请坛子里高手帮我看看这个MATLAB程序怎么修改，谢谢！

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这是程序原文：
%%%%% Matlab program (.m) file that gives solutions to "Problem Set Zero"
%%%%% You should play with this file to learn simple Matlab commands.
%%%%% Use "help" at the command window to figure out what various commands
%%%%% do. Eg, type "help clear" to figure out what the next line does.
clear all
%%%%% QUESTION 1: STEADY STATE
%%%%% Declare parameter values
g     = 0.03;
n     = 0.01;
s     = 0.20;
alpha = 0.33;
delta = 0.04;
parameters = [g;n;s;alpha;delta];
%%%%% Declare some symbols
syms kk gg nn ss aa dd
answer = solve('(1+gg)*(1+nn)*kk-ss*kk^aa-(1-dd)*kk=0',kk);
k_bars = double(subs(answer,{gg,nn,ss,aa,dd},parameters))
% Matlab treats "symbols" differently to "numbers"; the commands above
% declare some symbols then solve the difference equation for its
% steady-state values. I then substitute the vector of parameters into the
% answer to get "numbers" back. Use the Matlab syntax "help syms", "help
% solve" etc etc to investigate this more closely.
% Keep the non-trival (positive) steady-state for future reference     
         
k_bar  = max(k_bars);
ky = k_bar^(1-alpha);
mpk = 1+alpha*k_bar^(alpha-1);
display(['Non-trivial steady state          =   ',num2str(k_bar)])
display(['Capital/output ratio              =   ',num2str(ky)])
display(['Gross marginal product of capital =   ',num2str(mpk)])
%%%%% QUESTION 2: TRANSITION TO STEADY STATE
error = 10;
iter  = 1;
k0    = k_bar*exp(-0.5);
% This sets the initial condition while the next command sets up an empty
% vector that we will use to store the sequence of capital stocks.
kt    = [];
% The following loop iterates on the difference equation until the error is
% small (less than 10e-6)
while error > 10e-6
    if iter == 1
        k = k0;
    else
        k = kt(iter-1);
    end
   
    k1 = (s/((1+g)*(1+n)))*k^(alpha) + ((1-delta)/((1+g)*(1+n)))*k;
% The previsious line computes the subsequent capital stock k1 given
% today's capital stock k
   
    error = abs(k1-k_bar);
    kt = [kt;k1];
    iter  = iter+1;  
   
end
T = length(kt);
t = [0;cumsum(ones(T-1,1))];
figure(1)
plot(t,kt,t,k_bar*ones(T,1),'k--')
title('Figure 1: Time path of the capital stock')
xlabel('time, t')
ylabel('capital stock, k(t)')
legend('k(t)',0)
%%%%% Now for the phase diagram (this is a little tricky...)
kmax = 2*k_bar;
ks   = linspace(0,kmax,1000);
psis = (s/((1+g)*(1+n)))*ks.^(alpha) + ((1-delta)/((1+g)*(1+n)))*ks;
kt0 = kt(1:T-1);
kt1 = kt(2:T);
figure(2)
if k0 < k_bar
    plot(kt0,kt1,'b>-',ks,ks,'k--',ks,psis,'r-')
    title('Figure 2: Phase diagram')
else
    plot(kt0,kt1,'b<-',ks,ks,'k--',ks,psis,'r-')
    title('Figure 2: Phase diagram')
end
xlabel('k(t)')
ylabel('k(t+1)')
legend('k(t+1)','45-degree line','\psi(k(t))',0)
%%%%% PROBLEM 3: GROWTH RATE APPROXIMATIONS
%%%%% The next command produces a 1000-by-1 vector of evenly spaced "zs"
%%%%% between 0 and 1.
zs = linspace(0,1,1000)';
g_approximate     = log(1+zs);
g_actuals         = zs;
figure(3)
plot(100*zs,100*g_approximate,'k--',100*zs,100*g_actuals)
title('Figure 3: Accuracy of log-differences as aproximate growth rates')
xlabel('growth rate (%)')
legend('log-approximate growth rate','actual growth rate',0)

运行后显示如下错误：
??? Attempt to execute SCRIPT solve as a function.
Error in ==> D:\Matlab\toolbox\symbolic\@sym\solve.m
On line 49  ==> [varargout{1:max(1,nargout)}] = solve(S{:});
Error in ==> D:\Matlab\work\ps0_solutions.m
On line 25  ==> answer = solve('(1+gg)*(1+nn)*kk-ss*kk^aa-(1-dd)*kk=0',kk);

谢谢！！！

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关键词：MATLAB程序 MATLAB atlab matla 怎么修改 MATLAB 程序 高手 坛子

没看明白；
解决什么问题的？
我先拿到程序里跑一边试试。

在你的当前目录，Matlab的搜索路径里面找 solve.m ， 然后把除了Matlab系统里面自带的solve.m 都改名，然后看看

16行solve函数找不到显示解，因此answer没能被赋值
可以试试fsolve求数值解

cdconan 发表于 2011-4-16 13:00
16行solve函数找不到显示解，因此answer没能被赋值
可以试试fsolve求数值解

哥们，试过了，不行啊～这个问题很纠结，还请大家帮忙啊

5# wb123456

1. 
   
2. %%%%% Matlab program (.m) file that gives solutions to "Problem Set Zero"
   
3. %%%%% You should play with this file to learn simple Matlab commands.
   
4. %%%%% Use "help" at the command window to figure out what various commands
   
5. %%%%% do. Eg, type "help clear" to figure out what the next line does.
   
6. clear all
   
7. %%%%% QUESTION 1: STEADY STATE
   
8. %%%%% Declare parameter values
   
9. g     = 0.03;
   
10. n     = 0.01;
    
11. s     = 0.20;
    
12. alpha = 0.33;
    
13. delta = 0.04;
    
14. parameters = [g;n;s;alpha;delta];
    
15. %%%%% Declare some symbols
    
16. % syms kk gg nn ss aa dd
    
17. k_bars = fsolve(@(k)(1+g)*(1+n)*k-s*k^alpha-(1-delta)*k,4);
    
18. % k_bars = double(subs(answer,{gg,nn,ss,aa,dd},parameters))
    
19. % Matlab treats "symbols" differently to "numbers"; the commands above
    
20. % declare some symbols then solve the difference equation for its
    
21. % steady-state values. I then substitute the vector of parameters into the
    
22. % answer to get "numbers" back. Use the Matlab syntax "help syms", "help
    
23. % solve" etc etc to investigate this more closely.
    
24. % Keep the non-trival (positive) steady-state for future reference     
    
25.          
    
26. k_bar  = max(k_bars);
    
27. ky = k_bar^(1-alpha);
    
28. mpk = 1+alpha*k_bar^(alpha-1);
    
29. display(['Non-trivial steady state          =   ',num2str(k_bar)])
    
30. display(['Capital/output ratio              =   ',num2str(ky)])
    
31. display(['Gross marginal product of capital =   ',num2str(mpk)])
    
32. %%%%% QUESTION 2: TRANSITION TO STEADY STATE
    
33. error = 10;
    
34. iter  = 1;
    
35. k0    = k_bar*exp(-0.5);
    
36. % This sets the initial condition while the next command sets up an empty
    
37. % vector that we will use to store the sequence of capital stocks.
    
38. kt    = [];
    
39. % The following loop iterates on the difference equation until the error is
    
40. % small (less than 10e-6)
    
41. while error > 10e-6
    
42.     if iter == 1
    
43.         k = k0;
    
44.     else
    
45.         k = kt(iter-1);
    
46.     end
    
47.    
    
48.     k1 = (s/((1+g)*(1+n)))*k^(alpha) + ((1-delta)/((1+g)*(1+n)))*k;
    
49. % The previsious line computes the subsequent capital stock k1 given
    
50. % today's capital stock k
    
51.    
    
52.     error = abs(k1-k_bar);
    
53.     kt = [kt;k1];
    
54.     iter  = iter+1;  
    
55.    
    
56. end
    
57. T = length(kt);
    
58. t = [0;cumsum(ones(T-1,1))];
    
59. figure(1)
    
60. plot(t,kt,t,k_bar*ones(T,1),'k--')
    
61. title('Figure 1: Time path of the capital stock')
    
62. xlabel('time, t')
    
63. ylabel('capital stock, k(t)')
    
64. legend('k(t)',0)
    
65. %%%%% Now for the phase diagram (this is a little tricky...)
    
66. kmax = 2*k_bar;
    
67. ks   = linspace(0,kmax,1000);
    
68. psis = (s/((1+g)*(1+n)))*ks.^(alpha) + ((1-delta)/((1+g)*(1+n)))*ks;
    
69. kt0 = kt(1:T-1);
    
70. kt1 = kt(2:T);
    
71. figure(2)
    
72. if k0 < k_bar
    
73.     plot(kt0,kt1,'b>-',ks,ks,'k--',ks,psis,'r-')
    
74.     title('Figure 2: Phase diagram')
    
75. else
    
76.     plot(kt0,kt1,'b<-',ks,ks,'k--',ks,psis,'r-')
    
77.     title('Figure 2: Phase diagram')
    
78. end
    
79. xlabel('k(t)')
    
80. ylabel('k(t+1)')
    
81. legend('k(t+1)','45-degree line','\psi(k(t))',0)
    
82. %%%%% PROBLEM 3: GROWTH RATE APPROXIMATIONS
    
83. %%%%% The next command produces a 1000-by-1 vector of evenly spaced "zs"
    
84. %%%%% between 0 and 1.
    
85. zs = linspace(0,1,1000)';
    
86. g_approximate     = log(1+zs);
    
87. g_actuals         = zs;
    
88. figure(3)
    
89. plot(100*zs,100*g_approximate,'k--',100*zs,100*g_actuals)
    
90. title('Figure 3: Accuracy of log-differences as aproximate growth rates')
    
91. xlabel('growth rate (%)')
    
92. legend('log-approximate growth rate','actual growth rate',0)

复制代码

谢谢上楼的朋友

6楼很强啊。遇到几次了。

对了，你说这些程序是出自一本经济学的书籍。
可以推荐下不？
我找找看。

函数定义关键字是function，不是Function