the histgram of G is showing a straght line in this code

R10=2e3
dR=0.01
DeltaR=(rand(1000,1)-0.5)*2
R1=R10*(1+DeltaR*dR)
figure; hist(R1,20)
R20=18e3
dR=0.01
R2=R20*(1+DeltaR*dR)
figure; hist(R2,20) 
G=1+R2./R1
figure; hist(G,200)

[Walter Roberson]

format long g
R10=2e3

R10 = 
        2000

dR=0.01

dR = 
                      0.01

DeltaR=(rand(1000,1)-0.5)*2;
DeltaR(1:10)

ans = 10×1

0.456850403382957 0.829686060953362 -0.89642340724095 0.466597228156508 0.977663054020667 -0.87630487096742 0.888444975804023 -0.35566284936498 0.70593257404108 0.123539848371262

R1=R10*(1+DeltaR*dR);
R1(1:10)

ans = 10×1

2009.13700806766 2016.59372121907 1982.07153185518 2009.33194456313 2019.55326108041 1982.47390258065 2017.76889951608 1992.8867430127 2014.11865148082 2002.47079696743

figure; hist(R1,20)

R20=18e3

R20 = 
       18000

dR=0.01

dR = 
                      0.01

R2=R20*(1+DeltaR*dR);
R2(1:10)

ans = 10×1

18082.2330726089 18149.3434909716 17838.6437866966 18083.9875010682 18175.9793497237 17842.2651232259 18159.9200956447 17935.9806871143 18127.0678633274 18022.2371727068

figure; hist(R2,20) 

G=1+R2./R1

G = 1000×1

10 10 10 10 10 10 10 10 10 10

figure; hist(G,200)

Answer 1

[Walter Roberson]

DeltaR=(rand(1000,1)-0.5)*2
R1=R10*(1+DeltaR*dR)

So R1 is a constant times (1+DeltaR*dR)

R2=R20*(1+DeltaR*dR);

so R2 is a constant times (1+DeltaR*dR)

G=1+R2./R1

The (1+DeltaR*dR) are present in both R2 and R1, so the R2./R1 is going to be R20/R10 which is a constant. So G is going to be 1 + the ratio of the two constants, and so is going to be a constant.

[ZHuoyan Liu]

Thanks!

that means that if i create two seperate (1+DeltaR*dR) for R1 and R2 the result should be showing the randomised output right?

[Walter Roberson]

If you have two difference random DeltaR for R1 and R2, then Yes, the result will not be the same for all of the entries.