## SharePoint 2010 – Three-State ootb Workflow with multiple user for a step?

I want to use the three state out of the box workflow in SharePoint 2010. My problem is now, that I only can assign a task to one user for a step (It is not possible to add a second person in this field)

Is there any way to assign a task to a group?

p/s: I can’t use any other software such as InfoPath or SPD.

## real analysis – Finding a metric that contracts on average in one step

I am looking at different configurations of particles on a $$d$$-ary tree with $$n$$ nodes and $$lfloor frac{n}{2}rfloor$$ particles. If we start at a configuration $$x$$, an edge is selected uniformly at random and if that edge has one particle and one empty space, then the particle and empty space are swapped. If the selected edge has two particles or no particles, then we remain at $$x$$. In this case, I am trying to find a metric $$rho (x,y)$$ between configurations $$x$$ and $$y$$ that differ along a single edge such that $$(1)$$ is satisfied.

$$mathbb{E} (rho(X_1,Y_1))-rho(x,y) leq 0 tag{1}$$

where $$X_1,Y_1$$ correspond to one move starting from $$x,y$$ respectively and $$mathbb{E}$$ is the expectation. The following figure gives an example for $$d=2$$. Notice that $$x,y$$ differ only along one edge.

For the above figure,

$$mathbb{E} (rho(X_1,Y_1))-rho(x,y) = -frac{1}{n-1}rho(x,y)+frac{d-1}{n-1}(rho(x,y)+rho(y,y^{+}))+frac{d}{n-1}(rho(x,y)+rho(x,x^{+}))+frac{1}{n-1}(rho(x,y)+rho(y,y^{-})) tag{2}$$

Because we decrease the distance in one move by $$rho(x,y)$$ only when we select the edge along which $$x,y$$ differ, which happens with probability $$frac{1}{n-1}$$ and in all other cases we increase the distance after one move. Let $$x,y$$ differ at height $$h$$ from the root. If we select another edge at height $$h$$ apart from the edge along which $$x,y$$ differ, which happens with probability $$frac{d-1}{n-1}$$, then $$x$$ remains at $$x$$ and $$y$$ moves to $$y^{+}$$ and that explains the second term in $$(2)$$. If we select an edge at height $$h+1$$, which happens with probability $$frac{d}{n-1}$$, then $$x$$ moves to $$x^+$$ and $$y$$ remains at $$y$$ and that explains the third term in $$(2)$$. Similarly, if we select an edge at height $$h-1$$, which happens with probability $$frac{1}{n-1}$$, then $$x$$ remains at $$x$$ and $$y$$ moves to $$y^-$$ and that explains the fourth term in $$(2)$$.

My attempt: I have tried $$rho(x,y)=h$$ and $$rho(x,y)=alpha^{h}$$ for $$alpha>1$$ a constant but both of them did not satisfy $$(1)$$ mainly for the reason that the distance increases in too many ways while the distance decreases only when we select one particular edge i.e. the one along which $$x,y$$ differ. Any thoughts about how to find $$rho(x,y)$$? Or how to pose this as a different problem?

## Looking for clarity on step in fractional knapsack problem proof

I’m reading through the proof for the fractional knapsack problem presented here: https://www.cs.rice.edu/~nakhleh/COMP182/Knapsack.pdf

I follow the proof until the definition of the new set $$Q = langle q_1, q_2, …, q_n rangle$$

• for every $$1 leq ialso $$q_{i}=y_{i})$$ which is constructed as follows:
• $$q_j=y_j$$
• For $$i=j+1$$ to $$n$$

$$begin{array}{l} – d^{prime} leftarrow min left(q_{i}, dright) \ – q_{i} leftarrow q_{i}-d^{prime} \ – d leftarrow d-d^{prime} end{array}$$

I don’t understand where the $$q_{j+1…n}$$ elements come from. We only fill $$Q$$ up to $$q_j$$, but to fill $$q_{j+1}$$ we reference $$q_i$$ which is undefined — should it be 0? What am I missing?

## plotting – Step size is effectively zero; singularity or stiff system suspected

I am trying to solve four coupled nonlinear differential equations.
But, everytime I am getting
NDSolve::ndsz: At t == 2.0833315360868916`, step size is effectively zero; singularity or stiff system suspected.
I have tried all the possible ways to solve such kind of problem with similar kind of problems available in stack.
I want to get the value of g1,g2,g3 and g4 for large value of t i.e. 1 to 100 but I am getting only for small value of t?
Could you please suggest me some way to sort out the issue?
(Eta) = 0.125;
rg1s = Derivative(1)(u1)(t) – u1(t)u3(t) – 3u1(t)u4(t) + u2(t)u3(t) + 3u2(t)u4(t) + 2(1 + 2(Eta))u1(t)^2 == 0;
rg2s = Derivative(1)(u2)(t) – u2(t)u3(t) – 3u2(t)u4(t) + u1(t)u3(t) + 3u1(t)u4(t) + 2(1 + 2(Eta))u2(t)^2 == 0;
rg3t = Derivative(1)(u3)(t) – 0.5
((u1(t) – u2(t))^2 + u3(t)^2 – 3
u4(t)^2 + 6u3(t)u4(t)) == 0;
rg4t = Derivative(1)(u4)(t) – 0.5
((u1(t) – u2(t))^2 + u3(t)^2 + 5
u4(t)^2 – 2u3(t)u4(t)) == 0;
sol = NDSolve({rg1s, rg2s, rg3t, rg4t, u1(0) == -0.6
0.2, u2(0) == -0.6
0.2, u3(0) == 0.6*0.2, u4(0) == 0.2}, {u1, u2, u3, u4}, {t, 1, 10})

Thank you

## Name for data structure that keeps track of minimal/maximal element through unit step updates

There’s a data structure that has the following operations:

• $$text{init}(n)$$ – Initializes $$n$$ integer elements set to zero, $$O(n)$$.
• $$text{inc}(i)/text{dec}(i)$$ – Increments/decrements the element at index $$i$$ by one, $$O(1)$$.
• $$text{get}(i)$$ – Returns the element at index $$i$$, $$O(1)$$.
• $$text{min}()/text{max}()$$ – Returns an index containing a minimal/maximal value, $$O(1)$$.

One possible implementation of the above is a linked list of buckets, each bucket storing the elements of value $$k$$, combined with a vector storing which bucket an element $$i$$ is in, and a vector directly to the node inside the bucket.

Does this have a name?

## PowerShell Script Not Working on SQL Server Job Step

I hope you’re doing fine!
I’ve some .CSV files generated in a shared folder (SERVER01DIRCurrent) and would like to copy to another folder in same server01 (SERVER01DIRArchive). Script must to copy, rename files adding timestamp and then removing from source folder. PS script works very well on Powershell ISE or Windows Powershell, but, when I try to schedule same script on SQL Server Job, I’m receiving errors below.

``````\$source = "\SERVER01DIRCurrent" \$destination = "\SERVER01DIRArchive"

Get-ChildItem \$source -Recurse -Include *.csv | % {
\$name = \$_.Name.Split(".")(0) + "_" + (\$_.CreationTime | Get-Date -Format yyyymmdd) + "_" + (\$_.CreationTime | Get-Date  -Format hhmmss) + ".csv"
#\$name = "Finished_" + (\$_.CreationTime | Get-Date -Format yyyymmdd) + "_" + (\$_.CreationTime | Get-Date  -Format hhmmss) + ".csv"
#\$name = "Finished_" + \$_.Name.Split(".")(0) +  "_" + (\$_.CreationTime | Get-Date -Format yyyymmdd) + "_" + (\$_.CreationTime | Get-Date  -Format hhmmss) + ".csv"
Rename-Item \$_ -NewName \$name
Move-Item "\$(\$_.Directory)\$name" -Destination \$destination }
``````

Message
Unable to start execution of step 1 (reason: line(9): Syntax error). The step failed.

After receive error above, I’ve changed Line(9) as below:

``````\$source = "\SERVER01DIRCurrent"
\$destination = "\SERVER01DIRArchive"

Get-ChildItem \$source -Recurse -Include *.csv | % {
\$name = \$_.Name.Split(".")(0) + "_" + (\$_.CreationTime | Get-Date -Format yyyymmdd) + "_" + (\$_.CreationTime | Get-Date  -Format hhmmss) + ".csv"
#\$name = "Finished_" + (\$_.CreationTime | Get-Date -Format yyyymmdd) + "_" + (\$_.CreationTime | Get-Date  -Format hhmmss) + ".csv"
#\$name = "Finished_" + \$_.Name.Split(".")(0) +  "_" + (\$_.CreationTime | Get-Date -Format yyyymmdd) + "_" + (\$_.CreationTime | Get-Date  -Format hhmmss) + ".csv"
Rename-Item \$_ -NewName \$name
Move-Item "\SERVER01DIRCurrent\$name" -Destination \$destination
}
``````

*Message
The job script encountered the following errors. These errors did not stop the script:
A job step received an error at line 4 in a PowerShell script.
The corresponding line is ‘Get-ChildItem \$source -Recurse -Include .csv | % { ‘.
Correct the script and reschedule the job.
The error information returned by PowerShell is:
‘Cannot find path ‘SERVER01DIRCurrent’ because it does not exist. ‘.
Process Exit Code 0. The step succeeded.

After that, I’ve tried to change line(9) again, using \$source parameter:

``````Move-Item \$source + "" + \$name -Destination \$destination
``````

But, it didn’t work and return same error above: ‘Cannot find path ‘SERVER01DIRCurrent’ because it does not exist. ‘.

Please, could you help me how to solve it? It looks like “\$(\$_.Directory)…” is not working on SQL job. Not sure why and how to solve it.
Thanks!

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## differential equations – Specific problem with NDSolve step size, stiffness

I am trying to solve a non-linear second order boundary value problem in a finite interval. The differential equation is $$y”-frac{a}{b}y-frac{u_n}{b}y^3-frac{ge_0}{b}x=0,$$ with $$xin(0,L)$$, along with the boundary conditions, $$y'(0)-frac{o}{b}y(0)=0,qquad y'(L)+frac{l}{b}y(L)=frac{ge_0L}{b}.$$
In the code attached below, with the chosen parameters, the solution is regular and shows up until $$a=1.2$$. On using $$a=1.3$$ and beyond, I get the error “At x == 0.99505895226764`, step size is effectively zero; singularity or stiff system suspected.”

I am changing the input right after $$z$$ in Plot(Evaluate…

Working:

``````ClearAll;
Clear(Derivative)
fneeonnd(z_?NumericQ, a_?NumericQ, b_?NumericQ, o_?NumericQ,
l_?NumericQ, g_?NumericQ, L_?NumericQ, c_?NumericQ, u_?NumericQ,
un_?NumericQ, eo_?NumericQ) := (
ss = NDSolve({y''(x) - (a/b)*y(x) - (un/b)*(y(x))^3 - (g*eo/b)*x ==
0, y'(0) - (o/b)*y(0) == 0, y'(L) + (l/b)*y(L) == (g*eo*L/b)},
y, {x, 0, L}
, Method -> "StiffnessSwitching", WorkingPrecision -> 100);
y(z) /. ss)
Plot(Evaluate(
fneeonnd(z, 1.2, 0.05, 0.1, 0.1, 0.5, 1, 0.1, 0.1, 0.4, 1)), {z, 0,
1}, PlotStyle -> {Thickness(0.007)}, BaseStyle -> {FontSize -> 20},
Frame -> {True}, FrameLabel -> {"z", (Eta)},
FrameStyle -> {{Black, Directive(Thick)}, {Black,
Directive(Thick)}, {Black, Directive(Thick)}, {Black,
Directive(Thick)}})
``````

NOT Working:

``````ClearAll;
Clear(Derivative)
fneeonnd(z_?NumericQ, a_?NumericQ, b_?NumericQ, o_?NumericQ,
l_?NumericQ, g_?NumericQ, L_?NumericQ, c_?NumericQ, u_?NumericQ,
un_?NumericQ, eo_?NumericQ) := (
ss = NDSolve({y''(x) - (a/b)*y(x) - (un/b)*(y(x))^3 - (g*eo/b)*x ==
0, y'(0) - (o/b)*y(0) == 0, y'(L) + (l/b)*y(L) == (g*eo*L/b)},
y, {x, 0, L}
, Method -> "StiffnessSwitching", WorkingPrecision -> 100);
y(z) /. ss)
Plot(Evaluate(
fneeonnd(z, 1.3, 0.05, 0.1, 0.1, 0.5, 1, 0.1, 0.1, 0.4, 1)), {z, 0,
1}, PlotStyle -> {Thickness(0.007)}, BaseStyle -> {FontSize -> 20},
Frame -> {True}, FrameLabel -> {"z", (Eta)},
FrameStyle -> {{Black, Directive(Thick)}, {Black,
Directive(Thick)}, {Black, Directive(Thick)}, {Black,
Directive(Thick)}})
``````

## maximum likelihood – Confused looking at an answer key. How does the professor go from the second to last step to the \$>1\$ line?

Let $$X_1, …, X_n$$ be iid with one of two pdfs. If $$theta = 0$$, then

$$begin{equation*} f(x|theta) = begin{cases} 1 text { if } 0 < x< 1\ 0 text{ otherwise} end{cases} end{equation*}$$

while if $$theta = 1$$, then

$$begin{equation*} f(x|theta) = begin{cases} 1/(2sqrt{x}) text { if } 0 < x < 1\ 0 text{ otherwise} end{cases} end{equation*}$$

Find the MLE of $$theta$$.

First, we need to find the density and likelihood functions for each potential value of $$theta$$. We will ignore the interval outside $$0 < x < 1$$.

For $$theta = 0$$ we have the density function $$f(x|theta) = 1$$ therefore the likelihood function is $$L(0|x) = 1$$ over our interval $$0 < x_i < 1$$.

For $$theta = 1$$ we have the density function $$f(x|theta) = frac{1}{2sqrt{x}}$$ and therefore the likelihood function is $$L(1|x) = frac{1}{2Pi_{i=1}^n sqrt{x_i}}$$ over our interval $$0 < x_i < 1$$.

Next, we use some algebraic manipulation to turn $$x_i < 1$$ into $$L(1|x) = frac{1}{2Pi_{i=1}^n sqrt{x_i}}$$. Simplifying these to match will show us the point at which our MLE of $$theta$$, $$hat{theta}$$ changes. We find:

begin{equation*} begin{aligned} & x_i < 1 \ & sqrt{x_i} < 1 \ & frac{1}{sqrt{x_i}} > 1\ & Pi_{i=1}^n left(frac{1}{2sqrt{x_i}} right) > left( frac{1}{2}right)^n \ & > 1 end{aligned} end{equation*}

How did he get this result? I’m confused on how to interpret the >1 as well. Any advice on the meaning here is appreciated

So we say if $$Pi_{i=1}^n left(frac{1}{2sqrt{x_i}} right) > 1$$ then the MLE of $$theta = 1$$. The reverse is also true therefore:

$$begin{equation*} hat{theta} = begin{cases} 0 text{ if } Pi_{i=1}^n left(frac{1}{2sqrt{x_i}} right) leq 1\ 1 text{ if } Pi_{i=1}^n left(frac{1}{2sqrt{x_i}} right) > 1 end{cases} end{equation*}$$

## linux – openvpn startup error: Failed at step CHDIR spawning /usr/bin/openvpn

I’m trying to start an openvpn v2.4.9 server with running

``````systemctl start openvpn-server@server
``````

as root on Arch Linux.

This results in the following error:

`````` The job identifier is 176005 and the job result is failed.
Nov 03 13:42:35 hostname audit[1]: SERVICE_START pid=1 uid=0 auid=4294967295 ses=4294967295 msg='unit=openvpn-server@-etc-openvpn-server.onf comm="systemd" exe="/usr/lib/systemd/systemd" hostname=? addr=?>
Nov 03 13:42:35 hostname systemd[100169]: openvpn-server@-etc-openvpn-server.service: Changing to the requested working directory failed: No such file or directory
Nov 03 13:42:35 hostname systemd[100169]: openvpn-server@-etc-openvpn-server.service: Failed at step CHDIR spawning /usr/bin/openvpn: No such file or directory
Subject: Process /usr/bin/openvpn could not be executed
Defined-By: systemd
Support: https://lists.freedesktop.org/mailman/listinfo/systemd-devel

The process /usr/bin/openvpn could not be executed and failed.
``````

I am not quite sure what is meant with “Failed at CHDIR spawning”.

/usr/bin/openvpn is existing and executable for everyone:

``````root@hostname > ls -l /usr/bin/openvpn
-rwxr-xr-x 1 root root 788544 Apr 20  2020 /usr/bin/openvpn
``````

/usr and /usr/bin are enterable by anyone as well.

My config is located at /etc/openvpn/server.conf and looks like this:

``````user nobody
group nobody

persist-key
persist-tun

proto udp
proto udp6

dev tun

ca /etc/openvpn/easy-rsa/pki/ca.crt
cert /etc/openvpn/easy-rsa/pki/issued/cert.crt
key /etc/openvpn/easy-rsa/pki/private/key.key
dh /etc/openvpn/easy-rsa/pki/dh.pem

cipher AES-256-CBC
auth SHA512
comp-lzo
reneg-sec 36000

server 192.168.100.0 255.255.255.0
push "route 192.168.100.0 255.255.255.0"

keepalive 10 36000

status openvpn-status.log

log /var/log/openvpn.log

verb 6
``````

None of the two logfiles list any valuable information.