Canker sores are also known as mouth ulcers and as that latter moniker might suggest, these are ulcers that appear in the mouth.
There are actually different types of canker sore. These are the 'simple canker sore' and the 'complex canker sore'. They might both seem pretty simple to you though: both of them are just painful ulcers in your mouth.
The difference is in the way these conditions present. A simple canker sore is likely to appear two to three times a year and will generally last up to a week. They generally last for about a week.
Complex canker sores meanwhile are less common and these generally appear in those that have previously had canker sores.
Although they have similarities, it is important to distinguish between canker sores and cold sores, which are not the same thing. Cold sores are caused by viruses and will appear on the outside of the mouth, usually under the nose or around the lips. Canker sores by contrast appear inside the mouth and are not caused by the herpes simplex type 1 virus.
So, what does cause canker sores? The unfortunate answer is: all kinds of things! Canker sores can be caused by everything from stress, to illness, to direct trauma. In some cases, they may be the result of irritation in the mouth, or of vitamin deficiencies. In other cases, canker sores might appear to emerge for no apparent reason whatsoever. They just felt like appearing.
The good news though is that there are also lots of ways you can squash a canker sore. In this post, we're going to take a look at some of those methods and you can then try using them yourself to remove the canker sore quickly and start enjoying your food again!
Yes, the best treatment of all for canker sores and the one that is nearly always guaranteed to work is to play the old 'waiting game'. As mentioned, canker sores tend to be self-limiting and will go away on their own after about a week.
So, while it might be very frustrating right now, it will pass in time. And before the canker sore completely goes away, it will first shrink and become a lot less angry and painful. Time is the great healer of all wounds! Well… most anyway.


text:
Q: Multi threaded task not utilizing CPU? I have a method in C# that I want to run from multiple threads for lets say 20 times in a console app, on a 4 core machine. The problem I am facing is that the results are not what I expect them to be. When I run the method 25 times sequentially, the time taken is X and I can see in perfmon that max CPU is 100% (it tells me that it is using 1 core). Running the same method using multiple threads, I expect the execution time to be X/4 and also I expect the max CPU usage in perfmon to be 400% (since it is a 4 core machine). However, I can only see that the execution time is X/2 and the max CPU usage has never been more than 275%. I have tried various things such as creating my own threads, using threasdpool etc. and nothing seems to work. Can somebody please explain/help me understand it better?
Other interesting thing is that if I replace my method with a dummy task using Thread.SpinWait(x) the execution time is X/4 and I can see max CPU going to 400%. This tells me that something is wrong in my method and I don't understand what it is. I have no locks/sleep anywhere in my method. Following is the code I use to execute:
public static void DoWorkParallel()
{
    var s = new List<string> { "a", "b", "c", "d", "e", etc. };
    s.ParallelForEach2(x =>
                       {
                           MyTask(x);
                       });
}

public static void DoWorkSequential()
{
    var s = new List<string> { "a", "b", "c", "d", "e", etc. };
    foreach (var ss in s)
    {
        MyTask(x);
    }
}

public static void ParallelForEach<T>(this IEnumerable<T> collection, Action<T> action)
{
    var results = collection.Skip(1).Select(item => new
    {
        action,
        res = action.BeginInvoke(item, null, null)
    }).ToArray();

    action(collection.First()); /* let the mainthread do a job too*/

    foreach (var r in results)
    {
        r.action.EndInvoke(r.res); /*then wait the rest of time */
    }
}

A: Thread.SpinWait is running in a tight loop, performing no I/O and accessing no memory -- it's doing no useful work, so comparing its CPU usage to that of your task is a fallacy.
Without details of what your task is doing (whether it accesses files, waits on mutexes, etc.), it's difficult to determine what's going on.
One thing you can do is to add a Stopwatch instance to your task, and use it to print the time elapsed for each invocation. Compare the results of running tasks with DoWorkSequential and ParallelForEach -- each invocation should take the same amount of time regardless of the method. If the tasks run by ParallelForEach are taking longer, it could indicate you have, say, a mutex that is in contention and needs to be replaced with another locking method.

A: You may want to set the MaxDegreeOfParallelism to 4 in a ParallelOptions parameter to your Parallel.Foreach invocation.  If you don't, dotnet4 uses a lot of worker threads; their management and synchronization overhead can be a little wasteful. I had a similar problem and this worked for me.
http://msdn.microsoft.com/en-us/library/system.threading.tasks.paralleloptions.aspx
Also, keep things simple here. Remember that when you use tightly-wrought parallelism, you're trading off records per second (cpu core throughput) against records per quarter (programmer productivity).

redpajama_set_name: RedPajamaStackExchange


Courtesy of Dr jude onyenakazi.
Check this website daily, I will be posting preparatory questions from past internship exams which I participated.
U guys are really trying.
Comment:Good morning. Pls, I need past questions on physiotherapy internship exams. Thanks.


text:
namespace BestSportsStore.Web.Controllers
{
    using AutoMapper;
    using Infrastructure.Mapping;
    using Microsoft.AspNet.Identity;
    using Models.Product;
    using Services.Contracts;
    using System.Linq;
    using System.Web.Mvc;

    public class BaseController : Controller
    {
        protected const string PartialViewProducts = "_ProductsPartial";
        protected const int PageSize = 3;
        protected const string Underscore = "_";
        protected const string SortByDefault = "Likes";
        protected const string SortOrderDefault = "Desc";

        private readonly IProductsService productsService;

        public BaseController(IProductsService productsService)
        {
            this.productsService = productsService;
        }

        protected IMapper Mapper
        {
            get
            {
                return AutoMapperConfig.Configuration.CreateMapper();
            }
        }

        public IProductsService ProductsService
        {
            get
            {
                return this.productsService;
            }
        }

        public string CurrentUserId
        {
            get
            {
                return this.HttpContext.User.Identity.GetUserId();
            }
        }

        protected IQueryable<ProductViewModel> Sort(IQueryable<ProductViewModel> products, string sortBy, string sortOrder)
        {
            if (sortBy == SortByDefault)
            {
                if (sortOrder == SortOrderDefault)
                    return products.OrderByDescending(p => p.Likes.Count);
                else
                    return products.OrderBy(p => p.Likes.Count);
            }

            if (sortBy == "Price")
            {
                if (sortOrder == SortOrderDefault)
                    return products.OrderByDescending(p => p.Price);
            }

            return products.OrderBy(p => p.Price);
        }
    }
}

redpajama_set_name: RedPajamaGithub


text:
Chef Stockholm HB är ett handelsbolag baserat i Stockholm som ger ut tidningen Chef (tidskrift).

Chef Stockholm HB ägs av Ledarna.

Förutom att ge ut tidningen Chef anordnar företaget även utbildningar för chefer på olika nivåer, samt diverse evenemang, bland annat Stora Chefsdagen i oktober och Chefgalan (tidigare Kompetensgalan) i mars.

Externa länkar
 Officiell webbplats
 Chef på Twitter
 Chef på Facebook

Referenser

Fotnoter

Svenska tidskrifter

redpajama_set_name: RedPajamaWikipedia


When it comes to buying something, one of the Most Important Concerns is usually the "Price". Many of us usually ask for cheaper and cheaper prices.
However, for domain registration, it's simply more than prices! And this is also why one should register his/her domain name for multiple years.
1.Save time, effort and lower the risks of losing your domain.
Multi-year domain registration prevents your domain from expiring too soon. This can save a lot of hassle as registering an expired domain can be very complicated, and there are risks that your domain might be used for unethical activities.
2.Save more with one lump sum payment.
New domain registration can be a lot cheaper (for the 1st year). But when it comes to domain renewal, the fee goes back to normal. Registering a domain for a long period of time can avoid paying for high renewal fees. Moreover, some domain registrars/hosting providers even offer Big Discount for multi-year domain registration so that customers can save more.
Register a .MY Domain for 1st yr and renew year after year.
Total: RM480 Register a .MY Domain for 4 years.
Google has higher confidence towards those who are serious about keeping their business around for a long time.
Google sees domain registered for a longer time as something more reliable (lower chances of being a spammer). This can influence Google's sandbox effect for newer sites.
Act wise, and never lose your domain!
Previous articleMobile Friendly or Mobile Responsive?
Next article5 Things to Consider Before Signing Up for Web Hosting.
Internet Marketing Executive of Exabytes, Love the way Internet can be creative and grow in marketing.


text:
Strategic Ways To Partner With One Schoolhouse
One Schoolhouse is an organization that partners with independent schools to provide innovative student learning and school operations best practices. It is comprised of a consortium of more than 100 independent schools from across the country. Our consortium schools engage with us to strategically enhance their curricular objectives and offerings. Learn more
Students Recognized for Academic Achievements
Congratulations to One Schoolhouse students for academic achievements. For their recognitions by the National Center for Women and Information Technology (NCWIT) Aspirations in Computing Awards:
Lizzie Baratta of Garrison Forest School, Maryland, received an Honorable Mention in the Maryland Affiliate of NCWIT. Lizzie takes AP Computer Science A.
Cassie DeVera of Marymount School in New York received an Honorable Mention in the Greater NYC Affiliate of NCWIT. Cassie takes AP Computer Science A.
Sarah Greisdorf of Holton-Arms School in Maryland, received a National Honorable Mention of the NCWIT Award for Aspirations in Computing. Sarah takes both Creating Tomorrow: Designing for a Digital World and AP Computer Science A.
Cinthia Hernandez of Flintridge Sacred Heart Academy, California, is a 2017l winner in the Los Angeles and Orange County Metro NCWIT. Cinthia takes AP Computer Science A with One Schoolhouse.
Eunji Lee of Sacred Heart Academy Hamden in Connecticut is a 2017 winner of the Connecticut Affiliate of NCWIT. Eunji takes AP Computer Science A.
Audrey Moore of Laurel School in Ohio is a 2017 winner of the Ohio Affiliate of NCWIT. Audrey takes AP Computer Science A.
Kennedy Singleton of Louise S. McGehee School in Louisiana is a 2017 winner of the Mid-Gulf Coast Affiliate of NCWIT. Kennedy takes Creating Tomorrow: Designing for a Digital World.
Congratulations to Mackenzie Barnes of St. Mary's Episcopal in Tennessee for being accepted to Tennessee Governor's School for International Studies program this summer. Mackenzie takes Chinese with One Schoolhouse.

redpajama_set_name: RedPajamaCommonCrawl


text:
Napkin Folding
Food & Drink Issues
A Butler's Joke
The Butler did it!
Private Service Survey
Preparing an Event
Visitors » Extras » The Butler did it!
THE BUTLER REALLY DID DO IT HERE!
The dangers inherent in falling out with Butlers are not to be underestimated. Two woman who lived in the country house in the UK, now occupied by Prince and Princess Michael of Kent had a Butler for 40 years. When they announced after dinner that they would have to "let him go", he served coffee as usual, selected a 12-bore from the gunroom ...... and shot them.
DUKE OF MARLBOROUGH
Working in private service can be very confusing. People want someone who is part of the family, but who is still their staff. Younger employers may find the balance particularly hard to strike. Not that old money always treat private staff with respect. At Blenheim Palace, near Oxford/UK, almost a quarter of a mile from the staff quarters, a former Duke of Marlborough still rang for the Butler to change television channels.
BUTLERS HURT AFTER FLEEING PRINCE
CAIRO (Reuters, November 18, 1998) - Two Egyptian butlers were hospitalized in serious condition Tuesday after being injured while trying to escape from a high-rise hotel where they had worked for a Saudi Prince, the police said. Ahmed Noureddin and Ahmed Abdel-Satar tied bed sheets together and used them to lower themselves from the 29th floor of the Ramsis Hilton to the 26th, and were injured when jumping to the 24th floor, the police said. The police said Abdel-Satar claimed the Prince's American bodyguards had forced them to stay in the hotel for two years. Abdel-Satar also asked authorities to "save'' nine other butlers working for the Prince, who lived in the hotel. Hotel officials were not immediately available for comment. The police said Abdel-Satar spoke of ``torture and humiliation'' by the guards and complained that he had not been paid for six months. The police said they had a file on the bodyguards, who in 1996 had beaten up an Egyptian who was parking his car near the Prince's.


text:
Q: Iterating objects in React JS What is the Error in the below code? I tried to iterate over the array of objects using the map. but I am receiving an error. Please find attached the code for reference.
import "./styles.css";

export default function App() {
const sample = [{
samplePath : "/website",
 par:{
  abc:"123",
  def: "678",
  ghi:"456"}
}];

const createURL = (re)=> {

const q = Object.entries(re.par)
    .map(([key, value]) => key +'='+value).join("&");

return "?"+ q;
}

console.log(createURL(sample));

return (

Hello CodeSandbox

Start editing to see some magic happen!

);
}

I am receiving the following error

A: You're trying to call Object.entries on re.par.  In this case, re is an array, so calling .par on it returns undefined.  I'm not sure what your desired result is, but if you access the first element in the array and then .par, maybe that's what you're after?

const sample = [{
samplePath : "/website",
 par:{
  abc:"123",
  def: "678",
  ghi:"456"}
}];

const createURL = (re)=> {

const q = Object.entries(re.par)
    .map(([key, value]) => key +'='+value).join("&");

return "?"+ q;
}

console.log(createURL(sample[0]));

A: In your example sample is an array, containing one object.
So, either, remove the [] such that sample is an object like you are referring to it in your code. Or, Object.entries(re[0].par)


See the image: I got above error when trying to send a RestAPI request.
were you able to run the test using the play button in object repository?
See the attachment, that is the way I am sending a request.
In my case, I am taking test data from excel file.
I have 85 test data records, and for the first 76 test records, everything is OK.
I just got the issue from the record 77th to the last one.
Also, If I only run test data of record 77th --> the latest record (9 rows of data in excel file), there is NO Socket error issue.
i had this issue when i run test data from excel file after records 180+.
I am running into this as well, right at the same 77 number by the look of it.
It's a data driven test reading data from a CSV file. The first 70 requests go through without issue.
While not 100% the same, I made a Postman Runner, for that single test and was able to send it 100 times without issue. The problem appears to be within Katalon and a certain threshold for the number of tests it can send, not with the API itself.
I recreated my test to use a POST request from the Object Repository. I have again set this up as a Data Driven test using a CSV file with 100 items. It consistently fails after processing 70 items with the same Socket Closed error.
Is there something within this process that is actually opening 70 sockets?
Is there some additional piece needed to clean up after each API request?


text:
\section{Introduction}

\setcounter{equation}{0}

During the last two decades there has been an increasing interest in the
study of quantum effects in cosmological backgrounds. Since the
appearance
of the pioneer works by Parker \cite{Parker1} and Grib {\it et al} \cite
{Grib} among others, a great body of papers have been published on the
problem of scalar and spin 1/2 particle creation in different
cosmological
configurations. More recently there have been some attempts to analyze
in
this framework the problem of particle creation in the vicinity of
cosmic
strings \cite{Parker2,Sahni}

Since the problem of particle creation in curved space-times is closely
related to the definition of particle in a curved background, different
approaches and interpretations have appeared during these last two
decades,
being the most popular the adiabatic and the corpuscle interpretations,
an
overview of the most important results and approaches can be found in
\cite
{Birrel} and \cite{Grib2}. Despite the great effort devoted to a better
understanding of quantum phenomena in expanding cosmological universes,
the
number of articles discussing concrete models is relatively scarce. This
is
due to the fact that, in most of the cases, the relativistic wave
equations
in curved backgrounds are very difficult to solve and therefore is
necessary
to reduce the original problem to a simpler one.

Recently, the exact solution of Klein-Gordon equation in an expanding
(asymptotically flat) cosmological universe filled with radiation has
been
reported \cite{Costa}, the authors obtain a thermal planckian
distribution
of particles created when they compare the ''in'' and ''out'' vacua via
the
Bogoliubov coefficients. The model presented in \cite{Costa} encourages
us
to analyze more complex configurations.

A good scenario for discussing the problem of scalar and spin 1/2
particle
creation in a cosmological background, is an isotropic universe with
zero
cosmological constant $(\Lambda ~=~0)$ filled with a mixture of
radiation
and dust-like matter \cite{Chernin,Kharbediya} in a spatially open
Friedmann
Robertson-Walker model of the form
\begin{equation}
\label{Fried}ds^2~=-dt^2+a^2(t)\left( d\chi ^2+\sinh {}^2(\chi
)(d\vartheta
^2+\sin {}^2\vartheta d\varphi ^2)\right)
\end{equation}
from (\ref{Fried}) we obtain
\begin{equation}
\label{state}\frac{d\epsilon }{p+\epsilon }=-3\frac{da}a,
\end{equation}
\begin{equation}
\label{state2}\quad \kappa \epsilon =\frac 3{a^2}\left[ (\frac{da}{dt}%
)^2-1\right]
\end{equation}
where $\kappa $ is the constant appearing in the Einstein's Equations,
$%
\epsilon $ is the energy, and $p$ is the pressure. Assuming that there
is
not interaction between radiation and dust, we have $\epsilon =\epsilon
_d+\epsilon _r$ where, \cite{Chernin}
\begin{equation}
\epsilon _d=\frac A{a^3},\quad \epsilon _r=\frac B{a^4},
\end{equation}
after introducing the conformal time $\eta $ related to $t$ by
\begin{equation}
\label{time}\eta =\int \frac{dt}a
\end{equation}
and integrating $a(\eta ),$ and imposing as initial condition $a(\eta
)\rightarrow b$ when $\eta \rightarrow -\infty ,$ being $b$ a constant,
we
obtain the line element

\begin{equation}
\label{metric}ds^2~=~b^2(e^\eta ~+~1)^2%
\pmatrix{-d\eta ^{2}~+~d\chi ^{2}~+\sinh ^{2}\chi (d\vartheta
^{2}~+~\sin ^{2}\vartheta &d\varphi ^{2})}
\end{equation}

\noindent Among the advantages of the line metric (\ref{metric}) we can
mention that, this line element is regular for any value of the time
parameter $\eta .$ . The homogeneous and isotropic space-time
(\ref{metric})
evolves in permanent expansion towards and asymptotically flat region at
$%
\eta \rightarrow +\infty .$ This asymptotic behavior resembles the one
obtained in the perfect fluid cosmological model \cite{Costa}. In
addition, $%
\partial /\partial \eta $ approaches a timelike Killing vector when
$\eta
\rightarrow -\infty $. Also, we have that, after separating variables in
the
Klein Gordon and Dirac equations, the equations governing the time
dependence are solvable in terms of special functions. It is the purpose
of
the present paper to obtain the quantum spectral distribution of scalar
and
spin 1/2 particles created in the cosmological model associated with the
line element (\ref{metric}) The paper is organized as follows: In \S 2
we
solve the Klein-Gordon equation in the background field (\ref{metric})
and,
in \S 3 we also solve the Dirac equation in this metric. Finally, in \S
4,
based on the results obtained in \S 2 and \S 3, we compute, via the
Bogoliubov coefficients, the probability distribution of the number of
created particles \cite{Parker3}

\section{Solution of the Klein-Gordon equation}

\setcounter{equation}{0}

The covariant generalization of the Klein-Gordon equation reads:

\begin{equation}
\label{2.1}%
\pmatrix{g^{\alpha \beta }\nabla _{\alpha }\nabla _{\beta }-\xi
R-m^{2}}\Phi
{}~=~0
\end{equation}
\noindent where $R$ is the scalar curvature

\begin{equation}
\label{R}R~=~\frac 6{b^2(e^\eta +1)^3}
\end{equation}

\noindent and $\xi $ is a dimensionless parameter which for a
conformally
coupling takes the value of $\xi ~=~1/6.$

\noindent Introducing a solution of the form

\begin{equation}
\label{2.3}\Phi ~=~T(\eta )X(\chi)Y(\vartheta ,\varphi )
\end{equation}
\noindent and substituting (\ref{2.3}) into, (\ref{2.1}) we find that,
after
separating variables, Eq. (\ref{2.1}) reduces to the following set of
ordinary differential equations

\begin{equation}
\left( \frac{d^2}{d\vartheta ^2}+\cot \vartheta \frac d{d\vartheta
}+\frac
1{\sin {}^2\vartheta }\frac{d^2}{d\varphi ^2}\right) Y~=~l(l+1)Y
\end{equation}

\begin{equation}
\label{ecua}\left( \frac{d^2}{d\eta ^2}+2\frac{\dot a}a\frac d{d\eta
}+\frac{%
\ddot a}a+m^2a^2+\sigma ^2-1\right) T=0
\end{equation}
\begin{equation}
\left( \frac{d^2}{d\chi ^2}+2\coth \chi \frac d{d\chi
}+\frac{l(l+1)}{\sinh
{}^2\chi }+\sigma ^2\right) X=0
\end{equation}
where $l$ and $\sigma $ are constants of separation, with $l\geq 0$ and
$%
\sigma ^2>1.$

After introducing the auxiliary function $Z(\eta )$,
\begin{equation}
T(\eta )~=~{\frac{Z(\eta )}{a(\eta )}}\qquad
\end{equation}

Eq. (\ref{ecua}) takes the form:

\begin{equation}
\label{second}\left( {\frac{d^2}{d\eta ^2}}+m^2a^2(\eta )+\sigma
^2-1\right)
Z~=~0
\end{equation}
\noindent substituting the expansion factor $a(\eta )~=~b(e^\eta ~+~1)$
into
(\ref{second}), we obtain the equation,

$$
\left( {\frac{d^2}{d\eta ^2}}+m^2b^2e^{2\eta }~+2m^2b^2e^\eta
+b^2m^2+\sigma
^2-1\right) Z~=~0
$$
\noindent whose solution can be expressed in terms of confluent
hypergeometric functions $M(a,b,z)$ and $U(a,b,z)$ \cite{Abramowitz} as
follows:

$$
Z(\eta )~=~e^{\eta c}\exp (-imbe^\eta )(c_1M({\frac
12}~+~c~+ibm,2c+1,2ibme^%
\eta )
$$
\begin{equation}
+~c_2U({\frac 12}~+~c~+ibm,2c+1,\ 2ibme^\eta )
\end{equation}
\noindent where $c_1$ and $c_2$ are arbitrary constants and $c$ is given
by
the expression

\begin{equation}
\label{c}c~=~-i(\sigma ^2~+~m^2b^2~-~1)^{1/2}
\end{equation}
The solutions of (2.4) are given in terms of the harmonic polynomials:
$%
Y=Y_{lm}(\vartheta ,\varphi )$, also we have that the solution of the
equation (2.6) reads

\begin{equation}
X~=~{\frac 1{\sqrt{\sinh \ \chi }}}P_{i\nu -1/2}^{-l-1/2}(\cosh \chi )
\end{equation}
\noindent where $P_\beta ^\alpha (\cosh x)$ are the torus functions
\cite
{Gradshtein}. and $\nu ^2~=~\sigma ^2-1$.

\section{Solution of the Dirac equation}

\setcounter{equation}{0}

The covariant generalization of the Dirac equation in curved space-time
reads \cite{Brill}

\begin{equation}
\label{3.1}\left\{ \tilde \gamma ^\mu (\partial _\mu -\Gamma _\mu
)+m\right\} \Psi ~=~0
\end{equation}
\noindent where $\tilde \gamma ^\mu $ are the curved gamma matrices
satisfying the relation,

\begin{equation}
\left\{ \widetilde{\gamma }^\mu ,\widetilde{\gamma}^\nu \right\}
_{+}=2g^{\mu \nu }
\end{equation}
\noindent and $\Gamma _\mu $ are the spin connections \cite{Brill}.
Choosing
to work in the diagonal tetrad gauge for $\tilde \gamma ^\mu $,

\begin{equation}
\tilde \gamma ^0=~a^{-1}\gamma ^0,\qquad \tilde \gamma ^1=a^{-1}\gamma
^1,\tilde \gamma ^2~=a^{-1}(\sinh \chi )^{-1}\gamma ^2,
\end{equation}
\begin{equation}
\tilde \gamma ^3~=~a^{-1}(\sinh \chi )^{-1}(\sin \vartheta )^{-1}\gamma
^3
\end{equation}
\noindent where $\gamma ^\mu $ are the standard Dirac flat matrices, we
have
that the Dirac equation (3.1) takes the form

\begin{equation}
\label{Dirac}\left\{ {\frac 1a}\gamma ^0\partial _\eta +{\frac 1a}\gamma
^1\partial _\chi +\frac 1{a\sinh \chi }\gamma ^2\partial _\vartheta
+\frac
1{a\sinh \chi \sin \vartheta }\gamma ^3\partial _\varphi +m\right\} \Psi
=0
\end{equation}
\noindent where we have introduced the spinor $\Psi $,

\begin{equation}
\label{3.5}\tilde \Psi =a^{-2}(\sinh \chi )^{-1}(\sin \vartheta
)^{-1/2}\Psi
\end{equation}
\noindent Applying the algebraic method of separation of variables \cite
{Andrush,Shishkin,Victor} it is possible to write eq. (\ref{Dirac}) as a
sum
of two first order differential operators ${\bf \hat K}_1$ $\ {\bf \hat
K}_2$
satisfying the relation

\begin{equation}
[{\bf \hat K}_1,{\bf \hat K}_2]_{-}=0,\ \left\{ {\bf \hat K}_1+{\bf \hat
K}%
_2\right\} \Phi =0
\end{equation}
\begin{equation}
\label{3.7}{\bf \hat K}_1\Phi =\lambda \Phi =-{\bf \hat K}_2\Phi
\end{equation}
\noindent where

\begin{equation}
\label{3.8}\Psi =\gamma ^1\gamma ^2\gamma ^3\Phi ,
\end{equation}
\begin{equation}
\label{3.9}{\bf \hat K}_1=\pmatrix{\gamma ^{0}\partial _{\eta }+ am
}\gamma
^1\gamma ^2\gamma ^3,
\end{equation}
\begin{equation}
\label{bf}{\bf \hat K}_2=(\gamma ^1\partial _\chi +{\frac 1{\sinh \chi
}}%
\gamma ^2\partial _\vartheta +{\frac 1{\sinh \chi \sin \vartheta
}}\gamma
^3\partial _\varphi )\gamma ^1\gamma ^2\gamma ^3\ {}
\end{equation}
in this way, we have separated the time variable $\eta $ from the
spatial $%
\chi ,\vartheta ,$ and $\varphi $ variables. The problem arises when we
try
to reduce (\ref{bf}$)$ to the form (\ref{3.7}). It is not difficult to
see
that there are not separating matrices allowing that step. In order to
go
further we write equation $({\bf \hat K}_2~+~\lambda )\Phi ~=~0$ as
follows:

\begin{equation}
\label{3.11} \pmatrix{\hat{\bf K}_{3}+\hat{\bf K}_{4}\gamma ^{1}\gamma
^{2}}%
\Phi ~=~0
\end{equation}
\noindent where ${\bf \hat K}_3$ and ${\bf \hat K}_4$ are two commuting
differential operators given by the expressions

\begin{equation}
{\bf \hat K}_3~=~(\sinh \chi )%
\pmatrix{\gamma ^{2}\gamma ^{3}\partial _{\chi }+\lambda },\quad {\bf
\hat K}%
_4~=~\left( \gamma ^2\gamma ^3\partial _\vartheta +\frac 1{\sin
\vartheta
}\partial _\varphi \right) ,
\end{equation}
After introducing the auxiliary spinor $\Sigma $

\begin{equation}
\label{3.13}\Phi ~=~%
\pmatrix{\hat{\bf K}_{4}~+~\gamma ^{1}\gamma ^{2}\hat{\bf K}_{3}}\Sigma
\end{equation}
\noindent we reduce equation (\ref{3.11}) to the form;

\begin{equation}
\left[ {\bf \hat N}_1+{\bf \hat N}_2\right] \Sigma ~=~0,\quad \left[
{\bf %
\hat N}_1,\ {\bf \hat N}_2\right] _{-}~=~0
\end{equation}
\noindent with

\begin{equation}
{\bf \hat N}_2\Sigma =-{\bf \hat N}_1\Sigma =k^2\Sigma
\end{equation}
\noindent where

\begin{equation}
{\bf \hat N}_1~=~(\sinh \chi )^2%
\pmatrix{\partial ^{2}_{\chi }+\coth \chi \partial _{\chi }-\lambda
\gamma ^{2}\gamma ^{3}\coth \chi +\lambda ^{2}}
\end{equation}
\begin{equation}
\label{ene}{\bf \hat N}_2~=~%
\pmatrix{\partial ^{2}_{\vartheta }+i\gamma ^{2}\gamma ^{3}k_{\varphi
}(\sin \vartheta )^{-2}\cos \vartheta -k^{2}_{\varphi }(\sin \vartheta
)^{-2}}
\end{equation}
\noindent $k_\varphi $ in (\ref{ene}) is the eigenvalue of the operator
$%
-i\partial _\varphi $.

\noindent Choosing to work in the following representation of the Dirac
matrices:
$$
\gamma ^0=\left(
\begin{array}{cc}
-i & 0 \\
0 & i
\end{array}
\right) ,\quad \gamma ^1=\left(
\begin{array}{cc}
0 & \sigma ^3 \\
\sigma ^3 & 0
\end{array}
\right) ,
$$

$\quad $%
\begin{equation}
\gamma ^2=\left(
\begin{array}{cc}
0 & \sigma ^2 \\
\sigma ^2 & 0
\end{array}
\right) ,\quad \gamma ^3=\left(
\begin{array}{cc}
0 & \sigma ^1 \\
\sigma ^1 & 0
\end{array}
\right) ,\quad
\end{equation}

\noindent we have that the spinor $\Sigma $ takes the form

\begin{equation}
\label{3.19}\Sigma ~=~\left(
\begin{array}{c}
p(\chi )A(\vartheta ) \\
q(\chi )B(\vartheta ) \\
r(\chi )C(\vartheta ) \\
s(\chi )D(\vartheta )
\end{array}
\right)
\end{equation}
\noindent the functions $p(\chi ),q(\chi ),r(\chi )$ and $s(\chi )$
satisfy
the differential equation

\begin{equation}
\pmatrix{\sinh \chi (d_{\chi }+\mp i\lambda )}%
\pmatrix{\sinh \chi (d_{\chi }+\pm i\lambda )}X(\chi )+k^2X(\chi )~=~0
\end{equation}
\noindent where the upper signs correspond to $p(\chi ),r(\chi )$, and
the
lower signs are related to $q(\chi )$ and $s(\chi )$.

\noindent The functions $A(\vartheta ), B(\vartheta ), C(\vartheta ),
D(\vartheta )$ are solution of the equation

\begin{equation}
\pmatrix{d_{\vartheta }\pm k_{\varphi }(\sin \vartheta )^{-1}}%
\pmatrix{d_{\vartheta }\mp k_{\varphi }(\sin \vartheta
)^{-1}}Y(\vartheta
)-k^2Y(\vartheta )~=~0
\end{equation}
\noindent where the upper signs correspond to $A(\vartheta )$ and $%
C(\vartheta )$, and the lower ones correspond to $B(\vartheta )$ and $%
D(\vartheta )$. After substituting the expression (\ref{3.19}) into
(\ref
{3.13}) we arrive at

\begin{equation}
\label{3.22}\Phi ~=~ik%
\pmatrix{p(\chi )B(\vartheta )(1+i)\alpha \cr q(\chi )A(\vartheta
)(1-i)\alpha \cr r(\chi )D(\vartheta )(1+i)\beta \cr s(\chi )C(\vartheta
)(1-i)\beta\cr }%
e^{ik_\varphi \varphi }
\end{equation}
\noindent where $\alpha $ and $\beta $ are functions depending on the
variable $\eta $. The functions $A,B,C,D$, and $p,q,r,s$ and are related
by
the following systems of coupled partial equations:

\begin{equation}
\label{3.23}(d_\vartheta -{\frac{k_\varphi }{\sin \vartheta }})%
\pmatrix{A\cr C}~=~-k\left(
\begin{array}{c}
B \\
D
\end{array}
\right) ,\quad (d_\vartheta +{\frac{k_\varphi }{\sin \vartheta }})%
\pmatrix{B\cr D}~=~k\pmatrix{A\cr C}
\end{equation}
\begin{equation}
\label{3.24} \sinh \chi (d_\chi +i\lambda )\pmatrix{p\cr r}~=~k%
\pmatrix{q\cr s},\quad \sinh \chi (d_\chi -i\lambda )\pmatrix{q\cr
s}~=-k%
\pmatrix{p\cr r}
\end{equation}
\noindent where $\epsilon $ and $\omega $ are constants. From
(\ref{3.22})
and (\ref{3.9}) we can put $p(\chi )~=~r(\chi )$, $q(\chi )~=~s(\chi )$,
$%
A(\vartheta )~=~C(\vartheta )$, and $B(\vartheta )~=~D(\vartheta )$. The
functions $\alpha $ and $\beta $ satisfy the system:

\begin{equation}
\label{3.25} (\partial _\eta +ima)\beta ~=~-\lambda \alpha ,\qquad
(\partial
_\eta -ima)\alpha ~=~\lambda ~\beta
\end{equation}
\noindent from which we obtain:

\begin{equation}
(d_\eta ^2~+~m^2b^2e^{2\eta }~+e^\eta ~(2m^2b^2-imb)~+~m^2b^2~+~\lambda
^2)\alpha ~=~0
\end{equation}
\begin{equation}
(d_\eta ^2~+~m^2b^2e^{2\eta }~+e^\eta ~(2m^2b^2+imb)~+~m^2b^2~+~\lambda
^2)\beta ~=~0
\end{equation}
\noindent whose solution can be written in terms of confluent
hypergeometric
functions as follows

$$
\alpha (\eta )~=~e^{\eta c}e^{-imbe^\eta }(-c_0{\frac{(c+imb)}\lambda }%
M(c+imb+1,\ 2c+1,2imbe^\eta )
$$
\begin{equation}
+~c_1{\frac{(c^2~+~m^2b^2)}\lambda }U(c+imb+1,2c+1,\ 2imbe^\eta ))
\end{equation}
\begin{equation}
\beta (\eta )~=e^{\eta c}e^{-imbe^\eta }\ \left(
c_0M(c+imb,2c+1,2imbe^\eta
)~+~c_1U(c+imb,2c+1,2imbe^\eta )\right)
\end{equation}

\noindent where $c$ is given by the expression:

\begin{equation}
\label{3.30}c~=~-i(\lambda ^2~+~m^2b^2)^{1/2}
\end{equation}
\noindent The solution of the coupled system of equations (\ref{3.23})
can
be expressed in terms of Jacobi polynomials as follows:

\begin{equation}
A(\vartheta )~=~\sin {}^k\varphi (\vartheta )\cos (\vartheta
/2)P_n^{(k}\varphi ^{+1/2,}{}^k\varphi ^{-1/2)}(\cos \vartheta )
\end{equation}
\begin{equation}
B(\vartheta )~=~\sin {}^k\varphi (\vartheta )\sin (\vartheta
/2)P_n^{(k}\varphi ^{-1/2,}{}^k\varphi ^{+1/2)}(\cos \vartheta )
\end{equation}
\noindent where $n$ is given by the expression

\begin{equation}
n~=~\mid k\mid -\mid k_\varphi \mid -{\frac 12}
\end{equation}
\noindent The system (\ref{3.24}) can be rewritten in the following
matrix
form

\begin{equation}
\left( {\bf I\ }d_\chi +i\lambda \sigma ^3-\frac{i{\it k}}{\sinh \chi }%
\sigma ^2\right) \left(
\begin{array}{c}
p \\
q
\end{array}
\right)
\end{equation}
\noindent after applying the matrix transformation $S$

\begin{equation}
S^{-1}\pmatrix{p\cr q}~=~\pmatrix{P\cr Q},S~=\frac 1{\sqrt{2}}%
\pmatrix{1+i\sigma ^{1}}
\end{equation}
\noindent we reduce (\ref{3.24}) to the form

\begin{equation}
\label{3.36}(d_\chi -{\frac{ik}{\sinh \chi }})Q~=~\lambda P\qquad
(d_\chi +{%
\ \frac{ik}{\sinh \chi }})P~=~-\lambda Q
\end{equation}
\noindent with,

\begin{equation}
P+iQ=~p,\quad Q~+iP~=~q
\end{equation}
\noindent the solution of the system (\ref{3.36}) can be expressed in
terms
of Gauss hypergeometric functions \cite{Gradshtein},

$$
P(\chi )~=~\sinh {}^{ik}(\chi )\cosh (\chi /2)\ {\tt x}
$$
\begin{equation}
F({\frac 12}~+i\lambda ~+ik,{\frac 12}~-i\lambda ~+ik,{\frac
32}+ik,(1-\cosh
\chi )/2)
\end{equation}
$$
Q(\chi )~=~-\pmatrix{({1\over 2}+ik)/\lambda }\sinh {}^{ik}(\chi )\sinh
(\chi /2)\ {\tt x}
$$
\begin{equation}
F({\frac 12}~+i\lambda ~+ik,{\frac 12}~-i\lambda ~+ik,{\frac
12}+ik,(1-\cosh
\chi )/2)
\end{equation}
\noindent Finally, from (\ref{3.5}), (\ref{3.8}) and (\ref{3.22}) we
find
that the solution of the Dirac equation (\ref{3.1}) reads

\begin{equation}
\label{spinor}\tilde \Psi =a^{-2}(\sinh \chi )^{-1}(\sin \vartheta
)^{-1/2}\left(
\begin{array}{c}
p(\chi )B(\vartheta )(i+1)\beta (\eta ) \\
q(\chi )A(\vartheta )(1-i)\beta (\eta ) \\
p(\chi )B(\vartheta )(1+i)\alpha (\eta ) \\
q(\chi )A(\vartheta )(1-i)\alpha (\eta )
\end{array}
\right) e^{ik_\varphi \varphi }
\end{equation}
\noindent Here some comments are in order. The election of a diagonal
tetrad
for separating variables and solving the Dirac equation in the
background
field (\ref{metric}) was done for the sake of simplicity. If we want to
reobtain an angular dependence of the spinor (\ref{spinor}) in terms of
spherical harmonics, it is enough to achieve the transformation relating
the
diagonal tetrad gauge with the Cartesian one \cite{villalba} Obviously,
this
transformation does not affect the time dependence of the spinor
solution.

\section{Particle creation}

\setcounter{equation}{0}

In this section we proceed to analyze the phenomena of scalar and Dirac
particle creation in the cosmological background associated with the
metric (%
\ref{metric}). The model in question begins at $\eta ~=-~\infty $ and
evolves to $\eta \rightarrow +\infty $. Among the particularities of our
model we have to mention that, like \cite{Costa}, the universe
(\ref{metric}%
) is asymptotically flat, but does not present a period of contraction
during its evolution toward the future.

In order to establish the asymptotic behavior of the solutions, at $\eta
{}~=-\infty $ and at $\eta ~=+\infty $, of the corresponding wave
equations,
we write the Hamilton-Jacobi equation

\begin{equation}
g^{\alpha \beta }S,_\alpha S,_\beta ~+~m^2~=~0
\end{equation}
\noindent in the background field (\ref{metric}). Then we obtain:

\begin{equation}
\label{HJ}-(S,_\eta )^2~+~(S,_\chi )^2~+~{\frac 1{\sinh {}^2\chi
}}\left(
(S,_\vartheta )^2~+~{\frac 1{\sin {}^2}}_\vartheta (S,_\varphi
)^2\right)
+~m^2a^2~=~0
\end{equation}
We can separate in Eq. (\ref{HJ}) the time dependence from the spatial
one
as follows:

\begin{equation}
S~=~{\sf T}(\eta )~+~F(\chi ,\vartheta ,\varphi )
\end{equation}
\noindent then, we obtain

\begin{equation}
({\sf T},_\eta )^2~=~m^2a^2~+~\Lambda ^2,\ \rightarrow {\sf T}~=~\pm
\int
(m^2a^2~+~\Lambda ^2)^{1/2}d\eta
\end{equation}
\noindent where $\Lambda $ is a constant of separation. Then, the
quasiclassical asymptotes of the solutions are

\begin{equation}
\label{asy}\Psi \rightarrow F(\chi ,\vartheta ,\varphi )\exp
\pmatrix{\pm i\int (m^{2}a^{2}+\Lambda ^{2})^{1/2}d\eta }
\end{equation}
The asymptotic behavior of the function $T(\eta )$ is

\begin{equation}
\lim _{\eta \rightarrow -\infty }{\sf T}~\rightarrow ~\pm
(m^2b^2+\Lambda
^2)^{1/2}\eta ,\quad \lim _{\eta \rightarrow \infty }{\sf T}\rightarrow
\pm
mbe^\eta ~
\end{equation}
The relation between the conformal time $\eta $ and the cosmic time $t$
can
be obtained from (\ref{time}) giving as result,
\begin{equation}
\label{relation}b(e^\eta +\eta )=t
\end{equation}
notice that the functions $b(e^\eta +\eta )$ monotonically increases,
and $%
t\rightarrow \pm \infty $ when $\eta \rightarrow \pm \infty ,$ and, when
$%
\eta $~$\rightarrow $~+$\infty $  the metric (\ref{metric}) becomes a
Milne
universe \cite{Birrel}

We have that the solutions of the Klein-Gordon equation with the
asymptotic
behavior (\ref{asy})can be written as,

\begin{equation}
\label{4.7}Z_{-\infty }(\eta )~=~(2c)^{-1/2}e^{\eta c}\exp (-imbe^\eta
)M({%
\frac 12}~+~c~-ibm,2c+1,2ibme^\eta )
\end{equation}
\noindent for $\eta $~$\rightarrow $~$-\infty $, and

\begin{equation}
\label{4.8}Z_\infty (\eta )~=~e^{i\pi c/2}e^{\eta c}\exp (-imbe^\eta
)U({%
\frac 12}~+~c~-ibm,2c+1,2ibme^\eta )
\end{equation}
\noindent for $\eta $~$\rightarrow $~$\infty $. where, in order to
analyze
the form (\ref{4.7}) takes at infinity, we have used the formula: \cite
{Abramowitz}

\begin{equation}
\label{4.9}U(\alpha ,\ \gamma ,\ z)\rightarrow z^{-\alpha }\left(
1+O(\frac
1z)\right) ,
\end{equation}
and the relation (\ref{relation})

\noindent Notice that, for the constant $c$ given by (\ref{c}),
$T_{-\infty
} $ and $T_\infty $ can be associated with a positive frequency mode at
$%
\eta ~\rightarrow ~-\infty $ and $\eta ~\rightarrow ~\infty $
respectively.
The next step is to compute the Bogoliubov coefficients relating the two
different vacua at $\eta ~=~-\infty $ and $\eta ~\rightarrow ~\infty $
\cite
{Birrel}. In the present case it is possible to obtain $\beta $ without
performing the integration. Indeed, using the recurrence formula between
the
confluent hypergeometric functions $U(\alpha ,\gamma ,z)$ and $M(\alpha
,\gamma ,z)\ $\cite{Abramowitz},

\begin{equation}
\label{4.10}M(\alpha ,\gamma ,z)~=~{\frac{\Gamma (\gamma )}{\Gamma
(\gamma
-\alpha )}}e^{i\pi \alpha }U(\alpha ,\gamma ,z)~+~{\frac{\Gamma (\gamma
)}{%
\Gamma (\alpha )}}e^{i\pi (\alpha -\gamma )}e^zU(\gamma -\alpha ,\gamma
,-z)
\end{equation}
\noindent substituting $\alpha ~=~{\frac 12}~+~c$~$+ibm$, $\gamma
{}~=~2c+1$
and\ \ $z~=~2ibme^\eta \ $ into (\ref{4.9}) and taking into account the
relation

\begin{equation}
e^zU(\gamma -\alpha ,\gamma ,-z)~=~e^zz^{1-\gamma }(-1)^{1-\gamma
}U(1-\alpha ,2-\gamma ,-z)
\end{equation}
\noindent we arrive at,

$$
Z_{-\infty }(\eta )~=~{\frac{\Gamma (2c+1)}{(2c)^{1/2}\Gamma ({\frac
12}%
{}~+~c~-ibm)}}e^{i\pi /2}e^{i\pi c/2}e^{-\pi mb}Z_\infty (\eta )~+~
$$
\begin{equation}
\label{4.12}+~{\frac{\Gamma (2c+1)}{{(2c)^{1/2}}\Gamma ({\frac
12}~+~c~-ibm)}%
}e^{-i\pi /2}e^{-i\pi c/2}e^{-\pi mb}(2mb)^{-2c}Z_\infty ^{*}(\eta )
\end{equation}
\noindent then, using the relations for the $\Gamma $ function, \cite
{Gradshtein}

\begin{equation}
\label{4.13}2^{2z-1}\Gamma (z)\Gamma (z~+~{\frac 12})~=\sqrt{\pi }\Gamma
(2z)
\end{equation}

\begin{equation}
\mid \Gamma (iy)\mid ^2=\frac \pi {y\sinh (\pi y)}
\end{equation}

\noindent and considering the limit case when $b$ is small, we arrive at

\begin{equation}
\mid \beta \mid ^2~=~{\frac 1{e^{2\pi \nu }-1}}
\end{equation}
\noindent where is $\nu ~=~(\sigma ^2~-~1)^{1/2}$

\noindent Analogously, we have that the Bogoliubov coefficients for the
Dirac case can be obtained by means of the functions $a(\eta )$ and
$\beta
(\eta )$ determining the time dependence of the spinor (\ref{spinor}).
Then,
considering the corresponding positive frequency solutions at the origin
and
at infinity,

\begin{equation}
\beta _{-\infty }(\eta )~=~e^{\eta c}e^{-imbe^\eta
}M(c+imb,2c+1,2imbe^\eta
)
\end{equation}
\begin{equation}
\beta _\infty (\eta )~=~e^{i\pi c/2}e^{\eta c}e^{-imbe^\eta
}U(c+imb,2c+1,2imbe^\eta )
\end{equation}

\noindent where the constant $c$ is given by (\ref{3.30}). Taking into
account the relation (\ref{4.9}) and (\ref{4.10}) we obtain

$$
\beta _0(\eta )~=~{\frac{\Gamma (2c+1)}{\Gamma (c+1+imb)}}e^{i\pi
(c+imb)}e^{-i\pi c/2}\beta _\infty (\eta )~+
$$
\begin{equation}
+~{\frac{\Gamma (2c+1)}{\Gamma (c-imb)}}e^{-i\pi (c+imb+1)}e^{i\pi
c/2}(2mb)^{-2c}\beta _\infty ^{-}(\eta )~
\end{equation}
\noindent then, for small values of $b$, using the relation for the
gamma
function $\Gamma (z),$ \cite{Gradshtein}

\begin{equation}
\mid \Gamma (iy+\frac 12){}\mid ^2={{{\frac \pi {\cosh (\pi y)}}{}{}{}}\
}
\end{equation}

\noindent and the expression (\ref{4.13}) we arrive at:

\begin{equation}
\label{Fermi}\mid \beta \mid ^2~=~{\frac 12}{\frac 1{e^{2\pi \lambda
}+1}}
\end{equation}
\noindent where $\lambda $ \ in (\ref{Fermi}) is the separation constant
appearing in (\ref{3.7})

\section{Concluding remarks}

In this paper we have analyzed the process of creation of scalar and
Dirac
particles in an isotropic and homogeneous cosmological universe filled
by
non interacting radiation and dust-like matter. It is worth noticing
that,
also in this model we have obtained that the average number of particles
created satisfies a thermal distribution law. This result show that the
equation of state does not determine the quantum spectrum but the time
behavior of the line element associated with the cosmological universe.
{}From
a mathematical point of view, this paper shows the capabilities of the
algebraic method of separation of variables and how it works when first
order separation is not possible. We hope that the results presented in
this
article will be of help in understanding quantum phenomena in
cosmological
backgrounds. \vspace{1.0 cm}

redpajama_set_name: RedPajamaArXiv


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text:
Q: How to list certificates, trusted by OpenSSL? As I understand, any software working with X.509 certificates may have own basis to decide, whether a certificate is trusted or not.
AFAIK OpenSSL just consults a list (such as, for example, /etc/ssl/certs) and checks if the certificate is present there.
Is there a way for OpenSSL to list all certificates which it trusts? I know that I can consult that file myself (on my particular installation of OpenSSL), but is there a (installation-independent) way to get the trusted list from OpenSSL itself?

A:
AFAIK OpenSSL just consults a list (such as, for example, /etc/ssl/certs) and checks if the certificate is present there.

No, OpenSSL trusts nothing by default. You have to instruct it what to trust. There's even a FAQ topic covering it: Why does <SSL program> fail with a certificate verify error?:

This problem is usually indicated by log messages saying something
  like "unable to get local issuer certificate" or "self signed
  certificate". When a certificate is verified its root CA must be
  "trusted" by OpenSSL this typically means that the CA certificate must
  be placed in a directory or file and the relevant program configured
  to read it. The OpenSSL program 'verify' behaves in a similar way and
  issues similar error messages: check the verify(1) program manual page
  for more information.

You can also test your connection to Google to see how OpenSSL behaves:
$ openssl s_client -connect google.com:443
CONNECTED(00000003)
depth=2 C = US, O = GeoTrust Inc., CN = GeoTrust Global CA
verify error:num=20:unable to get local issuer certificate
verify return:0
...
Start Time: 1407377002
Timeout   : 300 (sec)
Verify return code: 20 (unable to get local issuer certificate)

Notice the above fails because OpenSSL does not trust GeoTrust Global CA by default. Actually, there's another trust point in the chain and that's Google Internet Authority G2.
You can remedy the situation by telling OpenSSL what to trust. Below, I use -CAfile option with Google Internet Authority G2:
$ openssl s_client -connect google.com:443 -CAfile google-ca.pem
CONNECTED(00000003)
depth=3 C = US, O = Equifax, OU = Equifax Secure Certificate Authority
verify return:1
depth=2 C = US, O = GeoTrust Inc., CN = GeoTrust Global CA
verify return:1
depth=1 C = US, O = Google Inc, CN = Google Internet Authority G2
verify return:1
depth=0 C = US, ST = California, L = Mountain View, O = Google Inc, CN = google.com
verify return:1
...
Start Time: 1407377196
Timeout   : 300 (sec)
Verify return code: 0 (ok)

Next, you can act like a browser by going to cURL and download cacert.pem. cacert.pem has lots of CAs in it:
$ openssl s_client -connect google.com:443 -CAfile cacert.pem
CONNECTED(00000003)
depth=3 C = US, O = Equifax, OU = Equifax Secure Certificate Authority
verify return:1
depth=2 C = US, O = GeoTrust Inc., CN = GeoTrust Global CA
verify return:1
depth=1 C = US, O = Google Inc, CN = Google Internet Authority G2
verify return:1
depth=0 C = US, ST = California, L = Mountain View, O = Google Inc, CN = google.com
verify return:1
...
Start Time: 1407377356
Timeout   : 300 (sec)
Verify return code: 0 (ok)

You're not quite as bad as a browser with its hundreds of CAs and subordinate CAs, but you're getting close:
$ cat cacert.pem | grep -o "\-\-\-\-\-BEGIN" | wc -l
     153

OpenSSL security model is in contrast to the web app/browser security model, where the browser carries around a list of trust anchors or trust points known as Certificate Authorities (CAs). Note: in this model, the wrong CA could claim to certify a site, and the browser would be no wiser.
This has happened in the past, and it will likely happen again in the future. For a good history of PKIX funny business, see CAcert's Risk History. For example, you know Google Internet Authority G2 and GeoTrust Global CA certify Google's sites. There's no reason for a Dutch CA called Diginotar to claim to certify them, or a French Cyberdefense Agency to claim to certify them.
Related to security models: another problem with the web app/browser model is you cannot package the one trust anchor or CA needed for your app and use it (assuming you have a trusted distribution channel). Your certificates gets tossed in the pile with the CA Zoo. Others can still claim to certify your site, and you can claim to certify other sites.
The security model one of the reasons web apps are relegated to low value data. Web apps should not handle medium value or high value data because we can't place the needed security controls.

Is there a way for OpenSSL to list all certificates which it trusts?

No need since the list has 0 members :)

Also see How to find out the path for openssl trusted certificate?.

A: I'm wonder if this has changed in some way since jww's response.
If I submit:
$ openssl s_client -connect google.com:443
It works successfully, retrieves 4 total certs, and returns:
Start Time: 1484661709
Timeout   : 300 (sec)
Verify return code: 0 (ok)

I believe this is because servers should be setup to send, along with the certificate, any intermediate and root certificates that are needed to verify the full chain, right?

A: I recently looked into this, and found no way to get OpenSSL to list the certificates in its trusted set. The best way I found was to, as you point out, "consult that file [/etc/ssl/certs] myself (on my particular installation of OpenSSL)".
You can be more installation-independent about finding the directory which OpenSSL consults.  openssl version -d prints the path to it.
% openssl version -d
OPENSSLDIR: "/opt/local/etc/openssl"

OpenSSL looks here for a file named cert.pem and a subdirectory certs/. Certificates it finds there are treated as trusted by openssl s_client and openssl verify (source: the article, What certificate authorities does OpenSSL recognize?).
So, you can do something like:
% find -H `openssl version -d | sed -E 's/OPENSSLDIR: "([^"]*)"/\1/'`/(cert.pem|certs) \
-type f -exec cat {} \+

This prints out the entire contents of the files which OpenSSL expects to contain certificates.  If you want less than the entire file, then replace cat with the appropriate commands.

A: Self-answer: this text suggest calling SSL_CTX_set_default_verify_paths() to trust the certificates in a system-dependent directory.


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Madden NFL '95 is a football video game. It features John Madden on the cover.Madden NFL '95 (for the Sega Genesis only) was the first Madden game to have both the NFL team logo and NFLPA licenses. This version of Madden NFL was also the first Madden to showcase no passing windows, though passing windows can be retained using the options menu. Previously included "Past Championship Teams" and "All-Franchise" teams are not included. Player stats are tracked individually both in-game, and throughout a season. The cover has John Madden on it, but it also features two actual NFL players, Dallas Cowboys left tackle Erik Williams and San Francisco defensive lineman Karl Wilson who has his number edited to show the fictional number 70 which was not used on the 49ers roster.It was released to both the Game Boy and Sega Game Gear.The game also features the theme song from the NFL on Fox.
Madden NFL '96 is a football video game designed for the 1995 NFL season, licensed by the NFL. The AI has been boosted and can now hurry in two-minute drill situations, spike the ball, and cover the receivers with better efficiency.It was the last to explicitly be endorsed by the NFL on Fox, although a knock-off/rendition of the NFL on Fox's iconic theme would continue to be used in Madden for several years afterward.
Madden NFL 97 is the first Madden NFL football game to be created in the 32-bit gaming era. It was originally created for the PlayStation and Sega Saturn, but other versions were also released, including the 16 bit Super Nintendo and Sega Genesis.This was the first edition of Madden to implement the use of a salary cap when customizing team rosters. It also featured the new Baltimore Ravens, the expansion franchise that absorbed the roster of the Cleveland Browns for the 1996 season. The game also adds a team of real free agents (as opposed to the blank slate used in Madden 96) which can be traded and added to other teams' rosters (led by recently semi-retired quarterback Randall Cunningham), and could be played by using a cheat code.
Madden NFL 98 is a football video game. It features John Madden on the cover. It was the last version of the Madden series to be released for the SNES, Sega's Genesis, and Saturn platforms as well as the last Madden game to utilize the 2D sprite based in-game models for the players/referee, on 3D playing fields.The Green Bay Packers had the best team overall in the game with a 95. The worst team overall was the Baltimore Ravens with a score of 71. A 3 way tie between the Dallas Cowboys, the New England Patriots and the Green Bay Packers for the best offense with a perfect score of 100. There is also a 3 way tie for the best defense in the game between the San Francisco 49ers, the Seattle Seahawks and the Green Bay Packers with a score of 97. The San Diego Chargers had the best special teams in the game with a score of 96.
Magic Boy is a platform video game developed by Blue Turtle and published for the Amiga, Atari ST and DOS by Empire Interactive and the Super Nintendo Entertainment System in North America and Europe by JVC.
Magical Drop (マジカルドロップ?), sometimes referred to by native Japanese players as MagiDro (マジドロ?), is a series of puzzle games originally released in the arcade, and later primarily released for several platforms such as the Neo Geo, Super Famicom, Sega Saturn and Neo Geo Pocket Color developed by Data East. The games are notable for being extremely fast-paced.
Magical Pop'n (マジカル ポップン?) is a two-dimensional video game developed by Polestar and published by Pack-In-Video for the Super Famicom. The game tells the story of a girl referred to as "Princess" who sets out to retrieve a magic gem that was stolen by the Demon King when her kingdom was invaded.
Majyūō (魔獣王 King of Demons?) is a 1995 Japanese platformer video game developed and published by KSS for the Super Famicom. The game was released exclusively in Japan on August 25, 1995.
Makeruna! Makendou 2: Kimero Youkai Souri is an Action game, developed by Success and published by Datam Polystar, which was released in Japan in 1995.
Manchester United Championship Soccer is a 1995 soccer video game for the Super NES.


text:
Amid sustainability criticism, poll finds public prefers planning
Andria Krewson | Jun 14, 2012
Amid a political environment attacking global sustainability initiatives, a new national poll by the American Planning Association released Thursday finds bipartisan support for planning, as well as belief that community planners play a key role in economic recovery.
Some political groups have attacked planning and are crafting laws that would bar cooperative planning or ban the word "sustainability." Results of the new poll show those efforts don't reflect the primary concerns of U.S. residents, who are focused on jobs and the economy and believe a combination of market forces and planning – not the market alone, or planning alone – best support the economy.
To read more about the survey, including charts and graphics, click here to visit the APA's site.
"The results confirmed what we already believed: that it was a very vocal minority concerned about Agenda 21," said Mitchell Silver, president of the American Planning Association and Raleigh's chief planning and development officer.
"Agenda 21" is a voluntary global environmental initiative backed by President George H.W. Bush in 1992. Silver said the APA has spent time this year discussing the initiative and the opposition to it, because it has surfaced in political discussion. In North Carolina it has come up in the N.C. General Assembly and in the N.C. gubernatorial campaign, after GOP candidate Pat McCrory tweeted support for a resolution opposing Agenda 21.
The new poll results, though, will help the planning organization focus efforts on economic issues like jobs, Silver said.
The most important planning priorities for those surveyed? Jobs, safety, schools, neighborhoods and water quality, according to the poll.
"The public in general is speaking loud and clear: 'We need you to help us plan for economic recovery,' " Silver said in an interview. The APA will use results to refocus its mission on those priorities, he said.
The poll surveyed 1,300 U.S. residents in March to help determine what the public wants from planners and what perceptions exist. Planners wanted a clear picture amid a squeeze on government budgets and vocal claims by those who see planning as part of a global conspiracy.
An example of a question on the issue of planning vs. "market forces" and the results (graphed) are below:
Some people believe that community planning is a necessary part of improving the U.S. economy and encouraging job growth, while others believe that "market forces" alone will help the economy and bring more jobs. Which of the following statements comes closest to your belief?
Some state and local governments, including Gaston County, have adopted laws that criticize or outlaw any action seen to support Agenda 21. Those critics see the U.N. initiative at minimum as increasing government regulation and at maximum as a U.N. conspiracy to use planners to bring Marxism to the United States.
In the N.C. General Assembly, House Joint Resolution 983 calls on North Carolina to not comply with Agenda 21. In Alabama, a new law opposing Agenda 21 bans any local or state government entity from doing business with any organization linked to the initiative.
Amid that charged political environment, the new APA poll found bipartisan support for planning, and found a belief that planning can help tackle lingering economic challenges. Amid the poll's other findings:
Less than a third of respondents believe their communities are doing enough to address economic issues.
Two thirds believe their communities need more planning and that it takes a combination of market forces and planning to help the economy create more jobs.
Half of respondents want to be involved in community planning in the future.
Almost half the respondents say living in their community is worse for most people than it was five years ago.
Respondents describe ideal communities as having locally owned businesses nearby (55 percent); the ability to grow old in the same neighborhood (54 percent); availability of sidewalks (53 percent); energy-efficient homes (52 percent); and availability of transit (50 percent).
Most respondents (85 percent) don't know enough about Agenda 21 to hold an opinion; 6 percent oppose it and 9 percent support it.
The vocal Agenda 21 opponents weren't the sole reason for the poll but were an important driver, said Jason Jordan, legislative director of the national planning organization.
"In many cases, it has proven to be enormously disruptive and confrontational," he said. In addition, it had been 10 years since the organization last conducted a poll, he said, and given the economic and community changes, the group wanted to check in on attitudes toward planning. Planning, in and of itself, has also come under attack in some quarters.
The most surprising finding?
"We were very heartened at how broad-based the support for planning was across demographic groups and party affiliation," Jordan said. Only 22 percent of Republicans surveyed believed market forces alone were enough to improve the economy and encourage job growth, while 14 percent of independents and 6 percent of Democrats believed market forces alone were enough.
"That runs directly counter to what we're hearing from the critics," Jordan said.
Silver echoed Jordan's interpretation. "I was surprised at the level of support across demographic groups," for planners, he said, as well as geographic groups in rural areas, small towns and big cities.
It was refreshing to see that when the poll asked specific questions about safety and schools, he said, people responded positively to planning.
"When you explain what planners do, people have a better reaction," he said.
The survey has an error margin of plus or minus 3 percentage points. It was designed by Collective Strength in Austin, Texas, and reviewed and fielded by Harris Interactive.
For more on the survey, including charts and graphics, visit the APA's site.
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The programmes and products that I provide will always enable you to produce the results that you are seeking as long as you follow them through in sequence and take the necessary actions as agreed.
Working together with one of the products this will enable us to have one session each month for three months, to work on any issues that arise and make sure you are keeping on track with your progress and the results you want to achieve.
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Call me on 01295 272263 / 07783 107 236 to book your sessions today. If you are uncertain as to what would best suit your unique situation, call me as above or email me [email protected]. I look forward to helping you soon.


The Buffalo National River begins as a trickle in Boston Mountains of Arkansas, where it flows north and then eastward through the Ozark Mountains, merging 150 miles later with the White River at Buffalo City.
The Ponca "headwaters" area (and where we're located) offers the Buffalo National River's finest scenic beauty. With its massive towering bluffs, countless waterfalls, grazing elk herds, pioneer homesteads, early Native American bluff shelters, and remote unspoiled location, the upper Buffalo National River wilderness is considered a prime outdoor destination by adventure enthusiasts.
In fact, the river is such a special place that in 1972, Congress designated it as America's first national river. This unique status protects the Buffalo from commercial or residential development, along with other destructive operations such as damming and gravel mining, thus preserving its scenic beauty and natural wild state forever.
Buffalo Outdoor Center outfits for the Buffalo's upper district, which begins at Ponca and continues downriver 52 miles to a remote river access called Woolum. The river's most scenic stretch is the 24-mile Ponca-to-Pruitt section, offering more than 50 Class I and II rapids in the first 10 miles, as well as numerous hiking opportunities along the way. Two of the river's most prominent geological features are located on this stretch of river: Hemmed-In Hollow and Big Bluff, respectively, the tallest waterfall (250 feet) and bluff face (550 feet) between the Appalachians and the Rockies.
Canoes and kayaks are the preferred boats of choice on the Buffalo National River, although higher water levels permit rafting from time to time. All three boat types can be rented at BOC and vehicle shuttle services are also provided. The typical day-trip is 8 to 10 miles in length, depending upon the launch site. Multi-day trips can be arranged for as many days as you like and your takeout can be anywhere on the Buffalo, including the middle and lower river. Use our helpful interactive Buffalo National River map to learn more about river access points, mileage, and scenic highlights.
March through June is the season for the upper Buffalo, as our section of river is highly rainfall dependent. Weekends in May and June are very popular, so advance reservations are recommended if your trip includes a Saturday or Sunday. Otherwise, a reservation isn't necessary for weekday trips.
Buffalo Outdoor Center has been outfitting on the river for well over 30 years. We'd love to put our knowledge and expertise to work for you, helping you plan a Buffalo National River adventure that makes the most of your valuable vacation time.


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Speed dating in India
Bret Thorn | Apr 11, 2011
"Did you see that guy's watch? That thing's worth $300,000," Jeff Sinelli, founder of the Which Wich? sandwich chain told me after a couple of potential franchisees left his table.
I hadn't seen the watch, but I'd noticed his stylish cufflinks made out of string.
We had both observed his and his partner's complete lack of a sense of humor.
What do you expect from financiers whose pride and joy seemed to be a distribution network for PVC pipes?
I'm in India, covering a trade mission organized by the United States Commerce Department's Commercial Service. This particular mission is focused on franchisors, mostly restaurant companies, looking for Indian partners.
The mission is centered around what are essentially speed dates. They're meetings between the franchisors and potential franchisees to do a combination of pitching themselves and getting acquainted with one another — almost exactly like first dates, really.
I've been sitting in on some of the meetings to see what they're like — the pictures in this blog entry show you what they look like.
What I've seen so far indicates that finding business partners through blind dating — even with dates who have been vetted by the Commercial Service — is about as tough and almost as grueling as finding a love match that way. I'm not sure what the equivalent of a one-night stand would be in this case, but please let me know if you can think of one.
The would-be franchisors, being Americans working in foodservice, would talk about having "passion" for the business. The potential franchisees — real estate agents, retailers, PVC pipe distributors with lots of money of unknown origin — would blink once or twice and stare back blankly.
But you never know. The McDonald's executive in charge of western and southern India comes from the lubricant business, so maybe Mr. Fancy Watch will be the man behind Which Wich? India.
Personally, I liked the owner of The Devil's Workshop, a bakery chain with the slogan "Food you hate to love."
After the speed dates, we all went to the Residence of the American Consul General for a reception, and I learned that India is by far the world's largest consumer of whisky. So after a glass of perfectly acceptable red wine I switched to nice single malts and chatted with Subway franchisees and politicians, commercial attachés and Assistant Commerce Secretary Nicole Lamb-Hale, who is on the trip with us.
Marketing consultant Jagdeep Kapoor explained to me the challenges of presenting the right message in India. He pointed out that standing in line to order food is insulting to many middle class Indians, because it seems like begging. Besides, they're accustomed to having servants and like to be waited on.
But then for other middle class Indians, who studied in the United States, American fast food gives them a sense of nostalgia and they want to stand in line for it ("They feel like they're in New York") and then get exactly the same food that they would have in the U.S.
Others want food that reminds them of their mother's home cooking, so it should taste completely Indian.
I asked Kapoor how you could do all of that with one brand while maintaining a single identity.
"Give the people what they want," he said, which didn't really answer the question, but it is indeed what a restaurant should do.
He also said that Indians love bossing people around, so that the model at Subway, or any chain where your food is assembled in front of you, appeals to them because they can tell the restaurant worker exactly how they want their food to be made.
Indeed, that seems to appeal to everyone.
TAGS: Archive Marketing


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San Jose mayor healing broken bones after New Year's Day bike crash
Ashley McBride Jan. 2, 2019 Updated: Jan. 2, 2019 3:59 p.m.
San Jose mayor healing broken bones after New Year's...
1of5San Jose Mayor Sam Liccardo arrives to attend a City Council meeting at San Jose City Hall in San Jose, Calif. on Tuesday, Dec. 11, 2018.Photo: Jessica Christian / The Chronicle
2of5San Jose Mayor Sam Liccardo is expected to make a full recovery following a bike accident on New Year's Day in northeast San Jose.Photo: San Jose Mayor's Office
3of5From left to right, San Francisco Mayor Ed Lee, Metropolitan Transportation Commission Chairman Dave Cortese and San Jose Mayor Mayor Sam Liccardo stand with bicycles during Ford GoBike's launch event on Friday, Sept. 9, 2016, in San Francisco.Photo: Noah Berger, Special to The Chronicle
4of5San Jose Mayor Sam Liccardo attends the opening of the Continental Silicon Valley Research and Development Center for automotive technology in San Jose, Calif. on Wednesday, April 12, 2017.Photo: Paul Chinn / The Chronicle 2017
5of5San Jose Mayor Sam Liccardo listens in to San Jose police chief Larry Esquivel during a press conference following the shooting death of a San Jose police officer earlier today, Tuesday, March 24, 2015, in San Jose, Calif.Photo: Santiago Mejia, The Chronicle
San Jose Mayor Sam Liccardo is expected to make a full recovery following a bike accident on New Year's Day in northeast San Jose.
He has months of physical therapy ahead, but Liccardo was up and walking around at San Jose Regional Medical Center on Wednesday morning, spokesman David Low said.
A photo released by the mayor's office showed him walking between hospital staff while wearing a back brace over hospital garb.
"I expect to be working from home this week, and back at City Hall doing the job I love next week," Liccardo said in a statement.
By Steve Rubenstein
San Jose Mayor Sam Liccardo hit by car while bicycling,...
Michelle Robertson
San Jose mayor will get tattoo of city's logo — under one...
By Ashley McBride
Woman dies after early morning hit-and-run in San Jose
San Jose police officer injured in early morning car crash
A driver struck Liccardo while he was riding his bike around 12:30 p.m. Tuesday in the 600 block of Salt Lake Drive, about 3 miles northeast of downtown San Jose. He suffered two fractured vertebrae and a fractured sternum, but is expected to be released from the hospital in a day, the mayor's office said.
The driver who struck Liccardo remained at the scene and received a citation for failing to yield to another motorist, police said.
A frequent cyclist, Liccardo was believed to have been on his way to Sierra Road, a popular, challenging climb into the foothills for experienced bicyclists.
Ashley McBride is a San Francisco Chronicle staff writer. Email: ashley.mcbride@sfchronicle.com Twitter: @ashleynmcb
Ashley McBride
Follow Ashley on:
ashleynmcb
Ashley McBride covers breaking news on The Chronicle's metro desk as a Hearst Journalism Fellow. Previously, she worked as an interactive learning producer and editor at the Poynter Institute in St. Petersburg, Fla., and as a web producer at the Palm Beach Post in West Palm Beach, Fla. She graduated with a master's degree in journalism from Syracuse University and completed her bachelor's degree at the University of Miami. A Tampa, Fla., native, Ashley enjoys cooking, reading and listening to podcasts in her spare time.
Greek man charged with murdering former Bay Area scientist in Crete


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text:
Q: pycharm shell issue that doesn't occur in terminal I am currently working on a project that calls the Word2Vec class from gensim. I have downloaded gensim via:
pip install -U gensim

If I open a python shell and type:
from gensim.models import Word2Vec

I get the following error:
Traceback (most recent call last):
  File "<input>", line 1, in <module>
  File "/Applications/PyCharm.app/Contents/helpers/pydev/pydev_import_hook.py", line 21, in do_import
    module = self._system_import(name, *args, **kwargs)
  File "/Library/Python/2.7/site-packages/gensim/__init__.py", line 6, in <module>
    from gensim import parsing, matutils, interfaces, corpora, models, similarities, summarization
  File "/Applications/PyCharm.app/Contents/helpers/pydev/pydev_import_hook.py", line 21, in do_import
    module = self._system_import(name, *args, **kwargs)
  File "/Library/Python/2.7/site-packages/gensim/models/__init__.py", line 13, in <module>
    from .word2vec import Word2Vec
  File "/Applications/PyCharm.app/Contents/helpers/pydev/pydev_import_hook.py", line 21, in do_import
    module = self._system_import(name, *args, **kwargs)
  File "/Library/Python/2.7/site-packages/gensim/models/word2vec.py", line 100, in <module>
    from gensim.models.word2vec_inner import train_sentence_sg, train_sentence_cbow, FAST_VERSION,\
  File "/Applications/PyCharm.app/Contents/helpers/pydev/pydev_import_hook.py", line 21, in do_import
    module = self._system_import(name, *args, **kwargs)
  File "__init__.pxd", line 155, in init gensim.models.word2vec_inner (./gensim/models/word2vec_inner.c:9234)
ValueError: numpy.dtype has the wrong size, try recompiling

After uninstalling and reinstalling many scipy, numpy, and gensim variations. And after using the following commands based on what I've read about the above error:
pip uninstall numpy
pip uninstall pandas
pip install pandas

easy_install --upgrade numpy

pip install -U gensim

Now I get a different error:
Traceback (most recent call last):
  File "<input>", line 1, in <module>
  File "/Applications/PyCharm CE.app/Contents/helpers/pydev/pydev_import_hook.py", line 21, in do_import
    module = self._system_import(name, *args, **kwargs)
  File "/Library/Python/2.7/site-packages/gensim/__init__.py", line 6, in <module>
    from gensim import parsing, matutils, interfaces, corpora, models, similarities, summarization
  File "/Applications/PyCharm CE.app/Contents/helpers/pydev/pydev_import_hook.py", line 21, in do_import
    module = self._system_import(name, *args, **kwargs)
  File "/Library/Python/2.7/site-packages/gensim/models/__init__.py", line 13, in <module>
    from .word2vec import Word2Vec
  File "/Applications/PyCharm CE.app/Contents/helpers/pydev/pydev_import_hook.py", line 21, in do_import
    module = self._system_import(name, *args, **kwargs)
  File "/Library/Python/2.7/site-packages/gensim/models/word2vec.py", line 92, in <module>
    from gensim import utils, matutils  # utility fnc for pickling, common scipy operations etc
ImportError: cannot import name utils

If I type:
from gensim import utils

I get the same error as above. For some reason utils is not being imported properly from gensim but I don't know why this would be happening. I am fairly new to python and this is rather confusing to me but what I've noticed after all the previous installations one BIG FIX!
If I call:
from gensim import utils

The import utils error does not occur when I invoke python2.7 in my terminal. But when I use the python interpreter in pycharm (the following version):
2.7.10 (/System/Library/Frameworks/Python.framework/Versions/2.7/bin/python2.7)

Python 2.7.10 (default, Jul 14 2015, 19:46:27)
[GCC 4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.39)] on darwin

The error still occurs.
When I invoke pycharm in my terminal it calls python from the following library:
Python 2.7.10 (v2.7.10:15c95b7d81dc, May 23 2015, 09:33:12)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin

I don't know if the python shell in python vs the terminal python shell are different but I don't know why one shell works fine but the other brings up the import utils error. Any help would be much appreciated!


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VMware, a global leader in cloud infrastructure and business mobility, helps customers realise possibilities by accelerating digital transformation. With VMware solutions, organisations are improving business agility by modernising data centres and integrating public clouds, driving innovation with modern apps, creating exceptional experiences by empowering the digital workspace, and safeguarding customer trust by transforming security. The VMware NSX Network Virtualization and Security Platform delivers a completely new operational model for networking that forms the foundation of the Software-Defined Data Centre (SDDC). NSX enables the creation of entire networks in software and embeds them in the hypervisor layer, abstracted from the underlying physical hardware while micro-segmentation and granular security are delivered to the individual workload. With 2016 revenue of $7.09 billion, VMware is headquartered in Palo Alto, CA and has over 500,000 customers and 75,000 partners worldwide.
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www.bcx.co.za
SMEs require a high level of trust in their cloud service provider
9 Jan 2019 - Kirsten Doyle
Dr Nkosi Kumalo, managing executive of sales at BCX.
Cloud is not a tool to save costs, but a tool to drive efficient utilisation of IT investments, says BCX.
According to statistics, 63% of SMEs don't survive past their second year. There are several factors that inhibit the growth of these entities in SA, including government red tape, lack of skills, labour regulations, lack of innovation, crime and access to funding.
ITWeb Cloud Summit 2019
Is your organisation ready for the cloud (r)evolution? Find out at the ITWeb Cloud Summit 2019, to be held on 12 February at The Forum in Bryanston. During the one-day event, we intend to provide a glimpse into the future of cloud, by delving into topics that go beyond the usual tools and solutions. Expert speakers will share information and insight on all aspects of cloud computing, including cloud services, delivery models, benefits, security, adoption, strategies and evolutions. Attendance is free to qualified end-users, click here to find out more.
Cloud computing is an area that could address most, if not all these challenges as it provides IT resources that are scalable, on demand and pay-per-use, convenient, independent of location and cost effective.
So says Dr Nkosi Kumalo, managing executive of sales at BCX, who will be presenting on 'The role of Cloud Computing in addressing SME challenges in South Africa' at the ITWeb Cloud Summit 2019, to be held on 12 February at The Forum in Bryanston.
During the course of his research, Dr Kumalo surveyed 265 mature SMEs, or SMEs that have passed the survival stage and are classified as small or medium companies that employ between 50 to 200 employees by the National Small Business Act. The research revealed that a high percentage of the respondents do not have an in-house IT department, and more than 62% have experienced server downtime over the past 12 months.
An enabler of growth
"Cloud computing becomes a strong enabler of growth as it takes away the pain of IT outages and the need to onboard expensive skills to support downtime. Information technology downtime has a big impact on both staff morale and customer service, which are unique selling points for SMEs. The nimbleness and agility of SMEs makes them to be attractive partners in the business environment and IT is key to giving them an edge in this regard," says Kumalo.
There's an enormous opportunity in making SMEs active contributors in how cloud is served to address their needs.
- Dr Nkosi Kumalo, BCX
According to him, cloud software usage is weighing heavily on non-core applications like messaging, office, payroll processing and suchlike. "Thirty percent of matured SME companies surveyed are using ERP applications, but the rest are not. All companies surveyed are aware of cloud computing and its benefits, but the transition towards mission-critical services is slow and a major concern for adoption of cloud."
He says it's also common to see that SMEs do not have a deliberate approach towards cloud computing adoption, and have a preference for using ready-made solutions that are available in an on-premises environment. He further states that there's an enormous opportunity in making SMEs active contributors in how cloud is served to address their needs. "SMEs appreciate cloud solutions, but protection of IP and relinquishing of control is a top concern, requiring a high level of trust between the service provider and the cloud user."
Kumalo adds that data growth is also another major issue when it comes to resource forecasting, as it becomes extremely difficult to know the capacity needed in the following year for budgeting purposes.
"Cloud computing provides a perfect solution to scale up and down, depending on usage requirement."
Above the hype
Delegates attending Kumalo's talk will get a clear definition of cloud, above the hype, and will learn how cloud is not a tool to save costs, but a tool to drive efficient utilisation of IT investments.
Cloud is not a tool to save costs, but a tool to drive efficient utilisation of IT investments.
"SMEs need to understand business policies and industry regulations so that they can embrace cloud computing in an informed and compliant manner. It's unwise to generalise and be emotional about a journey towards the cloud, they need to look beyond the noise," he concludes.
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