Post 9nVDRCaDbOd8SogF84 by angristan@mstdn.io
(DIR) More posts by angristan@mstdn.io
(DIR) Post #9nVDRCaDbOd8SogF84 by angristan@mstdn.io
2019-10-02T03:31:55Z
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1.a. 2 mbps / 1 mbps = 2 user at the same time, or 2 / 1 / 0,2 = 10 users average1.b. 0.2 (20%)1.c. 0.2^3 = 0.008correct?
(DIR) Post #9nVDT1g97DVxDvK4Ya by angristan@mstdn.io
2019-10-02T03:32:14Z
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that seems easy but I'm afraid of having missed something
(DIR) Post #9nVDeRnOLzzXPQ9r96 by loke@functional.cafe
2019-10-02T03:34:21Z
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@angristan With circuit switching, you can only have two users in this case. Circuit switching gives you a dedicated link that cannot be shared with other users.In practice, circuit switching isn't really used anymore though. All circuits you're getting is going to be packet switched under the hood, since dedicated circuits are quite wasteful.
(DIR) Post #9nVDiW80BiDC0JjsG0 by angristan@mstdn.io
2019-10-02T03:35:04Z
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@loke ok thank you 🙏
(DIR) Post #9nVETVIh5AcN3292qe by d599f84e@mastodon.xyz
2019-10-02T03:43:33Z
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@angristanI think for c) it's rather increasing than decreasing. The more users, the more probable that someone is transmitting.Does that make sense?
(DIR) Post #9nVFCssmFfutDXf58K by angristan@mstdn.io
2019-10-02T03:51:43Z
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@d599f84e but the more users, the less probable that *all* users are transmitting at the same time though?
(DIR) Post #9nVMMM5BkrtO47vjF2 by dot@mastodon.mg
2019-10-02T05:10:39Z
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@angristanWhy isn't the probability that 3 users are transmitting at the same time equal to 0?
(DIR) Post #9nVQJmZnCBSBBLH1SC by angristan@mstdn.io
2019-10-02T05:56:15Z
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@dot oh