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VikaD [51]
3 years ago
5

For an 18 Mbps LTE data stream with a symbol time of 66.67 μs, how many subcarriers are created?

Computers and Technology
1 answer:
vampirchik [111]3 years ago
8 0
<span>1/.0000667 = 14,992.5;
18,000,000/14,992 = 1,200 subcarriers
</span><span>For an 18 Mbps LTE data stream with a symbol time of 66.67 μs, 1,200 subcarriers are created.</span>
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It takes 2 seconds to read or write one block from/to disk and it also takes 1 second of CPU time to merge one block of records.
Alexxx [7]

Answer:

Part a: For optimal 4-way merging, initiate with one dummy run of size 0 and merge this with the 3 smallest runs. Than merge the result to the remaining 3 runs to get a merged run of length 6000 records.

Part b: The optimal 4-way  merging takes about 249 seconds.

Explanation:

The complete question is missing while searching for the question online, a similar question is found which is solved as below:

Part a

<em>For optimal 4-way merging, we need one dummy run with size 0.</em>

  1. Merge 4 runs with size 0, 500, 800, and 1000 to produce a run with a run length of 2300. The new run length is calculated as follows L_{mrg}=L_0+L_1+L_2+L_3=0+500+800+1000=2300
  2. Merge the run as made in step 1 with the remaining 3 runs bearing length 1000, 1200, 1500. The merged run length is 6000 and is calculated as follows

       L_{merged}=L_{mrg}+L_4+L_5+L_6=2300+1000+1200+1500=6000

<em>The resulting run has length 6000 records</em>.

Part b

<u><em>For step 1</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{2300}{100} \times 2 sec\\T_{I.O}=46 sec

So the input/output time is 46 seconds for step 01.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 2300 for step 01
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{2300}{100} \times 1 sec\\T_{CPU}=23 sec

So the CPU  time is 23 seconds for step 01.

Total time in step 01

T_{step-01}=T_{I.O}+T_{CPU}\\T_{step-01}=46+23\\T_{step-01}=69 sec\\

Total time in step 01 is 69 seconds.

<u><em>For step 2</em></u>

Input Output Time

Input Output Time is given as

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{I/O per block} is 2 sec

So

T_{I.O}=\frac{L_{run}}{Size_{block}} \times Time_{I/O \, per\,  block}\\T_{I.O}=\frac{6000}{100} \times 2 sec\\T_{I.O}=120 sec

So the input/output time is 120 seconds for step 02.

CPU  Time

CPU Time is given as

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}

Here

  • L_run is 6000 for step 02
  • Size_block is 100 as given
  • Time_{CPU per block} is 1 sec

So

T_{CPU}=\frac{L_{run}}{Size_{block}} \times Time_{CPU \, per\,  block}\\T_{CPU}=\frac{6000}{100} \times 1 sec\\T_{CPU}=60 sec

So the CPU  time is 60 seconds for step 02.

Total time in step 02

T_{step-02}=T_{I.O}+T_{CPU}\\T_{step-02}=120+60\\T_{step-02}=180 sec\\

Total time in step 02 is 180 seconds

Merging Time (Total)

<em>Now  the total time for merging is given as </em>

T_{merge}=T_{step-01}+T_{step-02}\\T_{merge}=69+180\\T_{merge}=249 sec\\

Total time in merging is 249 seconds seconds

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 Preamble is a 7- octet pattern of alternating the 0's and 1's and it allow devices in the network to easily synchronized the receiver clock. It also provide the bit level synchronization.

It basically allow the start of frame delimiter to provide the byte synchronization and assign a new frame. The bits are transmit orderly from left to right in the frame.  

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Giga usually means 10^9

We're trying to work giba, which is 2^30 into the vocabulary to differentiate them because of the confusion and difference that they create.
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names = ["Kevin", "Joe", "Thor", "Adam", "Zoe"]

names.sort()

for x in names:

   if x == "Thor":

       break

   else:

       print(x)

I made up my own names for the sake of testing my code. I wrote my code in python 3.8. I hope this helps.

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