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Comment by greenbit | Hacker News Reader

Does this mean that probably somewhere below that tower, there were operator stations that would have allowed any of the 5500 lines to connect to any of the other lines? How many simultaneous calls would that even be? Correct me if I borked this, but 5500!/(2^(5500/2)), perhaps?

It seems plausible that if the phone had only just been invented, you'd initially set up small systems that would in fact allow any line to connect to any line. That'd be fine for maybe even a few dozen lines. But as the image shows, that doesn't scale too well.

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The total number of connections would all but certainly be far greater than the maximum number of simultaneous calls which could be supported.

Under PSTN, public-switched telephone networks, a not-infrequent occurrence, especially when calling long-distance, was to get a message "all circuits are busy". When each call was literally a circuit, and the "switch" (the central telephone exchange) made and broke those circuits as calls began and ended, my understanding (not my area of expertise, but one of some interest) is that this meant that all available interchange connections were occupied. For long-distance, this was typically far lower than for local calls (most phone traffic is local), and for international calls, lower still. The first transatlantic telephone cable could support only 36 simultaneous calls, in 1956. Calls were short, expensive, and all but exclusively for business and government subscribers.

<https://hamhistory.org/first-transatlantic-telephone-cable/>

I'd expect that the Stockholm exchange probably supported a few hundred simultaneous calls, probably a few (a dozen or so perhaps) per operator, who had to physically connect each call.

Indeed, I'm old enough to remember "all circuits are busy."

You lived through covid too?

A lot of phone systems saturated themselves when things first shut down. Probably not physical copper constraints but the virtual interconnects between providers:

https://productioncommunity.publicmobile.ca/t5/Get-Support/A...

https://archive.ph/D9qQ4

Through the 1990s (and prior) this was not uncommon, particularly on high-traffic days, often holidays (in the US: Mother's Day, Thanksgiving, and Christmas), when calling long-distance. Also during natural disasters, when local calls might also be affected.

There were also systems outages, such as the 1988 Hinsdale switching station which took out phone service to most of the Chicago area (both local and long-distance):

"1988 PHONE CRISIS TIED TO 1 BROKEN POWER LINE" (1988)

Turns out that that switch was a SPOF:

To make sure such a crisis doesn`t happen again, Illinois Bell Telephone Co. announced Friday it is embarking on a $80-million, five-year program to construct a complete duplicate telephone network system throughout its Chicago suburban operation and to redesign its fire protection systems.

<https://www.chicagotribune.com/1989/03/11/1988-phone-crisis-...>

Even a large confluence of connections like this would likely still have had local switchboards.

When you made a call, your local operator would have connected you either to a local number on their own board, or to another local board as needed. That second operator would have then connected you to the desired number.

Each board would only have limited connection lines to each other board (or to a branch exchange). So if all the connections from board A to board B (or to the branch exchange) were in use, the caller would have to try again later.

If you have n nodes, each one needs to connect to n-1 other nodes. But the connection from A to B is the same as the one from B to A. Therefore, there are n(n-1)/2 total connections.

Sure, that's true for single connections, the state of the switchboard when you have exactly one patch plugged in. If there are 5000 phones, there would be 5000 possible starting points and 4999 possible ending points, and as you say, A-B and B-A being equivalent, there are two ways to get there, so divide by 2.

But what if you then connect a second pair of phones, leaving the first patch in place? 4998 choices for A by 4997 for B again divided by two. And so on, until you place the 2500th patch on the board. But then there's more division to do, because there are 2500! different sequences to fill the board with that exact same pattern. It's a fun little combinatorics exercise.

Well, on further consideration, I arrived at N!/((2^(N/2)*(N/2)!)

Still, pretty astronomical.

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