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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. 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

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