with your host….

Dave Hy, the rocket scientist guy

(Not to be confused with the untalented and scientifically-challenged Bill Nye, the stupid guy)

489 Lecture #11Reliability Analysis and Cost Modeling

Readings:L&W Section 19.2L&W Chapter 20L&W Chapter 22

Reliability = The science of forestalling failureWhy is this needed?

Ans.: The vulnerability and inaccessibility of space systems(Things have to be kept working right on their own)

Penalty for lack of reliability: Economic Loss = How much a failure costs

“Reliability of components or systems” = the probability that they will not fail.

Reliability analysis uses the laws of probability.Probability has two meanings:

Relative frequency interpretation

“State of knowledge” (Laplace)Pr(A)

1.0A is certain to

happen

0.5Don’t know one way or the other

0Sure that A won’t

happen

Probability – What you need to know

Suppose: Pr[A] = Probability that event “A” will happenThen: Pr[Not A] = 1 - Pr[A]

If A and B are independent: Pr[A and B] = Pr[A]Pr[B]

If Fs = system failure probability, then: VF = Expected cost of failure=

FsVs

(Vs = economic resources needed to compensate for the loss of the s/c and its launch)

Using these simple facts, we can address:

Given the reliability of components, the reliability of the system

How to estimate component reliability

How to maximize reliability cost-effectively

Given the reliability of components, find the reliability of a subsystem

Example: A cargo drop system (CDS) for LBSS (Rk = probability of success for component “k”)

1. Carrier s/c exits lunar orbit, hovers at drop altitude

R1= 0.99

2. Start maneuver engines

R2= 0.98

3. Terrain sensors enabled

R3= 0.99

5. Hazard avoidance system enabled

R5= 0.95

4. Air bags deploy R4= 0.98

6. Cargo pallet released R6= 0.99

7. Carrier ascends R7= 0.98

9.Safe touchdown event R9= 0.99

Successful landing

8. Air bags survive impact R8= 0.99

Let RCDS denote the probability that the Cargo Drop System works. Because each of the 9 components has to work:

RCDS = Pr{ Component 1 works and Component 2 works and ……Component 9 works}

We can assume in this case that the success of each component is independent of all the others – CDS success is compounded of

independent events. Therefore:RCDS = Pr{ Component 1 works} x Pr{Component 2 works} x…

Pr{Component 9 works}= R1 x R2 x R3 x R4 x R5 x R6 x R7 x R8 x R9

= (0.99) (0.98) (0.99) (0.98) (0.95) (0.99) (0.98) (0.99) (0.99)= 0.850

Finally, the probability of failure, FCDS, is:FCDS = 1- RCDS = 1-0.850 = 0.15

Note: Despite just a few percent failure probability for each component, the failure probability for the whole CDS builds up to

15%

More Complex Situations …

Estimating component reliability: Use past statistics…

Methods to reduce Fs

±Fault avoidance: Use design margins, high quality parts, close inspection and

testing, etc. “Do it right the first time”

ΔFault tolerance: Design in the ability to operate after the failure of some

components.

ΩFunctional redundancy:After failure of a component, another component performs

the functions of the failed unit, even though its primary function is something different.

πGround support robustnessThe vehicle allows ground support to perform “workarounds”

to solve problems (frequently involves software manipulations)

1. Carrier exits lunar orbit, hovers at drop altitude

R1= 0.99

2. Start maneuver engines R2= 0.98

3. Terrain sensors enabled R3= 0.99

6. Cargo pallet released R6= 0.99

4. Air bags deploy R4= 0.98

7. Carrier ascends R7= 0.98

8. Air bags survive impact R8= 0.99

9.Safe touchdown event R9= 0.99

Successful landing

5.B Hazard avoidance system enabled

R5.B= 0.95

5.A Altitude reduction maneuver

R5.A= 0.95

Robust Hazard Mitigation

Subsystem (HMS)

Methods to reduce Fs – Example of Functional Redundacy

To calculate reliability, first concentrate on the “Robust HMS” block:

5.B Hazard avoidance system enabled

R5.B= 0.95

5.A Altitude reduction maneuver

R5.A= 0.95

When we have components in parallel, the whole block works if any one of the components works. To find the reliability in such cases, first find the probability of failure – call it FRobust HMS

Clearly, the whole block can’t fail unless all the components fail:FRobustHMS = Pr{ Component 5.A fails and Component 5.B fails}But the failure of each component is independent of the other component, so:FRobustHMS = Pr{ Component 5.A fails}x Pr{Component 5.B fails}But the probability of Component 5.A failing is 1-R5.A, etc.FRobustHMS = (1-R5.A)( 1-R5.B)

= (0.05)x(0.05) = 0.0025

Now, the reliability of this block, call it RRobustHMS is just:RRobustHMS = 1 - FRobustHMS = 0.9975

Note: By means of redundancy, we’ve decreased the likelihood of failure in the robust HMS from 5% to 0.25%.

To finish up our calculation of reliability for the new (partially redundant) CDS subsystem, just compute RCDS as before, but substitute RRobustHMS in place of R5:

REDL = R1 x R2 x R3 x R4 x RRobustHMS x R6 x R7 x R8 x R9 = (0.99) (0.98) (0.99) (0.98) (0.9975) (0.99) (0.98) (0.99) (0.99)

= 0.893So, we’ve raised the overall reliability by over 4%

____________________________________________________________

Note: When you have a calculation such as R1 x R2 x R3 x R4 x R5 x R6 x R7 x R8 x R9, where there are many factors and each component reliability is close to unity, you can get a quick (but good) approximation by simply

subtracting the sum of the failure probabilities from unity;REDL = R1 x R2 x R3 x R4 x R5 x R6 x R7 x R8 x R9

= (1 – F1) (1 – F2) …(1 – F9)= 1 – (F1+ F2+… +F9) + products of small numbers

≈ 1 – (F1+ F2+… +F9)… which is a lot simpler (and less prone to roundoff error) than multiplying many

numbers, each one very slightly less than 1.0.(In the present case, the approximation gives RCDS ≈ 0.888)_______________________________________________

Cost concepts showing optimum reliability budget [Hecht, 1973]. F0 is the probability of failure for the baseline

system.(From L&W, 2nd Edition)

Cost Estimation ProcessCost Estimation Process(L&W Ch(L&W Chapterapter 20) 20)

××First compose a WBS (Work Breakdown Structure),First compose a WBS (Work Breakdown Structure), using Fig. 20-2 as a using Fig. 20-2 as a modelmodel

÷÷L&W distinguish three life cycle phases:L&W distinguish three life cycle phases:

>>Research, Development, Test & Evaluation (RDT&E)Research, Development, Test & Evaluation (RDT&E)

==Production Production Theoretical First Unit (TFU) Theoretical First Unit (TFU)

<<Operations and Maintenance Operations and Maintenance (O&M)(O&M)

We We emphasize here : RDT&Eemphasize here : RDT&E and TFU. and TFU. RDT&E as unRDT&E as undderstood erstood in thein the context of L&Wcontext of L&W includes the development of includes the development of reasonably maturreasonably maturee technology technology elements up to elements up to the level required for the the level required for the flight.flight. It does not refer to the basic technology develo It does not refer to the basic technology developmentpment and flight and flight

validatiovalidationn demos demos

Cost Estimating Methods

1)Detailed, bottom-up estimating

2)Analogy-based estimating (find a similar item then try to adjust for differences)

3)Parametric estimating. Use math relations between design parameters and cost that are compiled from statistics of previous programs. These relations are

called “CERs” (Cost Estimating Relationships)

For preliminary design, (3) is best. But (3) is subject to caveats:

CER’s only applicable to the range of historical data

Parametric estimating not satisfactory for estimating items involving major technological advancements or

fundamental paradigm shifts. (see L&W, p.788)

Recommended Process

First, use (3) for initial estimates

Then revisit system elements that are new or innovative and use method (2) (or even (1))

For the new and innovative elements, we will request detailed “bottom up” data on similar items from JSC

B) Apply the factors given in Table 20-8 to the RDT&E CERs to allow

for development heritage.

C) Compute software costs using Table 20-10

D) Next estimate ground segment and operations costs using Table

20-11

For the Parametric estimating stage, take the following steps:

A)Use Table 20-6 (for small sats.)*, combined with Table

20-9 (that breaks out the fractions due to nonrecurring costs versus recurring costs, to give the RDT&E versus TFU

costs)

E)Communications equipment: Table 20-13

F)Launch costs: Table 20-14

G)Finally, be sure to apply the inflation factors relative to the year 2000 in Table 20-1.

______________________* We use CERs for small sat.s because, although the

statistical data base is smaller, it pertains to programs wherein consistent efforts were made to reduce costs

Cost-Risk Analysis

MLE = “Most likely estimate” = mean value of cost

SE = “Standard Error” = The standard deviation about the mean

Cost risk due to uncertainties arising from technical innovations are estimated by the “Technology readiness level” (TRL) – a graduated scale introduced by NASAFor this project, we need to produce two main outputs:

1)Costs for RDT&E and for the TFU, with uncertainty estimates.

2)Identify the basically new technologies that will be needed to implement the design and estimate their (relatively low) TRLs.

By 1985, OSCAR 10 and others demonstrated store and forward communications. Satellites could become “flying bulletin boards”. A

single small satellite could provide global mail service!

Low Cost, Small Satellite Design(L&W Chapter 22)

Before 1990, the trend was a small number of very capable and

expensive spacecraft

Cheaper access to space:NASA’s Get Away Special program for small (<68kg)

payloadsAriane rocket’s ASAP (Ariane Structure for Attached

Payloads). Up to 6 small sats. Carried on the ASAP ring (each up to 50 kg)

New trend: Miniature satellite technology or “microspace” (see L&W Table 22-1) Factors in emergence;

Low cost access to spaceDigital comm. Equipment that can fit into a GAS

containerDigital store and forward comm..

(“An 8.5 kg microsatellite’s ability was unavailable in half ton satellites of 15 years ago”)