3 ms·
Surely if a connection erodes it is either still connected and working or not connected and not working? It wouldn't get slower.
by jemmyw 3y ago
Surely if a connection erodes it is either still connected and working or not connected and not working? It wouldn't get slower.
- ozr 3y agoThe process of recognizing that a subset of a component has failed, recovering from the failure, and resuming would result in a slow-down. I'm not sure that's a full explanation for the experience though.
- PaulHoule 3y agoFor something like an Ethernet connection a cable can certainly be half bad, good enough that you can transfer some data but with speed and reliability problems particularly involving weirder protocols that involve broadcast on the LAN and such.
- btgeekboy 3y agoThe connection itself wouldn't get slower, but the outer mechanism using that connection may retry, and the retry may be successful. The appearance to the user is one of slowness. For example, one of the symptoms of networking packet is slowness. Assume 2% of packets are lost. A packet is lost, the receiver re-requests it, and the retry has a 98% chance of being successful. End result: the transmission completes, but at a slower pace.
- whoopdedo 3y agoSystems are built to be fault tolerant because hardware errors are inevitable. Bit-flips in RAM and disks, line noise in transmission circuits, voltage fluctuations, etc. And these problems are statistical. In a new machine the chance of error may be 1%, then as it ages the error rate increases to 10%. Repeat something a billion times and those percentages mean you have a million errors. Not only in time but density; a chip with a thousand transistors is going to have fewer problems than one with 10 million. That's why a much older system can be faster than an old but more recent one. But also because the older system will be less fault tolerant. Back then the cost of handling a bit-flip or low-voltage wasn't higher so they ignored the errors and continued to run at full speed as it corrupted your data.