 Solution

The standard conjugate gradient convergence bound isHere $\kappa \le 5$, so the factor in parentheses is at most $q=(3-\sqrt{5})/2\simeq 0.381966$. Since $2q^{25}<10^{-10}$, 25 iterations suffice to reduce the energy-norm error by a factor exceeding $10^{10}$.

There are at most $m/20$ nonzero entries of $t$, and each contributes to at most two off-diagonals. Thus $A-I$ has at most $m^2/10$ nonzero entries. Initialize a matrix-vector product with the diagonal contribution $p$, which costs no arithmetic because the diagonal is one. Each off-diagonal term needs one multiplication and one addition, giving at most $m^2/5$ floating-point operations per product.

A standard conjugate gradient step adds two inner products, three vector updates and two scalar divisions, at most $10m$ further operations. Starting from $x_0=0$ makes $r_0=b$ and avoids an initial matrix-vector product; its squared norm needs $2m-1$ operations. The total for 25 steps is at mostThis supplies an explicit meaning of the paper's large-$m$ qualification. The quoted conjugate gradient convergence bound is the usual exact-arithmetic result.