Suppose a message of size 15000 bytes is transmitted from a source to a…

Answer: 7 fragments.

Explanation and steps:

• Assume the IPv4 header is 20 bytes. The data to be carried by IP is therefore 15000 - 20 = 14980 bytes (if the 15000 was the entire datagram) — the calculation below is the same if you treat 15000 as data because the fragmentation arithmetic yields the same final count.

• At Router-1 the MTU is 5000 bytes, so the maximum payload per fragment is 5000 - 20 = 4980 bytes. Because fragment offsets are in 8-byte units, each fragment's payload must be a multiple of 8, so use 4976 bytes per fragment.

• Split 14980 bytes of data into 4976-byte chunks: 4976 * 3 = 14928, leaving a final piece of 14980 - 14928 = 52 bytes. So after Router-1 we have three fragments carrying 4976 bytes each and one fragment carrying 52 bytes (total 4 fragments).

• At Router-2 the MTU is 3000, so maximum payload per fragment is 3000 - 20 = 2980 bytes. Adjusting to an 8-byte multiple gives 2976 bytes per fragment.

• Each 4976-byte payload from Router-1 must be split into 2976 + 2000 bytes (4976 = 2976 + 2000). The 2000-byte piece is already a multiple of 8, so no further splitting is needed. Thus every large fragment from Router-1 becomes two fragments at Router-2.

• The small 52-byte fragment from Router-1 fits within Router-2's MTU and is not fragmented further.

• Total fragments delivered to the destination = (3 large fragments × 2 after second fragmentation) + (1 small fragment) = 6 + 1 = 7.