# How is wavelength assignment in Wavelength Division Muxing Rings determined? [closed]

I have this image that supposedly documents how to handle a 3-node wavelength division muxing ring:

I know an n-node ring will need wavelengths (n * (n-1) / 2).

But I have no idea how which wavelengths are received and which are transmitted is determined. It seems almost random from this picture. The only thing that seems to make more sense is the table, but I have no idea how it was derived.

Is this picture correct, or is there something I'm missing?

• The diagram looks fine to me, the wavelengths allow any node to reach any other node with the least waste of colours - Is there a specific part that you don't find clear?
– iTom
Dec 3 '17 at 20:06
• I suppose I'm just not seeing a pattern. I don't see how those wavelengths were assigned, or how you would expand this to a 4 (or larger) ring. Why does T1 transmit wavelength 1 and 3? Why does T2 transmit 2 and 3? Why does T3 transmit 1 and 2? In a 4-node (6 wavelength) ring, would there still only be 1 wavelength between the receiver and transmitter? And how is that wavelength chosen? Dec 3 '17 at 20:11
• "I know an n-node ring will need wavelengths (n * (n-1) / 2)." That is a full mesh. It is the same thing with three nodes, but a ring with four nodes will not need connections between nodes 1 and 3, and nodes 2 and 4. A full mesh requires connections between every node, while a ring only requires connections between a node and its two neighbors (upstream and downstream). Dec 3 '17 at 21:36
• @RonMaupin Apparently, they're building a mesh of links with an optical ring.
– Zac67
Dec 3 '17 at 21:43
• @Zac67, possibly, but it may be that the question is confusing with three nodes since a ring and a full mesh are the same thing with three nodes. They become different at four nodes. Dec 3 '17 at 21:48

ACTUAL SOLUTION:

I marked this post solved a little early. After some help from outside SE, I figured out how to go about this. You need to think of the problem like this. You need 1 wavelength to talk between any 2 nodes, back and forth. So A->B requires its own wavelength, which can also be used for B->A

So for every possible pair of nodes, you need 1 wavelength. Thus, n(n-1)/2 wavelengths. If you start out with a table, this can help a lot.

You don't need a wavelength for a node to talk to iteself. So let's start with allowing node 1 to talk to node 2, using the first wavelength, wavelength #1.

``````  |R1|R2|R3|R4|
-------------
T1|  |#1|  |  |
T2|#1|  |  |  |
T3|  |  |  |  |
T4|  |  |  |  |
``````

Now we need a wavelength to talk from node 1 to node 3:

``````  |R1|R2|R3|R4|
-------------
T1|  |#1|#2|  |
T2|#1|  |  |  |
T3|#2|  |  |  |
T4|  |  |  |  |
``````

And so on until we complete the table:

``````  |R1|R2|R3|R4|
-------------
T1|  |#1|#2|#3|
T2|#1|  |#4|#5|
T3|#2|#4|  |#6|
T4|#3|#5|#6|  |
``````

And with that table filled in, the connections are easy. Transmitter 1 needs to output #1-3. Transmitter 2 needs to output #1, #4, #5.

Just thinking of each wavelength as a way to communicate between any given pair of nodes was a massive help.

As it seems, each node splits off two λs from the fiber loop, each originating from one of the other nodes. Then it adds its data on (the same) two λs to the remains on the fiber and passes on the loop.

Each λ addresses one of the other nodes as per the table. Follow each λ from the transmitter to the receiver and you'll get the picture:

• N1 & N2 communicate using λ3
• N1 & N3 use λ1
• N2 & N3 use λ2

EDIT: With four nodes you'll need three more colors:

• N1 & N2 use λ1
• N1 & N3 use λ2
• N1 & N4 use λ3
• N2 & N3 use λ4
• N2 & N4 use λ5
• N3 & N4 use λ6

Each nodes terminates the three colors used by its links and lets the other three pass by.

In a nutshell, you need sum(1..(n-1)) colors, with Gauß's sum formula that's n(n-1)/2.

• Thanks. I believe that's what I was missing. Once I traced the lines from the receiver to the transmitter, and noticed that they increment and the transmitters and recievers move everything but that wavelength. However, it does get slightly confusing with 4 nodes. Would it look like this? imgur.com/a/Y846R There's only 4 wavelengths used between the nodes themselves, which makes it a little strange. Not sure if that's the intention. Dec 3 '17 at 20:24
• With four nodes, each node would receive and transmit 3 λs - not sure about that n(n-1)/2 formula at all...
– Zac67
Dec 3 '17 at 20:28
• The formula is something I was given, and I ended up running into it on some papers on the subject as well. So would I split the 6 wavelengths and have 3 moving from the receiver to the transmitter? Dec 3 '17 at 20:33
• I'm starting to see how n(n-1)/2 wavelengths doesn't make sense. However, it's what I was given. I can't not use it. (Even if it's possibly wrong) Though I don't see any way to squeeze 6 wavelengths into 4 nodes. My current idea imgur.com/a/Y846R doesn't make much sense. And I don't see any nCr way of getting 4 combinations from 6 items. Ugh. Dec 3 '17 at 20:43
• I'm just going to assume that the n(n-1)/2 from my professor must be wrong. I'm going with this solution. imgur.com/a/pdH94 Thanks for the assistance. Dec 3 '17 at 20:51