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# Steady-state analysis of single exponential vacation in a PH/MSP/1/∞ queue using roots

We consider an infinite-buffer single-server queue where inter-arrival times are phase-type (PH), the service is provided according to Markovian service process (MSP), and the server may take single, exponentially distributed vacations when the queue is empty. The proposed analysis is based on roots of the associated characteristic equation of the vector-generating function (VGF) of system-length distribution at a pre-arrival epoch. Also, we obtain the steady-state system-length distribution at an arbitrary epoch along with some important performance measures such as the mean number of customers in the system and the mean system sojourn time of a customer. Later, we have established heavy- and light-traffic approximations as well as an approximation for the tail probabilities at pre-arrival epoch based on one root of the characteristic equation. At the end, we present numerical results in the form of tables to show the effect of model parameters on the performance measures.

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03/25/2020

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## 1 Introduction

In recent times, queueing models with non-renewal arrival and service processes have been used to model networks of complex computer and communication systems. Traditional queueing analysis using Poisson processes is not powerful enough to capture the correlated nature of arrival (service) processes. The performance analysis of correlated type of arrival processes may be done through some analytically tractable arrival process viz., Markovian arrival process ( see Lucantoni et al. The has the property of both time varying arrival rates and correlation between inter-arrival times. To consider batch arrivals of variable capacity, Lucantoni [20] introduced batch Markovian arrival process ( . The processes and are convenient representations of a versatile Markovian point process, see Neuts [23] and Ramaswami [27]. Like the , Markovian service process is a versatile service process which can capture the correlation among the successive service times. Several other service processes, e.g., Poisson process, Markov modulated Poisson process () and phase-type () renewal process can be considered as special cases of . For details of , the readers are referred to Bocharov [3] and Albores and Tajonar [1]. The analysis of finite-buffer queue has been performed by Bocharov et al. [4]. The same queueing system with multiple servers such as has been analyzed by Albores and Tajonar [1]. Gupta and Banik [17] analyzed

queue with finite- as well as infinite-buffer capacity using a combination of embedded Markov chain and supplementary variable method.

During the last two decades, queueing systems with vacations have been studied extensively. For more details on this topic, the readers are referred to a recent book by Tian and Zhang [30] and references therein. An extensive amount of literature is available on infinite- and finite-buffer - and -type queueing models with multiple vacations, see first few chapters of [30], Karaesmen and Gupta [18] and Tian et al. [29]. However, limited studies have been done on queue with single vacation, see Chapter 4 of [30]. In the past few years, there is a growing trend to analyze queueing models with renewal or non-renewal arrival and service processes with server vacation, see, e.g., Lucantoni et al. [22] and Shin and Pearce [26]. The analysis of phase-type server vacation for the case of queue has been carried out by Chen et al. [7]. Baba [2] analyzes queue where the server is allowed to take working vacations as well as vacation interruptions. Samanta [28] discussed a discrete-time queue with single geometric vacation time. Recenly, Chaudhry et al. [9, 12] discussed queues with single and batch arrivals using the roots method, respectively.

In this paper, we carry out the analytic analysis of the queue with exponential single vacation through the calculation of roots of the denominator of the underlying vector generating function of the steady-state probabilities at pre-arrival epoch. In this connection, the readers are referred to Chaudhry et al. [9, 10, 12], Tijms [31] and Chaudhry et al. [11] who have used the roots method. The roots can be easily found using one of the several commercially available packages such as Maple and Mathematica. The algorithm for finding such roots is available in some papers, e.g., see Chaudhry et al. [11]. The purpose of studying this queueing model using roots is that we obtain computationally simple and analytically closed form solution to the infinite-buffer queue with the vacation time following exponential distribution. It may be remarked here that the matrix-geometric method (MGM) uses iterative procedure to get steady-state probabilities at the pre-arrival epochs. Further, it is well known that for the case of the MGM it is required to solve the non-linear matrix equation with the dimension of each matrix in this equation being the number of service-phases involved in a queue. In the case of the roots method, we do not have to investigate the structure of the transition probability matrices (TPM) at the embedded pre-arrival epochs. It may be mentioned here that the basic idea of correlated service was first introduced by Chaudhry [13]. Further, it may be remarked here that the analysis of the infinite-buffer queues with renewal input and exponential service time under exponential server vacation(s) has been carried out by Tian and Zhang [30], see Chapter 4. The queueing model that we are going to consider has non-renewal service () and exponential single vacation time. In addition, we discuss several other quantitative measures such as system-length distribution at a post-departure epoch and expected busy and idle periods. Later, we have established heavy- and light-traffic approximations as well as an approximation for the tail probabilities at pre-arrival epoch based on one root of the characteristic equation. Finally, some numerical results have been presented which may help researchers/practitioners to tally their results with those of ours.

## 2 Description of the model

Let us consider a single-server infinite-buffer queueing system with the server’s single vacation. The inter-arrival time of customers, the service time of a customer and the vacation time of the server are represented by the generic random variables (r.v.’s)

and , respectively. Let denote the distribution function (D. F.) of the random variable with and

the corresponding probability density function (p.d.f.) and Laplace-Stieltjes transform (LST), respectively. The inter-arrival time

is assumed to have a general distribution with p.d.f. , D. F. and LST .
Arrivals. The inter-arrival times are assumed to be independent and identically-distributed (i.i.d.) random variables and they are independent of the service process as well as vacation time. The inter-arrival time distribution

is an important special case of general distribution as the distribution possesses nice vector and matrix form representation. Several probability distributions such as Earlang, hyper-exponential, generalized Earlang, Coxian etc. can be treated as special cases of

-distribution. It may be noted here that -distribution is a special case of general distribution. If the inter-arrival times follow -type distribution with irreducible representation , where & are a vector and a matrix of dimension and , respectively, the p.d.f. and D.F. of inter-arrival times are given by

 FA(x) = 1−α eTxeη,for x≥0, (1) andfA(x) = −α eTxTeη=α eTxT0,for x>0, (2)

where is a non-negative vector and satisfies and is an vector with all its elements equal to 1. Throughout the paper we write a subscript as the dimension of the column vector and sometimes we write by dropping its subscript. The mean inter-arrival time during a normal busy period is given by

 1λ=α∫∞0xeTx dx(−T)eη = −α(T)−1eη. (3)

Services. The customers are served singly according to the continuous-time Markovian service process () with matrix representation . The is a generalization of the Poisson process where the services are governed by an underlying -state Markov chain. For more details on , the readers are referred to recent papers by Chaudhry et al. [9, 12]. Let denote the number of customers served in units of time and the state of the underlying Markov chain at time with its state space . Then is a two-dimensional Markov process with state space . Average service rate of customers (the so called fundamental service rate) of the stationary is given by , where with denoting the steady-state probability of servicing a customer in phase . The stationary probability row-vector can be calculated from with , where . The customers are served singly according to a with steady-state mean service time .

Now, let us define as the matrix whose th element is the conditional probability defined as

 Pi,j(n,t)=Pr{N(t)=n,J(t)=j|N(0)=0,J(0)=i},1≤i,j≤m.

Let be the matrices whose elements are . Then using Chaudhry et al. [9, 12], it may be derived that

 P∗(z,t)=eL(z)t,|z|≤1, t≥0, (4)

where and

Vacations. The server is allowed to take a single vacation whenever the system becomes empty. On return from a vacation if the server finds the system nonempty he will serve the customers present in the queue, otherwise the server waits for a customer to arrive and the system continues in this manner. For an exponential single vacation time represented by the r.v. , the LST, p.d.f. and D.F. are given as follows:

 f∗V(s)=γγ+s,   fV(x)=γe−γx,  FV(x)=1−e−γx. (5)

where is assumed as the mean vacation time. The Vacation times are independent of the arrival as well as of the service processes. The traffic intensity is given by which is also independent of the vacation process.

## 3 The vector generating function of the number of customers served during an inter-arrival and other related probability matrices

Let denote the matrix of order whose th element represents the conditional probability that during an inter-arrival period customers are served and the service process passes to phase , provided at the initial instant of the previous arrival epoch there were at least customers in the system and the service process was in phase . Then

 Sn=∫∞0P(n,t)dFA(t), n≥0. (6)

If is the matrix-generating function of , where are the elements of , then, using (6) and (4), we get

 S(z) = ∞∑n=0Snzn=∫∞0∞∑n=0P(n,t)zndFA(t) (7) = ∫∞0P∗(z,t)dFA(t)=∫∞0eL(z)tfA(t)dt=f∗A(−L(z)).

The evaluation of the matrices can be carried out along the lines proposed by Lucantoni [20]. For the sake of completeness, we have given the procedure of obtaining , see Lucantoni [20]. One may note that the computation of using Equation (7) may be cumbersome. However, the following scheme may be efficient and is given by

 S(z) = limN→∞N∑n=0Snzn, (8)

where may be obtained as proposed in Chaudhry et al. [10].

We further introduce a few more notations which are required for the rest of the analysis of the queueing model under consideration. Now from renewal theory of semi-Markov process, if we let and denote the remaining and elapsed times of an inter-arrival time, respectively, then

 FˆA(x)=F˜A(x)=∫x0λ(1−FA(y)) dy, (9)

which we use while deriving the expression for in Equation (12). Similar to the case of inter-arrival time, if we let denote the remaining vacation time, then

 FˆV(x) = 1−e−γx,[Using (???)] (10)

and

 fˆV(x) = γe−γx. (11)

As above, we introduce the matrices of order whose th element represents the limiting probability that customers are served during an elapsed inter-arrival time of the arrival process with the service process being in phase , given that there were at least customers in the system with the service process being in phase at the beginning of the inter-arrival period. Then, from Markov renewal theory as given in Chaudhry and Templeton [8, p. 74-77], we have

 Ωn = λ∫∞0P(n,x)(1−FA(x)) dx,n≥0. (12)

The matrices can be expressed in terms of the matrices and their relationship discussed in [12] is as follows:

 Sn = δn,0Im+1λΩnL0+1λΩn−1L1.1{n≥1},n≥0, (13)

where is an indicator function and takes value 1 if the the condition is satisfied, otherwise it takes value 0.

Further, let be the conditional probability that at least customers are served in and the service process is in phase at the end of the th service completion, given that there were customers in the system and the service process was in phase at time . The probabilities then satisfy the equations

 ˜Pij(n,t+Δt) = ˜Pij(n,t)+m∑k=1Pik(n−1,t)[L1]kjΔt+o(Δt),

with the initial condition Rearranging the terms and taking the limit as , it reduces to

 ddt˜Pij(n,t) = m∑k=1Pik(n−1,t)[L1]kj,n≥1

for , with the initial conditions . This system may be written in matrix notation as

 ddt˜P(n,t) = P(n−1,t)L1,n≥1, (14)

with .
Let denote the probability that exceeds an inter-arrival time , then

 ω = ∫∞0Pr(A<ˆV|A=x).fA(x) dx (15) = ∫∞0Pr(ˆV>x).fA(x) dx = ∫∞0(1−FˆV(x))fA(x) dx=f∗A(γ).

Similarly, if we let denote the probability that exceeds , then

 τ = ∫∞0(1−FˆV(x))fˆA(x) dx=f∗ˆA(γ). (16)

Remark 2.1:  and may be derived in slightly different way. In the following we present a slightly different derivation for and may be done similarly.

 ω = ∫∞0Pr(A<ˆV|ˆV=x).fˆV(x) dx (17) = ∫∞0Pr(A

One of the frequently used inter-arrival time is phase-type renewal process which also serves as a special case of several other inter-arrival time distributions and is well-known in the literature. Therefore, we state the above formulae (17) and (16) for the case of phase-type inter-arrival time by the following theorem.

###### Theorem 3.1

If inter-arrival time follows a -distribution with irreducible representation , where and are of dimension , then the expressions for and are as follows.

 ω = 1+γα.(T−γIη)−1.eη, (18) τ = 1−λγα.(T−γIη)−1.T−1eη. (19)

Proof: Using the definition of and , after little algebraic manipulation the the results (35) and (19) may be obtained.

In the following, we further define a few notations which are required to analyze the queueing model under consideration. If we let with , i.e., inter-arrival time is greater than remaining vacation time, then

 FA+(x) = ∫x0∫∞0fˆV(y)fA(y+s) dy dsPr{A>ˆV} (20) = ∫x0∫∞0fˆV(y)fA(y+s) dy ds1−ω,

where may be called excess inter-arrival time. Further, let us denote and as the remaining and elapsed times of the excess inter-arrival time random variable , respectively, then

 FˆA+(x)=F˜A+(x)=∫x0λ1(1−FA+(y)) dy, (21)

where is the mean of the random variable For an important special case of phase-type inter-arrival time the distribution of excess inter-arrival time is also phase-type which is proved in the following theorem.

###### Theorem 3.2

If inter-arrival times follow a -distribution with irreducible representation , where and are of dimension , then the distribution of is also phase-type with representation , where and are of dimension and are given by

 α1 = γ1−ωα(−T+γIη)−1, (22) T1 = T. (23)

Proof: Simple algebraic calculation will give the proof.

Further, we introduce a few more matrices which are required for the analysis of this queueing model. With vacation ending, let denote the matrix of order whose th element represents the conditional probability that customers are served during an excess inter-arrival time and the service process passes to phase , provided at the initial instant of the previous arrival epoch there were at least customers in the system and the server was on a vacation with the service phase . Then

 Vn = ∫∞0P(n,t)dFA+(t), n≥0. (24)

Now, with vacation ending, let denote the matrix of order whose th element represents the probability that at least customers are served during an excess inter-arrival period and the service process is in phase with the server going on vacation at the end of the -th service completion, provided at the initial instant of previous arrival epoch there were exactly customers in the system and the server was on a vacation with the service phase . Then similar to the results derived above, we obtain

 V∗n = ∫∞0˜P(n,t)dFA+(t),n≥1. (25)

Further, with vacation ending, let denote the matrix of order whose th element represents the limiting probability that customers are served during an elapsed excess inter-arrival time with the service process being in phase , given that there were at least customers in the system with the server being on a vacation with the service phase at the beginning of the inter-arrival period. Then we have the following expression for

 Δn = ∫∞0P(n,t)dF˜A+(t)=λ1∫∞0P(n,t)(1−FA+(t)) dt,n≥0. (26)

The relationships among the matrices , and can be derived as follows.

 Δ0 = λ1(Im−V0)(−L0)−1, (27)

and

 Δn = (Δn−1L1−λ1Vn)(−L0)−1,n≥1. (28)

Further, using [12], it can be shown that

 V∗n = 1λ1Δn−1L1, n≥1. (29)

Similarly, let denote the matrix whose th element represents the limiting probability that or more customers have been served during an elapsed excess inter-arrival time and the service process is in phase with the server going on vacation at the end of th service completion, provided at the previous arrival epoch the server was on a vacation with service phase and the arrival lead the system to state or more customers. Then we can write

 Δ∗n = ∫∞0˜P(n,t)dF˜A+(t)=λ1∫∞0˜P(n,t)(1−F˜A+(t)) dt,n≥1. (30)

Now using the procedure discussed in [12], one may derive the following relation:

 Δ∗n+1 = (Δ∗n−Δn)(−L0)−1L1,n≥1. (31)

Finally, we define a few notations which are required to analyze the queueing model under consideration. Let given that . It is needless to mention that since is exponentially distributed, the distribution of will be Erlang of order two, which is a phase-type distribution with two states. Let the phase type representation of be denoted as with To calculate the distribution function of , we need to define which denotes the probability that exceeds an inter-arrival time . Then,

 ω2 = ∫∞0Pr(A<ˆV+V|ˆV+V=x).fˆV+V(x) dx (32) = ∫∞0Pr(A

Similarly, if we let denote the probability that exceeds , then

 τ2 = ∫∞0FˆA(x)fˆV+V(x) dx. (34)

Further, if an inter-arrival time is following distribution with the above representation, then following the derivation as presented in Theorem 3.1, we have

 ω2 = 1+(γ2)(α⊗\boldmathβ).(T⊗I2+Iη⊗U)−1.(eη⊗e2), (35) τ2 = 1−λ(γ2)(α⊗\boldmathβ).(T⊗I2+Iη⊗U)−1.(T−1eη⊗e2). (36)

Now the distribution function of may be derived as

 FA++(x) = ∫x0∫∞0fˆV+V(y)fA(y+s) dy dsPr{A>ˆV+V} (37) = ∫x0∫∞0fˆV+V(y)fA(y+s) dy ds1−ω2.

Similarly, let us denote and as the remaining and elapsed times of the random variable , then

 FˆA++(x)=F˜A++(x)=∫x0λ2(1−FA++(y)) dy, (38)

where is the mean of the random variable For an important special case of phase-type inter-arrival time, the distribution of is also phase-type whose representation can be obtained following Theorem 3.2 as , where

 α2 = γ21−ω2(α⊗% \boldmathβ)(−T⊗I2−Iη⊗U)−1 (39) T2 = T⊗I2. (40)

Further, with another vacation ending, let denote matrix whose th element represents the conditional probability that customers are served during an excess inter-arrival time and the service process passes to phase with the server becoming idle (after completing service of the -th customer), provided at the initial instant of the previous arrival epoch there were at least customers in the system and the server was on a vacation with the service phase . Then

 Cn = ∫∞0P(n,t)dFA++(t), n≥0. (41)

Similarly, with the vacation ending, let denote the matrix of order whose th element represents the probability that at least customers are served during an excess inter-arrival period and the service process is in phase with the server becoming idle (after completing service of the -th customer), provided at the initial instant of previous arrival epoch there were exactly customers in the system and the server was on a vacation with the service phase . Then, we can define

 C∗n = ∫∞0˜P(n,t)dFA++(t),n≥1. (42)

Also, with the vacation ending, let denote matrix whose th element represents the conditional probability that customers are served during an elapsed excess inter-arrival time and the service process passes to phase with the server becoming idle (after completing service of the -th customer), provided at the initial instant of the previous arrival epoch there were at least customers in the system and the server was on a vacation with the service phase . Then,

 Φn = ∫∞0P(n,t)dF˜A++(t)=λ2∫∞0P(n,t)(1−FA++(t)) dt,n≥0. (43)

Similarly, with the vacation ending, let denote the matrix whose th element represents the limiting probability that at least customers have been served during an elapsed excess inter-arrival time and the service process is in phase with the server becoming idle (after completing service of the -th customer), provided at the previous arrival epoch the server was on a vacation with service phase with the arrival leading the system to state customers. Then we can write

 Φ∗n = ∫∞0˜P(n,t)dF˜A++(t)=λ2∫∞0˜P(n,t)(1−FA++(t)) dt,n≥1. (44)

One may note here that the matrices are required to obtain other matrices by using the above relations (27)-(28). The relationships among the matrices and can be similarly obtained as for the matrices and . These relationships are given below.

 Φ0 = λ2(Im−C0)(−L0)−1, (45) Φn = (46)
 Φ∗1 = (Im−Φ0).(−L0)−1L1, (47)
 Φ∗n+1 = (Φ∗n−Φn)(−L0)−1L1,n≥1. (48)
 C∗n = 1λ2Φn−1L1, n≥1, (49)

The matrices are calculated exactly the same way as we derive the matrices , see Chaudhry et al. [csa15]. For more information on , readers are referred to Bocharov [3], Albores and Tajonar [1] and Gupta and Banik [17].

## 4 Analysis of GI/MSP/1/∞ queue with single vacation

We consider a queueing system with single vacation as described above. In the following subsections we obtain steady-state distributions for this queueing system at different epochs considering .

### 4.1 Stationary system-length distribution at pre-arrival epoch

Consider the system just before arrival epochs which are taken as embedded points. Let  be the time epochs at which arrivals occur and the time instant before . The inter-arrival times are i.i.d.r.v.’s with common distribution function . The state of the system at is defined as where is the number of customers present in the system including the one currently in service. Whereas denotes phase of the service process and or 1 indicates that the server is on vacation or busy . In the limiting case, we define the following probabilities:

 π−j,0(n) = limk→∞P{Nt−k=n, Jt−k=j, ξt−k=0}, n≥0, 1≤j≤m, π−j,1(n) = limk→∞P{Nt−k=n, Jt−k=j, ξt−k=1}, n≥0, 1≤j≤m,

where represents the probability that there are customers in the system just prior to an arrival epoch of a customer when the server is on vacation with phase of the service process . Similarly, denotes the probability that there are customers in the system just prior to an arrival epoch of a customer when the server is in a busy (when ) or dormant (when ) state with phase of the service process . Let and be the row vectors of order whose -th components are and respectively.

Observing the state of the system at two consecutive embedded points, we have an embedded Markov chain whose state space is equivalent to Observing the system at two consecutive embedded Markov point, we have the following system of vector difference equations

 \boldmathπ−0(0) = ∞∑k=0\boldmathπ−0(k)((1−ω)V∗k+1−C∗k+1)+∞∑n=0% \boldmathπ−1(n)((1−ω)V∗n+1), (50) \boldmathπ−0(n) = \boldmathπ−0(n−1)ωIm,n≥1, (51) \boldmathπ−1(0) = ∞∑k=0\boldmathπ−0(k)C∗k+1+∞∑n=0\boldmathπ−1(n)(ω(V∗n+1+n∑j=0Vj)+(1−ω)n∑j=0Vj−n∑i=0Si), (52) \boldmathπ−1(n) = ∞∑k=n−1\boldmathπ−0(k)(1−ω)Vk−n+1+∞∑j=n−1\boldmathπ−1(j)Sj−n+1,n≥1. (53)

Multiplying (53) by , summing from to , after adding (52) and using the vector-generating function , we obtain

 \boldmathπ−∗1(z)[Im−zS(z−1)] = ∞∑j=0∞∑i=j+1\boldmathπ−0(j)Vizj−i+1


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# Solve $x\frac{dy}{dx}=x-y$ using the integrating factor method and by using the substitution $y=vx$

I have been asked to show that the solutions to this equation are identical regardless of the method used. I did both methods and I think I may have got it right but I am not sure. Please check:

Using the integrating factor method I found that the factor was $x$. Therefore $\frac{d}{dx}(xy)=x$ so the result was $y=\frac12 x+\frac cx$

Using the substitution $y=vx$ enabled me to get the separable equation: $x\frac{dy}{dv}+2v=1$ I then solved to give: $\frac1{\sqrt{1-2v}}=Ax$ where $A=e^c$. Rearranging to try and get the form of the answer that I got using the other method gave me: $y=\frac12x-\frac{1}{2A^2x}$. It looks similar to my last answer and would be the same if it is permissible to simply say that $c=-\frac{1}{2A^2}$. Is this allowed or have I done something wrong?

• they are equivalent. – abel Mar 19 '15 at 0:23
• So I have done it right? – RobChem Mar 19 '15 at 0:23
• Yes it is allowed. – science Mar 19 '15 at 0:24
• yes. you did it right. – abel Mar 19 '15 at 0:24

i don't know if you care for another way to do this. that is a third way. but here it is. we write the differential equation as $$\frac{dy}{dx} = \frac{x-y}x$$ split this into equations $$\frac{dy}{dt} = x - y, \, \frac{dt}{dx} = \frac 1 x$$ we solve the second one and get $$x = Ce^t, \frac{dy}{dx} + y=Ce^t$$ the solution for $y$ is $$y = Be^{-t} + \frac C 2 e^t$$ now we get rid $t$ by subbing $\frac x C$ for $e^t$ and arrive at the same solution you got $$y = \frac{BC} x + \frac 1 2 x$$
$$xy'+y=x \implies (xy)'=x.$$

format: plain


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# Mass problem please check my work.

1. Jul 17, 2006

### 62646264

I don't know how to use LaTex or anything yet, so I'll just have to word this right.

Question: If the density of a fluid is given by

p = exp ^ -(x^2+y^2+z^2)^3/2

what is the total mass inside the unit sphere.

Since mass = integral (p dA)
where dA is the element of area

I switched to spherical coords...

M = [int(0..2*Pi)]:[int(0..Pi)]:[int(0..1) : [exp ^ -(p) ^ 3/2] * p^2*sin(phi) dp d(phi) d(theta)

where this is the triple integral of the function designated, with integration bounds listed in parantheses and the order of integration shown at the end of the equation.

I got the answer 2.213...if anyone sees any mistakes, please let me know. Sorry it's not easier to read...

2. Jul 17, 2006

### 0rthodontist

What you use here actually is mass = the integral of p with respect to V, volume.

You did make a calculation error: in your integral for M, the expression for the mass should have a p2 term, not a p term, in the exponential (which will reduce to e-(p3))


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# 2015 CAP Congress / Congrès de l'ACP 2015

13-19 June 2015
University of Alberta
America/Edmonton timezone
Welcome to the 2015 CAP Congress! / Bienvenue au congrès de l'ACP 2015!

## Status of Long-Baseline Neutrino Experiments

16 Jun 2015, 15:45
30m
CAB 235 (University of Alberta)

### CAB 235

#### University of Alberta

Invited Speaker / Conférencier invité Particle Physics / Physique des particules (PPD)

### Speaker

Dr Nicholas Hastings (University of Regina)

### Description

The current generation of long-baseline neutrino oscillation experiments employ an off-axis $\nu_\mu$ (or $\bar{\nu}_\mu$) beam produced by the decay of pions created when a proton beam strikes a target. The beam is monitored at detector facilities near the production point before travelling hundreds of kilometres to a far detector. Aiming the beam centre slightly away from the far detector provides the off-axis configuration which selects a narrow energy band beam tuned to maximize the oscillation probability. The status of these experiments will be presented. The Tokai to Kamioka (T2K) experiment consists of a $\nu_\mu$ beam produced at the Japan Proton Accelerator Research Centre (J-PARC) in Tokai on the East coast of Japan, which is monitored by a suite of detectors before travelling 295 km to the Super-Kamiokande (SK) water Cerenkov detector. T2K has been in operation since 2010 and has been continually releasing new and exciting neutrino oscillation results. The most recent precision $\nu_\mu \to \nu_e$ appearance and $\nu_\mu$ disappearance oscillation measurements as well as initial results running the experiment in the $\bar{\nu}_\mu$ beam configuration will be presented. The NO$\nu\hspace{-0.11ex}$A experiment, utilizing the NuMI beam and a near detector at Fermilab and a far detector at a distance of 810 km, began operation in 2014. The current status of NO$\nu\hspace{-0.11ex}$A will also be shown.

### Primary author

Dr Nicholas Hastings (University of Regina)

 Slides


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# China's Water Shortage and Power Plants (their power plants definitely have a drinking problem)

In the previous post, I described how thermal power plants use a massive amount of water. This time we are going to explore a specific case. As usual, it's China.

Power plant water use can be a problem in a water-stricken area. Let's look at a case-study. China is a water-stricken area, and has a lot of thermal power plants. In fact, China uses more primary energy than any other country in the world. Unfortunately, their power plants are far less efficient than they should be. So they are wasting water, and this is unsustainable. Moreover, China has 1,350 million people. The US has 314 million.

First, let's look at the rainfall of China, compared to the US:

Rainfall in China, in inches

Rainfall in the US, in Inches

Looks pretty similar, right? Now recall that the US has 1/4 the population of China. And pretty much the exact same amount of area. Keep that in mind while we look at China's powerplant locations:

China's water stressed areas, compared to where power plants are planned. Source,

So. The places that have the most people and need the most power are the same as the dry places. In other words, China is building the bulk of its thermal power plants in the area that can't provide sufficient water to cool the power plants.

Before coming to the complete picture, let's check out the water use:

Fresh Water Use in the US.
source

In the US, 80% of water use is to grow food and to make electricity.

Finally, where is all this water coming from? Rain alone isn't enough, it comes from the ground. Fresh water from the ground is not unlimited, and we are running out of it. It's called Fossil Water, and here is what the situation looks like in the US:

Water withdrawals in the US

In other words, a huge chunk of our country is relying on water that will not exist in a few decades.

And looking at China:

China's groundwater depletion rate

In the US, the scale of groundwater depletion tops out around 400 cubic kilometers. In china, it tops out at 3,000 in regions. That's not to say that the US won't run out. It just says that China is in serious trouble.

Again, 80% of water use is for electricity and agriculture. And China has 4x the people of the US. There is not sufficient water. Would you rather run out of electricity, or run out of food? It's not an easy choice, but food can be imported. That being said, someone has to grow the food, and that country better have a robust water supply. Moreover, food growth is a low income industry. A country that marries itself to being a food supplier, unless it charges gouging levels of prices, is marrying itself to never being a high-income country. But charging price-gouging levels is a bad idea.

While this mental exercise was fun, let's look at some examples.

First, while Californians probably shouldn't have been growing water-intensive almonds in a dessert in the first place, running out of water has imperilled the world supply of all sorts of nuts and things. They are tearing up their farms because of lack of water.

That's only the start. Drought in Syria helped bring about war there. Syria is a tiny country that doesn't matter on the world scheme. India, China, and Pakistan face water shortages. Combined, they have 1/3 the world population. They also happen to hate each other. As climate change progresses, and some countries face droughts, people may not want to choose between food and electricity. They may try to divert water supplies, sparking tensions and even war.

So. Does your power plant have a drinking problem? If you live in China, it definitely does, and it's causing all sorts of strife.

Wrapping it all together: Yes, a country can import food. But you know how much of the world relies on the middle east for oil, and we talk about energy security? That's just stuff that makes your cars move. Remember how Russia threatens to shut off natural gas to Europe if they don't get in line with Russia's plans, and so much of Europe is cowed? That stuff keeps homes warm, but it isn't as important as food. Imagine a powerful country that is mostly reliant on other countries for food to stay alive. That's a really bad situation. The country in this situation has to either take dictations from whoever feeds them (not really a problem if you are getting your food from non-powerful nations, but still irksome), or has to take over a food-producing country.

One potential solution: Chinese power plants are notoriously inefficient. If you have a 25% thermodynamically efficient powerplant, it uses 30% more water than a 37.5% efficient power plant. China should either shut down inefficient plants and require new construction that is efficient, or require retrofits of old plants. It would be very expensive, but less expensive than the social and political cost of running out of water too soon. What about the US? Most of our plants are pretty efficient already. Especially our Natural Gas plants that much of the country runs on. We probably spend too much water on watering desserts to make food, but that's another story.

An almost-final note. While solar power and wind power use water in construction, their water use is minimal compared to that of thermal power plants. Barring solar-thermal (it's thermal, it uses water), these renewable resources are the only answer to the reducing the choice between electricity and food. In other words, expansion of wind power and solar PV is the only cheat code we have to deal with this impending water shortage.

One last thing. Why did I single out China? Only because I know a lot about China. Pakistan will have water shortage issues, but they already don't have electricity. In the summer, they have blackouts for up to 20 hours a day cause they can't produce enough electricity. This is a country of 180 million people, bordering India, and sharing a strong mutual resentment with India. More on this later, though.

- Jason Munster

# The President's State of the Union Address. Geopolitics of Oil: Expanded US (and Russian) Oil Drilling and the Middle East's Bane.

An oil rig in the Bakken. This represents a massive shift of primary energy resource production power in the world. picture source

What do the price of oil, the president's state of the union address, and middle eastern stability have in common? In the address, the President talked about fighting climate change, but the US is going full-tilt towards more drilling. While this sounds like hypocrisy, it actually puts the US and the world in a better position to deal with climate change. Sounds crazy? Keep reading. Here's a hint though: What would happen if the price of oil dropped to $40 overnight? Much of the Middle East power structure would collapse. The Science! (skip this if you only care about what oil prices do for the Middle East) Because this is a science blog, I am writing science stuffs here. Hokay, the background. The world consumes around 90 million barrels of oil a day. How much of this does the US produce? Check out this graph: Historic US oil production. Source is EIA I want to point out three things. First, in the 70s and 80s the US was one of the world's most prolific oil producing countries. A lot of this came from huge Texas fields, like Eagle Ford. And then the gigantic basins like Eagle Ford ran out of easily accessible oil, and US oil production collapsed. Now look at the ramp rate of production in the most recent years. The rate of increase in production is unprecedented. It's going up fast. Next, lets focus on Texas and North Dakota: North Dakota and Texas Oil Production Notice the massive rate of increase of production. From 2008 to 2012, Texas alone increased production so much that it provided an additional 2% of the world's oil. North Dakota is producing nearly 1 million barrels a day, or slightly more than 1% of the world oil. Together, they produce 3.75 million barrels per day, 4% of the world's oil. Let's put this in perspective. Iran produces 4 million barrels a day. North Dakota and Texas alone produce nearly this much. Look at those growth rates. Are they showing any signs of slowing down? No. In other words, the US is rapidly becoming one of the world's most prolific producers of oil. How'd this even happen? Hydrofracking and horizontal drilling. Earlier I said that Eagle Ford and such were played out. In reality, all the easy oil in it was pulled out. The remaining oil is like the Bakken: tight oil. Let me emphasize this: Every single major play of the 70s and 80s is about to become a new Bakken. That means going back to the days when the US was the largest oil producer in the world. That means Russia is also going to be able to ramp up production, once they figure out how to hydrofrack. So the price of oil is going to drop in the future. What This Means for Politics and the Middle East; Expansion of drilling in the US and Russia will have two major effects. The first is that we will no longer rely on oil from the Middle East to supply the world markets. In this case, the world will care less about stability in the middle east. To the point where the world would just let the Middle East burn, just like we let happen in Africa today (except for Israel, I would guess) This would mean less military spending, which would in turn mean more domestic investment (or lower taxes, but our ailing infrastructure and gutted R&D budget really could stand to be brought back up to where it was when the US rose to become the world's only superpower). The second implication of this glut of oil is much more far-reaching. It means is lower oil prices worldwide. If oil drops below$75 a barrel, even Saudi Arabia struggles. It'll be hard for the Middle East to make trouble when they cannot afford to. Now let's say hypothetically that Iran funds terrorist groups (I haven't researched this and don't know whether it is true, so it's a hypothetical). If the price of oil drops to the point where they can no longer profitably produce, then suddenly our hypothetical country cannot fund terrorism. And we save moneys from no longer needing as much anti-terrorism programming.

In other words, we save money because we won't be sending military to the middle east, and because terrorism will potentially be more poorly funded.

Let's sum up: produce more oil, the price of oil drops, countries and companies make less money from oil (mostly countries, companies have a way of maximizing profits pretty well), since countries have less money, they can't push their state agendas as much.

Wrapping This Up

So, pretty much, drill more in the US (where we can regulate emissions), save a ton of American (and European) money by no longer having to make sure there is peace around oil resources, use that money to fix all the problems we've created with the environment. In other words, more drilling is a potential long-term solution because it will eventually free up more funds in the federal budget.

It's my guess that the president can't say, "We need to drill in the US to make it so we don't have to spend money stabilizing the Middle East. Once that is no longer a problem, we can use the extra money to address climate change."

Some criticisms: If the decrease in the price of oil results in more oil consumption, this would be bad for the environment. We need to continuously improve efficiency of vehicles and industry, and decrease our demand for oil and fossil fuels. Any money saved from military intervention happening less should be driven towards this goal.

That's all for now! Thanks for reading.

- Jason Munster

# Geoengineering

So. Science can fix anything, right? Only if we have lots of time and money. And grad students that function as indentured servants in a pyramid scheme to get tenure.

Back to the point. The truth is that science can't fix everything on short time scales. Climate is one of them. Geoengineering can help to a degree, but it will only get us part of the way there to avoid the worst consequences of climate change. Let's discuss some.

White roofs, white roads, white buildings.

Two articles back, we discussed albedo, or reflecting sunlight. Ice reflects 90%, water reflects 90%. Whatever is reflected tends to go to space and not stay in the Earth system and warm it up. In fact, whatever is absorbed then gets in the greenhouse trapping loop, warming up the Earth a good bit. Dark surfaces (our roofs, our roads, most of our buildings) reflect little and absorb a lot. So, paint them all white, and more light is reflected. Excellent!

"But Jason," you say, "Cities are only a small percentage of land area. How could this possibly help? I mean, the rest of the Earth will still absorb just as much heat. Right?"

And to you I say, "Excellent, sir! That is true. Making all our stuff white won't do much for the overall heat budget of the Earth. I am so proud of you for reading most of my website so you quickly figure stuff like that out."

So what does it do?

The heat island effect is based on the fact that cities are covered in dark buildings and pavement, and have a very low albedo, so they absorb heat

Cities are fucking warm. They suffer from this thing called the "heat island effect." That is a fancy way of saying that they are so dark, they absorb the sunlight and are easily 10 degrees F (around 5 degrees C) warmer than they should be. Turn everything white, and you can cool the city. This will actually have a very large effect on how hard our AC units have to work in the summer. Imagine if your city was suddenly 10 degrees F cooler. How sweet would that be? I posit that it would be pretty rad.

This one seems to help a bit, but we will still be using tons of energy and producing CO2 in all other ways. Moreover, it won't solve the problem of the agriculture, ice caps, and acidifying ocean.

Putting CO2 in the ground

There are two ideas of sequestering CO2 in the ground. The first is capturing it at the source. Like power plants. This sounds like an easy idea, but the first problem is the energy it takes to capture it. Thermal power plants take in atmospheric air. Which is 78% nitrogen, and 21% O2. Even if all the O2 were converted to CO2, what comes out of the power plant stack is still 78% nitrogen. Separating the two to store the CO2 takes more energy. In fact, the power plant is roughly 30% less efficient. So it needs to burn a lot more coal or natural gas to produce the same amount of power, and will cost a lot more to build. And any fancy idea you have to get around this 30% efficiency hit won't work. No matter what, you either have to pre-concentrate O2 to get a pure stream of CO2 on the other side, or separate the CO2 on the emission side.

The next problem is where to store it once you get it. Gases like to leak out of things. Some companies are trying to store the CO2 underground, much like petroleum is stored underground in a lot of places. This is why you need to separate it from the nitrogen in the air. There just isn't enough space to store both the CO2 and the nitrogen, and also it is expensive to pump stuff underground. Another issue is that it is unclear how long storing CO2 will last in the ground, since it more or less needs to be done indefininately.

Finally, since 35% of our energy use is from cars driving down the road, and it is impossible to capture that CO2. So Carbon Capture and Storage (CCS) from the source still won't do everything we need.

Direct Capture
The next idea is to capture CO2 directly from the air. We have increased CO2 in the atmosphere from 280 parts per million (.028%) 400ppm. The idea of direct capture is to do the opposite. Draw down the CO2 and then store it somewhere. Some might suggest we store it in trees, but that is an awful lot of trees, and unless we bury them trees somewhere underground, they are just gonna get consumed by bacteria and become CO2 again. Other options are to mechanically and chemically separate CO2 from the air, and them store it underground as above. This is very expensive. It might work in the future, but for now it won't.

The bonus of this, if it ever works, is that it is the best way to reverse our issues from an engineering standpoint. We can turn back the clock.

Stratospheric Injection

Injecting small sulfur or other particles into the atmosphere cools the entire globe by reflecting some small portion of sunlight before it hits the rest of the Earth. We know this cause when mountains like Pinatubo and St. Helens explode, they launch particles into the stratosphere and we get a cold year.

SO2 increase in the stratosphere by exploding volcano

Some people have suggested that we could do this. Just inject stuff into the stratosphere to reflect sunlight. The problem? It turns out that everything small enough to cause the proper scattering just happens to be the right size to promote adsorption of water particles. Which then allows for rapid recycling of CFCs in the stratosphere.

"But Jason," you say, "I thought recycling was good!"

Recycling plastics is good. Stratospheric recycling of CFCs is bad. Cause what happens is a CFC reacts with ozone, breaking it apart, wrecking the ozone layer, and then usually is all like, "Man, I am exhausted from catalyzing that reaction, I am gonna take a break." But that water that adsorbed onto our reflective particle provides an excellent place for it to re-radicalize. Which means it is ready to take out another Ozone particle. That's right, our CFC goes to chill out on some water droplets, effectively taking a restful timeout at a pool, and gets ready for work again destroying the ozone layer.

Let's pull this all back together. We try to put stuff in the upper stratosphere, if could make CFCs more effective at destroying the ozone layer, and then we are all screwed in a much much larger way than climate change. Cause the ozone layer is what protects us from getting fried by a lot of UV rays.

Here's where things get fun. Imagine you are a small country of 1 million people living on an island. And that island is going to get inundated with water in 20 years unless climate change is reversed. You don't give a damn about a chance of destroying the ozone layer. You only care about saving your people and your country. Stratospheric injection isn't exactly nuclear science. We aren't going to have rogue nations stumbling through how to do this, and failing all the time.

I'll leave you to ponder what all that means, cause it is more fun that way, and we are already at 1200 words.

The upshot of this is that it also fails to solve the acidifying of the ocean, we don't know how well it will work, and we don't know what will go wrong.

Solar Reflector

Another idea is to put huge mirrors in space and reflect a chunk of the sunlight coming in. This could work. Wasn't this a plot in some Bond movie, though? Also, it would be mad expensive. Probably much more expensive than some other options. And much like the option directly above, we still acidify the ocean.

Review

Hokay, so. Most of the technologies for fixing our problem don't exist, don't work, are too expensive, or could kill us all. And if they do work in the future, they won't solve all the problems we are creating. Even the one that does solve all the problems, direct capture from the atmosphere, won't do crap for our plight if we rely on that alone. As a species, we can easily outstrip any CO2 removal measures just by burning more things. Even if after rigorous testing proved all these work, we would need to some combination together to get anywhere. And even with that, we need to reduce the continued growth of emissions worldwide, otherwise no science or engineering solution will stop climate change.

Depressing, eh?

- Jason Munster

# Climate Change 2

I am not expert on different effects of climate change. But I do know a good smattering of random things. More importantly, several of my coworkers in grad school are at the forefront of the research of a lot of things here.

Here are some events relating to climate change, with indications of how much I think I know about it. So, for these things, I will have a title, than a 5 star rating for my level of confidence in the material I am presenting. 5/5 means I think I know a whole lot, 4/5 means I know what a grad student in a related field should know, and I probably am friends with one of the experts in the field, 3/5 means I am conversant in it, 2/5 means I understand it a little and have seen the math, 1/5 means I have heard of it and think it is worth mentioning.  It is important to note that anything rated 1 or 2 should be taken with a grain of salt, and should absolutely not be cited. I don't really know much about these things, other than they are possible.

Melting Ice Sheets -  3/5

A snapshot of the Arctic sea ice extent from June 2013. Area of sea ice has decreased over time

It seems like every summer, the news programs get all abuzz over the Arctic ice extent. No matter which way it goes, they get excited. The extent is literally the surface area that this ice covers. But as we discussed on an early post about thermodynamics, the amount of heat energy you have to pump into a system does not relate to its surface area, but instead to its volume, since volume is directly related to mass. And the story of ice volume yearly minimum is more telling: the minimum ice volume in the summer has decreased by a factor of nearly 50% over the past 5 years. In other words, half of the summer ice is gone.

The areal extent of ice seems somewhat erratic. The volume measurement of arctic ice over the last 5 years is a much more important measurement

What happens when the ice goes away?

Albedo changes - everyone knows about this, so I won't rate my knowledge here. In the last post, I mentioned that ice reflects 90% of light energy, and water absorbs 90%. If the sea ice disappears, more heat can be absorbed and trapped by the Earth, causing warming to happen more quickly.

Shortwave radiation is high-energy radiation from the sun. Longwave is infrared that comes off from Earth. Ice reflects shortwave (sun) radiation.

Stronger temperature changes in high latitudes

As the planet warms, the warming will be more felt in the high latitudes (ie the Arctic and Antarctic). As you can guess, this will have feedbacks with the ice melting and albedo changes.

Projected temperature increases show that the high latitudes will have far more profound temperature increases under climate change.

The habitats of the Arctic will present another positive feedback - 5/5

This is what I study directly. I don't model this, I measure it. Well, my team does. I am a small part of that. In normal biomes, plants pull CO2 from the atmosphere and turn it into plant matter. A lot of this is leaves or grass and such. They then die, fall to the ground, and get consumed by bacteria or oxidized to become CO2 again. So most of the CO2 consumed by plants and such is recycled back into the atmosphere.

In cold places, it is different. Moss and grass grow in the summer (no trees, permafrost prevents them from ever taking root). Much of this after it dies does not get recycled to CO2 again,cause the freeze already happened and it is too cold for the stuff to become CO2. This has happened in the Arctic for 300,000 years or more. In the first 3 meters of Arctic soil, there is enough undigested carbon to double the amount of CO2 in the atmosphere. Obviously it won't all release at once, and most of it may not release. But even if a part of a percent started being released per year, it would match mankind's CO2 emissions. This hasn't started happening yet, but if it did, we'd want to work fast to reverse it if we hope to prevent climate change from jumping into a strong positive feedback loop that we cannot control.

More on this later, when I describe my actual research and what I do day to day.

Weather patterns change - 1/5

I can barely even hand-wave at this one. The ocean strongly influences atmospheric circulation patterns. Hurricanes, for instance, always form over the ocean. This is because the ocean has a ton of thermal momentum (it doesn't change temperature at the same rate as the atmosphere) and the top layer of it is well-mixed, so even if the top few inches warm up, it will rapidly be cooled off by the water beneath. The atmosphere has much less thermal momentum, mostly because it is far less dense than water. So what happens when you have an ice cap? The water-atmosphere interaction is cut off. The water is sealed away from the atmosphere, and suddenly the ocean stops controlling wind patterns and such. And then very large-scale atmosphere-driven wind patterns can develop without ocean waters impeding it. This leads to wacky weather. Like increased snow in winter at mid-latitudes, and much more variable weather. This is why we now call it climate change instead of global warming. Some places will get cooler, but the variability of weather patterns will increase because of this sort of event. Like in Boston on May 2th where we broke the record low, and then on may 29th we broke the record high. Yay more climate variability.

Drought in the US. Much of the west coast is short of water.

In addition to weather variability, some trends will be more pronounced. Dry seasons will be more dry and last longer. Rainy seasons will have more intense storms. This can be a problem, cause droughts prevent agriculture from working.

Increases in Floods and Droughts - 2/5

There are floods called 100-year floods, cause they should only happen once every 100 years. Areas of Australia had two 100-year floods in a decade. This is because climate change will make large weather events, like floods and droughts, a lot more frequent.

Torrential rains flood Australia pretty frequently these days. Expect more of this in many parts of the world as climate change takes hold.

Melting Glaciers 4/5 (I hang out with the world experts on this all the time, cause they are cool)

Did you know that everything with mass exerts a gravitational force. Yes, hard to believe, but it is true! And it turns out that mountains and glaciers exert a sideways gravitational force. One that is strong enough to pull water from the oceans towards them. In other words, if the Greenland ice sheet melts, the sea level Greenland would actually drop. And the sea level around India, Africa, and South America would rise a more than you would expect. So instead of seeing 7m of sea level rise from all of Greenland melting, they might see 8m. In other words, all the poor countries that didn't put the GHGs in the atmosphere, and also cannot afford to prepare for the rise, will take the brunt of this one.

Disease - 1/5

Many people predict that certain diseases will become more rampant. Like how trees are getting destroyed all over California, because certain tree-eating bacterias and insects can survive in the slightly warmer weather. More trees and plants will die, yes. The disease part is a bit questionable how it will work. Diseases of many times will shift where they work, but it won't necessarily expand it. But just think about how much fun most of my readers (predominantly American) will have if Malaria creeps north into a bunch of our states. Overall, though, the jury is still out on this one.

Food Production difficulties - 2/5

Many staple grains, like corn and wheat, won't grow as easily if the temperature rises even 2 or 3 C. The world food supply could easily run short, especially with the combination of increasing population from 7 to 9 billion over the next 40 years or so, and the fact that as much of the world gets wealthier, they want more meat. Why is the meat thing an issue? It takes about 40 lbs. of grains and such to make 1 lbs. of cow meat. For pigs, it is much better, with a ratio of about 8. Cause pigs are excellent at turning calories into food for us. Yet another reason to like bacon, eh?

Anyways, food supplies will become more strained. It could be a very serious issue. People might fight over it. By people, I mean countries.

Also interesting, I sometimes brew beer with a guy who is one of the experts on this.

Wrapping up

I have only touched on a few things here. As more come up, I will update this post and tell people to check it out. Before leaving, let's review some of this stuff.

Wealthy countries by and large have pushed a ton of greenhouse gases into the atmosphere. It is causing climate change. Because of how gravity and glaciers work, climate change is going to effect predominantly Southern hemisphere countries. In other words, South America, Africa, areas around India, etc. Pretty much, it is going to have a more profound effect on the countries that can't afford to build walls around their cities to hold back water, and can't throw money and science at the problems as easily. Climate change already punishes poor countries cause they cannot afford to deal with the changes, but the melting glaciers problem exacerbates their situation.

One great example: If emissions of greenhouse gases are not somewhat arrested and sea level rises 1m, at least 17 million people in Bangladesh will find themselves inundated. Where are they going to go? They are surrounded by an ocean, India, Burma, and a whole slew of mountains called the Tibetan Plateau (think Himalayan mountains). India doesn't want them, they are already crowded. Burma is rather hostile. Sending 17 million climate refugees anywhere is likely to cause a problem. And that is just one country.

Hokay, that was depressing to write. To end on a cheery note, climate change will make the weather in both Canada and Siberia much nicer. Also, when the ice caps melt in the Arctic, international trade will have all new sorts of inexpensive ways to move around! This will prove useful.

Oh, one the thing.

The Arctic has a ton of resources that can be mined / produced. So when that ice melts, there will be a wealth a resources. And probably a lot of fighting over said resources.

- Jason Munster

# Oil exports and imports

I am going to be lazy this week and post a very short one.

We discussed the Bakken before. It is producing nearly a million barrels a day of oil. This is $100 million in oil sales a day. It costs up to$45/barrel to lift the oil out of the bakken. The Bakken makes profits of about $50 million a day, or about$18 billion a year.

Let's compare this to Saudi Arabia. They pull oil out of the ground for $1 per barrel (it is very efficient there). They make profits of roughly$100/barrel. They produce 12 million barrels a day. In other words, they make $1.2 billion per day. The US currently imports 11 million barrels of oil per day, and we use a total of 20 million. We transfer about$1.1 billion out of the US per day to drive cars.

What is the end product of this money? Many countries with oil have what is called the "resource curse." It is also known as Dutch disease. It turns out if your country has a ton of natural resources, it becomes inefficient and actually has less growth than it would have otherwise. This has several reasons. First, if a government makes a ton of oil money (or copper or gems), they tend to provide stuff for free to the populace. They take the money they make and just use it to pay for social things, like schools, hospitals, etc. Citizens pay less or no taxes. Corruption and handouts tend to be rife, but since the citizenry gets stuff for free, they don't even care. It is highly inefficient, and the country does not develop a real economy. They often will not use the money to diversify the economy, and once the resource runs out, the country is literally worse off than it was before they found the resource. Finally, having resources tends to allow despotic regimes to thrive. Iraq, Iran, Venezuela,

Interesting, eh? Maybe is a good reason to drive a fuel efficient car. Except that China and India and other developing countries with increasing numbers of cars will ensure every barrel of oil finds a buyer.

One last thing. The US gets its oil imports almost entirely from North America, from Mexico and Canada. The Middle East oil goes to China and the rest of the east. Why does the US spend so much money to maintain peace in the Middle East? Cause if the oil produced there came off the market, the former buyers will have to go elsewhere to buy their oil. Despite the fact that we don't get oil from the middle east, the global price (and our price) is much lower than it would be if war caused the oil to stop flowing from there. So why doesn't China start also enforcing peace in the Middle East? How come the US is the major country to fund this peace keeping? Frankly put, no other country has the experience of the US with having troops on foreign soil. China doesn't yet have the experience to effectively do this sort of thing. But they are already practicing. Eventually, as China continues to rise, they will likely begin to shoulder responsibility of ensuring the flow of oil.

# Nuclear Reactors Final

I am getting bored of this topic, and I want to get to wind-solar-hydro so I can finish up with the energy technology stuff and write varied stuff. So I am going to compress it a lot. If anyone wants to see it expanded, let me know and I will take care of it later.

This post is about breeder reactors, thorium reactors, preventing nuclear proliferation, how the electricity costs stack up to other power plants, and why nuclear power is so expensive.

Nuclear power is expensive because the up-front costs are massive. The cleanest and most efficient coal fired power plants might cost a billion dollars to build. It might take 5 years to build it. Nuclear power plants seem to cost about $8 billion to build with all the safety features they use to prevent nuclear meltdowns (seriously, the new tech is very safe, and it shows in the cost). And they seem to take 8-15 years to build, depending on how much Greenpeace or pretty much every other group tries to stop construction via litigation. In other words, they take out an$8B loan and accrue interest for 8-15 years before they can start paying it back. Stuff costs a lot. Why do it? Cause nuclear waste can be contained, unlike the NOx and CO2 from natural gas and coal plants. Also, South Korea thinks it can build a nuclear plant a nuclear power plant in a short amount of time for only $5 billion. United Arab Emirates decided this was a good idea, and is buying four South Korean nuclear reactors to desalinate water. Schematic of the South Korean nuclear power plant to be built in the United Arab Emirates, from link above. The US Energy Administration Administration agrees that nuclear power is now less expensive than it used to be. I have ripped a table straight off their web page that shows it (see, I really am getting lazy in this post). Table 1. Estimated levelized cost of new generation resources, 2018 U.S. average levelized costs (2011$/megawatthour) for plants entering service in 2018
Plant type Capacity factor (%) Levelized capital cost Fixed O&M Variable O&M (including fuel) Transmission investment Total system levelized cost
Dispatchable Technologies
Conventional Coal 85 65.7 4.1 29.2 1.2 100.1
Advanced Coal 85 84.4 6.8 30.7 1.2 123.0
Advanced Coal with CCS 85 88.4 8.8 37.2 1.2 135.5
Natural Gas-fired
Conventional Combined Cycle 87 15.8 1.7 48.4 1.2 67.1
Advanced Combined Cycle 87 17.4 2.0 45.0 1.2 65.6
Advanced CC with CCS 87 34.0 4.1 54.1 1.2 93.4
Conventional Combustion Turbine 30 44.2 2.7 80.0 3.4 130.3
Advanced Combustion Turbine 30 30.4 2.6 68.2 3.4 104.6
Advanced Nuclear 90 83.4 11.6 12.3 1.1 108.4
Geothermal 92 76.2 12.0 0.0 1.4 89.6
Biomass 83 53.2 14.3 42.3 1.2 111.0
Non-Dispatchable Technologies
Wind 34 70.3 13.1 0.0 3.2 86.6
Wind-Offshore 37 193.4 22.4 0.0 5.7 221.5
Solar PV1 25 130.4 9.9 0.0 4.0 144.3
Solar Thermal 20 214.2 41.4 0.0 5.9 261.5
Hydro2 52 78.1 4.1 6.1 2.0 90.3

Note that last column is \$ per megawatt hour. It is the bottom line cost of producing power from that plant. First, what is dispatchable vs non-dispatchable? Dispatchable means you get it whenever you want it. You can ramp it up or down however you please. Non-dispatchable means that you depend on external factors, like the fickle winds of.. well.. winds?

Tangent! Winds are really just redistribution of energy from the equator to the poles. The sun shines more at the equator, heating it up, and then energy likes to move from areas of high energy to areas of low energy, so it does it using wind. And sometimes hurricanes. So really, wind power is just really inefficient solar power. You know what else is really inefficient (and slow) solar power? Hydrocarbons and coal. Cause they are really just buried plant and algae matter and such. That is tens to hundreds of millions of years old. So, coal and oil are really just really old, slow, and dirty solar power. Tangent done!

Tangent picture? This shows how the equator heats more than the rest of the earth. These extra heat has to redistribute to be more even. Hurricanes start near the equator cause of the heat there, then move away from it. from: http://oceanworld.tamu.edu/resources/oceanography-book/oceansandclimate.htm

Nuclear power is almost as cheap as coal power, and cheaper than clean coal (note, clean coal still produces a ton of CO $_2$ and NO $_X$ )! What gives? How is nuclear so inexpensive? Well, we haven't built a nuclear power plant in the US in years. We don't know what it will actually cost. Those are just estimates. Also, people are quite scared of nuclear power. The cost of building nuclear power rises when you have environmentalists and NIMBY folks suing the pants off nuclear power developers. But let's make one thing clear: if the new generation of nuclear power plants are as inexpensive as they are supposed to be, the power is less expensive that all other power plants other than natural gas (note: the US does not have capacity to build more hydro power), and less expensive than even that if you account for NO $X$ produced by and methane leaks associated with natural gas power (methane production and transport will always have leaks, and it is 23x as powerful a greenhouse agent as CO2).

Let's look at a few more things on the chart above. Remember when I said natural gas got cheap? Look at how cheap it is to produce power from natural gas on the chart above. Think anyone is building nuclear, solar, or offshore wind when you can build and deploy reliable natural gas power? Somehow, the answer to this is yes. Yes, people are building all these things, despite being expensive. Which is kind of cool.

Before moving away from costs, look at the variable costs. The variable costs are high for everything except renewables and nuclear. Why is this? Cause renewables and nuclear don't really use fuel. Yes, a nuclear plant uses fuel, but it costs almost nothing relative to the labor and the capital costs. All the cost of these is upfront CAPEX (capital expense), and then you get free power.

Finally, lets take a really close look at the variable costs. This link is pretty sweet for those of you interested. It contains variable costs for each power source. You can see that fuel is the bulk of cost of fossil steam plants, but less than a quarter of the total cost of nuclear, and nuclear fuel is 1/4 the price per energy unit than even dirt cheap natural gas.

Enough about costs! Onto breeder reactors!

In the first post I mentioned that one part of nuclear reactions is to give off neutrons. Sometimes instead of a neutron splitting an atom, the atom absorbs it. U-235 is the uranium we use in nuclear reactors. U-238 doesn't produce as much heat, cause it doesn't like to decay as fast, so it isn't viable nuclear fuel. Or is it?! U-238 is like a catcher in baseball. Except it catches neutrons. And then it incorporates them into its nucleus to become U-239. In other words, it really isn't like a catcher in baseball.

The breeder reaction series. From: http://nuclearpowertraining.tpub.com/h1019v1/css/h1019v1_76.htm

What's special about U-239? It decays rapidly through a special type of decay to become neptunium-239 and eventually plutonium 239! This process can extract up to 100x the energy from nuclear fuel. You know what's magic about that? Less nuclear waste produced. Also, you produce a ton of nuclear fuel this way. You can also use thorium-232, which then becomes uranium-233 after absorbing a neutron, which can in turn be used for nuclear fuel. Thorium is very cheap and very abundant. So the plutonium and uranium that is magically created through awesomely manipulating nuclear forces is then used in nuclear thermal reactors to produce power.

Nuclear proliferation!

Having a nuclear power plant does not mean you can make nuclear bombs. Nuclear bombs required U-235 enriched to a very high level. What exactly is enrichment? Natural uranium is less than a few percent U-235, the rest is U-238. The uranium comes as a solid, and is processed by making it dance with a bunch of flourine. UF $_6$ is produced, which is gasified uranium. The U-238 is slightly heavier than U-235, so it very very slowly settles to the "bottom" if you spin it very fast in centrifuges. Once you have enriched it to somewhere between 5 and 8%, U-235, it is good to go into a reactor and make energy. To make a bomb, you have to enrich it to around 90%. Enriching it further gets exponentionally more difficult. Getting from 50% to 70% is much more difficult than getting it from 10% to 50%. So making bombs is hard.

What about plutonium? Seems like any fool can make plutonium. And in fact, they can! All you have to do it get some U-235, wrap it in U-238, and you have make a breeder reactor in your back yard! This seriously happened. Someone made a breeder reactor in their garage at 17. And this wasn't one of those kids that comes from a brilliant family with a ton of money that goes to work in a world famous lab and 'discovers' a new technique under the watchful eye of one of the most brilliant researchers in the world. This was your every day kid who was just really interested in something.

Except that building a nuclear weapon from plutonium is even more difficult than from uranium. Cause when you make the plutonium, you always get a large amount of another plutonium isotope. The other isotope loves to go critical much earlier than Pu-239. Remember what happens to a potentially critical nuclear reaction when the fuel gets split up? No? Remember what happened to Chernobyl when a small gas explosion spread the core out everywhere? It completely shuts down the critical reaction. In other words, plutonium loves to accidentally blow up to early and just spread itself around without going critical. Not much of a weapon there. How do we have plutonium bombs then? Really smart people made special triggering mechanisms to make this happen. How do they do this? I dunno. If I did, I wouldn't be out in public writing a blog, I'd be doing super secret awesome research somewhere.

Turns out that only the US and a handful of other countries have figured this one out. So while any old fool can make a breeder reactor, the combined science of most nations is not good enough to figure out how to do it.

One last thing. If nuclear power is so difficult, how did so many countries get it? Well, US and Russia developed it. The US gave it to China at some point to balance some power issues with the Soviet Union. The US gave it to several other allies as well. The Soviet union gave it out some, too. China then distributed to crazies like North Korea years later, and Pakistan and India were given it through similar pathways. In other words, it is still pretty difficult to develop.

Hey, I just covered 4 whole things in one post, and managed to get more terrible jokes in. Awesome.

Aww darn, I forgot to include the small amount of original research I did on this topic. Next article


text:
### Still have math questions?

Algebra
Question

If $$v = m ^ { ( 6 - \frac { n } { r } ) } ,$$ express $$n$$ in terms of $$r , m$$ and $$v$$ by taking logarithms to base $$10 .$$

$$n = \frac { r } { \log m } ( 6 - \log v )$$

$$n = r \log m ( \log v - 6 \log m )$$

$$n = r ( 6 - \frac { \log v } { \log m } )$$

$$n = \frac { 1 } { m } ( 6 m r - r v )$$

solve for $$n , v = m ( 6 - \frac { n } { r } ) : n = - \frac { r \ln ( v ) - 6 r \ln ( m ) } { \ln ( m ) }$$


text:
PROGRAMS

# ps

## Overview

The ps command (a.k.a process status) provides a snapshot of the current processes running on the system.

For repetitive updating of the process information (e.g. updates on how much CPU and memory each process is using, similar to Task Manager in Windows) see the top command instead.

## Default Implementation

By default, ps only prints processes owned by the current user, AND are processes that are associated with the terminal that called ps. Normally this will result in quite a small amount of output, perhaps only 2-5 processes.

gbmhunter@ubuntu:~$ps PID TTY TIME CMD 71811 pts/3 00:00:00 bash 73771 pts/3 00:00:00 ps gbmhunter@ubuntu:~$

You will always be guaranteed these two above when running from a bash terminal, obviously bash is running, and so is the ps command (it includes itself).

## More Complete Information

More complete information on the running processes of the system can be found by providing arguments, such as the UNIX-style ps -e:

[ghunter@ubuntu ~]$ps -e PID TTY TIME CMD 1 ? 01:15:50 systemd 2 ? 00:00:26 kthreadd 3 ? 00:02:24 ksoftirqd/0 8 ? 00:01:55 migration/0 9 ? 00:00:00 rcu_bh 10 ? 06:22:40 rcu_sched 11 ? 00:00:51 watchdog/0 12 ? 00:01:00 watchdog/1 ... If you want more detail on the command that started each process, use ps -ef. [ghunter@ubuntu ~]$ ps -ef
UID         PID   PPID  C STIME TTY          TIME CMD
root          1      0  0 Jan02 ?        01:15:51 /usr/lib/systemd/systemd --system --deserialize 22
root          2      0  0 Jan02 ?        00:00:26 [kthreadd]
root      45147      1  0 Jun06 ?        00:00:19 /usr/sbin/crond -n
root      45697      1  0 Jun06 ?        00:06:39 /usr/sbin/irqbalance --foreground
root      45718      1  0 Jun06 ?        00:00:00 /usr/sbin/atd -f
ntp       45742      1  0 Jun06 ?        00:00:18 /usr/sbin/ntpd -u ntp:ntp -g

Remember that PID is the process ID and PPID is the parent process ID.

If you use BSD style options (no dash), ps will print the command and the provided options for each process, rather than the executable name:

gbmhunter@ubuntu:~$ps -aux USER PID %CPU %MEM VSZ RSS TTY STAT START TIME COMMAND root 1 0.0 0.0 33896 4020 ? Ss Jan27 0:02 /sbin/init root 2 0.0 0.0 0 0 ? S Jan27 0:00 [kthreadd] root 3 0.0 0.0 0 0 ? S Jan27 0:01 [ksoftirqd/0] root 5 0.0 0.0 0 0 ? S< Jan27 0:00 [kworker/0:0H] root 7 0.0 0.0 0 0 ? S Jan27 0:30 [rcu_sched] root 8 0.0 0.0 0 0 ? S Jan27 0:00 [rcu_bh] root 9 0.0 0.0 0 0 ? S Jan27 0:00 [migration/0] root 10 0.0 0.0 0 0 ? S Jan27 0:00 [watchdog/0] root 11 0.0 0.0 0 0 ? S Jan27 0:00 [watchdog/1] root 12 0.0 0.0 0 0 ? S Jan27 0:00 [migration/1] root 13 0.0 0.0 0 0 ? S Jan27 0:01 [ksoftirqd/1] root 15 0.0 0.0 0 0 ? S< Jan27 0:00 [kworker/1:0H] ... This is only a snapshot of the total number of processes it will print! ## Supported Options ps can support a large and confusing amount of different option styles, including UNIX options (one dash), BSD options (no dash) and GNU long options (two dashes). ## ps With grep The output of ps can be piped to grep to filter the results. For example, if you wanted to only look for processes with the word hocus_pocus in it: $ ps -aux | grep hocus_pocus

Note: grep will match anything on the line printed by ps -aux. That means that hocus_pocus will be matched against the username column and any paths in the process name.

However, aside from having to use two commands, there are other disadvantages to using ps with grep. A completely new program, pgrep was built to try and provide a better process-searching tool.


text:
# How to choose data for training a predictive model for attrition prediction

Trying to build a predictive model for attrition prediction at service desk/call center.

Have daily data on the following parameters:

1.Call quality - QTM (0-100%), 2.No. of calls - Calls(Number) 3.Attendance 4.Customer feedback(1/0) Q1,Q2 (0-100%) for both, agents who left the job and for the ones who are still there, for a duration of 6 months.

Aim: to predict agents tendency/probability of staying/leaving based on his/her daily performance.

Doubts i have, 1. how should i use the data to train the model(logistic regression)

should it be trained based on the avg of the parameters taken over a duration of 6 months.

**if so can we test the daily metrics based on a model which is trained using mean of the parameters for 6 months.

this is my first attempt at making a predictive model,i have gone thru various case studies/models such as the titanic survival model using logistic regression,Wisconsin DEWS model.

I decided to model using the weekly aggregates of the the two populations(attrites and Non-attrites).

The Data Set (approx 5 months data,with weekly aggregates of the two populations i.e Attrites and Non-Attrites.) AW1 : Week1 Aggregates of the performance metrics for Attrites NAW1: Week1 Aggregates of the performance metrics for Non-Attrites

Post this i ran a logistic Regression on 80% of this data-set and kept aside the other 20% for testing. Results of the logistic regression:

and then i used the predict function on the 20% of the data which contained 3 data points for both attrites and Non-attrites,so to be 100% accurate the model should have predicted 3 as attrites and 3 as Non-attrites but the correct prediction is 5/6 that is one wrong prediction out of 6.

Please help me in interpreting the meaning of the results of the model all the z values are zero im not sure what that signifies.

Googled a little regarding the z values = 0 issue and came across some posts on stackoveflow that suggested using "bayesglm" instead of "glm" did that and the results are good at the first look but being a newbie in the field i would like you to guide me with respect to the statistical significance of the issue and is the model really as good as the results of the "bayesglm" or is it just by fluke.

the model gives a 100% accurate prediction on the test set now 6/6.

I would aggregate the data to weekly aggregate numbers, assuming that great / bad agents have some what consistent call center performance over the six months. Sometimes aggregating erases the effects of outliers before they can be classified as such. This would account for shifts in performance across the total 6 month period as well.

When it comes to sampling using 80% of data points to develop model and 20% to validate would be a good start. Can adjust those numbers depending on how big a data set you are dealing with.

I utilize Iowa State papers some times. Here is a good one on the basics (pdf).

Hope You have fun!!

Update: Just so we are clear you are aggregating by week per customer service rep right?

Both models don't fit good. You can tell variable fits using the coefficients section of the results. Significant variables have the stars next to there P value (more stars equals more significant typically and lower P value). Based on that none of your variables are actually

It's good that you are comparing the model vs actual results. ROC curves capture the model differences pretty well. Try running this and post what you get.

library(pROC)
g <- roc(admit ~ prob, data = mydata)
plot(g)

Update: Its weekly aggregates, population wise(i.e the attrite population and the active population),didn do it agentwise because we will have cases when a agent leaves when he was at his peak performance but those are exceptional cases so i thought it would be better to compare the two populations, please advise if that's not the correct way of thinking

SO AW1 is first weeks performance metric aggregates for Attrites, similarly NAW1 is first weeks performance metric aggregates for Non-attrites/Active agents.

Ran the "step" fuction(Selects a formula-based model by AIC) on the bayesglm model and the results are as below;

Aggregating all the agent results together will mean you essentially are over fitting to match the total population metrics and not the agent's performance. Recommend that you tie in the agent level results. You mentioned there being a chance that a great agent leaves unexpectedly but for a well run unit that should be a rarity. Also, recommend you change model family parameter to

family = binomial(link = "probit")

This should give you probability of default for each agent. This would

• great point Ken, so basically in order to avoid the bias that might be introduced due to shifts in performance over such a long duration(6 months).....it will be better to aggregate over a weeks performance and train based on the weeks aggregate values and later test using week aggregates. i have a silly doubt : can we use a model which is trained using the aggregate values(over a week)....to test daily performance values,will it be a viable option in terms of the prediction accuracy. or should we test the same way i.e over week aggregate values. Sep 3, 2014 at 8:52
• It's definitely possible. In your case, a solid model should perform well in peak and minimal periods of call volume Sep 6, 2014 at 4:10
• Hey Ken thanks for the motivation, i tried out the whole idea and the results look like below, please check and advise as im having difficulties in interpreting the results of the model. Sep 9, 2014 at 8:11
• Hey Gung, i have added the ROC curve and the info regrading the aggregates,please check and guide me. Sep 10, 2014 at 9:27

In engineering terms, the "fit an equation" approach alone often becomes voodoo. It does not answer "why". It does not inform effective actions.

So if I manually entered the data:

Subject Calls   QTM Attend  Q1_rev  Q2_Rev  Status
AW1 24.5    98.5    71.21   94.44   90.35   1
AW2 28  96.06   70.96   97.22   93.08   1
AW3 30  95.23   64.51   97.7    94.02   1
AW4 27  96.4    65.57   97.14   93.9    1
AW5 29  97.14   76.47   97.22   94.23   1
AW6 27  90.48   81.79   97.22   94.33   1
AW7 29  97.89   59.34   97.1    94.38   1
AW8 28  92  97.78   97.28   94.5    1
AW9 29  97.17   91.11   98.08   94.72   1
AW10    27  86.49   97.72   98.14   94.9    1
AW11    26  89.8    95.74   98.17   94.79   1
AW12    27  93.95   97.78   98.44   95.08   1
AW13    27  82.14   92.85   98.65   95.49   1
NAW1    26  95.49   87.8    99.27   94.74   0
NAW2    30  94.19   84.83   98.3    94.65   0
NAW3    32  94.17   80.44   98.66   94.67   0
NAW4    29  96.41   81.05   98.71   95.06   0
NAW5    30  96.49   85.46   98.6    95.03   0
NAW6    28  94.84   88.69   98.69   95.24   0
NAW7    29  94  91.22   98.82   95.81   0
NAW8    29  95.47   88.75   98.57   95.84   0
NAW9    29  96.84   84.88   98.47   95.82   0
NAW10   29  93.8    93.56   98.38   95.96   0
NAW11   27  94.02   87.07   98.35   95.96   0
NAW12   28  97.82   84.65   98.33   96.01   0
NAW13   29  96  86.62   98.36   96.09   0

And then use the following code to look at the variables in terms of "Staus":

#housekeeping
rm(list = ls())

library("vioplot")

#set working directory
setwd("C:/Users/mrmunroe/Desktop/")

ind1a <- as.numeric(subset(data$Calls,data$Status==1))
ind2a <- as.numeric(subset(data$Calls,data$Status==0))

ind1b <- as.numeric(subset(data$QTM,data$Status==1))
ind2b <- as.numeric(subset(data$QTM,data$Status==0))

ind1c <- as.numeric(subset(data$Attend,data$Status==1))
ind2c <- as.numeric(subset(data$Attend,data$Status==0))

ind1d <- as.numeric(subset(data$Q1_rev,data$Status==1))
ind2d <- as.numeric(subset(data$Q1_rev,data$Status==0))

ind1e <- as.numeric(subset(data$Q2_Rev,data$Status==1))
ind2e <- as.numeric(subset(data$Q2_Rev,data$Status==0))

par(mfrow=c(1,5))
vioplot(ind1a,ind2a,
names=c("Calls | Status1","Calls | Status0"),
col="Red")

vioplot(ind1b,ind2b,
names=c("QTM | Status1","QTM | Status0"),
col="Orange")

vioplot(ind1c,ind2c,
names=c("Att | Status1","Att | Status0"),
col="Green")

vioplot(ind1d,ind2d,
names=c("Q1r | Status1","Q1r | Status0"),
col="Blue")

vioplot(ind1e,ind2e,
names=c("Q2r | Status1","Q2r | Status0"),
col="Violet")

Then I get the following image. When I, as a human look at it, and I consider that if "Status 0" is someone who attritioned, then it looks like those who leave are the better and brighter. If that is the case then it is diagnostic. My sincere hope is that this is not the case, but unlike voodoo it makes a difference between data and understanding.

The fact that we have a predictive model doesn't mean that we understand why. Both descriptive models (answer what, like the table) and predictive models (answer how, like the equation)are things computers can do. Diagnostic tools, tools that actually empower change and improvement have to have "why" as the answer. Currently the only such tool is a human mind.

There is a program called JMP that has an amazingly powerful tool called a "Variability plot". It is straightforward to use and, at Intel, is the primary graph used in the company. It is a strong and effective tool for reducing very complex seeming problems to their basic roots. On of the things that they do is handle multi-level nesting. I need to learn more about lattice graphics and plotting in R to be able to make a 3-level variability plot in R.

I understand that you've already checked an answer, but I suggest an alternative approach all the same. Why not use random survival forests to predict the conditional cumulative hazard of attrition? This method has the benefit of having few assumptions about the underlying hazard function, and does not rely on the proportional hazards assumption. In addition, The method is implemented in randomForestSRC (formerly within randomSurvivalForests). The limitation of this method is that you would need both tenure of the workers as well as an indicator for whether there was attrition even for each worker. I would prefer this method over logistic regression because time is not really discrete in this process; it is at least approximately continuous. Another benefit of random survival forests is that it automatically considers complex interactions among your model features.

• Tried using rf(random forests) and its giving good results as of now,but i would like to try the approach that you have suggested using randomsurvivalforest,would be a good start for me if you could explain the basic funda behind ramdomsurvivalforests or if could point me in the right direction in terms of its application and the constraints of the function in R. Sep 19, 2014 at 8:35
• The basic idea is that random survival forests builds forests of trees based on splits on the estimated cumulative hazard function. You can find links to helpful tutorials and primers at ccs.miami.edu/~hishwaran/ishwaran.html Sep 19, 2014 at 16:30


text:
# How do you solve sqrt(8-2x) = x-4?

Oct 2, 2017

$x = 2$
$x = 4$

#### Explanation:

$\sqrt{8 - 2 x} = x - 4$

Square both sides to remove radical:

$8 - 2 x = {\left(x - 4\right)}^{2}$

Expand right hand side:

$8 - 2 x = {x}^{2} - 8 x + 16$

Collect like terms on one side of the equation:

${x}^{2} - 6 x + 8 = 0$

Factor ${x}^{2} - 6 x + 8 = 0$:

$\left(x - 2\right) \left(x - 4\right) = 0 \implies x = 2$ and $x = 4$


text:
# Microeconomics: Calculating and Comparing MRS for Various Utility Functions [closed]

I am having a problem determining "which utility function is not like the rest." The following utility functions are:

$$U_1(x,y)=x^{2/3}y^{1/3}$$

$$U2(x,y)=3x^2*y+2$$

$$U_3(x,y)=\frac{1}{3}\ln (x)+\frac{2}{3}\ln( y)$$

$$U_4(x,y)= 4\ln(x)+2\ln(y)-12$$

So my general thought in approaching this question was to obtain the MRS for each of the functions. After doing this, however, I found U1 and U2 to have an MRS of -2y/x. However, I am having a much bigger problem differentiating U3 and U4. My intuition is that U4 would be equal to U1 and U2 because we value x twice as much as we value y. however, can anyone help out with the math?

The rule for differentiating a natural logarithm is

$$\frac{d}{dx}a\ln(x)=\frac{a}{x}$$

so \begin{align}\frac{\partial{}}{\partial{}x}U_3(x,y)&=\frac{\partial{}}{\partial{}x}\left(\frac{1}{3}\ln(x)+\frac{2}{3}\ln(y)\right)\\&=\frac{1}{3}\frac{1}{x}\\&=\frac{1}{3x}.\end{align}

\begin{align}\frac{\partial{}}{\partial{}y}U_3(x,y)&=\frac{\partial{}}{\partial{}y}\left(\frac{1}{3}\ln(x)+\frac{2}{3}\ln(y)\right)\\&=\frac{2}{3}\frac{1}{y}\\&=\frac{2}{3y}.\end{align}

$$\text{MRS}=-\frac{\frac{\partial{}}{\partial{}x}U_3(x,y)}{\frac{\partial{}}{\partial{}y}U_3(x,y)}=-\frac{\frac{1}{3x}}{\frac{2}{3y}}=-\frac{y}{2x},$$

I guess that by following this template you should be able to work out the other derivatives and the MRSs that you are looking for.

• So when I take this partial derivative, my final answer here would look like this: (-1/3x+2lny)/(1/3lnx+2/3y)? I appreciate your help! – Zach Smith Feb 13 '16 at 17:00
• @ZachSmith No, I finished off the example for $U_3$ so you can see how the whole process works. The method for calculating the MRS for $U_4$ is similar. – Ubiquitous Feb 13 '16 at 18:15


text:
# SNR relation with modulation index and noise power

What is the relation between the AM modulation index and the SNR? (I plotted this and it came as the shape of the exponential graph)

Also what is the relation between the channel noise power and the SNR? I plotted this and I got the shape as this graph:

(ignore the labels of the axis)

• This question is unanswerable because you give no indication of where the noise is and how it is affecting the signal. – Dilip Sarwate Apr 2 '14 at 13:44

If $x(t)$ is your message signal, then you can write the AM signal as

$$s(t)=A[1+mx(t)]\cos\omega_0 t$$

where $A>0$ is a real-valued constant, $m$ is the modulation index, and $\omega_0$ is the carrier frequency (in radians). If you assume that the noise added by the channel is white with power spectral density $N_0$, then the SNR after demodulation is

$$SNR=\frac{A^2m^2\overline{x^2(t)}}{4N_0B_x}$$

where $B_x$ is the bandwidth of the message $x(t)$ and $\overline{x^2(t)}$ is its average power.

• But SNR is not an exponential function of modulation index $m$ as the OP claims his result shows, so the OP must be using a different definition of SNR. – Dilip Sarwate Apr 2 '14 at 14:07
• @DilipSarwate You're right, but this is the most basic case and if the OP is not happy with it, this will hopefully at least start a discussion revealing the exact details of the question. – Matt L. Apr 2 '14 at 15:03
• For SNR I used power of signal/power of noise – user1930901 Apr 2 '14 at 18:55
• Yes, that's what's usually done. – Matt L. Apr 2 '14 at 20:21
• so what you posted is applicable or not? – user1930901 Apr 3 '14 at 6:55


text:
#### Chapter 5 Geometry

Section 5.1 Elements of Plane Geometry

# 5.1.2 Points and Lines

In geometry, a place or a position in a plane is idealised to the most basic object, namely a point. A single point itself cannot be characterised any further.
For several points, relations between these points can be considered in different ways - and points can be used to define new objects such as line segments and lines (see figure below). Mathematically, these objects are sets of points.

First, we consider a line segment and the distance between points. To do this, we need a comparison tool for measuring distance. In mathematics, this tool is a comparative length called the unit length. For applications, appropriate length units such as metres or centimetres are chosen, depending on the task in hand.
##### Line Segments and Distances 5.1.1
Given two points $A$ and $B$, the line segment $\stackrel{‾}{AB}$ between $A$ and $B$ is the shortest path between the two points $A$ and $B$.

The length of the line segment $\stackrel{‾}{AB}$ is denoted by $\left[\stackrel{‾}{AB}\right]$. The line length equals the distance between the two points $A$ and $B$.

A ray of light emitted by a distant star or by the sun is an appropriate notion of a ray starting at the initial point $A$ and proceeding through a second point $B$ indefinitely. A ray is also called a half-line.

Continuing the path of a line segment $\stackrel{‾}{AB}$ on both ends indefinitely results in a line.
##### Line 5.1.2
Let $A$ and $B$ be two points (i.e. point $A$ is different from point $B$). Then, $A$ and $B$ define exactly one line $AB$.

Considering, beside $A$ and $B$, an additional point $P$, we can ask for the distance $d$ of the point $P$ from the line $AB$, which is defined as the shortest path between $P$ and one of the points of the line $AB$.

Given three points $P$, $Q$, and $S$ in the plane, the lines $SP$ and $SQ$ can be defined.
The two lines have the point $S$ in common. If the point $Q$ is also on the line $SP$, then $SQ$ and $SP$ denote one and the same line. If the point $Q$ does not belong to the line $SP$, the line $SQ$ is different from the line $SP$. Then, the two lines have only the point $S$ in common. The point $S$ is called an intersection point.
If any two lines $g$ and $h$ do not have any points in common, the smallest distance between points on $g$ and $h$, respectively, is called the distance between the lines $g$ and $h$. Hence, $g$ and $h$ do not have any point in common if they have a distance larger than $0$. Two lines are called parallel if every point on one of the two lines has the same distance from the other line.
A single line can be described by the distance of two points $M$ and $M\text{'}$ as well: The set of all points with the same distance from two points $M$ and $M\text{'}$ is a line.
In geometry, it is a typical approach to define new objects by means of certain properties such as the distance. In this way, a circle can also be described very easily.
##### Circle 5.1.3
Let a point $M$ and a positive real number $r$ be given.
 Then, the set of all points at distance $r$ from point $M$ is a circle around $M$ with radius $r$.