Trident reliability

“De Omnibus Dubitandum Est”. René Descartes

Abstract

This paper describes various approaches for estimating Trident I and Trident II SLBM launch reliability using binomial model both in frequentist’s and Bayesian math paradigms. The results and opinions expressed by author do not reflect any official point of view.

Framework for research

Mysterious light

One of November nights in 2015 there was something unusual in the sky: A mysterious light that streaked across San Diego’s night sky Saturday, visible as far away as Nevada and Arizona, was a Trident missile test-fired by the Navy. The test was part of a scheduled, on-going system evaluation test, said Cmdr. Ryan Perry with the Navy’s Third Fleet. ¹ So what’s it all about? Is Trident missile so reliable?

Trident family

Here are some interesting facts about Trident program. The Trident was built in two variants: the I (C4) UGM-96A and II (D5) UGM-133A; however, these two missiles have little in common. While the C4, formerly known as EXPO (Extended Range Poseidon), is just an improved version of the Poseidon C-3 missile, the Trident II D5 has a completely new design (although with some technologies adopted from the C4). The C4 and D5 designations put the missiles within the “family” that started in 1960 with Polaris (A1, A2 and A3) and continued with the 1971 Poseidon (C3). Both Trident versions are three-stage, solid-propellant, inertially guided missiles, and both guidance systems use a star sighting to improve overall weapons system accuracy. ²

The Trident II D5 missile system, which is manufactured in the US by Lockheed Martin, is a three-stage solid-fuel inertially-guided rocket approximately 13m long, nearly 2m in diameter and weighing 60 tonnes. It has a range of between 6,500km and 12,000km, dependent upon payload, and its accuracy is measured in metres. The precision of ballistic missiles is measured by what is known as the circular error probability or CEP, which is the radius of the circle within which half the strikes would impact. The CEP for Trident is reported to be around 90 metres: thus, each warhead would impact within 90 meters of the target point with a probability of 50%. Each missile is technically capable of carrying up to 12 warheads. The missile’s own inertial guidance system then takes over. After the third rocket motor has separated, the warhead carrier takes a star sighting to confirm the missile’s position and then manoeuvres to a point at which the warheads can be released to free-fall onto their targets. Each missile has a MIRV (multiple independently targetable re-entry vehicle) capability which enables it to engage multiple targets simultaneously. ³

Problem

We want to compare two US Navy programs connected with testing submarine-launched ballistic missiles (SLBM): TRIDENT I (C4) and TRIDENT II (D5). The main reason to do such comparison is trivial and plain as obvious too. Here is a quote from CBO report ⁴: “The current strategic nuclear forces—consisting of submarines that launch ballistic missiles (SSBNs), land-based intercontinental ballistic missiles (ICBMs), long-range bombers, and the nuclear weapons they carry—are reaching the end of their service lifetimes. Over the next two decades, the Congress will need to make decisions about the extent to which essentially all of the U.S. nuclear delivery systems and weapons will be modernized or replaced with new systems. CBO estimates that over the 2015–2024 period, the Administration’s plans for nuclear forces would cost $348 billion, an average of about $35 billion a year.”

Also UK as a member of NATO and US partner inspite of BREXIT is facing real problems with nuclear deterence strategy where Trident D5 is the conerstone of nuclear arsenal. Here we quote BASIC (British American Security Information Council) conclusions: “There have been several recent attempts to estimate the lifetime cost of the follow-on system. The Liberal Democrats in September 2006 estimated a capital and operating cost for the Successor system as £76 billion over its development and 30 year deployment lifetime. In 2009 Greenpeace estimated it to be at least £97 billion. The BASIC Trident Commission’s second briefing authored by defence economist Keith Hartley in March 2012 estimated a total cost of some £89.6 billion.” ⁵

So money talks for both US and UK as far as Trident D5 is concerned!

Main assumptions

Now we should do some assumptions to narrow the scope of research:

1. We consider random experiment with exactly two possible outcomes “success” and “failure”, in which the probability of success is the same every time the experiment is conducted ⁶.
2. We need some example of a real experiment. Let’s see it here: ⁷, ⁸.
3. It is known from the book by Robert C. Aldridge ⁹ that reliability of Trident missile is 80% i.e. probability for successful launch from underwater submarine. We’ll try to prove this estimate.
4. There are four possible events during test flight: success, partially success, failure and non-test. We consider “partially success” event as a “failure” and do not count any “non-test” event.

R packages

Here we use the following packages of R language for required calculations:

library(binom)
library(pwr)
library(ggplot2)
#From Rasmus Baath 
#Bayesian First Aid Alternative to the Binomial Test
#http://www.sumsar.net/blog/2014/06/bayesian-first-aid-prop-test/
library(rjags)

## Loading required package: coda

## Linked to JAGS 4.1.0

## Loaded modules: basemod,bugs

library(mcmc)
library(stringr)
source(file = "bayes_prp_tst.R")
source(file = "binom_test.R")
source(file = "r_jags.R")
source(file = "generic.R")
#From Survivability/Lethality Analysis Directorate, ARL
#"Binomial Distribution: Hypothesis Testing, Confidence Intervals (CI),
#and Reliability with Implementation in S-PLUS"
source(file = "bino.rel.size.R")
source(file = "bino.ci.R")
source(file = "bino.rel.tab.R")

Frequentist paradigm

Here we produce some point estimates and confidence intervals for Trident C4 and D5 test flights based on frequentist paradigm.

TRIDENT I (C4)

We begin with Trident C4 - the first option of modern US SSBN nuclear force. Here are some quotes from FAS reports. The C4 missile development flight test program commenced on 18 January 1977 with the successful launching of C4X-1 from the flat pad (25C) at Cape Kennedy. This was followed by 17 additional C4X launches from pad 25C. Of these 18 flights, 15 were successful, 1 was a partial success, 1 was a failure and 1 was a no test (due to ground support equipment error). The C4X program completed on 23 January 1979. This was followed by the firing of 7 PEMs (Performance Evaluation Missiles versus Production Evaluation Missiles) from the SSBN-657 during the period 10 April 1979 through 31 July 1979. PEM-1 had a first stage motor failure but PEMs 2 through 7 were successful. It was this successful flight test program that lead to SECNAV James Woolsey to comment in January 1980 that TRIDENT I (C4) was “the most successful submarine launched ballistic missile development program to date.”

Moreover, the development flight test program was progressing so satisfactorily that after the 12th flight test of the C4X was successfully conducted, Lockheed on 19 May 1978 proposed that the total number of development (C4Xs and PEMs) be reduced from 30 to 25. Following the 16th flight test which was successful, the Director of SSPO determined on 27 November 1978 that the technical objectives of the C4 development program had been met and that the development flights could be reduced from 30 to 25 flight tests (18 C4Xs and 7 PEMs).¹⁰

So we got for C4 flight test sample of 24 trials (without 1 “non-test”) and 21 events as “success” together with 3 events as “failure” counting “partially successful” event as “failure” too. Here is our first point estimate of the probability of success for C4: P_c4=21/24=0.875. Not so bad! But why SECNAV James Woolsey was so confident about the development flight test program for c4? Let’s see.

Null hypothesis

Here we postulate our null hypothesis for C4 H₀| P=0.9 and alternative - H₁| P<0.9. To prove it or not we use Z-test and binomial test. ¹¹

Normal approximation Z-test

Frequentists presume that there is a knowable state of the world that corresponds to some estimated parameter or quantity in a model ¹². For frequentists, a probability is a measure of the the frequency of repeated events where parameters \(\theta\) are fixed (but unknown), and data \(D\) are random. For normal ¹³ approximation the binomial distribution ¹⁴ we use the following rule ¹⁵:

\(\theta=N(\mu,\sigma)\)

\(\mu \pm 3\sigma = np\pm 3{\sqrt {np(1-p)}}\in [0,n]\)

21*0.875+3*sqrt(24*0.875*(1-0.875))

## [1] 23.23556

21*0.875-3*sqrt(24*0.875*(1-0.875))

## [1] 13.51444

So everything within 3 standard deviations of its mean is within the range of possible values and we have Z-test ¹⁶,¹⁷:

\(Z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\)

Let’s do it!

z.test <- function(phat,p,n ){
q=1-p
z=(phat-p)/sqrt(p*q/n)
p.value=pnorm(z,p*q,n*p*q)
return(p.value)
}
z.test(21/24,0.9,24)

## [1] 0.4087854

So Z-test gives \(p.value=0.4087854\) that is we can’t reject H₀| P=0.9. What about binomial test?

Binomial test

The main idea of classical binomial test is to estimate 95% (for instance) confidence interval for population probability of \(X\) events during \(N\) trials \(P(X|N,Pe)\) given a priori probability of elementary event \(Pe=X/N\) for sample as \(P(X|N,Pe)=\sum_{K=1}^XC_{N}^{K}P^{K}(1-P)^{N-K}\), where \(C_{N}^{K}=\frac{K!}{N!(N-K)!}\).¹⁸ For binomial test we use binom.test function.

graph <- function(n,p){
  x <- dbinom(0:n,size=n,prob=p)
  barplot(x,names.arg=0:n, main=sprintf(paste('Binomial distribution:','N=', n, ', Pe=', p)),
          col = "lightblue",xlab = "Number of events, X", ylab = "P(X|N,Pe)")
}
graph(24,0.9)

binom.test(21,24,0.9,alternative = "less") #C4 null (P=0.9) versus alternative (P<0.9) hypothesis test

## 
##  Exact binomial test
## 
## data:  21 and 24
## number of successes = 21, number of trials = 24, p-value = 0.4357
## alternative hypothesis: true probability of success is less than 0.9
## 95 percent confidence interval:
##  0.0000000 0.9650467
## sample estimates:
## probability of success 
##                  0.875

#The same result can be produced by pbinom or dbinom
#pbinom(21,24,0.9)
#sum(dbinom(1:21,24,0.9))

As we see the result of exact binomial test for C4 is inspiring: p-value = 0.4357. And what about the successor D5?

TRIDENT II (D5)

Now we come up to Trident D5 - the successor of Trident C4 and the current option of US SLBM nuclear force. Here are some quotes from FAS reports. The D5 Development Flight Test Program originally consisted of 20 D5X missile flat pad flights and 10 PEM flights from a TRIDENT SSBN. Flight testing began in January 1987 and in 1988. The program was reduced to 19 D5Xs and 9 PEMs. Of the 15 tests conducted as of September 30, 1988, 11 were successful, 1 was partially successful, 2 were failures, and one was a “no-test” [the 15th flight test was destroyed by command destruct early in its flight while the missile was performing normally at the time the decision was made to destruct: therefore, the flight was a “no-test”]. ¹⁹

The first submarine launch was attempted by USS Tennessee (SSBN-734, the first D5 ship of the Ohio class) in March 1989. The latter failed because of problems with the first stage engine nozzle, and these problems delayed the IOC (Initial Operational Capability) of the Trident II until March 1990.²⁰

The U.S. Navy conducted successful test flights Nov. 7 and 9 2015 of two Trident II D5 Fleet Ballistic Missiles built by Lockheed Martin (NYSE: LMT). The world’s most reliable large ballistic missile, the D5 missile has achieved a total of 157 successful test flights since design completion in 1989. To support the U.S. Navy Strategic Systems Programs, Lockheed Martin is incorporating modernized electronics technology to cost effectively prolong the service life of the D5 missile design on current and next-generation submarine platforms. These two missile flights formally qualify the new flight control and interlocks electronics packages for deployment in 2017. The modernized avionics subsystems, which control key missile functions during flight, enable missile life extension through 2042²¹.

The U.S. Navy’s Fleet Ballistic Missile program flew into the future this week with its first 3-D-printed missile component. A part called a connector backshell made its debut as part of the Navy’s successful test flights March 14 to 16 of three Trident II D5 Fleet Ballistic Missiles built by Lockheed Martin (NYSE: LMT).The D5 missile, the world’s most reliable large ballistic missile, now has achieved a total of 160 successful test flights since design completion in 1989²².

The last flight test of D5 conducted by US Navy. ²³

So we got for D5 flight test sample of 27 trials (without 1 “non-test”) and 23 events as “success” together with 4 events as “failure” counting “partially successful” event as “failure” too. The estimate of the probability of success for D5 flight test sample looks like: P_D5=23/27=0.852. We see little decrease in D5 reliability compared to C4 according to flight tests. Such things happen with new missiles: D5 is larger than C4 while the latter is Poseidon C3 modification in fact.

Null hypothesis

As mentioned above we postulate null hypothesis for D5 the same way like for C3: H₀| P=0.9 and alternative - H₁| P<0.9. To prove it or not we use Z-test and binomial test.

Normal approximation Z-test

z.test(23/27,0.9,27)

## [1] 0.3518886

So Z-test gives \(p.value=0.3518886\) that is we can’t reject H₀| P=0.9. What about binomial test?

Binomial test

graph(27,0.9)

binom.test(23,27,0.9,alternative = "less") #D5 null (P=0.9) versus alternative (P<0.9) hypothesis test

## 
##  Exact binomial test
## 
## data:  23 and 27
## number of successes = 23, number of trials = 27, p-value = 0.2821
## alternative hypothesis: true probability of success is less than 0.9
## 95 percent confidence interval:
##  0.0000000 0.9477668
## sample estimates:
## probability of success 
##              0.8518519

#The same result can be produced by pbinom or dbinom
#pbinom(23,27,0.9)
#sum(dbinom(1:23,27,0.9))

Not so bad result for successor: p-value = 0.2821! Our H₀| P=0.9 can’t be rejected or we can reject it as true making error type-I with probability 0.2821.

Confidence intervals for flight tests

There is one more interesting estimate - confidence intervals for both SLBM flight tests. We can use both confidence level and reliability tables ²⁴, ²⁵ and binom.test function for this purpose too.

C4.fit<-binom.test(21,24,0.9,alternative = "two.sided") # C4 confidence interval
C4.fit

## 
##  Exact binomial test
## 
## data:  21 and 24
## number of successes = 21, number of trials = 24, p-value = 0.7282
## alternative hypothesis: true probability of success is not equal to 0.9
## 95 percent confidence interval:
##  0.6763886 0.9734407
## sample estimates:
## probability of success 
##                  0.875

D5.fit<-binom.test(23,27,0.9,alternative = "two.sided") # D5 confidence interval
D5.fit

## 
##  Exact binomial test
## 
## data:  23 and 27
## number of successes = 23, number of trials = 27, p-value = 0.3403
## alternative hypothesis: true probability of success is not equal to 0.9
## 95 percent confidence interval:
##  0.6626891 0.9581126
## sample estimates:
## probability of success 
##              0.8518519

We see 95% confidence intervals according to the flight test samples for C4 reliability (0.6763886 < P_C4 < 0.9734407) and for D5 reliability too (0.6626891< P_D5 < 0.9581126). Very good news for both SLBMs! Next we’ll try some new method approved by US DOD.

Power of test

We estimated above the probability of error type-I (p-value or \(\alpha\)) i.e. rejecting null hypothesis when it is true. But what if it is not true at all? Now we must estimate the probability of error type-II or \(\beta=1 - power.\)

C4

binom.power(p.alt = 0.5,n = 24,p = 0.9,alpha = 0.05,phi = 1,alternative = "less")

##     alpha 
## 0.9973643

D5

binom.power(p.alt = 0.5,n = 27,p = 0.9,alpha = 0.05,phi = 1,alternative = "less")

##     alpha 
## 0.9992586

pw<-binom.power(p.alt = 0.5,n = 1:30,p = 0.9,alternative = "less")
pwd<-data.frame(pww=pw,nn=1:30)
ggplot(pwd, aes(nn, pww)) + geom_line() + labs(title ="Power of binomial test, Pe=0.9", x = "Number of trials", y = "Power") 

Well, both test flights (C4, D5) have 99% power of test i.e. the probability of error type-II \(\beta\)=1% only. Good news for decision makers in Pentagon and tax payers all over US.

ARL-TR-5214

Army Research Laboratory published some interesting paper on binomial test ²⁶. Let’s try to use this official guide for the determination of sample size and maximum permissible number of failures (nf) required to establish a specific reliability (probability of success) with given probability (confidence).

dd21<-as.data.frame(bino.rel.tab(n = 30,r = 1:30,type = 1))
dd22<-as.data.frame(bino.rel.tab(n = 30,r = 1:30,type = 0))

plot(x=dd21$r,y = dd21$`80%`,type = "l",main = "ARL-TR-5214 upper and low bounds for P(r|n), n=30",xlab = "Number of success, r", ylab = "Probability of success, P", col="red") 
lines(x=dd21$r,y=dd21$`90%`,type="l", col="green")
lines(x=dd21$r,y=dd21$`95%`,type="l", col="blue")
lines(x=dd21$r,y=dd21$`97.5%`,type="l", col="yellow")
lines(x=dd21$r,y=dd21$`99%`,type="l", col="magenta")
lines(x=dd21$r,y=dd21$`99.5%`,type="l", col="pink")
lines(x=dd22$r,y=dd22$`80%`,type="l", col="red")
lines(x=dd22$r,y=dd22$`90%`,type="l", col="green")
lines(x=dd22$r,y=dd22$`95%`,type="l", col="blue")
lines(x=dd22$r,y=dd22$`97.5%`,type="l", col="yellow")
lines(x=dd22$r,y=dd22$`99%`,type="l", col="magenta")
lines(x=dd22$r,y=dd22$`99.5%`,type="l", col="pink")
legend("topleft",  c("80%","90%",  "95%", "97.5%", "99%", "99.5%"), fill = c("red","green", "blue","yellow", "magenta", "pink"))

grid()

ARL-TR-5214 procedures are based on the method of Clopper-Pearson (CP) ²⁷ and the likelihood ratio (LR) technique properties of the binomial distribution to derive interval estimates, which are in turn used in inference.

bino.rel.tab(n = 24,r = 21,type = 0)# low limit quantiles for C4 flight test sample

##       n  r     p      80%       90%       95%     97.5%       99%
## [1,] 24 21 0.875 0.782116 0.7424618 0.7077268 0.6763886 0.6388304
##          99.5%
## [1,] 0.6127353

bino.rel.tab(n = 24,r = 21,type = 1)# upper limit quantiles for C4 flight test sample

##       n  r     p       80%       90%       95%     97.5%       99%
## [1,] 24 21 0.875 0.9353829 0.9531772 0.9650467 0.9734407 0.9812071
##          99.5%
## [1,] 0.9854078

bino.rel.tab(n = 27,r = 23,type = 0)# low limit quantiles for D5 flight test sample

##       n  r         p       80%      90%       95%     97.5%       99%
## [1,] 27 23 0.8518519 0.7635656 0.725464 0.6923737 0.6626891 0.6272696
##          99.5%
## [1,] 0.6027344

bino.rel.tab(n = 27,r = 23,type = 1)# upper limit quantiles for D5 flight test sample

##       n  r         p       80%       90%       95%     97.5%       99%
## [1,] 27 23 0.8518519 0.9137514 0.9337829 0.9477668 0.9581126 0.9681962
##          99.5%
## [1,] 0.9739553

bino.rel.size(nf = (24-21),p = 0.95)# C4 required additional sample size after test flight

## $par
## $par$nf
## [1] 3
## 
## $par$p
## [1] 0.95
## 
## $par$g
## [1] 0.95
## 
## 
## $n
## [1] 153
## 
## $p
## [1] 0.950105
## 
## $g
## [1] 0.9505552
## 
## $n1
## [1] 152
## 
## $p1
## [1] 0.9497819
## 
## $g1
## [1] 0.9488355

bino.rel.size(nf = (27-23),p = 0.95)#D5 required additional sample size after test flight

## $par
## $par$nf
## [1] 4
## 
## $par$p
## [1] 0.95
## 
## $par$g
## [1] 0.95
## 
## 
## $n
## [1] 181
## 
## $p
## [1] 0.9501447
## 
## $g
## [1] 0.9508374
## 
## $n1
## [1] 180
## 
## $p1
## [1] 0.9498717
## 
## $g1
## [1] 0.9492507

bino.rel.tab(n = 161,r = 154,type = 0)# low limit quantiles for C4 program

##        n   r         p       80%       90%       95%     97.5%       99%
## [1,] 161 154 0.9565217 0.9370832 0.9279715 0.9198894 0.9124721 0.9033755
##          99.5%
## [1,] 0.8968935

bino.rel.tab(n = 161,r = 154,type = 1)# upper limit quantiles for C4 program

##        n   r         p       80%       90%       95%     97.5%       99%
## [1,] 161 154 0.9565217 0.9704807 0.9756481 0.9794192 0.9823434 0.9853585
##         99.5%
## [1,] 0.987187

bino.rel.tab(n = 164,r = 160,type = 0)# low limit quantiles for D5 program so far

##        n   r         p       80%       90%       95%     97.5%       99%
## [1,] 164 160 0.9756098 0.9593581 0.9518503 0.9450583 0.9387311 0.9308645
##          99.5%
## [1,] 0.9251953

bino.rel.tab(n = 164,r = 160,type = 1)# upper limit quantiles for D5 program so far

##        n   r         p       80%       90%       95%     97.5%       99%
## [1,] 164 160 0.9756098 0.9859649 0.9893201 0.9916269 0.9933154 0.9949465
##          99.5%
## [1,] 0.9958718

Very interesting result! C4 flight test program had passed all necessary additional 152 trials, while D5 had not already passed required 181 trials. We got for C4 program estimated 95% confidence interval in terms of low and upper limits as 0.9198894 < P_C4 < 0.9794192 while estimated 95% confidence interval in terms of low and upper limits for D5 program so far looks like 0.9450583 < P_D5 < 0.9916269. If ARL procedures are true we can come to the conclusion that there is vital progress in reliability of Trident II program as compared to Trident I in terms of upper limit 95% quantiles: P_c4,95%=0.9794192 and P_D5,95%=0.9916269.

Bayesian paradigm

“The probability of any event is the ratio between the value at which an expectation depending on the happening of the event ought to be computed, and the value of the thing expected upon its happening”. Thomas Bayes

Bayesians are more interested in the connection between evidence and a pattern of beliefs than in evidence and the true state of the world, which is unknown and possibly unknowable ²⁸. For Bayesians, a probability is a measure of the degree of certainty about values where parameters are random and data are fixed. So we apply Bayesian approach (model) with fixed data \(D\) while parameters \(\theta\) can vary: \(P(\theta|D) = P(D|\theta)P(\theta)/P(D)\) ²⁹, ³⁰, ³¹ that is \(Posterior = Prior * Likelyhood/Evidence\)

Suppose that we choose a \(Beta(\alpha, \beta)\) prior for \(\theta\)

where \(Beta(\alpha, \beta)={\dfrac {\Gamma (\alpha+\beta)}{\Gamma (\alpha)\,\Gamma (\beta)}}\)

\(\theta \in [0,1]\) , \(\alpha, \beta >0\) and \(\Gamma(Z) = \int\limits_{0}^\infty t^{Z-1}e^{-t}dt\)

\(P(\theta|\alpha,\beta) = Beta(\alpha, \beta)\theta^{\alpha-1}(1-\theta)^{\beta-1}\)

Posterior is now

\(P(\theta|k,n) \propto \theta^{k+\alpha-1}(1-\theta)^{n-k+\beta-1}\)

Posterior is again proportional to a Beta:

\(P(\theta|k,n) = Beta(\alpha +k, \beta +n-k)\)

\(P(\theta|k,n) = {\dfrac {\Gamma (\alpha+\beta +n)}{\Gamma (\alpha+k)\, \Gamma (\beta + n - k)}}P^{\alpha+k-1}(1-P)^{n-k +\beta -1}\)

where \(k\) is a number of outcomes as “success” and \(n\) - number of all trials.

Beta is the conjugate prior for binomial model: posterior is in the same form as the prior. To choose prior parameters, we think as follows: observe \(\alpha\) successes in \(\alpha + \beta\) prior “trials”. Prior “guess” for \(\theta\) is \(\alpha / (\alpha+\beta)\).³² Posterior distributions of parameters can be estimated using Markov chain Monte Carlo (MCMC) algorithms. Let’s do it!

Example without MCMC

Here is an example ³³ of using prior \(Beta(\alpha, \beta)\) for posterior binomial distribution of success outcomes given sample with 150 events.

set.seed(324807)
alpha=21 #number of prior successes 
beta = 3 #number of prior failures 
x<- rbinom(n = 150,size = 1,prob = 0.9) #the data

post.samp<-rbeta(n = 10000,shape1 = (sum(x==1)+alpha-1),shape2 = (sum(x==0)+beta-1)) #sample out of posterior
mean(post.samp) # mean probability for sample

## [1] 0.8897769

sqrt(var(post.samp)) #st.deviation for sample

## [1] 0.02361603

hist(post.samp, col = "grey",main = "Histogram of posterior sample distribution",xlab = "Posterior sample")
legend("topleft",  c("alpha=23, beta=3", "alpha=0, beta=0"), fill = c("red","blue"))
theta.plot<-seq(from=0.7,to=0.99,by=0.01) #values of theta to plot

post.plot<-dbeta(x=theta.plot,shape1 = (sum(x==1)+alpha-1),shape2 = (sum(x==0)+beta-1)) #posterior density values

lines(100*post.plot~theta.plot,type="l",col="red")

#compare to flat prior
post.plot.2<-dbeta(x=theta.plot,shape1 = sum(x==1),shape2 = sum(x==0)) #posterior density values
lines(100*post.plot.2~theta.plot,type="l",col="blue")

Credible intervals for flight tests

Given observed data \(D\), there is a 95% probability that the true value of \(\theta\) falls within the credible region (interval).

# C4
bstc4<-bayes.binom.test(21,24,0.9)
summary(bstc4)

##   Data
## number of successes = 21, number of trials = 24
## 
##   Model parameters and generated quantities
## theta: the relative frequency of success
## x_pred: predicted number of successes in a replication
## 
##   Measures
##          mean    sd  HDIlo  HDIup %<comp %>comp
## theta   0.847 0.069  0.709  0.964  0.764  0.236
## x_pred 20.343 2.367 16.000 24.000  0.000  1.000
## 
## 'HDIlo' and 'HDIup' are the limits of a 95% HDI credible interval.
## '%<comp' and '%>comp' are the probabilities of the respective parameter being
## smaller or larger than 0.9.
## 
##   Quantiles
##         q2.5%   q25% median   q75% q97.5%
## theta   0.689  0.806  0.855  0.897  0.955
## x_pred 15.000 19.000 21.000 22.000 24.000

plot(bstc4)

# D5
bstd5<-bayes.binom.test(23,27,0.9)
summary(bstd5)

##   Data
## number of successes = 23, number of trials = 27
## 
##   Model parameters and generated quantities
## theta: the relative frequency of success
## x_pred: predicted number of successes in a replication
## 
##   Measures
##          mean    sd  HDIlo  HDIup %<comp %>comp
## theta   0.829 0.068  0.694  0.949  0.853  0.147
## x_pred 22.396 2.625 18.000 27.000  0.000  1.000
## 
## 'HDIlo' and 'HDIup' are the limits of a 95% HDI credible interval.
## '%<comp' and '%>comp' are the probabilities of the respective parameter being
## smaller or larger than 0.9.
## 
##   Quantiles
##         q2.5%   q25% median   q75% q97.5%
## theta   0.675  0.788  0.835  0.878   0.94
## x_pred 17.000 21.000 23.000 24.000  27.00

plot(bstd5)

So we got Bayesian 95% credible intervals for Trident I C4 and Trident II D5 test flights:

Probability	Trident I C4	Trident II D5
Lower	0.709	0.694
Mean	0.847	0.829
Upper	0.964	0.949

Compare these results with ARL estimates:

Probability	Trident I C4	Trident II D5
Lower	0.707	0.785
Mean	0.875	0.925
Upper	0.965	0.987

Compare these results with binom.test estimates:

Probability	Trident I C4	Trident II D5
Lower	0.676	0.662
Mean	0.875	0.852
Upper	0.973	0.958

C41.dat<-data.frame(prob=c(0.709, 0.847, 0.964, 0.707, 0.875, 0.965, 0.676, 0.875, 0.973),
                names=c("Bayes","Bayes","Bayes","ARL","ARL","ARL","binom","binom","binom"))

D51.dat<-data.frame(prob=c(0.694, 0.829, 0.949, 0.785, 0.925, 0.987, 0.662, 0.852, 0.958),
                  names=c("Bayes","Bayes","Bayes","ARL","ARL","ARL","binom","binom","binom"))

par(mfrow=c(1,2))
boxplot(horizontal = TRUE, data=C41.dat, prob~names,col=c("red","green","blue"),main="Trident I C4",boxwex=0.3)
boxplot(horizontal = TRUE, data=D51.dat, prob~names,col=c("red","green","blue"),main="Trident II D5",boxwex=0.3)

ARL is very close to Bayesian estimate in terms of lower and upper bounds of probability value for C4 but not for D5! There is one more interesting result for D5 - lower and upper bounds of probability value produced by Bayes and binom.test are very close. Now let’s compare C4 and D5 test flights reliability using bayes.prop.test function.

suc <- c( 21, 23)
all <- c( 24, 27)

fit<-bayes.prop.test(suc,all)
plot(fit)

So we got 95% credible interval for D5 reliability (0.69 - 0.95) compared to C4 reliability (0.71 - 0.97) - very close intervals for two test flights samples.

Credible intervals for both programs

Now we want to compare reliability for both programs (C4, D5) using Bayes!

# C4
bstc4.all<-bayes.binom.test(154,161,0.95) 
summary(bstc4.all) 

##   Data
## number of successes = 154, number of trials = 161
## 
##   Model parameters and generated quantities
## theta: the relative frequency of success
## x_pred: predicted number of successes in a replication
## 
##   Measures
##           mean    sd   HDIlo   HDIup %<comp %>comp
## theta    0.951 0.017   0.918   0.981  0.434  0.566
## x_pred 153.151 3.786 145.000 159.000  0.000  1.000
## 
## 'HDIlo' and 'HDIup' are the limits of a 95% HDI credible interval.
## '%<comp' and '%>comp' are the probabilities of the respective parameter being
## smaller or larger than 0.95.
## 
##   Quantiles
##          q2.5%    q25%  median    q75%  q97.5%
## theta    0.913   0.941   0.953   0.963   0.979
## x_pred 145.000 151.000 154.000 156.000 159.000

plot(bstc4.all)

# D5
bstd5.all<-bayes.binom.test(160,164,0.95)
summary(bstd5.all) 

##   Data
## number of successes = 160, number of trials = 164
## 
##   Model parameters and generated quantities
## theta: the relative frequency of success
## x_pred: predicted number of successes in a replication
## 
##   Measures
##           mean    sd   HDIlo   HDIup %<comp %>comp
## theta    0.970 0.013   0.944   0.993  0.079  0.921
## x_pred 159.096 3.027 153.000 164.000  0.000  1.000
## 
## 'HDIlo' and 'HDIup' are the limits of a 95% HDI credible interval.
## '%<comp' and '%>comp' are the probabilities of the respective parameter being
## smaller or larger than 0.95.
## 
##   Quantiles
##          q2.5%    q25%  median    q75% q97.5%
## theta    0.939   0.962   0.972   0.979   0.99
## x_pred 152.000 157.000 160.000 161.000 164.00

plot(bstd5.all)

So we got Bayesian 95% credible interval for Trident I C4 and Trident II D5 programs:

Probability	Trident I C4	Trident II D5
Lower	0.918	0.944
Mean	0.951	0.970
Upper	0.981	0.993

Compare these results with ARL estimates:

Probability	Trident I C4	Trident II D5
Lower	0.919	0.933
Mean	0.956	0.967
Upper	0.979	0.987

Compare these results with binom.test estimates:

Probability	Trident I C4	Trident II D5
Lower	0.912	0.939
Mean	0.957	0.975
Upper	0.982	0.993

C4.dat<-data.frame(prob=c(0.918, 0.951, 0.981, 0.919, 0.956, 0.979, 0.912, 0.957, 0.982),
                names=c("Bayes","Bayes","Bayes","ARL","ARL","ARL","binom","binom","binom"))

D5.dat<-data.frame(prob=c(0.944, 0.970, 0.993, 0.933, 0.967, 0.987, 0.939, 0.975, 0.993),
                  names=c("Bayes","Bayes","Bayes","ARL","ARL","ARL","binom","binom","binom"))

par(mfrow=c(1,2))
boxplot(horizontal = TRUE, data=C4.dat, prob~names,col=c("red","green","blue"),main="Trident I C4",boxwex=0.3)
boxplot(horizontal = TRUE, data=D5.dat, prob~names,col=c("red","green","blue"),main="Trident II D5",boxwex=0.3)

Pay attention for increased D5 reliability (0.944 < P_D5 < 0.993) compared to C4 one (0.918 < P_C4 < 0.981) making difference in means as \(\mu\) = \(\mu(P_{D5})\) - \(\mu(P_{C4})\) = 0.970 - 0.951= 0.019 and standard deviations respectively \(\sigma\) = \(\sigma(P_{C4})\) - \(\sigma(P_{D5})\) = 0.017 - 0.013 = 0.004. One more interesting result is about the probabilities of the respective parameter \(\theta\) being smaller or larger than 0.95: P_D5,0.95 = 0.921 while P_C4,0.95 = 0.566. It means the estimate for P_D5 is more reliable and credible than for P_C4 that is we should be 92% confident that P_D5 > 0.95 while we can be only 57% confident about P_C4 to this end.

We see very close estimates for C4 reliability produced by ARL and Bayes. It’s not the case for D5 reliability where ARL estimates are lower than Bayesian ones. And there is one more interesting result: ARL and binom.test means are very close for C4, while upper bounds for C4 and D5 are similar both by Bayes and binom.test! So we may trust ARL procedures but make preference to Bayesian estimates as more robust ³⁴. In fact ARL procedures are based on priors similar to Bayesian except MCMC iterations.

The difference between the Bayesian values, and the classical approach, disappears as n increases. When the sample sizes become larger, the posterior distribution becomes less dependent on the subjective prior information and more dependent on the objective sample information. ³⁵

suc2 <- c( 154, 158)
all2 <- c( 160, 164)

fit2<-bayes.prop.test(suc2,all2)
plot(fit2)

Using Bayesian proportion test we see a little shift of 95% credible interval for D5 program (0.93 < P_D5 < 0.99) compared to C4 one (0.93 < P_D5 < 0.98) so far!

High alert probability

SLBM reliability and nuclear deterence

Hereby we produce some official points of view on US nuclear arms posture contained in open sources.

This means credibly underwriting U.S. defense commitments with tailored approaches to deterrence and ensuring the U.S. military continues to have the necessary capabilities across all domains—land, air, sea, space, and cyber. ³⁶

Credible deterrence results from both the capabilities to deny an aggressor the prospect of achieving his objectives and from the complementary capability to impose unacceptable costs on the aggressor. ³⁷

Deterrence, or more precisely, the theoretical construct of strategic deterrence, describes an ongoing interaction between two parties. In a deterrent relationship, one or both parties seek to persuade the other to refrain from harmful or dangerous actions by threatening or promising the other nation that the costs of acting will far outweigh the benefits. This can be done by threatening to impose high costs on the acting nation, threatening to deny the benefits the other nation may seek through its actions, and promising to withhold the costs if the nation forgoes the expected action.³⁸

Deterrence is a complex matter that depends on the credibiliity of retaliatory threats, the advesary’s perception of the consequences of retaliation, and in the case of nuclear retaliatory options, the role the United States would like nuclear weopons to play - and not to play - in world politics. ³⁹

The principle that nuclear deterrence (or any form of deterrence, for that matter) must be based on a high-confidence capability for second-strike retaliation even in the aftermath of a well-executed surprise attack is now well established. ⁴⁰

In a stable nuclear deterrent situation, neither the United States nor Russia could deprive the other of its capacity to inflict severe punitive damage in retaliation. Instability would exist if either side possessed a credible capability to strike without fear of reprisal, which could also be wielded as a threat. ⁴¹

The submarine force would offer a high degree of survivability for many decades – no peer competitor currently has any effective anti-submarine warfare capability against U.S. SSBNs at sea and technological breakthroughs that could threaten this survivability are several decades away. ⁴²

The US must maintain, upgrade, and procure capabilities to ensure survivability of sea-based ballistic missile forces. US ballistic missile submarines are currently superior to any adversary and rely on stealth for their survivability.⁴³

Trident system improvements make significant and substantial changes to the nuclear options available to the President. The advantages of sea-basing are numerous. Weapon systems are more survivable than land-based systems. Most importantly, sea-basing better supports the SIOP in both its ability to deter and strike if required. ⁴⁴

Reliability can be taken to mean any of several things, including the probability that a) the warhead will be delivered to within some range of the desired target, b) the warhead will detonate, or c) the warhead will detonate within some fraction of the design yield. These probabilities have different meanings when applied to single warheads versus classes of weapons. ⁴⁵

How much is enough?

Here we apply “reliability” as probability of successful launch of SLBM and delivery of warhead to within some range of the desired target in context of non-first strike tailored deterrence and without consideration of potential adversary’s BMD. ⁴⁶. So why it is so important to have as much reliable SLBM Trident D5 as possible? Let’s see!

par(mfrow=c(2, 4))
for(p in seq(0.5, 0.95, 0.05))
{
  x <- dbinom(0:24, 24, p)
  n<- 24
  barplot(x, names.arg=0:24, space=0, main=sprintf(paste('bin.dist.',n,p,sep=',')), col = "lightblue")
}

As we can see every 5% increase in reliability of one SLBM gives increase for overall readiness probability of 24 SLBM on board SSBN. That is why 85% reliability for one SLBM ⁴⁷ is the low limit providing minimum required readiness for SSBN! Some close estimates by experts are presented below ⁴⁸:

No less than 20 and no more than 24

What is the probability that no less than 20 and no more than 24 D5 missiles \(P(20\leqslant{X}\leqslant{24})\) would have successful launch outcome i.e. what is the probability of high readiness for Trident II SSBN on patrol given each D5 missile had 94%, 97% and 99% reliability (chance for success) - \(P\)?

\(P(20\leqslant{X}\leqslant{24})=C_{24}^{20}P^{20}(1-P)^{4}+C_{24}^{21}P^{21}(1-P)^{3}+C_{24}^{22}P^{22}(1-P)^{2}+\\+C_{24}^{23}P^{23}(1-P)^{1}+C_{24}^{24}P^{24}(1-P)^{0}\\\)

, where \(C_{n}^{k}=\frac{k!}{n!(n-k)!}\)

PD5L=sum(dbinom(20:24,24,0.944)) #low 
PD5M=sum(dbinom(20:24,24,0.970)) #mean
PD5U=sum(dbinom(20:24,24,0.993)) #upper
PD5L

## [1] 0.9904415

PD5M

## [1] 0.9993594

PD5U

## [1] 0.9999994

Y<-c()
X=seq(0.5,0.999,length.out = 100)
for (i in 1:100) {
  
  Y[i]<-sum(dbinom(20:24,24,X[i]))
}

PD52=data.frame(X,Y)
gr=ggplot(PD52,aes(x=X,y=Y)) + geom_line() + labs(title="Probability for 20-24 SLBM launch", x="SLBM Reliability", y="Overall launch probability")
gr2=gr+ geom_hline(aes(yintercept=PD5L), colour="#990000", linetype="dashed")
gr3=gr2+ geom_hline(aes(yintercept=PD5U), colour="#990000", linetype="dashed")
gr3

Here we apply some kind of hybrid “frequentest-Bayesian” approach for estimating probability \(P(20\leqslant{X}\leqslant{24})\). In fact we can apply pure Bayesian estimation to produce histogram for probability distribution. Let’s do it!

bstd5.all<-bayes.binom.test(160,164,0.95)
d5.all<-as.data.frame(bstd5.all$mcmc_samples[1:5000,])
pp<-d5.all$theta
dd5<-c()
for (i in 1:5000) {dd5[i]=sum(dbinom(20:24,24,pp[i]))}
hist(dd5,freq = T,xlim = c(0.99,1.0),breaks = 100,main="Histogram for probability distribution P(20<X<24) SLBM launch", xlab = "Probability",col = "lightblue")

summary(dd5)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.8693  0.9983  0.9995  0.9980  0.9999  1.0000

Exactly 24

What is the probability that exactly 24 D5 missiles would have successful launch outcome given 94%, 97% and 99% reliability chance for success? \(P(X=24)=C_{24}^{24}P^{24}(1-P)^{0}=P^{24}\)

plow=dbinom(24,24,0.944) #low 
pmean=dbinom(24,24,0.970) #mean
pupp=dbinom(24,24,0.993) #upper
plow

## [1] 0.2508002

pmean

## [1] 0.4814172

pupp

## [1] 0.8448546

X2=seq(0.8,0.999,0.001)
Y2=X2^24
PD522=data.frame(X2,Y2)
gr=ggplot(PD522,aes(x=X2,y=Y2)) + geom_line() + labs(title="Probability for exactly 24 SLBM launch", x="SLBM Reliability", y="Overall launch probability")
gr2=gr+ geom_hline(aes(yintercept=plow), colour="#990000", linetype="dashed")
gr3=gr2+geom_hline(aes(yintercept=pupp), colour="#990000", linetype="dashed")
gr3

Here we apply the same Bayesian estimation for \(P^{24}\). Let’s do it!

for (i in 1:5000) {dd5[i]=sum(dbinom(24:24,24,pp[i]))}
hist(dd5,freq = F, breaks = 30,main = "Histogram for probability distribution P(X=24) SLBM launch", xlab = "Probability",col = "lightblue")
x=dd5
curve(dnorm(x, mean=mean(dd5), sd=sd(dd5)), col="blue", add=TRUE, lwd=1) 
abline(v=0.5, col="red")

summary(dd5)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
## 0.05467 0.39980 0.50020 0.50460 0.61060 0.94760

We got 95% credible interval ( 0.2508002 < P_24D5,95% < 0.8448546) with 88.5% probability for SSBN to launch all 24 D5 missiles on high alert (P_D5=0.993) as an upper bound for this interval. Not so bad! So that is the US president’s confidence level as a commander in chief for his high alert nuclear asset - 14 SSBNs with 24 D5 missiles each (here we assume that each warhead has 100% reliability - very strong assumption for DOE device).

Exactly 4

In fact US Navy only once had launched 4 Trident II D5 missiles in one salvo from SSBN as a result of 97% probability of success for such event \(P(X=4)=P^4\) - dbinom(4,4,0.993)=0.9722926.

for (i in 1:5000) {dd5[i]=sum(dbinom(4:4,4,pp[i]))}
hist(dd5,freq = F, breaks = 20,main = "Histogram for probability distribution P(X=4) SLBM launch", xlab = "Probability",col = "lightblue")
x=dd5
curve(dnorm(x, mean=mean(dd5), sd=sd(dd5)), add=TRUE, col="blue",lwd=1)

summary(dd5)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.6161  0.8583  0.8910  0.8862  0.9211  0.9911

But how can it be? Is it a miracle or some technical innovation?

A little secret of D5 reliability

Here we can find some interesting clue to the high reliability of Trident II D5 missile. Let’s look through the GAO report ⁴⁹.

Each Trident II carries 24 D5 missiles, and each missile is equipped with the MK-6 guidance system, which is comprised of an inertial measurement unit and an electronics assembly. The Navy maintains spare MK-6 guidance systems on board each submarine and in its logistics pipeline for test and maintenance purposes. The Navy carries six spare MK-6s on board each patrolling submarine. These inventory objectives are based on maintaining the same high levels of readiness and reliability that were originally established between 1986 and 1987 during the Cold War era.

The Navy carries six spare MK-6s onboard each patrolling submarine. Our analysis indicates that having three onboard spares would decrease the guidance system’s operational readiness by only 3 percent (from 0.99979 to 0.96935) and having four onboard spares would result in only a 0.66 percent decrease (from 0.99979 to 0.99318). These decreases in MK-6 guidance system operational readiness would have a minimal effect on the overall D5 missile system’s operational readiness and reliability levels.

As we can see according to GAO report our estimate of upper bound for Trident II D5 reliability fits the official record as a result of maintaining MK-6 spare parts on board SSBN!

Probability for retaliation strike

The real probability for SSBN retaliation strike looks like \({P}_{RS}={P}_{EAM}{P(20\leqslant{X}\leqslant{24}|{EAM})}\), where \({P(20\leqslant{X}\leqslant{24}|{EAM})}\) is conditional probability for successful launch no less than 20 and no more than 24 SLBM given SSBN received correct EAM code from US National Command Authorities (NMCC, USSTRATCOM, E-4B NAOC, E-6A TACAMO) corresponding to the particular SIOP option ⁵⁰, ⁵¹, ⁵², ⁵³, ⁵⁴, ⁵⁵. We can’t estimate the probability \({P}_{EAM}\) of receiving correct EAM code by SSBN on patrol from TACAMO E-6A as main naval backbone of MEECN ⁵⁶ in stressed environment (EMP, ionization, etc ⁵⁷) during crisis scenario no matter what procedures are used (frequentest or Bayesian). It is beyond the scope of this paper (see ⁵⁸).

But we can guess somehow. What do we know about EAM? Here is a quote: Emergency Action Message (EAM) is a preformatted message that directs nuclear-capable forces[1] to execute specific Major Attack Options (MAOs) or Limited Attack Options (LAOs) in a nuclear war. Individual countries or specific regions may be included or withheld in the EAM, as specified in the Single Integrated Operational Plan (SIOP). ⁵⁹

Minimum Essential Emergency Communications Network (MEECN) is the Tri-Service transmission system which ensures delivery of Emergency Action Messages (EAM) to our strategic platforms. ⁶⁰

MEECN includes the Emergency Action Message (EAM) dissemination systems and those systems used for integrated Tactical Warning/Attack Assessment (TW/AA), presidential decision-making conferencing, force report back, re-targeting, force management, and requests for permission to use nuclear weapons. Efforts assure positive control of nuclear forces and connectivity between the Secretary of Defense, strategic and theater forces, and an informed decision-making linkage between the President, the Secretary of Defense, and the Combatant Commands. MEECN ensures our national leadership has proper command and control of our forces during times of national emergency, up to and including nuclear war.⁶¹

The format of EAMs starts with a six letter readout that is read in phonetic letters and the typical message is 30 characters long, although there have been many that do not stick to the 30 characters and can be much longer, or even less than that. There has even been an EAM as long as 238 characters long transcribed. Here is an example of an EAM recorded on January 13th 2014: “G5RTZN G5RTZN G5RTZN stand by, message follows G5RTZNMERFOBIV4EDDS3V6MZTHWMTX I say again G5RTZNMERFOBIV4EDDS3V6MZTHWMEX this is mainsail out”. ⁶²

Note. Nowadays NAOC as well as TACAMO can tramsmit EAM independently, using VLF long-trailing wire antenna and 200 kW transmitter both, but TACAMO VLF communication system is more reliable when flying within SSBN patrol location.

Suppose the length of EAM is 128 characters (letters and numbers) - something mean between short (30) and long (238) message. In the benign environment the probability of correctly receiving one character of EAM is 0.99999. Suppose EAM is transmitted 3 times by TACAMO and SSBN radioman must receive it correctly no less than 2 times. So we got the probability for correctly receiving EAM no less than 2 times out of 3 as \(P(2\leqslant{X}\leqslant{3})=C_{3}^{2}{P}_{EAM}^{2}(1-{P}_{EAM})^1 + C_{3}^{3}{P}_{EAM}^{3}(1-{P}_{EAM})^0\), where \({P}_{EAM}={P_{0}}^{128}\) and \({P_{0}}=0.99999\).

P0=0.99999
PEAM=P0^128
PEAM

## [1] 0.9987208

P23=sum(dbinom(2:3,3,PEAM))
P23

## [1] 0.9999951

Excellent result: \({P}_{EAM}=0.9999951\)! But what if environment is full of degradation along time due to EMP or ionization or something else, maybe natural magnetic field phenomena? What if radio waves paths are stressed by attenuation and absorption? What if the level of atmospheric noise is too high and signal to noise ration is too small? Will NEACP and TACAMO survive bolt-out-of-blue attack?

Here is a quote from S.Glasstone book ⁶³ The duration of the blackout for systems operating below frequencies about 30 megahertz is generally long in comparison with that of reduced performance. Absorption may also affect received noise levels if the noise reaches the receiver via the ionosphere. The major effect of nuclear detonations is to cause ionization i.e., an increase in electron density, which may lower the the ionospheric reflection altitude. The major consequences are phase anomalies and changes in signal strength and in noise from distant thunderstorms.

And this is a quote from thesis ⁶⁴: The ability of the command and control structure to withstand a nuclear blast and its attendant effects will directly impact how the war will be fought and eventually its outcome. A C3I system has many areas of vulnerability including susceptibility to blast and shock effects, thermal radiation, and electromagnetic pulse (EMP).

Now let’s produce some theoretical diagram for \({P}_{EAM}\) in case of severe degradation in stressed environment.

P0=seq(from=0.99999, to = 0.99, by = -0.00005)
PEAM=P0^128
P23<-function(X) sum(dbinom(2:3,3,X))
PRES=c()
for (i in 1:length(PEAM) ) PRES[i]=P23(PEAM[i])

DRES<-data.frame(P0,PRES)
dr=ggplot(DRES,aes(x=P0,y=PRES)) + geom_line() + labs(title="Probability for correctly receiving EAM", x="Probability of correctly receiving one character P0", y="Overall PEAM(2:3)")
dr2=dr+ geom_hline(aes(yintercept=0.5), colour="#990000", linetype="dashed")
dr3=dr2+geom_vline(aes(xintercept=0.9946), colour="#990000", linetype="dashed")
dr4=dr3+geom_hline(aes(yintercept=0.9), colour="#990000", linetype="dashed")
dr5=dr4+geom_vline(aes(xintercept=0.9983), colour="#990000", linetype="dashed")
dr5

As we can see for \(P0\leqslant{0.9946}\) the value of \({P}_{EAM}\leqslant{0.5}\) i.e. there is no chance for authorized retaliation strike by SSBN, while for \(P0\geqslant{0.9983}\) the value of \({P}_{EAM}\geqslant{0.9}\) i.e. there are good chances for authorized retaliation strike by SSBN. That is the decrease of \(P0\) value by 0.9983 - 0.9946 = 0.0037 or 0.37% would drastically reduce capabilities of NCA for authorized retaliation strike.

To be honest we must repeat once more - we don’t know how to estimate \({P}_{EAM}\) under degradation in stressed environment: too many factors (day time, season, sunspot number, transmitter and receiver locations, radiowave frequencies, transmitter power, antenna length, communications structure, command and control procedures, etc.) must be included in the model for calculating this value!

For example, one can use for such model Espenschied-Anderson-Bailey (EAB) empirical formula to estimate electromagnetic field ⁶⁵ \[E(\mu{V/m})=\sqrt{P}\cfrac{298000}d{e^{{-0.005d}(\cfrac{f}{300})^{1.25}}}\] where \[P= power, kw\] \[d= distance, km\] \[f= frequency, kHz\]

The real communications problem with SSBN is described in “FY 2000 President’s Budget Estimates”: This project (0204163N) develops communications systems elements which provide positive command and control of deployed ballistic missile submarines (SSBNs). This program provides enhancements to the shore-to-ship transmitting systems, shipboard receiver systems, and development of the Submarine Low Frequency (LF)/Very Low Frequency (VLF) Versa Module Eurocard (VME) Receiver (SLVR) System (formerly the Advanced VLF/LF VME Receiver (AVR) System). Continuing evaluation of this communications system is provided via the Strategic Communications Assessment Program (SCAP). Fixed VLF/LF develops an energy efficient, solid state, power amplifier replacement (SSPAR) for the VLF shore based transmitters of the Submarine Broadcast System, investigates improvement of the radio frequency high voltage insulators, bushings and antenna components used in these stations through the High Voltage Insulator Program (HVIP) and measures and signal propagation through the Coverage Prediction Improvement Program (CPIP). ⁶⁶

Pay attention that MEECN has been udergoing modernization twice since Ronald Reagan proposed increase in C3I budget for nuclear forces in 1981.

Presidential Directive (PD-59) and National Security Decision Directive (NSDD-13) have set the stage for the United States to place its nuclear forces and their supporting command and control structures in a position, where they have the capability to carry out a protracted nuclear conflict and “win” by denying the Soviet Union its war aims.⁶⁷

Strategic stability in an era of essential equivalence of C3I capabilities depends as much on survivability, endurance reconstitutability of C3I capabilities as it does on the size and character of strategic arsenal. ⁶⁸

Meanwhile communications with submerged submarine is a vital problem for NCA so far.

Here is one more quote from CBO report ⁶⁹, making our estimates for SSBN communications problem in stressed environment more comprehensible: CBO projects that the amounts budgeted for DoD’s nuclear command, control, communications, and earlywarning systems over 10 years would be $52 billion.

It is about 15% of the budget for nuclear forces during 2015-2024. We should note that these 15% do matter no less than the rest 85% of the nuclear forces budget. It is the case for Pareto principle (also known as the 80–20 rule, the law of the vital few, and the principle of factor sparsity) which states that, for many events, roughly 80% of the effects come from 20% of the causes ⁷⁰.

Consequences of retaliation strike

On November 21, 1962 Secretary of Defense McNamara in memo ⁷¹ to President Kennedy made some interesting estimation of nuclear retaliation potential: “It seems reasonable to assume that the destruction of, say, 25 percent of its population (55 million people) and more than two-thirds of its industrial capacity would mean the destruction of SU as a national society. Such a level of destruction would certainly represent intolerable punishment to any industrialized nation and thus should serve as an effective deterrent.”

Though this estimation made by McNamara nowdays could be interpreted as obsolete nevertheless it keeps the minds of all US presidents as decision makers and commanders in chief so far.

Needless to say, that nuclear war would have no winners: this undeniable fact proved by Hiroshima and Nagasaki in August 1945 as well as by Chernobyl in April 1986. One should see the ABC film “The Day After” (1983) to understand the consequences of retaliation strike no matter which side is engaged.

Conclusions

1. Bayesian estimates for Trident reliability are more preferable than classical binomial test and ARL procedures.
2. Using Bayes we should be 92% confident that P_D5 > 0.95 while we can be only 57% confident about P_C4 > 0.95.
3. We can be confident that no less than 20 and no more than 24 D5 missiles would have successful launch outcome in the 95% credible interval \(0.9904415 < P(20\leqslant{X}\leqslant{24}) < 0.9999994\) given 0.993 reliability for each SLBM.
4. There would be no more than 88.5% chance for all 24 D5 missiles to be launched \(P(X=24)=P^{24}\) given 0.993 reliability for each SLBM and only 25% chance with 0.944 reliability for each SLBM either. So spare parts do matter indeed!
   
5. In fact US Navy only once had launched 4 Trident II D5 missiles in one salvo from SSBN as a result of 97% probability of success for such event \(P(X=4)=0.9722926\).
6. The real probability for SSBN retaliation strike given SSBN received correct EAM code from US National Command Authorities in stressed environment \({P}_{RS}={P}_{EAM}{P(20\leqslant{X}\leqslant{24}|{EAM})}\) can’t be estimated no matter what procedures are used (frequentest or Bayesian).

Pax vobis et Dominus vobiscum!

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