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Cheap Engines PITA Analysis
Author: Mark Norton
Date: 05 April 2001
Cheap Engines are a pain in the ass. However, with IT, and IFE, CE will
give a player AD8 drives to start. Not too shabby on the whole, and good
tech to trade if need be. So as with so many other things in Stars! there's
a tradeoff to calculate. I didn't find this in the Advanced FAQ and while
it may have been done before, a large body of the former Stars! information
has been lost with various websites going down, so I took it upon myself to
figure it out myself (with some initial help from Dan Neely on #stars! on
irc.starlink.org. Thanks Dan!).
So, how much of a pain in the ass is Cheap Engines? A fleet with cheap
engines has a 10% chance of not moving for a given year. The payoff is
points in the RW, and 50% cheaper engines. That probability is for a single
year, but I may need to plan for a 4 or 5 year trip. How does Cheap Engines
affect long term travel?
What is the probability the fleet will get there in N years?
The fleet has a 10% chance of failing per year. Therefore it has a 90%
chance of success per year. The probability that the fleet will get there in
the nominal number of years is:
(0.9)^N
A more interesting question though is, for a given trip length, what is the
probability it'll take an extra year, two extra years, etc. For example,
a trip with a nominal length of 3 years might take 5 years. What is the
probability of that happening?
As a basis, there are at least 3 successes and 2 failure. Therefore we have
(0.9)^3 * (0.1)^2
However that is the probability of a *particular* trip. There are several
combinations of success and failure and that's where the binomial
distribution comes in. For a series of trials N, with M successes with a
probability P, the binomial distribution says that the probability of
exactly M successes is:
COMBIN(N, M) * P^M * (1-P) ^ (N-M)
Where the COMBIN function is N! / (M! (N-M)!).
In the case of the cheap engines travel, however, the combination function
counts too many. This is due to the fact that the last leg of a trip by
definition is a success. For example, take a trip that could have been 3
years, but ends up being 5 years. A M stands for Move and a F stands for
Failure.
Possible Combinations Impossible Combinations
--------------------- -----------------------
FFMMM MMMFF (because after 3 moves it's there)
MFFMM MMFMF (ditto)
MMFFM MFMMF
FMFMM FMMMF
MFMFM
FMMFM
The combination of 5 things taken 3 at a time would result in 10, while we
can see by simple inspection that the correct result is 6. However we can
also see that if we remove the last leg of the trip, the
combination function will once more work fine.
So we end up with a modified binomial distribution.
N = Length of actual trip in years
M = Nominal length of the trip in years
P = Probability of failure
COMBIN(N-1, N-M) * P^(N-M) * (1-P)^(M)
In English, the number of ways failures can be distributed times the
probability of the appropriate number of failures times the appropriate
number of successes. Note that for N - M = 0 you have to remove the
combination function in Excel.
If we put this in handy table format:
Nominal Trip Length
1 2 3 4 5 6 7 8 9 10
A 1 0.900 * * * * * * * * *
c 2 0.090 0.810 * * * * * * * *
t 3 0.009 0.162 0.729 * * * * * * *
u 4 0.001 0.024 0.219 0.656 * * * * * *
a 5 0.000 0.003 0.044 0.262 0.590 * * * * *
l 6 0.000 0.007 0.066 0.295 0.531 * * * *
7 0.001 0.013 0.089 0.319 0.478 * * *
T 8 0.000 0.002 0.021 0.112 0.335 0.430 * *
r 9 0.000 0.004 0.030 0.134 0.344 0.387 *
a 10 0.001 0.007 0.040 0.155 0.349 0.349
v 11 0.000 0.001 0.010 0.052 0.174 0.349
e 12 0.000 0.002 0.014 0.064 0.192
l 13 0.000 0.003 0.019 0.077
14 0.001 0.005 0.025
15 0.000 0.001 0.007
16 0.000 0.002
17 0.000
I rounded at the 1/1000th's place and then eliminated fields where the
value was zero after rounding (after the first one).
Another handy number to have is, "What is the probability that my fleet
will arrive in R years or less, with R being greater or equal to N, the
nominal trip length?". This should be the cumulative probability.
In Excel, it's easier just to sum the column up to that point.
Nominal Trip Length
1 2 3 4 5 6 7 8 9 10
A 1 0.900 * * * * * * * * *
c 2 0.990 0.810 * * * * * * * *
t 3 0.999 0.972 0.729 * * * * * * *
u 4 1.000 0.996 0.948 0.656 * * * * * *
a 5 1.000 0.991 0.919 0.590 * * * * *
l 6 0.999 0.984 0.886 0.531 * * * *
7 1.000 0.997 0.974 0.850 0.478 * * *
T 8 1.000 0.995 0.962 0.813 0.430 * *
r 9 0.999 0.992 0.947 0.775 0.387 *
a 10 1.000 0.998 0.987 0.930 0.736 0.349
v 11 1.000 0.997 0.981 0.910 0.697
e 12 0.999 0.996 0.974 0.889
l 13 1.000 0.999 0.994 0.966
14 1.000 0.999 0.991
15 1.000 0.998
16 0.999
17 1.000
The sums all add up to unity indicating that we've done this correctly.
I've eliminated further numbers past unity for clarity. Note that it IS
still possible to have a 1 year trip that takes 5, 10, even 100 years. It's
just very, very unlikely (and past 4 significant digits of precision.)
Using 95% as a confidence number (that is to say, the number at which we
feel confident we can rely on the number) from 1 year to 4 years, you can
usually expect to add a year. From 5 to 7 years, you can usually expect to
add two years. From 8 to 10 years, you can usually expect to add 3 years.
There are probably some other conclusions you can draw from the numbers as
well. If you have 100 fleets that need to travel 4 years away, about 66 of
them will arrive in 4 years, 92 ought to be there by 5 years and the rest
will trickle in afterwards.
Consider a trip that could take 2 years at W9 and 3 years at W8. We can see
from the second table that there is only a 73% chance that the fleet will
arrive in 3 years at W8, but there's a 97% chance that the fleet will arrive
in 3 years a W9. That percentage difference might mean having to plan extra
tankers to propel the fleet at W9 to get there in 3 years with more certainty.
CE will always be irritating and it seems to have the Murphy's Law effect
that you'll always fail when you need it the most, however with careful
planning, CE really isn't too much of a hardship (particular for IT where
quite a lot happens via gate, which CE doesn't affect.)
Regards,
Mark Norton
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