Scaling the flat organization
Posted on Sun 16 January 2022 in blog • 13 min read
There’s a common trope in management that goes something like “in order to better scale the organization as we grow, we need to keep it flat.” The thrust of the argument is that as the organization grows to meet customer and demand growth (and with it, growth in head count), additional levels of corporate hierarchy stifle that growth, and should thus be avoided.
For any knowledgedriven organization this is wrong, and just how wrong it is can be proven, numerically, with simple high school level maths. And in this context, “knowledgedriven organization” encompasses any technology company, any software engineering outfit, any technology services provider — in short, any organization that makes its money off the brains of its people.
Let’s establish a few selfevident facts about knowledgedriven organizations:
 The people who actually “make things happen” are the ones with no direct reports. The frontend designers, the infrastructure engineers, the backend specialists, the data analysts. Their managers (and their managers, and everyone else all the way up to the CEO) are charged with aggregating information, making decisions, removing obstacles to productivity, and perhaps providing some form of vision and guidance. But it’s individual, nonmanagerial contributors of all specializations that actually do things.
 In doing so, engineers work best in small teams with a great degree of autonomy. They will usually benefit from close working relationships with a small group of people.
 A manager’s role is thus twofold: remove any obstacles that stand in the way of the team accomplishing its goals, and act as an interface to other parts of the organization.
An example
With that in mind, let’s consider a hypothetical small company that is currently structured in teams of 5. There’s always 4 people reporting to one manager. Currently, that company is made up as follows:
 the CEO/founder, Alex (1 person),
 4 team leads (4 people),
 4 employees on each team, all of whom report to the respective team lead (in total, 16 people).
So, 21 people in all.
Management theory calls the number of reports per manager in an organization the span of control. I don’t like that term a great deal. For one thing, at four syllables it’s a bit of a mouthful, particularly if it needs to mentioned frequently. But more importantly, it’s not an accurate reflection of reality: in a knowledgedriven organization (like any technology or engineering company), it’s ludicrous to think that a manager “controls” their reports like puppets or robots. So, I’ll use a different term for the remainder of this article: I’m going to call the number of reports per manager the width of the company.
Also, I’ll use the term depth for the number of hierarchy levels that the company has. A sole proprietorship has a depth of zero. A company with a founderCEO and a few employees, but no other managers, has a depth of 1. Alex’ company, with one level of management reporting to Alex, and everyone else reporting to one of those managers, currently has a depth of 2.
So we can say that Alex’ company is currently narrow and shallow — it has small teams, and few management levels.
Now, the company has just closed a major funding round and several big customer deals, putting them on a solid growth trajectory. So, Alex expects the company to double in headcount on an annual basis for the foreseeable future.
So the question is: is it better for the organization to stick to the current width, and add depth as it grows, or should Alex increase its width, so that it can accomodate more people while retaining a shallower depth?
In other words, as the company scales, should it become deeper while staying narrow, or should it grow wider while staying shallow?
Fastforward five years
To look at that, Alex mentally fastforwards five years under the currently assumed growth model. After five years of doubling in headcount, the company now has $21 \cdot 2^5 = 672$ employees.
In this scenario, everyone in the company works still works in a 5person team, out of which one person is the leader. So every leader has 4 people that report to them. Let’s look at one employee, Sam. Sam works in a team with Joe, Jane, Harry and Ruth, and Ruth is the team lead. Let’s say her title is simply, “Manager”.
Ruth now has at most 3 peers of her own, and reports to someone who goes by “Senior Manager,” putting her in another team of no more than 5 at her management level. That Senior Manager has at most 3 peers again, all of whom report to a Director. A Director also, together with a maximum of 3 other Directors, reports to a VP, and the 4 VPs work together under Alex, who is still the CEO.
Now, I’ll tell you that for 672 people, you’ll not nearly have filled all those 5person teams. But try to intuitively guess, without doing the math, what organizational size this structure would accommodate. That is to say, with every person in the company being at most 5 hops away from the CEO, and everyone working in a group of 5, what’s the maximum company size this model can handle?
The answer is 1,365.
Let’s quickly break that down and see how we can plug other numbers in.
A gentle bit of maths, part I: team size and hierarchy levels
Say we take company’s width, that is the number of people working together in any group, excluding the leader, as $x$. In our example, that’s $4$.
Then, any team’s size (which we’ll call $n$, for reasons we’ll get to in a jiffy) is of course $x+1=5$.
The number of people any Senior Manager is reponsible for is $(x+1)x + 1 = x^2+x+1 = 21$ (that is, their Manager’s teams, and themselves).
The number of people any Director is responsible for is $((x+1)x+1)x+1 = x^3+x^2+x+1 = 85$.
You see where this is going. For any additional level of depth, we simply need to add another power of $x$.
And of course $1 = x^0$ — at the zeroth depth level there’s one single person: the CEO.
So we can express the number of people in an organization with a width of $x$ and a depth of $y$ as $$x^0 + x^1 + x^2 + … + x^y$$
or, more briefly:^{1} $$\sum_{i=0}^{y} x^{i}$$
And that, in turn, happens to work out to^{2} $${x^{y+1}1} \over {x1}$$
Plug in the numbers for $x=4$ and $y=5$, and we get 1,365.
A gentle bit of maths, part II: communications in complete graphs
Now, what’s our scaling constraint in a knowledge organization? The number of people you need to constantly be in touch with in order to accomplish your goals.
For Sam, those people are principally your Sam’s teammates team colleagues, including their manager, Ruth. That’s 4 people. However, it’s not enough for Sam to understand what he is exchanging with Jane, Joe, Harry, and Ruth; it’s also imperative for him to understand what they communicate about. So, Sam needs to keep himself appraisedof what Ruth told Harry, or what information Jane gave to Joe, and how Joe and Harry are coordinating their latest change (etc.).
That means that within a team, communications are a complete graph. And for a complete graph, the number of edges is given by $${n(n1)}\over 2$$
In our case, $n$ is our team size (including the leader), thus $x+1$ (the reports plus the leader).
So we can rewrite the completegraph formula as: $${{(x+1)(x+11)} \over 2} = {{x(x+1)} \over 2}$$
So in order for the team to be well informed of everyone’s actions at all times, a 5person team must keep track of 10 communications links between people. That’s absolutely doable, though we must keep in mind that the number of links does not grow linearly with the number of people in direct communications which each other, but it grows proportionally to the square of that number.
Sam’s manager Ruth, of course, works on two 5person teams: Sam’s, and Ruth’s team of fellow Managers reporting to a Senior Manager. That means Ruth needs to constantly keep in touch with the people on her team (including Sam), and also understand what everyone on her team of Managers is doing. Thus, she keeps track of 20 communications links. This is also true for her Senior Manager, that Senior Manager’s Director, and that Director’s VP. It’s only at the very top that the CEO has the luxury of directly managing only 4 VPs.^{3}
This should be flatter! Or should it?
Now, suppose someone tells Alex that in this growth plan the organization is much too hierarchical, and the organization must thus lose some of its projected hierarchy levels — that is, reduce its depth. Of course, the only way to do that while still being able to manage the same headcount growth is to make the company wider — in other words, have more people report to one manager than previously planned.
So Alex, being a good CEO, opens some spread sheet software and creates this handy table that simply plugs in values for $x$ and $y$, with $x$ (width) in columns and $y$ (depth) in rows.^{4}
2  3  4  5  6  7  8  9  10  11  

1  3  4  5  6  7  8  9  10  11  12 
2  7  13  21  31  43  57  73  91  111  133 
3  15  40  85  156  259  400  585  820  1,111  1,464 
4  31  121  341  781  1,555  2,801  4,681  7,381  11,111  16,105 
5  63  364  1,365  3,906  9,331  19,608  37,449  66,430  111,111  177,156 
6  127  1,093  5,461  19,531  55,987  137,257  299,593  597,871  1,111,111  1,948,717 
7  255  3,280  21,845  97,656  335,923  960,800  2,396,745  5,380,840  11,111,111  21,435,888 
8  511  9,841  87,381  488,281  2,015,539  6,725,601  19,173,961  48,427,561  111,111,111  235,794,769 
9  1,023  29,524  349,525  2,441,406  12,093,235  47,079,208  153,391,689  435,848,050  1,111,111,111  2,593,742,460 
10  2,047  88,573  1,398,101  12,207,031  72,559,411  329,554,457  1,227,133,513  3,922,632,451  11,111,111,111  28,531,167,061 
For our previous fiveyear plan, Alex can just look up the cell matching $x=4$, $y=5$ and finds our known outcome, a maximum head count of 1,365.
Now, Alex looks at what it takes to flatten the organization by eliminating one hierarchy level, or by two.

If we want to reduce depth by 1, we simply go up one row (thus, $y=4$) and find the value for $x$ that just accommodates 1,365 people or more. Alex sees that that’s $x=6$, which can accommodate 1,555 people. That is, increase the width by 2: reorganize from teams of 5 to teams of 7. Alex could also pick $x=5$, that is increase the width by only 1, which would land the company at a maximum head count of 781. That is well below what $x=4$ can handle, but it still lands Alex north of the original growth target of 672.

If we want to reduce depth by 2, we go up two rows ($y=3$) and do the same. We end up at $x=11$, which means to increase width by 7: reorganize from teams of 5 to teams of 12. Thus, we land at a maximum of 1,464 people, slightly exceeding the headcount we’re able to accommodate if we keep growing with the current structure. We could also do $x=10$ or $x=9$, landing us at maxima well below that (1,111 or 820), but still north of 672.
Now what does that mean in terms of communication channels each person has to maintain?
Again, what we want to keep in mind is the number of edges in a complete graph connecting $n$ (that is, $x+1$) points. For regular employees, we know that that’s $${x(x+1)}\over 2$$
And for any manager, who is effectively on two teams of size $x+1$ simultaneously, that’s $$2 \cdot {{x(x+1)}\over 2} = x(x+1)$$
Which means:

If we want to reduce depth by 1 and go from $x=4$ to $x=5$, every nonmanager employee now needs to be aware of 15 communications links (instead of 10), every manager, of 30 (instead of 20).

If instead we go from $x=4$ to $x=6$, every nonmanager employee now needs to be aware of 21 communications links, every manager, of 42.
So that’s a least a 50% increase, or even a doubling, of communications complexity.

For the elimination of two hierarchy levels (a depth reduction by 2), we’ll need to move from $x=4$ to at least $x=8$. At that point, every regular employee has at least 36 communications links on their teams to deal with; every manager deals with 72.

If instead we go to $x=9$, every nonmanager employee now needs to be aware of 45 communications links, every manager, of 90.

And for $x=10$, every nonmanager employee now needs to be aware of 55 communications links, every manager, of 110.
At this point Alex realizes that making the company wide and shallow, instead of narrow and deep, is painfully expensive in communication cost.
But what about all those managers we won’t have to pay?
A wellmeaning advisor interrupts Alex in the middle of planning. He interjects that Alex is missing a point, namely all the managers that the company will now no longer need, and the cost savings thus generated.
So Alex looks at the table again (width in columns, depth in rows):
2  3  4  5  6  7  8  9  10  11  

1  3  4  5  6  7  8  9  10  11  12 
2  7  13  21  31  43  57  73  91  111  133 
3  15  40  85  156  259  400  585  820  1,111  1,464 
4  31  121  341  781  1,555  2,801  4,681  7,381  11,111  16,105 
5  63  364  1,365  3,906  9,331  19,608  37,449  66,430  111,111  177,156 
6  127  1,093  5,461  19,531  55,987  137,257  299,593  597,871  1,111,111  1,948,717 
7  255  3,280  21,845  97,656  335,923  960,800  2,396,745  5,380,840  11,111,111  21,435,888 
8  511  9,841  87,381  488,281  2,015,539  6,725,601  19,173,961  48,427,561  111,111,111  235,794,769 
9  1,023  29,524  349,525  2,441,406  12,093,235  47,079,208  153,391,689  435,848,050  1,111,111,111  2,593,742,460 
10  2,047  88,573  1,398,101  12,207,031  72,559,411  329,554,457  1,227,133,513  3,922,632,451  11,111,111,111  28,531,167,061 
What’s handy here is that Alex can look at any one table cell, and the cell directly above it will contain the total number of managers (that is, people who have direct reports) for the same width. So,

for $x=4$, $y=5$ (our original scenario allowing the company to grow to 1,365 people), Alex would have to hire and pay a total of 341 managers.

for $x=6$, $y=4$ (the scenario that eliminates one level, and can accommodate 1,555 people), Alex’ company will need 259 managers. That’s 82 fewer managers, or a reduction by about 24%.

for $x=5$, $y=4$ (the scenario that eliminates one level, but accommodates only 781 people), Alex’ company will need 156 managers. That’s 185 fewer managers, or a reduction by about 54%.

for $x=11$, $y=3$ (the scenario that eliminates two levels, and can accommodate 1,464 people), the company will need 133 managers. That’s 208 fewer managers, or a reduction by about 61%.

for $x=10$, $y=3$ (the scenario that eliminates two levels, but accommodates only 1,111 people), the company will need 111 managers. That’s 230 fewer managers, or a reduction by about 67%.

for $x=9$, $y=3$ (the scenario that eliminates two levels, but accommodates only 820 people), the company will need 91 managers. That’s 250 fewer managers, or a reduction by about 73%.
A gentle bit of maths, part III: how much of our company will be managers?
It so happens that we can generalize this. If Alex looks at our table again, but considers the number of managers proportional to the number of people in the company, a pattern quickly emerges (again, width is in columns, depth is in rows):
2  3  4  5  6  7  8  9  10  11  

1  33.33%  25.00%  20.00%  16.67%  14.29%  12.50%  11.11%  10.00%  9.09%  8.33% 
2  42.86%  30.77%  23.81%  19.35%  16.28%  14.04%  12.33%  10.99%  9.91%  9.02% 
3  46.67%  32.50%  24.71%  19.87%  16.60%  14.25%  12.48%  11.10%  9.99%  9.08% 
4  48.39%  33.06%  24.93%  19.97%  16.66%  14.28%  12.50%  11.11%  10.00%  9.09% 
5  49.21%  33.24%  24.98%  19.99%  16.66%  14.28%  12.50%  11.11%  10.00%  9.09% 
6  49.61%  33.30%  25.00%  20.00%  16.67%  14.29%  12.50%  11.11%  10.00%  9.09% 
7  49.80%  33.32%  25.00%  20.00%  16.67%  14.29%  12.50%  11.11%  10.00%  9.09% 
8  49.90%  33.33%  25.00%  20.00%  16.67%  14.29%  12.50%  11.11%  10.00%  9.09% 
9  49.95%  33.33%  25.00%  20.00%  16.67%  14.29%  12.50%  11.11%  10.00%  9.09% 
10  49.98%  33.33%  25.00%  20.00%  16.67%  14.29%  12.50%  11.11%  10.00%  9.09% 
You’ll see that at a depth of 1, the share of managers is obviously $1 \over {x+1}$, but then as we increase in depth it quickly trends toward:^{5} $$1 \over x$$
The number of managers in Alex’ company is roughly the reciprocal of the company’s width. In other words, the number of managers is inversely proportional to width.
In contrast, the cost of communications is directly proportional to the square of the width.
At this point Alex realizes that while there are indeed savings to be made by the elimination of management in a wideandshallow company, they cannot possibly balance the added communication cost.
In other words: the cost in communications inefficiency grows much faster with width, so much so that it will eat up Alex’ company’s manager payroll savings several times over.
In summary
The “flat” (wide) organization scales poorly. Its growth in communication cost far outpaces its savings in payroll cost. And it scales progressively worse, the “flatter” (wider) it gets.

In this capitalsigma summation formula, $i$ doesn’t mean anything other than it being a counter. The formula is pronounced, in English, as “sum of $x$ to the $i$, from $i$ equals zero to $y$” (in other words, add up all wholenumber powers of $x$, from $x^0$ to $x^y$). ↩

You might notice that this expression is indeterminate for $x = 1$. Now I’d say the idea of a hierarchical company made up of oneonone teams (every manager has one report, who in turn is the manager of one report, and so on) is extremely unrealistic. But just for completeness’ sake, we can apply a limit to show that $$\lim_{x \to 1} {{x^{y+1}1} \over {x1}} = {y + 1}$$ In other words, such an organization could accommodate a number of people that is equal to its depth plus 1. ↩

This why they might also be able to appoint a CFO, CSO, CTO or whatever other Csuite functions are appropriate for the organization. So in the scenario we might end up with a handful more people than 1,365 for the Csuite and perhaps some number of staff in their offices. But for the purposes of this discussion those don’t make a big difference, so we’ll disregard them for now. ↩

I encourage you to compare the bottom rows and rightmost columns of this table to Wikipedia’s list of largest employers. ↩

If you’re curious, that is because the share of managers in relation to the total number of people in the company is $${\sum_{i=0}^{y1} x^{i}} \over {\sum_{i=0}^{y} x^{i}}$$ That works out to be $${x^y1} \over {x^{y+1}1}$$ Which, for $y=1$, is $${{x1} \over {x^21}} = {{x1} \over {(x+1)\cdot(x1)}} = {1 \over {x+1}}$$ And for larger values of $y$, both $x^y$ and $x^{y+1}$ become so large that the $1$ part barely matters, so it’s effectively: $${{x^y1} \over {x^{y+1}1}} \approx {x^y \over x^{y+1}} = {1 \over x}$$ In slightly more formal terms, we can consider $1 \over x$ the limit of the expression as $y$ goes to infinity: $$\lim_{y \to \infty} {{x^y1} \over {x^{y+1}1}} = {1 \over x}$$ ↩