Answer:
(Hope this helps can I pls have brainlist (crown)☺️)
Explanation:
The most crucial programme that runs on a computer is the operating system. It controls the memory and operations of the computer, as well as all of its software and hardware. It also allows you to communicate with the computer even if you don't understand its language.
An operating system is a piece of software that manages files, manages memory, manages processes, handles input and output, and controls peripheral devices like disc drives and printers, among other things.
This question is incomplete. The complete question is given below:
Your company has just brought a new 22-core processor, and you have been tasked with optimizing your software for this processor. You will run four applications on this system, but the resource requirements are not equal. Assume the system and application characteristics listed in table 1.1
Table 1.1 Four Applications
Application
A
B
C
D
% resources needed
41
27
18
14
% parallelizable
50
80
60
90
The percentage of resources of assuming they are all run in serial. Assume that when you parallelize a portion of the program by X, the speedup for that portion is X.
a. How much speedup would result from running application A on the entire 22-core processor, as compared to running it serially?
Answer:
Speedup = 50
Explanation:
- The speedup for that portion is x if we parallelize a portion of that program by X.
- If the whole program has no parallelize portion or in other words the whole program is running serially then the speedup will be zero.
- So in this scenario if parallelizable portion of A is 50% so according to above description the speedup is 50.
True, but the only way to get these constant place holders is of you program them in, in other words you have to tell the computer that you want them.
Answer:
The theoretical maximum capacity in bps for this link is 114.3 Gbps
Explanation:
The maximum capacity in bits per second can be calculated by Shannon's channel capacity formula:
<u>C = B*log₂(1 + SNR)</u>
Where B = bandwidth of the channel in Hz
SNR = Signal to Noise Ratio
We need to use SNR in linear scale not in dB scale so,
SNR = 10log₁₀ (20)
SNR = 13.01
Bandwidth = 60GHz - 30GHz = 30GHz
C = B*log₂(1 + SNR)
= (30*10⁹)*log₂ (1 + 13.01)
= 1.143*10¹¹ bps
C = 114.3 Gbps
