Hello everybody,

I want to determine the value for the variable S in 2 cases in the following function so that tfr=.85 is reached for both cases. First I tried to do this by solving the equation setting it equal to .85 which unfortunately did not work and then I tried to increase S successively until the minimum .85 would be reached. However, non of my attempts worked out as you can see below. I would be more than grateful if you could help me to solve this problem.

<code>

Target Fill Rate for all Bases

>tfr := 0.85

Average Base Pipeline for Minimum Lead Time

>basepipemin[1] := 0.2; basepipemin[2] := 1.8;

1st Approach

> for n to 2 do
RHS:=S[n]->sum((basepipemin[n]^j*evalf(exp(-basepipemin[n])))/factorial(j),j=0..S[n]);
S[n]:=solve(RHS(S[n])=tfr,S[n])
od;

2nd Approach

> for n to 2 do
for S to 10 while tfrbase <= tfr do
tfrbase := sum(basepipemin[n]^j*evalf(exp(-basepipemin[n]))/factorial(j), j = 0 .. S)
od od;

Thank you very much for any help in advance,

Fred

Please Wait...