If it costs $55,000 to put on a magic show for two weeks and the estimated revenue is $15,000 per night for 14 nights, what will
1 answer:
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Answer:
<u>The total time elapsed from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B is </u><u>25.11 ms</u>
Explanation:
Host A first converts the analog signal to a digital 64kbps stream and then groups it into 56-byte packets. The time taken for this can be calculated as:
time taken 1=
= (56 x 8) bits / 64 x 10³ bits/s
= 7 x 10⁻³s
time taken 1= 7 ms
The transmission rate of the packet from Host A to Host B is 4 Mbps. The time taken to transfer the packets can be calculated as:
time taken 2= (56 x 8) bits / 4 x 10⁶ bits/s
= 1.12 x 10⁻⁴ s
time taken 2= 112 μs
The propagation delay is 18 ms.
To calculate the total time elapsed, we need to add up all the time taken at each individual stage.
<u /><u> = Time taken 1 + Time taken 2 + Propagation Delay</u>
= 7 ms + 112 μs + 18 ms
= 0.025112 s
= 25.11 ms
Answer:
(a)Applications Time Stamp Events
(b)S=0.5(W-)+
, where
≤W≤
+8.
Explanation:
Some applications assume that clocks always advance, so they could timestamp events under this assumption.
In our case we have the wrong timed clock, say W and the hardware clock H which is supposed to advance at a perfect rate.
We proceed to construct a software clock such that after 8 seconds we can replace the wrong timed clock with the software clock in good conditions.
Let us denote the software clock with S.
Then, S=c(W-)+
where:
=The current Time(10:27:54) and;
c is to be found.
We already know that S=+4 when W=
+8,
So:
S=c(W-)+
+4=c(
+8-
)+
4=8c
c=0.5
We obtain the formula
S=0.5(W-)+
, where
≤W≤
+8.