Assign to avg_owls the average owls per zoo. Print avg_owls as an integer. Sample output for inputs: 1 2 4 3

In this situation, when the Adaptor or router gets the destination IP address (even if an incorrect MAC address is entered), it would strip the IP address from the Ethernet frame and using ARP, it will obtain the accurate MAC address of the destination.

Answer:

A) 2.56 ms

B) 128 μs

Explanation:

Time per bus cycle = 500 ns

Time taken to transfer bus control = 250 ns

Data transfer rate of the I/O device = 50 KB/s

Data is transferred one byte at a time.

A) Calculate the time the device occupies the bus during the transfer of a block of 128 bytes

Block size for transfer = 128 bytes

Bandwidth available = 50 KB/s

data transfer = (Block size ) / ( Bandwidth)

                     = ( 128 * 8 ) / ( 50 * 10^3 * 8 )

                     = ( 1024 ) / ( 50 * 10^3 * 8 )

                     = 2.56 ms

To compute the actual time taken for transfer, we incorporate the time required for control in both directions:

2.56 ms + 500 ns = 2.56 ms. This is because 500 ns is negligible.

B) cycle stealing mode

In this mode, each byte is transferred individually. The total control time for both directions will be twice the overall time, meaning 2 * 500 ns = 1000 ns. An extra control time of 250 ns is necessary at both ends.

Since 1 byte is transferred at once, the transfer time for 1 byte is 1 μs.

Thus, for 128 bytes = 128 * 1 μs = 128 μs.

Answer:

Vendor B

Explanation:

By using letters to signify each vendor and the symbol < to indicate which is cheaper, the problem can be summarized as follows: A<C, B<C, B<D, D<A.

These inequalities show that the vendors on the left are less expensive than those on the right, therefore, any vendor appearing on the right in these expressions cannot be the cheapest, ruling out C, D, and A. Thus, the least expensive vendor is B.