TITLE: Models of computation with decidable halting problem?
QUESTION [7 upvotes]: There are numerous examples of models of computation in which all programs halt, for example primitive recursion. 
Are there (non-trivial) examples of models in which only some programs halt, but the halting problem is still decidable? 
Does the decision procedure need to lie outside of the original model itself?
EDIT:
Carl Mummert gives very good example of a model of computation that has this property. But the example of polynomially clocked Turing machines is weaker than primitive recursion. 
Are there examples which are stronger than primitive recursion?

REPLY [5 votes]: Joel Hamkins points out that the decision procedure for any reasonable notion of "computability" is not going to be solvable by a function that is "computable" within that notion. 
Here is a contrasting example of a nontrivial model of computation in which the halting problem is solvable in the usual sense of computation. An index $e$ in our new system is a pair $(e_1, p)$ where $e_1$ is an index for a Turing machine and $p$ is a code for a polynomial over $\mathbb{N}$. In our new system, program $e$ is said to compute output $o$ on input $n$ (write $P_e(n) = o$) if and only if Turing machine $e_1$ computes $o$ on input $n$ in less than $p(|n|)$ steps. If $e_1$ runs for more then $p(|n|)$ steps then we say the computation of $P_e(n)$ is undefined (i.e. does not halt).  Here we assume the Turing machine uses binary coding for numbers and we let $|n|$ be the number of bits required to express $n$ in binary notation. 
This restricted model of computation is relatively common in the study of polynomial-time computability, where an index of the form $e = (e_1, p)$ is called a "polynomially clocked Turing machine". It's immediate from the definitions that a function is computable in the restricted model if and only if it is computable in polynomial time. Thus the model includes a very wide class of functions. However, because the time bound for index $e$ is already included in $e$, we can solve the halting problem for this model of computation with a normal Turing machine. (We cannot solve it with any polynomially clocked machine, of course.)