Why no swapping block in the last round of DES? Round 16 on the encryption side, and round 16 on the decryption side of DES have no swapping module.

Why is it like this? Can anyone provide any justification?
This is my homework question.
In &quot;Applied Cryptography&quot; — Chapter 12, , Bruce Schneier writes:

The final permutation is the inverse of the initial permutation and is described in Table 12.8. Note that the left and right halves are not exchanged after the last round of DES; instead the concatenated block $R$16$L$16 is used as the input to the final permutation. There’s nothing going on here; exchanging the halves and shifting around the permutation would yield exactly the same result.
This is so that the algorithm can be used to both encrypt and decrypt.