing on-chip parallelism
Static Scheduling of Synchronous Data Flow Programs for Digital Signal Processing
Exploiting Choice: Instruction Fetch and Issue on an Implementable Simultaneous Multithreading Processor
The Tera computer system
Related Papers (5)
Frequently Asked Questions (16)
Q2. How many registers are used in the 0-p case?
As the preemption latency increases, the number of dedicated registers decreases, the number of shared registers increases, and the total number of registers decreases monotonously.
Q3. What is the synthesis trajectory of the HYPER high level system?
Synthesis modules for hardware mapping and layout generation from HYPER high level synthesis system were used to complete the synthesis trajectory.
Q4. What is the way to evaluate the context switch overhead?
Since the peak usage of shared registers cannot be known a priori, edges are bound to registers (using PreemptionContextSynthesis()) to evaluate the context switch overhead, exactly.
Q5. What is the common problem associated with preemption context binding?
The optimization problem associated with preemption context binding can be de ned as: Given a scheduled task and a set of preemption points, bind the edges to the registers so that (i) the red edges are bound to dedicated registers, and (ii) the total number of registers is minimized.
Q6. What is the main idea behind the checkpoints approach?
The checkpoints (which incur some penalty in processorperformance) are used to divide the sequential instruction stream into smaller units to reduce the cost of resumption.
Q7. How does the scheme reduce the performance of the multi-task VLSI system?
The authors have also implemented a controller based scheme to eliminate the performance degradation by (i) partitioning the task states into critical sections, (ii) executing critical sections and (iii) preserving atomicity by rolling forward to the end of the critical sections on preemption.
Q8. What is the context switch overhead of a multitask VLSI system?
the context switch cost of a multi-task VLSI system with task set T is:j R j= Xt2Tj Rtd j +max t2T j Rts j (1)Performance degradation resulting from aborting a task is eliminated by (i) partitioning the task states into critical sections, (ii) executing critical sections and (iii) preserving atomicity of a critical section by rolling forward to the end of a critical section on preemption.
Q9. How do the authors get the preemption point set?
the best preemption point set one for each of the tasks is obtained by using the context switch cost function given by equation 1.
Q10. How many clock cycles does a task take?
Consider a system implementing two tasks A and B. Task A takes four clock cycles and task B takes six clock cycles for one iteration.
Q11. How is the preemption point insertion algorithm calculated?
Towards investigating preemption point insertion, consider a task with ve edges (e1; :::; e5), an application latency of eight clock cycles and an edge-toregister binding shown in gure 3.
Q12. What is the simplest way to represent a task?
Within this model, a task is represented as a hierarchical Control Data Flow Graph G(N;E; T ) (or CDFG), with nodes N representing the ow graph operations, and the edges E and T respectively the data and timing dependences between the operations.
Q13. how long does it take to get a new task to be active?
Notice that it has taken two clock cycles from the time a valid preemption request arrived (beginning of state A2) to the time the new task (task B) became active (end of state A3).
Q14. How many edges are accessed at the rst half cycle?
The register le has one input port and one output port which are accessed at the rst half cycle and the last half cycle, respectively.
Q15. What are the edges that do not straddle a preemption point?
The optimization problem can be de ned as follows: Given an underlying hardware model and N scheduled tasks, each with its own time bound ( ) and maximum preemption latency ( ), insert preemption points, and bind edges to registers, so that the context switch overhead is minimized.
Q16. What is the area overhead of the proposed scheme?
When compared to the background memory based schemes, the proposed scheme does not need (i) additional ports of the register les which are used to save/restore data to/from the background memories without stalling currently running task, (ii) additional buses to interconnect the register les to the background memories, and (iii) additional control logic to compute memory addresses.