Suppose the functions $F$ and $G$ can be computed in 5 and 3 nanoseconds by functional units $U_F$ and $U_G$, respectively. Given two instances of $U_F$ and two instances of $U_G$, it is required to implement the computation $F(G(X_i))$ for $1 \le i \le 10$. Ignoring all other delays, the minimum time required to complete this computation is _______ nanoseconds.

We need to evaluate $F(G(X_i))$ for 10 inputs ($i = 1, 2, \dots, 10$) using 2 units of $U_G$ ($3\text{ ns}$ each) and 2 units of $U_F$ ($5\text{ ns}$ each).

1. $G$-stage (first stage):

   • With 2 units of $U_G$, pairs of inputs are processed every $3\text{ ns}$.

   • $G(X_1), G(X_2)$ finish at $t = 3\text{ ns}$.

   • $G(X_3), G(X_4)$ finish at $t = 6\text{ ns}$.

   • $G(X_5), G(X_6)$ finish at $t = 9\text{ ns}$.

   • $G(X_7), G(X_8)$ finish at $t = 12\text{ ns}$.

   • $G(X_9), G(X_{10})$ finish at $t = 15\text{ ns}$.

2. $F$-stage (second stage - bottleneck stage):

   • The two $U_F$ units take $5\text{ ns}$ per pair of inputs.

   • At $t = 3\text{ ns}$, the first pair $G(X_1), G(X_2)$ is ready. Both $U_F$ units start processing and complete at $t = 3 + 5 = 8\text{ ns}$.

   • Subsequent pairs of $G$ results are already available when $U_F$ units finish each batch, so the $U_F$ units remain continuously busy after $t = 3\text{ ns}$:

     • Pair 1 ($X_1, X_2$): processed from $t = 3$ to $8\text{ ns}$

     • Pair 2 ($X_3, X_4$): processed from $t = 8$ to $13\text{ ns}$

     • Pair 3 ($X_5, X_6$): processed from $t = 13$ to $18\text{ ns}$

     • Pair 4 ($X_7, X_8$): processed from $t = 18$ to $23\text{ ns}$

     • Pair 5 ($X_9, X_{10}$): processed from $t = 23$ to $28\text{ ns}$

Thus, the total time required to complete all 10 computations is 28 nanoseconds.