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kipiarov [429]
2 years ago
12

POINT BOOST:

Computers and Technology
1 answer:
stiv31 [10]2 years ago
8 0

glenn

she's good what's your

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nikklg [1K]
You should use your Microsoft account because this will sync to all your windows devices.
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What is the purpose of the making of Nintendo Switch?
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To create a console that is suitable for children and families.

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Read 2 more answers
Assume a TCP sender is continuously sending 1,090-byte segments. If a TCP receiver advertises a window size of 5,718 bytes, and
Arturiano [62]

Answer:

for the 5 segments, the utilization is 3.8%

Explanation:

Given the data in the question;

segment size = 1090 bytes

Receiver window size = 5,718 bytes

Link transmission rate or Bandwidth = 26 Mbps = 26 × 10⁶ bps

propagation delay = 22.1 ms

so,

Round trip time = 2 × propagation delay = 2 × 22.1 ms = 44.2 ms

we determine the total segments;

Total segments = Receiver window size / sender segment or segment size

we substitute

Total segments = 5718 bytes / 1090 bytes

Total segments = 5.24587 ≈ 5

Next is the throughput

Throughput = Segment / Round trip

Throughput = 1090 bytes / 44.2 ms

1byte = 8 bits and 1ms = 10⁻³ s

Throughput = ( 1090 × 8 )bits / ( 44.2 × 10⁻³ )s

Throughput = 8720 bits / ( 44.2 × 10⁻³ s )

Throughput = 197.285 × 10³ bps

Now Utilization will be;

Utilization = Throughput / Bandwidth

we substitute

Utilization = ( 197.285 × 10³ bps ) / ( 26 × 10⁶ bps )

Utilization = 0.0076

Utilization is percentage will be ( 0.0076 × 100)% = 0.76%

∴ Over all utilization for the 5 segments will be;

⇒ 5 × 0.76% = 3.8%

Therefore, for the 5 segments, the utilization is 3.8%

4 0
3 years ago
The following is a sequence of undo-log records written by two transactions T and U: < START T>; ; < START U>; ; ; ;
tatiyna

Answer:

Giving that: The following is a sequence of undo-log records written by 2 transactions T and U:

< START T >;

< T,A,10 >;

< START U >;

< U, B, 20 >;

< T, C, 30 >;

< U, D, 40 >;

< Commit U >;

< T, E, 50 >;

< Commit T >;

1. < START U >

      Recovery action in this case will be undo(-1) and undo(0). All restored to its original Value

     log records < T, A, 10 >,  < T, abort >; as written out

2. < T, E, 50 >

      Recovery action in this case will be undo(8) and redo(0). A and C is restored to its original value, B and D are set to 20 and 40

      log records <T, C, 30 >,  < T, A, 10 >,  < T, abort >  are written out

3. < Commit T >

      Recovery action in this case will be redo(7) and redo(4). A and C are set to 10 and 30, B and D are set to 20 and 40

7 0
3 years ago
Explain what the hazard detection unit is doing during the 5th cycle of execution. Which registers are being compared? List all
kondaur [170]

Answer:

<em>The registers that are compared are instructions 3 and 4</em>

<em>Explanation:</em>

<em>From the question given,</em>

<em>Recall that we need to explain what the hazard detection unit is doing  during the 5th cycle of execution and which registers are being compared.</em>

<em>Now,</em>

<em>The instructions on the 5th cycle, at the stage ID/EX and IF/ID:</em>

<em>The instruction values are in ID/EX : sub $t2, $t3, $t6 (instruction 3)</em>

<em>The instruction values are in IF/ID: sub $t3, $t1 $t5 (instruction 4)</em>

<em>The register $t3 is compared in the instructions 3 and 4</em>

<em>The hazard detection unit between instruction 4 and 5t o be compared, it need to find out the values of $t1</em>

<em />

7 0
3 years ago
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