Add-on Percentage of total. I am not sure how to phrase this, but I have a problem where I need to adjust my total by a markup percentage. The resulting markup is included in the total, but I want the percentage of that markup to still match. I am calculating this today with a recursive algorithm, but I am looking for a more direct equation. 
For example:
Subtotal: 1000
Markup Goal: 10%

1) total   1100       actual markup  .090909
2) total   1110       actual markup  .099099
3) total   1111       actual markup  .09991
4) total   1111.1     actual markup  .099991
5) total   1111.11    actual markup  .099999
6) total   1111.111   actual markup  .1

Edit: Given a subtotal, I want to add a markup% such that:
markup / subtotal+markup = markup%

 A: If $m$ is the markup value, $s$ the subtotal and $p$ the markup percentage, then according to my interpretation your formula in the edit means
$$
\begin{equation*}
\frac{m}{s+m}100=p\tag{1a}
\end{equation*}
$$
or just
$$
\begin{equation*}
\frac{m}{s+m}=p,\tag{1b}
\end{equation*}
$$
if $p$ is the markup percentage expressed as a decimal value (in your example $.1$). You just have to solve
for $m$ (see computation below) to obtain
$$
\begin{equation*}
m=\frac{ps}{100-p},\qquad \text{or}\qquad m=\frac{ps}{1-p}.\tag{2a,b}
\end{equation*}
$$
If you prefer names instead of letters, the equivalent formulas are:
$$
\begin{eqnarray*}
\text{markup} &=&\frac{\text{markup%}\times \,\text{subtotal}}{100-\,\text{markup%}}\tag{3a} \\
\\
&&\text{or} \\
\\
\text{markup} &=&\frac{\text{markup% as decimal value }\times \,\text{subtotal}}{1-\,
\text{markup% as decimal value}}.\tag{3b}
\end{eqnarray*}
$$
In your example, $p=10\%$, $s=1000$. The first formula yields the markup
value
$$
\begin{equation*}
m=\frac{10\times 1000}{100-10}=\frac{1000}{9}\approx 111.11
\end{equation*}
$$
The second one for $p=.1$, $s=1000$ yields the same value
$$
\begin{equation*}
m=\frac{.1\times 1000}{1-.1}=\frac{100}{.9}=\frac{1000}{9}\approx 111.11
\end{equation*}
$$
Solution of $(1\text{a})$ for $m$
$$
\begin{eqnarray*}
\frac{m}{s+m}100 &=&p\Leftrightarrow 100m=p(s+m) \\
&\Leftrightarrow &100m=ps+pm \\
&\Leftrightarrow &100m-pm=ps \\
&\Leftrightarrow &(100-p)m=ps \\
&\Leftrightarrow &m=\frac{ps}{100-p}.
\end{eqnarray*}
$$
