I came across a curious and somewhat contentious Apple related tid-bit this morning. I was reading a piece from Macworld where Jeff Battersby notes that he recently purchased a new iPod touch some three weeks after the new 2.0 software became available but it still had the old firmware, and despite numerous calls to Apple it seemed he would have to pay for it. This seems out of sorts with Apple policy in the past. For example when I bought my iMac shortly after Leopard was released I was told I could get Leopard for free if it wasn’t on the system He had come to the conclusion that it was old inventory in the store, but I did a little digging and it looks like he iPod touch is not being sold with the 2.0 software at all.
On the Apple Store page for the iPod touch there’s this little gem:

Note the phrase: “Software Upgrade Required: Sold Separately”
So it would seem that Apple intends this to remain as a for purchase item for the iPod touch. This, in my opinion is pushing things a bit too far. I know some people will claim that I have “entitlement issues” for daring to suggest that Apple shouldn’t charge for something, but deliberately shipping a device with an old firmware just so they can charge you to upgrade to the latest is something I doubt even Microsoft would try to pull. In fact I’ve never heard of any other company ever doing something like this.
Are they within their rights to do so? Probably.
Is it the right thing to do? Absolutely not.
Sure it’s only $10 but that ads up pretty quickly considering the number of iPod touches being sold. It doesn’t really make any sense though. Surely they would want as many people using the App store as possible as the potential for revenue there is a lot stronger. The only reason I can think of to justify this apart from “because we can” is that there is an iPod touch refresh around the corner, because otherwise it makes Apple look incredibly greedy.
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I think it’s because the iPhone 2.0 firmware is a download software upgrade only. It isn’t available on a disc like Leopard.