issue with a while loop

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Adam Palmer
Adam Palmer el 3 de Ag. de 2014
Editada: dpb el 4 de Ag. de 2014
Hey everybody, got a simple question for ya. Im making a function file to determine when it is safe to drive for a person who has been drinking. The value for hours in the output is wrong, but the formula works well outside of the loop. The answer should be around 2-3 hours, but the while loop consistently returns .28 hours. Heres my code
BAC=.13;
Hrs=0;
Rt=.0140;
while BAC>.08
BAC=BAC-(Hrs*.0140);
Hrs=Hrs+.01
end
  1 comentario
Azzi Abdelmalek
Azzi Abdelmalek el 3 de Ag. de 2014
If the formula works well outside the loop, then don't use the loop

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dpb
dpb el 3 de Ag. de 2014
Editada: dpb el 3 de Ag. de 2014
You've got the while operating until BAC<=0.08 by adjusting Hrs but one can take the expression
BAC=BAC-(Hrs*.0140);
and solve for
0.08=0.13-Hrs*0.014
to find
>> (0.13-0.08)/0.014
ans =
3.5714
>>
What's wrong with your implementation that iterates where it isn't really needed is you're multiplying the difference/0.01 hr by the total hours every time instead of just the reduction/0.01 hr.
>> BAC=.13;
Hrs=0;
dHr=0.01;
Rt=.0140;
dHrRed=dHr*Rt;
while BAC>.08
BAC=BAC-dHrRed;
Hrs=Hrs+.01;
end
Hrs
Hrs =
3.5800
>>
This isn't quite as accurate as the direct solution but given the inherent nature of the problem the correlation probably isn't accurate to a tenth of an hour, either...
  2 comentarios
Adam Palmer
Adam Palmer el 3 de Ag. de 2014
Thanks dpb, Im still getting the hang of while loops as you can see lol
dpb
dpb el 3 de Ag. de 2014
Editada: dpb el 4 de Ag. de 2014
No problem; took me a few minutes to grok the issue, too.
OBTW, the above leaves Hrs at 0.01 greater than the actual solution because of the location of the increment of Hrs after the computation. The exit test doesn't occur until after that last increment but the final BAC value is based on the previous Hrs value.
To see this, try a modification...
BAC=.13;
Hrs=0;
dHr=0.01;
Rt=.0140;
dHrRed=dHr*Rt;
while BAC>.08
BAC=BAC-dHrRed;
if Hrs>=3.5, disp([Hrs BAC]), end
Hrs=Hrs+dHr;
end
3.5100 0.0807
3.5200 0.0806
3.5300 0.0804
3.5400 0.0803
3.5500 0.0802
3.5600 0.0800
3.5700 0.0799
>> disp([Hrs BAC])
3.5800 0.0799
>>
So, the answer is closer to the analytic for this particular set of values but your resolution is still only 0.01 hr, of course.
To fix this to get the nearest, rearrange it slightly --
BAC=0.13;
dHr=0.01;
Hrs=-dHr; % initialize to minus the delta
Rt=0.0140;
dHrRed=dHr*Rt;
while BAC>.08
Hrs=Hrs+dHr; % and increment first
BAC=BAC-dHrRed;
end

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Más respuestas (1)

Star Strider
Star Strider el 3 de Ag. de 2014
For zero-th order elimination kinetics, the loop should actually be:
BAC=.13;
Hrs=0;
Rt=.0140;
while BAC>.08
BAC=BAC-(0.01*.0140);
Hrs=Hrs+.01
end
This gives Hrs = 3.58.

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