A mathematical model is presented for computing total (spacecraft) subsystem cost including both the basic subsystem cost and the expected cost due to the failure of the subsystem. This model is then used to determine power subsystem cost as a function of reliability and redundancy. Minimum cost and maximum reliability and/or redundancy are not generally equivalent. Two example cases are presented. One is a small satellite, and the other is an interplanetary spacecraft

Patterson, Richard L.

Suich, Ronald C.

English

NASA Technical Reports Server

///-_3.._
NASA Technical Memorandum i04-453
w
7#==
, Balancing Reliability and Cost to
Choose the Best Power SUbsystem
Ronald C. Suich
California State University :
Fullerton, California
and .................
Richard L. Patterson
Lewis Research Center _- i i[ _ __ _[ i[ - _-[ _ i__ i ii
Cleveland, Ohio ..............
Prepared for the
26th Imersociety Energy Conversion Enginee_g Conference
"-- - cbgp6_-0red by the_ANS, SAE,ACS, AIAA, AS_, iEEE, andAIChE _
Boston, Massachusetts, August 4-9, 199i ...................
(NASA-TM-I04453)
COST TO CHOOSE THE
(NASA) 7 p
_ALANCING RELIABILITY AND Nqi-24952 m-_:_:;_'_'_:
BEST POWER SU_3SYSTEM
CSCL 05C
Unclas
G3/d3 0019929
==
https://ntrs.nasa.gov/search.jsp?R=19910015638 2020-03-19T18:04:59+00:00Z

BALANCING RELIABILITY AND COST
TO CHOOSE THE BEST POWER SUBSYSTEM
Ronald C. Suich
California State University
Fullerton, CA 92634
Richard L. Patterson
NASA Lewis Research Center
Cleveland, Ohio 44135
ABSTRACT
How does a design engineer or manager choose be-
tween a power subsystem with .990 reliability and a
more costly power subsystem with .995 reliability?
When is the increased cost of a more reliable power
subsystem justified?
A mathematical model is presented for computing total
(spacecraft) subsystem cost including both the basic
subsystem cost and the expected cost due to the failure
of the subsystem. This model is then used to determine
power subsystem cost as a function of reliability and
redundancy. Minimum cost and maximum reliability
and/or redundancy are not generally equivalent. Two
example cases are presented. One is a small satellite,
and the other is an interplanetary spacecraft.
INTRODUCTION
The methods described here can be applied to power
subsystems in launch vehicles, satellites, and on earth.
In addition, they can be utilized in many other types of
applications which do not necessarily involve power.
High reliability is not necessarily an end in itself. High
reliability may be desirable in order to reduce the statis-
tically expected cost due to a subsystem failure. Howev-
er, this may not be the wisest use of funds since the
expected cost due to subsystem failure is not the only
cost involved. The subsystem itself may be very costly.
We cannot consider either the cost of the subsystem or
the expected cost due to subsystem failure separately.
We therefore minimize the total of the two costs, i.e.,
the total of the cost of the subsystem plus the expected
cost due to subsystem failure.
We will be looking at subsystems which are part of a
larger main system, such as a power subsystem which is
a part of a main satellite system. In development of our
analyses, we will talk about subsystems and main
systems. In the examples, we will bring out power
subsystems and larger aerospace-related systems.
Expected value is an important ingredient in our quest
for finding the best power subsystem. Therefore, we'll
first consider the expected cost due to subsystem failure,
which is written as E{cost due to subsystem failure}.
As with all expected values, this number depends upon
both the dollar cost and the probability of its occur-
rence. Let c I be the dollar cost due to failure of the
subsystem, including all costs incurred by subsystem
failure (but not the cost of the subsystem itself). This
number could be the entire cost of the main system
(even greater in some circumstances) if failure of the
subsystem resulted in complete failure of the main
system. In other instances c1 would be less than the
cost of the main system, e.g., failure of the subsystem
resulted in only a partial failure of the main system.
Now the expected cost due to subsystem failure is c1
times the probabi_ty that this cost will be experienced.
Subsystem failure for us can only occur when the main
system is good. If the main system fails (for other than
failure of the subsystem), we'll not experience cost due
to subsystem failure. So, we discount the E{cost due to
subsystem failure} by multiplying by the reliability of
the main system. For example, let's consider a power
subsystem in a rocket. The rocket may explode on the
launch pad due to a fuel problem. Even if the power
subsystemwould have faded in _ght, we would not
experience this failure. Let rM be the reliability of the
main system (for other than failure of the subsystem),
and let rs be the reliability of the subsystem.
ThenE{costdueto subsystem failure}
= ciPr{subsystem failure J malnsystem
x Pr{main system good}
= cl(1-rs)r M **rMCl(1-rs).
good}
We can minimize this expected cost by building a sub-
system with an extremely low probability of failure, i.e.,
a subsystem with extremely high reliability. In this
situation it is not dear that we should build the most
reliable subsystem possible since this will minimize only
the expected cost due to subsystem failure but does not
consider the cost of building the subsystem itself. To
make this decision, we should not consider the two costs
separately. We therefore minimLze the total of the two
costs, i.e., the total of the lest of the subsystem plus the
exuected cost due to subsystem failure. The total cost
to be minimized is given by
C • cost of the subsystem + E{cost due to
subsystem failure}
= cost of subsystem + rMcl(1-rs).
In minimizing cost C we see that we are b01an_ng the
cost of the subsystem against the e_ected cost due to
subsystem failure.
NON-REDUNDANT POWER SUBSYSTEMS:
SELECTING THE BETTER OF
TWO ALTERNATIVES
Let's look at a microsatellite example. Suppose that we
have two possible power subsystems under consideration
for the microsatellite. Power subsystem 1, which costs
_200D00, has a .97 reliability. Power subsystem 2, with
a cost of $100,000, has a .94 reliability. Without further
information and analysis, there is no clear "best"power
subsystem, and the choice is often based upon the
amount budgeted for the power subsystem.
For further analysis, let Us say that the main micr0sat-
ellite system has a reliability (exclusive of the subsystem
under consideration) of rM = .96. We'll further assume
that failure of the power subsystem will result in a cost
of c1 = $10,000,000. Let us first compare the E{cost
due to subsystem failure} for each of the two power
subsystems. For power subsystem 1,
E{cost due to subsystem failure}
= rMqPr{subsystem failure}
= rMCl(l-rsl) -- .96 x $10,000,000 x .03 = $288,000.
For power subsystem 2,
E{cost due to subsystem failure}
= rMct(1-rsz ) = .96 x $10,000,000 x .06 = $576,000.
Since power subsystem 2 is less reliable than" power
subsystem 1, it has a higher _ed cost of failure.
However, since 2 is also less expensive to build, we
need to compare the overall cost, C., for 1 and for 2.
For power subsystem 1,
Csl = $'200,000 + $'288,000 = $488,000.
For power subsystem 2,
CS2 = $100,000+ $576,000= $676,000.
Since Csl < Cs2, we select power subsystem 1 over
power subsystem 2.
For further information on expected values or on
selecting the best subsystem in simple situations as
above, you may refer to [2]. We also note that the
methods contained in this paper do not consider time-
related functions, such as the cost of failure as a
function of mission time. Time-related functions are
covered in considerable depth in [2].
THE EFFECT OF REDUNDANCY:
K OUT-OF-N:G SUBSYSTEMS
In this article we'll direct our attention to a specific type
of subsystem, called a k-out-of-n:G subsystem. Such a
subsystem has n modules, of which k are required to be
good for the subsystem to be good. As an example
consider the situation where the engineer has a certain
power requirement. He may meet this requirement by
having one large power module, two smaller modules,
etc. The number of modules required is called k. For
example, the engineer may decide that k = 4. This
means that each module is 1/4 of the full required
power. Therefore, the subsystem must have 4 or more
modules for the full required power. The number of
modules used in the subsystem is called n. For
example, an n = 6 and k = 4 subsystem would have 6
modules each of 1/4 th power and thus would have the
output capability of 1.5 times the required power. The
engineer is free to choose n and k. Selection of the
different values of n and k results in diff_ent
subsystems, each with different costs and reliabililies.
Since each n and k yields a different subsystem with
different costs, we can therefore choose the subsyslem,
i.e., the n and k, which will minimize cost C.
Herewe'liassume perfect switching devices (if needed)
of negligible cost and independence of the modules of
the subsystem.
Although there are many variations of the k-out-of-n: G
model we'll present a few of these possibilities to give
some idea of its uses and potentialities.
MODEL 1 (k fixed and n variable)
The simplest k-out-of-n: G model is one where the
modules are independent and all have common
probability p of being good and common probability of
failure q = 1-p. Let X count the number of good
modules. Now E{cost due to subsystem failure}
= rMc1 Pr{subsystem failure}
+(:)= r,_c, Pr{ x < k} = ruc,__, p_q_-=
x=O
(1)
Recall that C = cost of subsystem + E{cost due to
subsystem failure}. We therefore need also to consider
the cost of the subsystem. First consider a simple situa-
tion where k is fixed. Here we are free to choose n.
Then n-k will be the redundancy or number of spares in
the subsystem. If each module costs c4 then the cost of
subsystem = nc4. Using this with (1) we obtain
C • cost of subsystem + E{cost due to subsystem
failure}
k-'l /31
-r,c,+r,,,:,.Eo[ +,,+e¢'"
We wish to fred the n which minimizes cost C.
As an example, let's look at an interplanetary
spacecraft. Consider the situation where k = 1, i.e.,
only one power module is required to be operational for
the power subsystem to be operational. Suppose that
the reliability of this inI.,,_gl.___]._ is .95, i.e., p = .95.
Let the reliability of the main spacecraft system for
other than failure of the subsystem be .9, i.e., rM = .9.
Suppose that the cost of one power module is 1
(hundred million dollars, for example) i.e., c4 = 1, and
that the cost due to failure of the power subsystem is 10
(hundred million), i.e., ca = 10.
Figure 1 Model 1 Example where ca = 10.
Figure I shows a plot of C for p ranging from .79 to .99
and n's of 1 through 4. If the reliability of a single
power module is .95, i.e., p = .95, note that n = 1 has
the lowest value of C. Therefore the best power
subsystem in this case is one with no spares. As a
matter of fact we can see from Figure 1 that the
subsystem of n = 1 has the lowest value of C for any p
> .87. Therefore, as long as the reliability of a single
power module is greater than .87, the best power
subsystem is one with no spares. If p < .87, then n =
2 has the lowest value of C. Therefore, if the individual
power module has reliability less than .87 (but greater
than .79), then the best power subsystem is one with
one spare. For values of p < .79, we should view the
graph over this range to find the best subsystem.
Now suppose that c I (cost due to failure of the
subsystem) increases to 50 (plot in Figure 2).
IIl.lS
g._
II .r._
7.1_[
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t :+-
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==s "-"-"-':_-"''"-':'"..-..-.7.7.7... .........
t.'=¢
$.dlB --l--l--l--l--l--l--l--l--l--
.lYre .lllJ .I_ Irll ._ .1911 .Sill ._ q_m .lJ'tll .qlgi
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Figure2 Model 1 Example where ca = 50.
We first note that if p = .95, then the n = 2 power
subsystem is the best. If we compare F'_ares 1 and 2
(at p = .95) we see that the larger value of c1 (in
Figure 2) requires a larger value of n. This prlndple
holds in general and makes sense. If the cost of
subsystem failure increases, then more redundancy is
required. Figure 2 reveals that if .83 < p < .98, then
the n = 2 power subsystem is the best. If p falls below
';83, then more redundancy is required (n = 3). If p >
.98, then no redundancy is required (n = 1).
MODEL 2 O:_othk and n variable)
Suppose in model 1 that we are also free to choose k in
our subsystem. If k is free to vary, then we*ll call this
model 2. Let c3 be the cost of a subsystem consisting
of exactly one module. Further suppose that the cost of
a subsystem with exactly k modules is c.3 g(k). Here
g(k) is the factor which measures the (generally)
increased cost of building a subsystem consisting of k
smaller modules rather than one large module. If g(k)
= 1 for all k, then a subsystem of k modules costs the
same as a subsystem consisting of a single module. Any
g(k) may be used. For example, if a subsystem of 2
smaller modules costs 4 times as much as a single
module subsystem then g(2) = 4. Therefore this
subsystem would cost c3 g(k) = c3 g(k) = 4<.3. If a
subsystem of 3 smaller modules costs 7 times as much
as a single module subsystem then g(3) =7, Other
values for g(k) may be defined in a similar manner.
Therefore, in the above example, g(1) = 1, 8(2) = 4,
g(3) = 7, etc. We also assume that each module in the
subsystem costs c3g(k)/k, which is 1/k th of the total
cost for k modules. Since we have a total of n modules
in the subsystem, then the cost of the subsystem =
nc3g(k)/k. Using this with (1) we obtain
C = cost of subsystem + E{loss due to subsystem
failure}
k-_[n /
For any particular situation with given values of c1, c.3,
rM, p and g(k) we select the n and k to minimize C as
given above. The n and k thus selected will be the
optimal subsystem.
Consider the example of a space electrical power
subsystem. A rough rule of thumb says that the cost of
smaller modules for a space electrical power subsystem
is proportional to the electrical power raised to the .7.
Thus, for this example g(k) ,, k(1/k) "7. Therefore, a
subsystem consisting of a single module ca.ft)able of full
power would cost c.3g(1) ,= c31(1/1)" = 1.0c3, a
subsystem consisting of 2 modules, each of 1/2 power,
would cost c.3g(2) = c37.(1/2)'7= 1.23c3 to build, etc.
An n = 3 and k - 2 subsystem, Le., one having 3
modules each of 1/2 power, would cost nc3 g(k)/k = 3
x 1.23%/2 = 1.85% to build.
Suppose for a small satellite that the cost due to power
subsystem failure, ¢t, is 100 (million dollars). Let the
reliability of the satellite (for other than failure of the
subsystem) be .99, i.e., rM= .99. Furthermore, the cost
of building a single module capable of full power is .5
million, i.e., c3 = .5. And last, let's say that each power
module has a reliability of .95.
From Figure 3 we see, at p = .95, that the n = 2, k =
1 power subsystem is the best (has the lowest value of
C). Note however, if p < .948, that the n = 4, k = 2
subsystem is the best. Additionally, note that this is a
much flatter curve. Unless we're fairly sure that p is
dose to .95, we might choose the n = 4 , k = 2
subsystem, since it gives us a relatively low value for C
over a wide range of p.
2
I1..11 r'm* I
Cl -140
¢I.'N r x
N-ill- I--
4.417 _ R. ,I II- 1' ....
'l.J _. N- | If* ':_oo
..tOo\ :" ;""
° o "\
c s.ml_ -o _x,. I
¢_ tt Ct OO %%
'P.p o ,.
_,12 I--1--1--1--1_
.l_@l .Ul@ ._Xl@ .tmm ._1@ ._11 .Ell .'Jill .qr/@l ._n .qpJ@l
P
Figure 3 Example for Model 2.
BASIC PROGRAMS
The authors will provide, upon request, copies of
BASIC programs (Ouickbasic 4.5) to both evaluate C
and also to search for an n and k which minimize C.
These programs are also appropriate for models other
than the two we_,e covered here (see [1] or [2] for more
detailed information on other models). If you wish a
copy on disk, please send a formatted disk floppy with
your request. We also note that all models may be
used when k is fixed by replacing nc3g(k)/k by nc4.
4 C;_t_,_L PgGE tSCF  ALWY
Additionally, the cost of launching the subsystem may
easilybe considered merely by including this cost in C
for the various models.
CONCLUSIONS
The methods brought forward in this paper can be used
to make very defmltive decisions in choosing the best
power (or other) subsystem from a number of
alternatives. Essentially we minimize the total of two
costs, Le., the total of the cost of the (power) subsystem
itself plus the expected cost due to the failure of the
(power) subsystem. A computer program which utilizes
the methods is available. Its output plots can yield very
dear, obvious, and straightforward decisions.
ACKNOWLEDGEMENTS
This work was supported by the NASA Office of
Aeronautics and Exploration Technology under Grant
Number NAG3-1100 from the NASA Lewis Research
Center. This manuscript represents U.S. Government
work and is not subject to copyright.
REFERENCES
[1] Suich, R. and Patterson, R., "K-out-of-n: G Systems:
Some Cost Considerations', to appear in the August
1991 issue of IEEE Transactions on Reliability.
[2] Suich, R. and Patterson, R., "How Much
Redundancy? - Some Cost Considerations, Including
Examples for Spacecraft Systems', NASA Technical
Memorandum 103197, 1990.
N/LSA
l_nJ #_wwAka She
Smm _a_ ,
1. _ No. 2. Ooveh_ AooMsk_ No.
NASA TM- 104453
4. Ttb end Sub_e
Balancing Reliability and Cost to Choose the Best Power Subsystem
7. _hoK=)
Ronald C. Suich and Richard L. Patterson
9. Pedorm_ _ Name scid_Addcm .....
National A_udcs and Space Administration
Lewis Research Center
Cleveland, Ohio 44135- 3191
12. Sponso_ngAger_ Narr_aridAddress
National Aeronautics and Space Administration
Washington, D.C. 20546-0001
Report Documentation Page
15.' _tary Notem
e. P.dorn_ _ Cod.
s. ewfom_ngO,_,/,mk_ _ No.
E-6294
10. Work Unit No.
506-41-4l
11. Contract 0¢ Grant No.
13. T_ of I_porl and Per_cl C_vemd
Technical Memorandum
14. spo_ J_r_y Code
ii
Prepared for the 26th Intersociety Energy Conversion Engineering Conference cosponsored by ANS, SAE, ACS, AIAA,
ASME, IEEE, and AIChE, Boston, Massachusetts, August 4-9, 1991. Ronald C. Suich, California State University,
Fullerton, California 92634; Richard L. Patterson, NASA Lewis Research Center. Responsible person, Richard L.
Patterson, (216) 433- 8166.
;le. Abstrad
How does a design engineer or manager choose between a power subsystem with .990 reliability and a more costly
power subsystem with .995 reliability? When is the increased cost of a more reliable power subsystem justified.'? A
mathematical model is presented for computing total (spacecraft) subsystem cost including both the basic subsystem
cost and the expected cost due to the failure of the subsystem. This model is then used to determine power subsystem
cost as a function of reliability and redundancy. Minimum cost and maximum reliability and/or redundmcy are not
generally equivalent. Two example cases are presented. One is a small satellite, and the other is an interplanetary
spacecraft.
17. Key Words (Suggested by Author(s))
Cost analysis; Reliability; Reundancy; Power; Space
18. Distrilx_m Statement
Unclassified - Unlimited
Subject Category 83
le..se_sty c,a,_. (o¢the mpo_) 20. SecurityCtaMif.(o¢_i, page) 21. No. ot
Unclassified Unclassified 6
| r
L
iASA FORM IIDQ, OCT M
"For sale by the National Technical Information Service, Springfield, Virginia 22161
22. Price"
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Balancing reliability and cost to choose the best power subsystem

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