Funny that you mention that but the comment you're replying to prompted me to do some back of the envelope calculations.
Things to note:
- About half Google's staff were technical
- There are ~80k employees?
- So assume 40k engineers
- Most engineers are ICs
- ICs are given levels from T3 (college grad) to T9 (can walk on water) with there being like a handful of Google Fellows beyond that (your Jeff Deans).
- Managers are M1s through M3s with M1s being baby managers. Most are M2s (equivalent to T6) who are managers of ICs and M3s (T7s) who are managers of managers (and higher level ICs).
- Higher level ICs (T7+) may well directly report to M2s or even directors or VPs but there's so few of those (~1%?) that they don't really change the math.
- In terms of rolled up head count you're looking at 5-20 for an M1/M2, 20-50 for an M3, 50-200 for a director, 150-1000+ for a VP.
- Directors have levels, D1 (T8) and D2 (T9). Sometimes you end up with 2 or even 3 directors within a management chain.
- The archetypal management chain is: CEO -> SVP -> VP -> Director -> M3 -> M1/M2 -> IC
So my opinion is the structure should ideally look something like this:
- M1/M2 average 10 ICs
- M3s average 5 M1/M2s
- Directors average 5 M3s
- VPs average 4 directors
with the following constraints:
- There is only 1 VP in a chain
- There is only 1 SVP in a chain
- There is only 1 director in a chain
- There are at most 2 managers in a chain
So 40K ICs need:
- 4K M1/M2s
- 800 M3s
- 160 directors
- 40 VPs
Now this "ideal" scenario obviously ignores some realities like a management structure built around products and infrastructure may not fit these numbers so neatly but then again some M1s could have 12 ICs while others have 8 and it all sort of works out.
I don't know what Google's numbers are here but I suspect it is MUCH MUCH higher than this.
It seems to me that as a back of the envelope estimation might be:
Suppose we say that an organization has n employees and that everyone in a position of leadership leads m people. The result would be a m-ary tree whose height is roughly:
h = log_m ((m - 1) * n))
If we say that there are l levels of seniority before one becomes a VP, then anyone not in the bottom l levels of the tree is a VP.
h_vp = # of levels of VPs = h - l
The number of VPs would then equal the number of nodes in a complete m-ary tree of height h_vp:
n = (m^h_vp) / (m - 1)
(Forgive me if I made any mistakes in my math, it's been a long time since I took algebra.)
Most companies have line managers. Line managers are not directors, they are managers. There are also senior managers. I think your schedule is good but I think you should add at least 1 layer maybe two.
Yeah, I just ran out of employees to cover and I worked backwards, so I didn’t get to use the term manager. I guess the system is good until the 1.000.000 mark ;)
I think line manager is a generic term for the person you explicitly report to though (e.g. the one who approves your time off), not a role in specific.
You can have higher ratios at the CXO and SVP level, and need lower ratios below director because at the higher level most people dont need as much day to day oversight.
Certainly an anomaly, but there's a distribution to the number of ICs assigned to the lowest level managers (M1/M2s). I've seen anywhere from 3 to 30. The people with 30 basically just jumped from 1:1 to 1:1 on repeat biweekly.
It's interesting how closely these estimates correspond to military TOEs. It's almost like we've figured out over the course of thousands of years what the most effective general-purpose hierarchical organization is.
Problem is that there aren't enough people to every VP, so they fight for headcount and projects leading to frequent reorgs or projects shuttering to make room for VP's pet projects.
