[gambiting]

Like I have posted elsewhere already: 5120x2880 = 14745600 pixels

14745600pixels x 3 colours, 1 byte each = 44236800 bytes

44236800/1024/1024 = 42.19MB each frame

42.19MB per frame x 60 frames a second = 2531.25MB.

TB2 has a bandwidth of 20Gbps = 2560MB/s

So yeah, theoretically TB2 has enough bandwidth. But it's a very tight fit.

[milen]

You're ignoring signalling and overhead. The actual usable bandwidth is about 17.3Gbps, thus not enough (for 60fps uncompressed stream).

DisplayPort v1.3 can deliver 25.92Gbps of usable bandwidth, thus required for 5k [1].

[1] http://www.extremetech.com/electronics/190130-displayport-1-...

[hamstergene]

You should divide 20Gbps by 9 bits (2275 MB/s) or even by 10 (2048 MB/s), not by 8, because in addition to useful payload, there are also packet headers and control packets being transferred. So it actually doesn't.

[mrb]

"Packet headers and control packets" are not the reason you should divide by 10. The reason is that Thunderbolt uses 8b-10b encoding so 1 byte of data is transferred as 10 physical bits. Therefore the maximum theoretical usable bandwidth is 20e9/10/1024/1024 = 1907 MiB/sec. Then on top of that you have to deduct the overhead from packet headers and control packets so the real-world usable bandwidth is even less than 1970 MiB/sec...

[aortega]

I believe 2560 is the post-coding bandwidth. Links are usually 3.125 Gbps before coding.

[jsheard]

Doesn't TB2 have two interleaved streams, one carrying DP1.2 and the other carrying 20Gbps PCI-E? Even if the PCI-E stream has enough bandwidth for 5K in theory, the ecosystem isn't in place to use it as a display output. That's what the DP part is for.

[richbradshaw]

For other people to play with this, just google "(5120 * 2880 * 3 * 60) bytes/s in Gib/s" to get the answer (just short of TB2s 20GiB/s assuming no overheads.