The document calculates the maximum bus speed as 44.5 ns per cycle, which is the sum of various delays. It also states that four clock cycles are needed to complete the input operation, with a new transfer initiated in the fourth cycle.

1. Solution
:
a) Maximum bus speed = bus driver delay + propagation delay + address decoder delay + time to
fetch the requested data + set up time
=2 +10 + 6 + 25 +1.5
= 44.5 ns per cycle
Ans: maximum speed of the bus: 44.5 ns
b) Number of clock cycles needed to complete the
input operation is 4.
Because a new transfer is started in clock cycle 4