In both the potential reduction and primal path following interior point methods for linear programming, a barrier function is constructed which contains the terms $-\sum \log x_j$ where $x_j$ are the variables. This is to keep the variables from being 0 and thus keep the current solution feasible.
However, in potential reduction, the new direction to move the current solution to is found by solving a linear program which contains as the single inequality constraint $||\mathbf{X^{-1}d}|| \leq \beta$ where $\beta < 1$ and $\mathbf{d}$ is the direction. This constraint is so that the new solution $\mathbf{x}+\mathbf{d}$ remains $> 0$ and thus feasible. $\mathbf{X}$ is a diagonal matrix with the elements of the variable vector $\mathbf{x}$ forming the diagonal.
However in the primal path method (related approach), one can just solve the Lagrangian dual of the barrier function problem and the new solution is guaranteed to be feasible - i.e. the property $\mathbf{||X^{-1}d}|| \leq \beta$ is guaranteed to be satisfied.
Is there some good intuition as to why primal path can get away with just solving the Lagrangian dual (and thus not have to deal with inequalities) while potential reduction cannot? The best I can think of is that the barrier function in the potential reduction algorithm contains the term $\log \mathbf{s'x}$ where $\mathbf{s'x}$ is the duality gap ($\mathbf{s}$ is the slack vector for slacks in the dual version of the problem). This term thus strongly pulls $\mathbf{x+d}$ into a potentially infeasible region. The barrier function for the primal path only has the term $\mathbf{c'x}$ (where $\mathbf{c'x}$ is the objective function of the original problem) in addition to the log terms in the barrier function so it does not exert as strong a pull (as it will not for instance go to $-\infty$ when $\mathbf{c'x}$ goes to 0 unlike $\log \mathbf{s'x}$ which will). However I don't know how to formalize this.