In this paper we show that the minimum number of comparisons necessary for the computation of the kth element of a totally ordered set of size n, $V_k (n)$, is bounded below by $n - k + (k - 1)\lceil \log _2 (n/(k - 1)) \rceil $. For $3 < k < n/4$, this bound is an improvement on the best lower bound presently known. A new algorithm which yields an upper bound that is better than the currently known bound for a large range of values of n will also be presented.