Question

The MTU on 1 GbE can be set as high as 9000 bytes using so-called "jumbo frames" at the data link layer. However, many 1 GbE interfaces still use MTUS of 1500. Let f be the number of bytes that need to be sent at the data link layer in a UDP transfer of a very large file with N bytes using an MTU of 1500 bytes, and left F be the number of bytes that would need to be sent if the same transfer were to happen using an MTU of 9600 bytes. Enter the value of f/F for verly large N as a decimal number with exactly two digits after the decimal point (using rounding to the nearest value). You may assume that the other parameters stay the same between the two MTUS: UDP header size: 8 bytes IP header size: 20 bytes Ethernet header size: 18 bytes

Answer?

Answer 1

Let us assume that 'f' is the number of bytes to be sent at the data link layer in a UDP transfer of a large file of 'N' bytes wiht an MTU of 1500 bytes.

Let us assume N = 65535 bytes (This is the maximum size for an IP Data-gram packet)

After the UDP header and IP header removal the packet size will be = (65535 - 28) Bytes = 65507 Bytes

For standard frame, the MTU is 1500 bytes.

The Overhead is given only for Data Link layer or Layer 2 which is = Frame Header + FCS = (14 + 4) Bytes = 18 Bytes,

Total Transmitted Size = 1500 + 18 Bytes = 1518 bytes (Payload size + overhead size)

So, f = 65507/1518 = 43.15 bytes

Let us assume 'F' be the number of bytes to be sent if the same transfer happens using an MTU of 9600 bytes

So, N = 65507 bytes

For the Jumbo frame type, the MTU is 9600 bytes.

Total Transmitted Size = 9600+18 = 9618 bytes

So, F = 65507/9618 = 6.81 bytes

So, f/F = 43.15/6.81 = 6.34 bytes

Hope this helps.