Which Of The Following Is An Example Of An Increment - T-Tees - Your Answers Await, Life's Questions Unraveled

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I know that the guide wants to emphasize the correct release/increment/sprint relationship, but my issue with the definition of increment is different.

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The way I see it (or used to see it) is this: Let’s say we’re at the beginning of the 8th Sprint, and our latest Increment is this:

I7 = { a, b, c, d, e, f }

Which means that it contains elements a, b, c, d, e, and f. We start working on a new item called g. After a while, it’s fully done, and when added to I7, would turn it into a new done Increment according to the definition of done:

I8-1 = { a, b, c, d, e, f, g }

Three days later, h and i are completely done, at the same time; therefore:

I8-2 = { a, b, c, d, e, f, g, h, i }

And close to the end of the Sprint, j is finished:

I8-3 = { a, b, c, d, e, f, g, h, i, j }

The team was also working on k and l, but they couldn’t finish them by the end of the Sprint, and therefore, those go back to the product backlog and not the increment.

Now it’s time for the Sprint Review meeting. What is the increment we’d show to the stakeholders? If the above scenario is what the Scrum Guide means, then what we’d show them is the latest increment (I8-3). Of course, we can consider the sum of all increments (I8-3 + I8-2 + I8-1 + I7 + …), but that would be the same as I8-3, and therefore, there’s no need to say the sum of increments when we can simply say the latest increment. As a result, I have some doubts that maybe the scenario I’ve described above is not compatible with the new Scrum Guide and it has something in mind that won’t exist in the latest increment and we have to add up all the increments. Is it so? Can someone give me an example or scenario in which the latest increment is different from the sum of all increments?

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The only thing I can think of, is for developers to work on multiple branches during the sprint, and have increments like the following:

I8b-1 = { a, b, c, d, e, f, g }

I8b-2 = { a, b, c, d, e, f, h, i }

I8b-3 = { a, b, c, d, e, f, j }

Then, those branches are merged into one and create the following at the end of the sprint:

I8 = { a, b, c, d, e, f, g, h, i, j }

Which can be referred to as the sum of the previous increments. However, I’m not comfortable considering I8b-1, I8b-2, and I8b-3 increments, because as long as they are not merged into one, I can’t consider them fully done, as no one can be sure what’s going to happen when they are merged, and some extra work may be required.

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