# How to multiply each value of a matrix by a number

6 views (last 30 days)
Dylan Springer on 16 Sep 2020
Answered: Sai Sri Pathuri on 22 Sep 2020
Hello, Right now I am getting a matrix at the end but all the values in each row are the same. Im not sure why this is happening. My results matrix is right but the unhedged and hedged are doing this error.
k_g = [0.66, 0.65, 0.64, 0.63, 0.62, 0.61, 0.60, 0.59, 0.55];
c_g = [0.085855 0.032191 0.020795 0.017001 0.013711 0.010851 0.008388 0.006291 0.001401];
k_b = [1.30, 1.25, 1.20, 1.15, 1.10, 1.05, 1.00, 0.95, 0.90];
c_b = [0.137213 0.082645 0.045060 0.028338 0.016146 0.007860 0.003277 0.001134 0.000245];
results_G = zeros(length(k_g), length(c_g));
results_B = zeros(length(k_b), length(c_b));
unhedged_G = zeros(length(k_g), length(c_g));
unhedged_B = zeros(length(k_b), length(c_b));
Eg = 643000000;
Eb = 272000000;
corr = 0.675;
n = 100000;
Rg = normrnd(0,9,[n,1]);
Rb = normrnd(0,11,[n,1]);
for j = 1:length(c_g)
for i = 1:length(k_g)
G = 0.6531*(1+Rg(i)/100);
netpayoff_g = (k_g(i)-G)-c_g(j);
results_G(i,j)= netpayoff_g;
unhedged_G(i,j)=Eg.*G;
end
j=j+1;
end
for b = 1:length(c_b)
for a = 1:length(k_b)
B = 1.234*(1+Rb(a)/100);
netpayoff_b = (k_b(a)-B)-c_b(b);
results_B(a,b)= netpayoff_b;
unhedged_B(a,b)= Eb.*B;
end
b=b+1;
end
hedged_G = unhedged_G + (9.*results_G)
hedged_B = unhedged_B + (9.*results_B)
hedged = hedged_G + hedged_B
results_G
results_B
best_G = max(results_G(:))
best_B = max(results_B(:))
jessupj on 16 Sep 2020
it might simply LOOK like the numbers are the same because Eg and Eb are large and the answer is displayed in scientific notation. try looking at the results after you run
format long g

Sai Sri Pathuri on 22 Sep 2020
You may try the following:
Create Rg and Rb as matrices with same length as results matrices
Rg = normrnd(0,9,[length(k_g), length(c_g)]);
Rb = normrnd(0,11,[length(k_b), length(c_b)]);
For the calculation of unhedged_G, define G as
G = 0.6531*(1+Rg(i,j)/100);
For the calculation of unhedged_B, define B as
B = 1.234*(1+Rb(a,b)/100);