Can a micro operation pass multiple pipeline stages with a single clock cycle?

Question

I'm learning a CPU architecture and currently found some high-level description of Intel x86_64 CPUs architecture. By the high level description I mean something like the following micro-op flow (might be wrong, this is how I currently see it):

1. Fetching code from memory
2. Caching the code in L1I cache
3. Activating Legacy Decode Pipeline
4. Put the decoded micro ops into the Decoded ICache and to the micro-ops queue (a.k.a IDQ)
5. Send micro-ops from the micro-op queue to RAT (a.k.a Renamer)
6. Allocate necessary resources and send micro-ops to the Reservation Station (a.k.a. Scheduler)
7. Dispatch micro-ops to the appropriate port
8. Write the result of the micro-ops to the writeback bus
9. Retire the micro-op

The question is if it is possible for micro-op to pass multiple pipeline stages for a single clock cycle?

For example:

1. A bunch of micro-ops are fetched from the Decoded ICache and put to the IDQ then the Renamer moves the micro-ops to the RS during the same clock cycle

2. Renamer moving a micro-op from IDQ to RS and then RS dispatches the micro-op to an issue port during the same clock cycle

3. When OOO core finishes execution of a micro-op its result is written to the writeback bus and then the micro-op is retired during the same clock-cycle.

Is it technically possible?

Answer 1

Yes and no.

You could make a processor core that might do the equivalent to your above sequence in one clock cycle, but it would happen at the cost of higher propagation delay, which would drive your clock rate down.

As a simplistic example, say that the propagation delay at each of your nine steps is the same. You've adjusted your clock so that each step is completed with some safe margin in propagation delay.

If you can keep your pipeline full all the time (which can't be done -- but let's say you can) then at each clock cycle an instruction starts and an instruction retires.

Now combine those pipeline steps in groups of three. Woo-hoo! Your pipeline is shorter! But your propagation delay has gone up -- so now you have to divide your clock rate by three, or nearly so.

So -- if you keep your pipeline full -- you issue an instruction each clock cycle, retire an instruction each clock cycle and your clock is three times slower than it was.

Which executes programs faster?