Question The maximum packet size (Maximum Transmission Unit or MTU) of an IP packet including IPv4 header on 100 Mbps Ethernet is usually set at 1500 bytes. A typical IPv4 header consists of 20 bytes, and a UDP header consists of 8 bytes. If we split up a file of 20,000,000 bytes so we can send it as a series of UDP payloads, how many IP packets do we have to send in order to transfer the entire file? Enter an integer number without formatting (no commas). Answer: The MTU for IP packets on 100 Mbps Ethernet is typically set to 1500 bytes. A typical IPv4 header consists of 20 bytes, and a UDP header consists of 8 bytes. If we split up a file of 26,000,000 bytes so we can send it as a series of UDP payloads, how many bytes do we have to send at the network layer in order to transfer the entire file? Enter an integer number (no commas). Answer: You are sending a 28,000,000 byte file using UDP over IP over an Ethernet with MTU 1500 bytes. The Ethernet header is 14 bytes and the frame checksum is 4 bytes. How many bytes do we send at the data link layer? Answer:

1] Total size of file =25,000,000 bytes\begin{array}{l}\text { UDP header }=8 \text { bytes } \\\text { IPV4 header }=20 \text { bytes } \\\text { packet size }=1500 \text { bytes }\end{array}payload + uDp header should be multiple of 8 . Hence it is 1480 bytes.\Rightarrow 1480+20 (ip headen) =1500 bytes which is exactly the packet size.\begin{aligned}\text { Actual data in } 1480 \text { bytes } & =1480-8 \\= & 1472 \text { bytes pen packet } .\end{aligned}\therefore The total no. of if packets to be sent is\begin{array}{l}=\frac{25000000}{1472} \\=16984 \text { packets. }\end{array}A]\begin{array}{l}\text { frame size }=1500 \text { bytes } \\\begin{aligned}\text { frame header } & =14 \text { bytes } \\\text { payload data } & =1500-14 \\& =1486 \text { bytes }\end{aligned}\end{array}Numben of frames needed to transfen\begin{aligned}28000000 \text { bytes } & =\frac{28000000}{1486} \\& =18842 \text { frames }\end{aligned}18842 frames will contain maximum payroad. While the last frame will Contains only 14 bytes header and remaining payload data.Number of bytes of payload data carry\text { by last frame } \begin{aligned}& =28000000-18842 \times 1486 \\& =788 \text { bytes }\end{aligned}\begin{aligned}\text { size of last frame } & =788+\text { headen size } \\& =788+14 \\& =802 \text { bytes }\end{aligned}Total amount of data send oven data\begin{aligned}\text { hink layen } & =1500 \times 18842+802 \\& =28263802 \text { bytes. }\end{aligned}Given M T U=1500 bytes\begin{aligned}\text { IPvy header } & =20 \text { bytes } \\\text { UDP header } & =8 \text { bytes } \\\text { size of file } & =26000000 \\\text { NDP pay 10ad } & =1500-(20+8) \\& =1500-28 \\& =1472 \text { bytes. }\end{aligned}Total number of ip packeks required to send 26000000 bytes\begin{aligned}\text { of data } & =\frac{26000000}{1475} \\& =17627 \text { bytes. }\end{aligned} ...