Answer:
<em>The visuals were not engaging.</em>
Explanation:
The answer is computer networking.
Complete Question:
Determine the number of cache sets (S), tag bits (t), set index bits (s), and block offset bits (b) for a 4096-byte cache using 32-bit memory addresses, 8-byte cache blocks and a 8-way associative design. The cache has :
Cache size = 1024 bytes, sets t = 26.8, tag bits, s = 3.2, set index bit =2
Answer:
Check below for explanations
Explanation:
Cache size = 4096 bytes = 2¹² bytes
Memory address bit = 32
Block size = 8 bytes = 2³ bytes
Cache line = (cache size)/(Block size)
Cache line =
Cache line = 2⁹
Block offset = 3 (From 2³)
Tag = (Memory address bit - block offset - Cache line bit)
Tag = (32 - 3 - 9)
Tag = 20
Total number of sets = 2⁹ = 512
He should make sure he has everything
Answer:
RS-485 2 wires, half-duplex for serial data transmission
Explanation:
The option mentioned in the answer section is certainly the correct option. And we do not need 4 wire RS-485 here as the half-duplex mode is going to work fine here, as the screen is never going to talk to the server, and hence at a time, it is only required for the receiver to receive and transmitter to transmit. However, you cannot certainly opt for fewer number receivers, and the multidrop option is required. And with RS-422A we are limited to 10, and hence, the best option is RS-485.
The first reason is it's noiseless due to the use of a twisted pair of cables. And the second reason is, the number of receivers can be as high as 32.