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baotong
一个报童从报刊发行中心订报后零售，每卖一份报纸可赚钱a元；若报纸卖不出去，则退回发行处，每退一份要赔钱b元。每天报童卖出的份数是随机的，但报童可以根据以往卖报情况统计得到每天卖k份报纸的概率密度p(k)。
（1） 求报童每天期望收益达到最大（或损失达到最小）的定报量z。
（2） 改变参数a/b的值，观察订报量的最优值变化，画出变化曲线。
试画出仿真流程图，进行程序实现，并对仿真结果进行分析。
(A newsboy subscribe from the press after the retail distribution centers, each sale of a newspaper can make money a million if newspapers can not be sold, then returned to Issue Department, each retire to a loss of a b element. Newsboy sold shares in a day is random, but the newsboy can sell based on past statistics reported to be selling daily newspaper k the probability density p (k). (1) seeking newsboy expected profit per day to maximize (or minimize loss) reported that the volume of fixed-z. (2) change the parameters a/b values, subscribe to observe changes in the volume of the optimal value, draw curve. Simulation test to draw a flow chart for program realization, and simulation results for analysis.)
- 2008-06-14 01:16:47下载
- 积分：1
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MNPBEM11
采用边界元方法求解电磁响应及系统本征模问题(boundary element method to investigate the eigmode and green function of the EM )
- 2012-06-05 18:52:52下载
- 积分：1
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ccamap
The code source of CCA-MAP loclaization algorithm using matlab
- 2013-03-21 05:53:32下载
- 积分：1
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LDPCcode
利用函数MadHG生成规则LDPC码的校验矩阵H，其行重为6,列重为3,行数为列数一半(行数越大越好)，H中任意两列没有围长为4的圈，并得到H对应的生成矩阵G，可以保证mod(G*H ,2)=0。使用方法为:[H,G] = MadHG(m,n,x),x= 1(得到的G左半部分为单位阵) or 2(G右半部分为单位阵),(use function MadHG Generation rules LDPC check matrix H, re-6, series of three, out for a few half of the number (A bigger the better), H 2 arbitrary no girth of the four circles, H counterparts and with the formation of matrix G, we can be assured of mod (G* H, 2) = 0. To be used : [H, G] = MadHG (m, n, x), x = 1 (the G part of the Left Front units) or 2 (G right half of the units RUF),)
- 2007-05-12 10:49:48下载
- 积分：1
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fractal-use
分形的练习一
①Koch曲线
用复数的方法来迭代Koch曲线
clear i 防止i被重新赋值
A=[0 1] 初始A是连接（0，0）与（1，0）的线段
t=exp(i*pi/3)
n=2 n是迭代次数
for j=0:n
A=A/3 a=ones(1,2*4^j)
A=[A (t*A+a/3) (A/t+(1/2+sqrt(3)/6*i)*a) A+2/3*a]
end
plot(real(A),imag(A))
axis([0 1 -0.1 0.8])
②Sierpinski三角形
A=[0 1 0.5 0 0 1] 初始化A
n=3 迭代次数
for i=1:n
A=A/2 b=zeros(1,3^i) c=ones(1,3^i)/2
A=[A A+[c b] A+[c/2 c]]
end
for i=1:3^n
patch(A(1,3*i-2:3*i),A(2,3*i-2:3*i), b ) patch填充函数
end
(Fractal
Exercise One
The ① Koch curve
Plural iteration Koch curve
clear i to prevent i is reassigned
A = [0 1] initial A is a connection (0,0) and (1,0) of the segments
t = exp (i* pi/3)
n = 2 n is the number of iterations
for j = 0: n
A = A/3 a = ones (1,2* 4 ^ j)
A = [A (t* A+ a/3) (A/t+ (1/2+ sqrt (3)/6* i)* a) A+2/3* a]
end
plot (real (A), imag (A))
axis ([0 1-0.1 0.8])
② Sierpinski triangle
A = [0 1 0.5 0 0 1] initialized A
n = 3 the number of iterations.
for i = 1: n
A = A/2 b = zeros (1,3 ^ i) c = ones (1,3 ^ i)/2
A = [A A+ [c b] A+ [c/2 c]]
end
for i = 1:3 ^ n
patch (A (1,3* i-2: 3* i), A (2,3* i-2: 3* i), b ) patch filled function
end)
- 2013-03-02 10:03:09下载
- 积分：1
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PID-CONTROL
基于水箱液位系统的三阶模型的阶跃响应测试，(three-tank )
- 2013-03-18 15:35:37下载
- 积分：1
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2210
matlab simulink implementation and analysis of Three-phase inverter2
- 2013-11-30 22:50:38下载
- 积分：1
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022d7ef2-7534-431e-bc7e-a1f183145d43
控制程序的简要程序，供大家参考，下午 大家能提出宝贵意见(l ddddfdggfffhhhhgfgfgg)
- 2009-04-18 17:49:42下载
- 积分：1
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Numerical_method_using_matlab
数值方法matlab版
numerical mnumerical method of matlabethod of matlab(numerinumerical method of matlabcal method of matlab)
- 2009-05-07 05:05:28下载
- 积分：1
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papr
采用映射法降低ofdm系统分均比的matlab仿真程序(Ofdm systems using sub-mapping method to reduce the matlab simulation program than)
- 2011-04-19 22:59:41下载
- 积分：1
