高性能计算机架构代写 Homework代写

ECE/CS 570 – High Performance Computer Architecture

Homework #1

高性能计算机架构代写 1.A multiprocessor machine has 1024 processors. On this machine we map a computation in which N iterate values must be computed and

1.

A multiprocessor machine has 1024 processors. On this machine we map a computation in which N iterate values must be computed and then exchanged between the processors. Values are broadcast on a bus after each iteration. Each iteration proceeds in two phases. In the first phase each processor computes a subset of the N iterates.

Each processor is assigned the computation of K = N/P iterates, where P is the number of processors involved. In the second communication phase each processor broadcasts its results to all other processors, one by one. Every processor waits for the end of the communication phase before starting a new computation phase. Let Tc be the time to compute one iterate and let Tb be the time to broadcast one value on the bus. We define the computation-to-communication ratio R as Tc/Tb. Note that, when P = 1, no communication is required.

At first, 高性能计算机架构代写

we use the premise of Amdahl’s speedup (i.e., the same workload spread across an increasing number of processors). Under these conditions:

(a) Compute the speedup as a function of P and R.

(b) Compute the maximum possible speedup as a function of P and R, for K = 1, 2, . . . , 1024.

Second,

we use the premise of Gustafson’s law, namely that the uniprocessor workload grows with the number of processors so that the execution time on the multiprocessor is the same as that on the uniprocessor. Assume that the uniprocessor workload computes 1024 iterates.

(d) What should the size of the workload be (as a number of iterates) when P processors are used, as a function of P and R? Pick the next closest integer value for the number of iterates.

2. 高性能计算机架构代写

This problem is about the difficulty of reporting average speedup numbers for a set of programs or benchmarks. We consider four ways of reporting average speedups of several programs:

take the ratio of average execution times, S1;
take the arithmetic means of speedups, S2;
take the harmonic means of speedups, S3;
take the geometric means of speedups, S4.

Consider the two improvements of a base machine, one improving floating-point performance and one improving memory performance: Two improvements are considered to a base machine with a load/store ISA and in which floating-point arithmetic instructions are implemented by software handlers. The first improvement is to add hardware floating-point arithmetic units to speed up floating-point arithmetic instructions. It is estimated that the time taken by each floating-point instruction can be reduced by a factor of 10 with the new hardware. The second improvement is to add more first-level data cache to speed up the execution of loads and stores. It is estimated that, with the same amount of additional on-chip cache real-estate as for the floating-point units, loads and stores can be speeded up by a factor of 2 over the base machine.

Three programs are simulated:

one with no floating-point operations (Program 1), one dominated by floating point operations (Program 2), and one with balance between memory accesses and floating point operations (Program 3). The execution time of each program on the three machines is given in Table 1:

(a) Compute the average speedups to the base machine for each improvement. Which conclusions would you draw about the best improvement if you were to consider each average speedup individually?

(b) To remove the bias due to the difference in execution times, we first normalize the execution time of the base machine to 1, yielding the normalized execution times in Table 2. Compute the four average speedups to the base machine for both improvements. Which conclusions would you draw about the best improvement if you were to consider each average speedup individually?

3. 高性能计算机架构代写

In this problem we evaluate the hardware needed to detect hazards in various static pipelines with out-of-order instruction execution completion. We consider the floating-point extension to the 5-stage pipeline, displayed in Figure 1.

Each pipeline register carries its destination register number, either floating-point or integer. ME/WB carries two instructions, one from the integer pipeline and one from the floating-point arithmeticpipeline.

Consider the following types of instructions consecutively:

integer arithmetic/logic/store instructions (inputs: two integer registers) and all load instructions (input: one integer register);
floating-point arithmetic instructions (inputs: two floating-point registers);
floating-point stores (inputs: one integer and one floating-point register).

All values are forwarded as early as possible. Both register files are internally forwarded. All data hazards are resolved in the ID stage with a hazard detection unit (HDU). ID fetches registers from the integer and/or from the floating-point register file, as needed. The opcode selects the register file from which operands are fetched (S.D fetches from both). Note that stores need both inputs in EX at the latest.

(a) 高性能计算机架构代写

To solve RAW data hazards on registers (integer and/or floating-point), hardware checks between the current instruction in ID and instructions in the pipeline may stall the instruction in ID. List first all pipeline registers that must be checked in ID. Since ME/WB may have two destination registers, list them as ME/WB(int) or ME/WB(fp). Do not list pipeline stages, list pipeline registers, and make sure that the set of checks is minimum.

(b)

To solve WAW hazards on registers, we check the destination register in ID with the destination register of instructions in various pipeline stages. List the pipeline registers that must be checked. Make sure that the set of checks is minimum. Important: remember that there is already a mechanism in the pipeline design to avoid structural hazards on the write register ports of both register files, so it is not possible for two instructions that write to the same register to reach WB in the same cycle.

Your solutions specifying the hazard detection logic should be written as follows for both RAW and WAW hazards:

if integer arithmetic/logic/store/load instruction in ID, check <pipeline registers>;
if FP load instruction in ID, check <pipeline registers>;
if FP arithmetic instruction in ID, check <pipeline registers>;
if FP store instruction in ID, check <pipeline registers >.4

4. 高性能计算机架构代写

In the pipeline of Figure 2, WAW data hazards on registers are eliminated and exceptions can be handled in the WB stage where instructions complete in process order as in the classic 5-stage pipeline. As always, values are forwarded to the input of the execution units.

(a) List all required forwarding paths from pipeline registers to either EX or FP1 to fully forward values for all instructions. List them as source → destination (e.g., FP2/FP1→FP1).

(b) Given those forwarding paths, indicate all checks that must be done in the hazard detection unit associated with ID to solve RAW hazards.

Your solutions specifying the hazard detection logic should be written as follows for RAW hazards on registers:

if integer arithmetic/logic/store or load instruction in ID, check <pipeline registers>;
if FP arithmetic instruction in ID, check <pipeline registers>;
if FP store instruction in ID, check <pipeline registers>.5

5. 高性能计算机架构代写

In this problem we compare the performance of four dynamically scheduled processor architectures on a simple piece of code computing Y = Y∗X + Z, where X, Y, and Z are (double-precision–8 bytes) floating-point vectors.

Using the core ISA of Table 0 in the Appendix, the loop body can be compiled as follows:

LOOP   L.D F0,0(R1)      /X[i] loaded in F0
       L.D F2,0(R2)      /Y[i] loaded in F2
       L.D F4,0(R3)      /Z[i] loaded in F4
       MUL.D F6,F2,F0    /Multiply X by Y
       ADD.D F8,F6,F4    /Add Z
       ADDI R1,R1,#8     /update address registers
       ADDI R2,R2,#8
       ADDI R3,R3,#8
       S.D F8, -8(R2)    /store in Y[i]
       BNE R4,R2,LOOP/   /(R4)-8 points to the last element of Y

The initial values in R1, R2, and R3 are such that the values are never equal during the entire execution. (This is important for memory disambiguation.) The architectures are given in Figures 3, 4, 5 and 6, and the same parameters apply. Branch BNE is always predicted taken (except in the basic Tomasulo, where branches are not predicted at all and stall in the dispatch stage until their outcome is known).

Keep in mind the following important rules (whenever they apply):

instructions are always fetched, decoded, and dispatched in process order;
in speculative architectures, instructions always retire in process order;
in speculative architectures, stores must wait until they reach the top of the ROB before they can issue to cache (i.e. a store issues to cache in the cycle after the previous instruction retires).
The solution should include two loads from the next iteration of the loop

(a) Tomasulo algorithm – no speculation. Please fill a table like Table 3 clock-by-clock for the first iteration of the loop. Each entry should be the clock number when the event occurs, starting with clock cycle 1. Add comments as you see fit.

(b) Tomasulo algorithm with speculation. Please fill a table like Table 4 clock-by-clock for the first iteration of the loop. Each entry should be the clock number when the event occurs, starting with clock cycle 1. Please be attentive to the fact that (contrary to Tomasulo with no speculation) stores cannot execute in cache until they reach the top of the ROB. Also branches are now predicted taken.

(c) ) Tomasulo algorithm with speculation and register fetch after issue. Please fill a table like Table 5 clock-by-clock for the first iteration of the loop. Each entry should be the clock number when the event occurs, starting with clock 1.8

(d) Speculative scheduling. Please fill a table like Table 6 clock-by-clock for the first iteration of the loop. Each entry should be the clock number when the event occurs, starting with clock 1.

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