Some fun with counting in logistics... and the square root law of distribution cost

Beyond the black box/ Motivation

When you think about optimizing your distribution network, mostly you are told to get some specialists of the number cruncher type with some 'bad a..' programs, collect lots of data, grind the figures and come out with some model that tells you which scenario is best.

It's a black box.

It may be needed after all but -I believe- only after you develop some solid understanding and intuition about the key drivers behind network design, notably, how the number of warehouses is driving your distribution cost.

In this short post, I will use counting and a bit of algebra to develop some closed, exact formulas that analytically describe the economies of scale in transportation/ distribution costs. I will use that in later posts to look at network design and optimization.

What drives transportation cost?/ Approach

When you increase the number of distribution centers (DCs)/ warehouses, what you really do, in addition to adding warehousing costs (*), is replacing some 'last mile' delivery aka 'milk run' costs with transportation cost to more warehouses; you reduce the transportation time and cost to reach your milk run distribution areas ('milk run feed') while increasing some transportation time and cost to reach the DCs ('DC feed' or replenishment).

So, let's look at how the distance/ time/ cost of 'feeding' or replenishing the DCs changes with the number of DCs. The same logic can then be applied to the milk runs feeds and a combined, comprehensive view on transportation cost. Throughout, we will assume that for 'feeding' we talk about point-to-point transportation; put differently, a DC will be fed from the regional DC with a point-to-point connection going to that specific DC and returning to the regional DC (without touching other DCs).

One Love/ Case of 'line land'

Let's start with a simple example: Chile (a long elongated distribution area) or 'line land' consisting of all DCs along a line (or on one single street).

If your line is very short (n=1), that is where you have your regional DC -which happens to be the DC- and the length of the DC feed is 0 (as the feed to the regional DC is out of scope).

If your line is 3 units long, then you put the regional DC in the middle and have to feed 2 more DCs at distance 1 each.

Looking at a small table for a few more cases (limiting to odd numbers just simplifies)

... it is easy to see that the total length of the one-way feeder path -point to point from central DC to all DCs- is (N2-1)/4, where the '^2' in the table is a fancy way of writing square or power of 2 or '2'. Even numbers of DCs will require a small correction to the formula which we will ignore.

If we include the return to the regional or central DC (factor 2) and add the units (feeder path was calculated in units of L/N), we find that “in line land of length L, we need transportation of the length L (N-1/N)/2 ≈ L N/2 (N>>1) to feed and replenish N DCs/ warehouses”.

Applying the same approach to M milk runs over the N DCs, we have M/N (assumed >> 1) milk runs per DC and a DC 'area' of length L/N; we find “in line land of length L with 1 out of N DCs, we need transportation of the length L/N (M/N-N/M)/2 ≈ L M/ (2 N2) to feed the M/N distribution milk runs for 1 DC”

Multiplying by N, we find the result for all N DCs “in line land of length L with N DCs, we need transportation of the length L (M/N-N/M)/2 ≈ L M/ (2 N) to feed the M distribution milk runs”.

Wow, so, while the transportation cost (assuming to grow 1-1 with the length of the transport) of the DC feed/ replenishment increases with the number N of DCs, the transportation cost of the milk run feed decreases with 1/N. Put differently

Doubling the number of DCs, doubles the transportation cost to feed the DCs but the transportation cost to feed the milk runs reduces by a factor of 2... in line land.

Song 2/ Case of 'square land' and the 'square root law of distribution cost'

We continue with a more realistic case: Manhatten or square land, where all roads are at a right angle to each other.

Again, if your area is very small (n=1), that is where you have your regional DC -which happens to be the DC- and the length of the DC feed is 0.

If your area is 3 blocks long, then you put the regional DC in the middle, you have to feed 4 more DCs at distance 1 each and then you have 4 more DCs at distance 2 each.

Looking at a small table for a few more cases (again, limiting to odd numbers just simplifies things)

... it is easy to see that the total length of the one-way feeder path -point to point from central DC to all DCs- is 1/2 sqrt(N) (N-1), where the 'sqrt' means the square root function. Again, allowing other numbers of DCs (that are not the square of an odd number) will just require small corrections which we will ignore.

If we include the return to the regional or central DC (factor 2) and add the units (feeder path was calculated in units of sqrt(A/N)), we find that “in square land of area A, we need transportation of the length sqrt(A) (N-1) to feed N DCs/ warehouses”.

Applying the same approach to M milk runs over the N DCs, we have M/N (assumed >> 1) milk runs per DC and a DC area of A/N; we find “in square land of area A with 1 out of N DCs, we need transportation of the length sqrt(A/N) (M/N-1) to feed the M/N distribution milk runs for 1 DC”.

Multiplying by N, we find the result for all N DCs “in square land of area A with N DCs, we need transportation of the length sqrt(A/N) (M-N) ≈ sqrt(A/N) M to feed the M distribution milk runs”.

Wow, so, while the transportation cost (assuming to grow 1-1 with the length of the transport) of the DC feed/ replenishment increases with the number N of DCs, the transportation cost of the milk run feed decreases with the factor sqrt(1/N). Put differently

Doubling the number of distribution centers (DCs), doubles the transportation cost to feed the DCs but the transportation cost to feed the milk runs reduces by a factor of 1/1.4142... = 71% in square land.

We call this the 'square root law of distribution cost'. It holds regardless of the specific choice of metrics (here: square land) as we will show in the next section (see also *3).

…Baby one more time / Case 3 'crow land'

Some may argue that notably for large distances, we should use a more realistic 'crow land' metric 'as the crow flies'.

Well, I'm not 100% sure it is more realistic but can be done; the respective table is

where for each step increasing the radius, we just cut the additional ring into 2n-1 partial ring segments of identical size/ angle (2 Pi/ (2n-1)).

If we include the return to the regional or central DC (factor 2), add the units (feeder path was calculated in units of sqrt(A/(N Pi))) and just focus on the leading term in N, we find that “in crow land of area A, we need transportation of the length sqrt(A/ Pi) 4/3 N ≈ sqrt(A) 0.752 N to feed N DCs/ warehouses”.

Etc.

The economies of scale or scaling view -how costs scale with the number of DCs- is identical to the one in square land:

Doubling the number of DCs, doubles the transportation cost to feed the DCs but the transportation cost to feed the milk runs reduces by a factor of 1/1.4142... = 71% in 'crow land'.

Have fun counting!

Notes and comments

(*) Yes, you are also increasing safety stocks, but I would argue that in most cases, the cost of transportation is more relevant.

(**) If you love playing with integer sequences, you may want to check out the respective entry in the online encyclopedia of integer sequences 0, 1, 5, 14... that you get when dividing the length of the feeder path by the common factor of 12

(*3) In a rhombus land (similar to square land but where with each additional step in radius just adds one layer) of area A, we need transportation of the length sqrt(A) (N-1) 2 sqrt(2)/3 ≈ sqrt(A) N 0.94 to feed N DCs.