Converting 2 bytes to Short in C#

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If you reverse the values in the BitConverter call, you should get the expected result:

int actualPort = BitConverter.ToUInt16(new byte[2] {(byte)port2 , (byte)port1 }, 0);

On a little-endian architecture, the low order byte needs to be second in the array. And as lasseespeholt points out in the comments, you would need to reverse the order on a big-endian architecture. That could be checked with the BitConverter.IsLittleEndian property. Or it might be a better solution overall to use IPAddress.HostToNetworkOrder (convert the value first and then call that method to put the bytes in the correct order regardless of the endianness).

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