Arman's stuff
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Coming to the End of- oh wait, hold on, one more thing

(Sun Jul 31 14:52:52 2011)

And as long as you're at it, would you mind...?

Just as a note: www.wolframalpha.com is very useful in determining what these graphs look like, if you can't visualize them in your head. Just replace m and b with actual values - say, 1 and 2.

When a job comes to an end, it follows a function, a graph if you will. Depending on various inputs, it can resemble many things. In the best of cases, the graph will be a simple linear function:
y = -m*x + b
Where y is the amount of work left to do, m is the rate at which you complete that work, and b is the amount you started with. Hopefully, you time it so that y=0 (there is no work left) by the time x = time to leave. Or a little before; calling it a day half way through because you've finished is a good thing (for you at least).

On the other hand, it may resemble the following instead:
y = m*x + b
In this case, the line on the graph is rising, rather than falling. Instead of crossing zero, it continues to rise, so that when you leave, there is a huge pile of work for the guy after you. Not so good for him, but you can rest assured that it's not your fault.

It may look like this:
y = b
This is where the number of things to do never changes. Probably because you are a bad worker, and never finish anything, or actually do any work at all. In this case, you should feel bad, probably because you've been fired.

Much better is the following case:
y = 0
In this case, you never had any work to do to begin with. In which case you probably should feel bad, because you were being paid to do nothing, which honestly is a pretty good gig. Bad for the company as a whole, maybe, but pretty good for you.

But those are linear; what about more complex graphs? Well, take a look at this:
y = m/x
While it starts big and goes down quickly, this graph doesn't linearly approach zero - in fact, it never really reaches zero at all. This is when the last task you have to complete stretches on forever; it doesn't change, but every time you think you're done, something else (slightly smaller) pops up in it, forcing you to go over it again. And again. And again.

This is even worse:
y = x^2 - 2m*x + m^2
At first, you think this is like the first graph - as time goes on, this drops towards zero. Good for you, good for the company. But suddenly, it curves away from zero into the infinite! The trick with this is to realize what's happening, and time it so you leave at the lowest point, before it starts to rise. Get out, while you still can!

But then there are the worst ones... the graphs that never die. Behold:
y = ABS(COS(x)) + COS(x)
You see the cosine of x, and think, sure, this is going to go up and down, up and down... no problem. Get out while it's going down! And you do. There is no work left; whew, you made it! But then... you get a call. Where did you leave those files? Do you remember what part of the program used the database? When you cleaned out your desk, did you find a paper with "important" written on it? Those keys you handed in... what were they for, again? You're in replay hell. You don't work there, but they still call you, asking for information, wondering if you'd mind bringing something in, even just trying to go to lunch with you. There is no way to get away from this, apart from ignoring their calls, moving, or both.

But that's all about the amount of work left; in some, you don't really care. [y=b] and [y=m*x + b] are going to leave work undone. However, sometimes you really need to finish something. Maybe it's because it's important to you. Maybe it's because it's important to your future employment. In those cases, leaving unfinished work is a bad idea, as it will inevitably come back to bite you. If your new job just so happens to be in the same building as your old job, well, "comes back to bite you" will happen much, much sooner than you could ever hope for. In those cases, you really, really want the amount of work left to be zero, or very close to zero.

Which brings me to my current situation.

My first jobs were like most first jobs; stocking shelves, watering plants, and packing boxes. Those jobs are easily in the [y = b] category. Before I started, there was a finite amount of work to do; after I left, there was the same finite amount of work to do. It never changed. Technically, the 'work growth rate' was x, and the 'work completion rate' was x (every day brought exactly as much work as I could do in a day), so it was more like [y = b + x - x]... but that's really the same thing. Leaving those jobs was nothing big; one day you're working, same as usual, and the next day you're done. Later jobs had other functions, but usually they ended up where I had either zero or near zero work left at the very end, and even if I did have more work, it wasn't something that the next drone couldn't do just as well.

This job is different. I care about how things turn out, and I also care about getting a good review from them for future employment. But, most importantly, I'm not actually leaving. I'm switching departments. My job is changing, but only in who my boss is - I'm still working with computers. Now, don't get me wrong; I enjoy working with computers. I will probably enjoy this job quite a bit. All except for the looming doom - I have unfinished work elsewhere in the company, and it will come back to haunt me. All the old baggage from the old job - things not completed, machines held together with a dirty hack, even grudges - is still there, waiting. The only real way to leave that behind is to move somewhere new. Instead of my equation stopping at x = time-to-leave, it keeps going, with bits of my old job leaking into my new job. Eh, who knows, maybe in a year or so I'll move back to where I was. It doesn't really matter. This isn't a new job, nor is it a promotion. It's what is known as a "lateral move" - you get more responsibility and more to do, but less bargaining power. You get a new job title, but only stack new things onto the old.

So how do you graph that?
y = m*x + F(x), where F(x) = x*y?



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