A faster way to replace variables than ReplaceAll

I noticed that for large matrix, the rule /. is very slow. In the example below, the combination of memorization and the replacement "/.sol[n-1]" inside the definition of sol[n] turns the code extremely slow. Just to calculate sol[1], it takes around 20 sec. So for sol[90] it will take 30 min! Is there any way to use a faster "replacement" technique than /. ?

ClearAll["Global*"]
sp = 100; ϵ = 0.2; β = 0.03; ρ = 0.5; α = 0.8; γ = 0.8;
V0 = 100;
J = 50; K = 50;
T = 100; Δt = 0.5;
Xmax = 300; Xmin = -200;
Vmax = 300; Vmin = 0;
ΔX = (Xmax - Xmin)/K;
ΔV = (Vmax - Vmin)/J;
F0[Xk_, Visc_] := Visc - sp;
F[k_, 0, n_] := 0;
F[k_, J, n_] := 0;
F[0, j_, n_] := 0;
F[K, j_, n_] := 0;
a[j_] := 0.5 β j Δt;
b[j_] := 1 + ( ϵ^2 j^2 + β) Δt;
c[j_] := -0.5 β j Δt;
d[j_] := -(0.25 ϵ ρ j/ ΔX) Δt;
sol[0] = Flatten[
Table[F[k, j, 0] -> F0[Xmin + k ΔX, j ΔV], {k, 1, K - 1}, {j, 1, J - 1}]];
sol[n_] :=
sol[n] = Module[{vars, eqns},
vars = Flatten[Table[F[k, j, n], {k, 1, K - 1}, {j, 1, J - 1}]];
(*THIS IS THE   PROBLEMATIC STEP!*)
eqns = Flatten[
Table[a[j] F[k, j - 1, n] + b[j] F[k, j, n] + c[j] F[k, j + 1, n] +
d[j] (F[k + 1, j + 1, n] + F[k - 1, j - 1, n] -
F[k - 1, j + 1, n] - F[k + 1, j - 1, n]) ==
F[k, j, n - 1], {k, 1, K - 1}, {j, 1, J - 1}]] /. sol[n - 1];
temp = Solve[eqns, vars][[1]];
FF = ArrayReshape[Values@temp, {K - 1, J - 1}];
Fc[k_, j_] := FF[[k, j]];
Flatten[Table[F[k, j, n] -> Fc[k , j ], {k, 1, K - 1}, {j, 1, J - 1}]]]
sol[1] // Timing
sol[900] // Timing (*This will take forever!*)
`
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• Dispatch is your friend... that said, the generation can be done more directly/efficently also. – ciao Feb 21 '15 at 6:05
• Thanks, rasher! The difference on speed is amazing. I wonder why isn't dispatch the default built in /. Do you know how can access the content of a Dispatch output without /. ? I've notice that the output is composed by "Length" and "Rules", but I would like to access "Rules" without creating another Table to use /. – MathematicaIsHard Feb 21 '15 at 12:01
• (1) It costs time to build the dispatch table. (2) Related: mathematica.stackexchange.com/questions/31613/… – Michael E2 Feb 21 '15 at 14:32