Performance Tab
The Performance tab allows you to see the current and historical CPU and memory utilization on your computer. This tab is useful if you are trying to diagnose why your computer or an application is running slower than normal. The CPU Usage and CPU Usage History boxes show how much CPU processing power your computer is currently using and has been using over time. The Memory and Physical Memory Usage History boxes display the amount of memory that is being used and how much has been used over time.
Physical Memory section: shows the information of RAM that is usable by the operating system.
Kernel memory: tells you how much memory is in use by the kernel and device drivers.
Networking Tab
The Networking tab allows you to have a general idea of how much traffic is flowing over a particular network interface on your computer. At the bottom of this screen you will see a list of all the network interfaces on your computer. Above that you will see graphs for each interface that show the network traffic flowing over them.
Users tab
In the Users tab, you'll find a list of all users who have an active session on the system.
1. Highlight a user and click Logoff to end that user's session.
2. Highlight a user and click Disconnect to end a user's session but preserve it in memory, so that user can later log on again and continue work.
Task 3. Amdahl's law
Amdahl’s law is a formula used to find the maximum improvement improvement possible by improving a particular part of a system. In parallel computing, Amdahl's law is mainly used to predict the theoretical maximum speedup for program processing using multiple processors.
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is the theoretical speedup -
is the time an algorithm takes to finish when running n threads -
is the fraction of the algorithm that is strictly serial (so 1- B is how much of the program can be run in parallel)
What this is basically saying is that the amount of speedup a program will see by using 
cores is based on how much of the program is serial (can only be run on a single CPU core) and how much of it is parallel (can be split up among multiple CPU cores).
- is the theoretical speedup
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is the fraction of the algorithm that can be made parallel - is the number of CPU threads
Example 1
If 30% of the execution time may be the subject of a speedup, p will be 0.3; if the improvement makes the affected part twice faster, s will be 2. Amdahl's law states that the overall speedup of applying the improvement will beS latency = 1 1 − p + p s = 1 1 − 0.3 + 0.3 2 = 1.18. {\displaystyle S_{\text{latency}}={\frac {1}{1-p+{\frac {p}{s}}}}={\frac {1}{1-0.3+{\frac {0.3}{2}}}}=1.18.}
Example 2
We are given a serial task which is split into four consecutive parts, whose percentages of execution time are p 1 = 0.11, p 2 = 0.18, p 3 = 0.23, and p 4 = 0.48 respectively. Then we are told that the 1st part is not sped up, so s 1 = 1, while the 2nd part is sped up 5 times, so s 2 = 5, the 3rd part is sped up 20 times, so s 3 = 20, and the 4th part is sped up 1.6 times, so s 4 = 1.6. By using Amdahl's law, the overall speedup is
S latency = 1 p 1 s 1 + p 2 s 2 + p 3 s 3 + p 4 s 4 = 1 0.11 1 + 0.18 5 + 0.23 20 + 0.48 1.6 = 2.19. {\displaystyle S_{\text{latency}}={\frac {1}{{\frac {p1}{s1}}+{\frac {p2}{s2}}+{\frac {p3}{s3}}+{\frac {p4}{s4}}}}={\frac {1}{{\frac {0.11}{1}}+{\frac {0.18}{5}}+{\frac {0.23}{20}}+{\frac {0.48}{1.6}}}}=2.19.} 
Notice how the 20 times and 5 times speedup on the 2nd and 3rd parts respectively don't have much effect on the overall speedup when the 4th part (48% of the execution time) is sped up only 1.6 times.
The point that Amdahl was trying to make was that using lots of parallel processors was not a viable way of achieving the sort of speed-ups that people were looking for. i.e. it was essentially an argument in support of investing effort in making single processor systems run faster.
1. Open created previously Word document Student Name_Lab№1.docx”
2. Write down the following data into second table titled “Lab1_task3”:
| 1. | What is the overall speedup if you make 10% of a program 90 times faster? | |
| 2. | What is the overall speedup if you make 90% of a program 10 times faster? | |
| 3. | The new CPU is 20 times faster on search queries than the old processor. The old processor is busy with search queries 70% of the time, what is the speedup gained by integrating the enhanced CPU? | |
| 4. | Assume 0.1% of the runtime of a program is not parallelizable. We are using supercomputer, which consists of 3,120,000 cores. Under the assumption that the program runs at the same speed on all of those cores, and there are no additional overheads, what is the parallel speedup on 30, 30,000 and 3,000,000 cores? B =0.001 | |
| 5. | The total time to execute a program is set to 1. The non-parallelizable part of the programs is 40% which out of a total time of 1 is equal to 0.4. The execution time of the program with a parallelization factor of 2 (2 threads or CPUs executing the parallelizable part) would be: |
