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.