Is this a proper alternative way for math model for TSP(Travelling Salesman Problem)? I have never seen a model that uses indexing in any article.So I have decided to publish it to be sure. I think indexing model is more suitable for generalizing the model than the subtour elimination method.I have tested the model using GAMS and it seems the model is fine.
For a TSP problem contains $n$ city datas are :
a.) $i$ and $j$ are cities
b.) $k$ is sequence order for cities
c.) $i = j = k = {1,2...n} $
d.) $d(i,j)$ equals the distance between city $i$ and city $j$
d.) $x(i,k)$ is binary variable that specifies whether city $i$ is travelled on $k^{th}$
e.) $t(i,j,k)$ is binary variable that gives information about $i$ and $j$ are connected at $k^{th}$ order
d.) $o(i,j)$ is binary variable that represents $i$ and $j$ are connected 
In base in this model we use connectedness between $i$ and $j$ as $o(i,j)$. If $i$ is on the $k^{th}$ order and $j$ is on ${(k+1)}^{th}$ , this means $i$ and $j$ are connected. We calculate this using $t(i,j,k)$. We find $o(i,j)$ summing all $t$ values for all $k$.
Minimize Total Distance $z$ $$z = \sum_{i=1}^n\sum_{j=1}^n o(i,j)*d(i,j)$$
Constraints 
1.) $i = j = k = {1,2...n}$ and $k + 1 = 1$ if $k = n$
$$x(i,k) + x(j,k+1) \ge 2 * t(i,j,k)$$
2.) $i = j = k = {1,2...n}$ and $k+1 = 1$ if $k = n$
$$x(i,k) + x(j,k+1) \le 2 * t(i,j,k) + 1$$
3.) $i = j = k = {1,2...n} $ $$o(i,j) = \sum_{k=1}^n t(i,j,k)$$
4.) $i = k = {1,2...n} $ $$\sum_{k=1}^n x(i,k) = 1$$
5.) $j = k = {1,2...n} $ $$\sum_{k=1}^n x(j,k) = 1$$
Total number of constraints is $2n^3 + n^2 + 2n + 1$
There is another model like this using mixed-integer nonlinear programming. It has $2n + 1$ equations but doesnt gives exact values for $n>10$
Finally does this model exists before or if not is this model correct? Please give a reference.

A bunch of formulations is described in 

A.J. Orman, H.P. Williams, A Survey of Different Integer Programming
  Formulations of the Travelling Salesman Problem, Link

I did not check if your version is in there.
