网络存储系统NSS基准性能测试(7.31)

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原创最新推荐文章于 2026-04-15 20:23:56 发布 1.3k 阅读 AI 写同款· GEO 优化 ›

STM32与M95M04 EEPROM数据存储方案详解嵌入式系统中的数据持久化存储是确保设备配置和用户偏好长期保存的关键技术。基于SPI接口的EEPROM芯片如M95M04，以其字节级擦写特性和百万次擦写寿命，成为替代内部Flash的理想选择。本文以STM32F100ZE与M95M04的组合为例，详细解析了从硬件连接到数据结构设计的全流程实现方案。通过SPI接口配置、存储分区管理、数据校验机制等核心技术，实现了智能家居场景下的温控设置、定时任务等配置数据的可靠存储。特别针对频繁写入场景，提出了写缓冲、磨损均衡等优化策略，并分享了实际项目中的故障排查经验与性能测阅读详情

传送门 ☞ Android兵器谱 ☞ 转载请注明 ☞ http://blog.youkuaiyun.com/leverage_1229

一、实验参数列表

二、MATLAB脚本（balanced job bounds.m）

clear;
 
N = input('SAN Performance N = ');
L = input('Queue length L = ');
Z = input('Thinking time Z = ');
 
%testing a group of datas
% N = 60;
% L = 32;
% Z = 0.23;
 
%model M/M/1
 
%host
E1 = 1/(64*33/8)+0.0002;
for n = 1 : L
    u(n) = (sqrt(n*n+4*n)-n)/2;    %utilization in percent
    rate(1,n) = u(n)/E1;   %I/O request rate in IOPS
    arate1 = sum(rate(1,n))/n;
end
 
    D(1) = E1 + arate1*(E1^2)/(2*(1-arate1*E1));
 
%fcf
E2 = 1/1062.5;
for n = 1 : L
    arate2 = rate(1,n);
    D(2) = arate2*(E2^2)/(2*(1-arate2*E2));
end
 
%dacc
h = 2.283;
v = 2.798;
E3 = 1/(64*66/8)+0.00016;
E4 = 1/(64*66/8)+0.00018;
ts = E3+(E3^2)*sqrt(1.39794);
for n = 1 : L
    u(n) = (sqrt(n*n+4*n)-n)/2;    %utilization in percent
    rate(3,n) = u(n)/E3;   %I/O request rate in IOPS 
    rate(4,n) = u(n)/E4;   %I/O request rate in IOPS 
    arate3 = sum(rate(3,n))/n;
    arate4 = sum(rate(4,n))/n;
    
D(3)=ts+((0.5*(h+(4*v-3*h-1)*0.5)+2*(1+h-2*v)*0.25)*(1/(E3^2))*arate3*E3*(2*(E3^2)))/((1/E3)-arate3);
D(4)=ts+((0.5*(h+(4*v-3*h-1)*0.5)+2*(1+h-2*v)*0.25)*(1/(E4^2))*arate4*E3*(2*(E4^2)))/((1/E3)-arate4);
end
 
Dmax = max(D(1:4));                             % maximum service demand per code              
Dsum = D(1)+D(2)+D(3)+D(4);           	% sum of total service demands
Davg = Dsum/4;                                  % average service demand per queue
 
for n = 1:N
    Rmin(n) = max(n * Dmax - Z, Dsum + ((n-1)*Davg*Dsum/(Dsum+Z))); % lower bound of response time 
    Rmax(n) = Dsum + ((n-1)*Dmax*(n-1)*Dsum/(((n-1)*Dsum)+Z));      % upper bound of response time    
end
 
% response time of MVA
for m = 2:4
    L(m) = 0;                    
end    
for n = 1:N
    R(1) = D(1);
    for m = 2:4
        R(m) = D(m) * (1 + L(m));
    end
    Tau = n / sum(R(:));
    for m = 2:4
        L(m) = Tau * R(m);          
    end
    
    Rn(n,1) = D(1);
    for m = 2:4
        Ln(n,m) = L(m);
        Rn(n,m) = R(m);
    end
    Taun(n) = Tau;
end
    
for n = 1:N    
    RTn(n) = sum(Rn(n,2:4));      % average response time    
end
 
t = 1:N;
 
% response time 
figure(1), plot(t, Rmin, 'g', t, RTn, 'r', t, Rmax, 'b'),xlabel('SAN performance (MB/s)'),ylabel('Response time(s)');

三、NSS（网络存储系统）边界性能