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If an address bus needs to be able to address four devices, how many conductors will be required? What if each of those devices also needs to be able to talk back to the I/O control device?
The solution of the previous question of the week can be seen below.
The maximum aggregate I/O transfer rate of the system is equal to:
700Kbystes/s + 700 Kbytes/s+ 2* 6.6 Kbytes/s+ 2* 1.2 Kbytes/s+ 10* 1Kbytes/s = 1,400 Kbytes/s+ 13.2 Kbytes/s+ 2.4 Kbytes/s+ 10 Kbytes/s= 1,425.6 Kbytes/s. This can be explained by the fact that a selector channel can’t service more than one device at a time.
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