SHA256 1 round equivance

[Balthazar2012]

My question is the following:

I have calculated the average number of operations in 1 round of SHA256 as follows:

Additions: 9.25
Bitwise Rotations: 9
Bitwise Shifts: 1.5
Bitwise AND: 5
Bitwise EXOR: 10

I have certain proposals that save the following number of operations in the SHA256 algorithm for Bitcoin mining:

Additions: 24
Bitwise Rotations: 12
Bitwise Shifts: 6
Bitwise AND: 0
Bitwise EXOR: 12

My question is, How can I represent the above in terms of a SHA256 round?

Do I just consider the additions? So that the saved operations above are equivalent to 24/9.25 = 2.5946 SHA256 rounds?

Any help is greatly appreciated.