Yes, there exists an effective bound on $D$. I am not sure who first found such a bound, but here is a nice reference :
The minimal number $e=e(I)$ such that $I=(\sqrt{I})^e$ I \supset (\sqrt{I})^e$is called the Noether exponent of$I$. The above article gives an effective bound for$e(I)$. More precisely, in the situation at hand, one may assume$n>N$and also$d_1 \geq d_2 \geq \cdots \geq d_n$. Then Jelonek proves that$e(I) \leq (d_1 \cdots d_N) \cdot d_n$(see Corollary 1.4 with$X=\mathbf{P}^N$). Thus the function$D(d_1,\ldots,d_n)=(d_1 \cdots d_N) \cdot d_n$(with$d_1 \geq \cdots \geq d_n$) works. 1 Yes, there exists an effective bound on$D$. I am not sure who first found such a bound, but here is a nice reference : MR2198324 : Jelonek, Z. On the effective Nullstellensatz. Invent. Math. 162 (2005), no. 1, 1--17. The minimal number$e=e(I)$such that$I=(\sqrt{I})^e$is called the Noether exponent of$I$. The above article gives an effective bound for$e(I)$. More precisely, in the situation at hand, one may assume$n>N$and also$d_1 \geq d_2 \geq \cdots \geq d_n$. Then Jelonek proves that$e(I) \leq (d_1 \cdots d_N) \cdot d_n$(see Corollary 1.4 with$X=\mathbf{P}^N$). Thus the function$D(d_1,\ldots,d_n)=(d_1 \cdots d_N) \cdot d_n$(with$d_1 \geq \cdots \geq d_n\$) works.