Each Ethernet frame can carry a max of 1500 bytes. A UDP segment with 7488 bytes of payload is transmitted. Find the total number of fragments and the size of the last fragment including IPv4 header (assume no options).

The correct option is 6 fragments, 108 bytes.

Let's break down the step-by-step derivation to find the total number of fragments and the size of the last fragment.

1. Understand the constraints and sizes:
- Maximum Transmission Unit (MTU) of Ethernet frame: 1500 bytes. This is the maximum size of the IP packet (including IP header and IP payload) that can be carried in an Ethernet frame.
- Size of IPv4 Header (with no options): 20 bytes.
- Maximum payload that a single IP fragment can carry:
$$\text{Max Payload}=\text{MTU}-\text{IP Header}=1500-20=1480\text{bytes}$$ Note that in IPv4 fragmentation, the payload size of each fragment (except the last one) must be a multiple of 8 bytes because the Fragment Offset field in the IP header is measured in units of 8-byte blocks. Since 1480 is divisible by 8 (1480 / 8 = 185), 1480 bytes is indeed the maximum payload size per fragment.

2. Determine the total payload to be transmitted:
- UDP segment payload = 7488 bytes.
- UDP Header size = 8 bytes.
- Total data to be carried by the IP layer (UDP Header + UDP Payload):
$$\text{Total IP Payload}=7488+8=7496\text{bytes}$$

3. Calculate the number of fragments:
Each of the first fragments will carry the maximum allowed payload of 1480 bytes.
- Number of full fragments needed:
$$\text{Number of full fragments}=\lfloor \frac{7496}{1480}\rfloor =5$$ - Payload carried by these 5 fragments:
$$5\times 1480=7400\text{bytes}$$ - Remaining payload for the last fragment:
$$\text{Remaining Payload}=7496-7400=96\text{bytes}$$ Since there is a remaining payload of 96 bytes, we need one more fragment.
Thus, the total number of fragments is:
$$5+1=6\text{fragments}$$

4. Calculate the size of the last fragment:
The last fragment contains the remaining IP payload and its own 20-byte IPv4 header.
- Size of the last fragment including IPv4 header:
$$\text{Last Fragment Size}=\text{Remaining Payload}+\text{IP Header}=96+20=108\text{bytes}$$

Therefore, we have a total of 6 fragments, and the size of the last fragment is 108 bytes.