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melomori [17]
3 years ago
14

If we have 6 processes in a round-robin scheduler with a quantum of 20ms a piece, and we increase the priority of one process su

ch that it now runs for 100ms, how much time of the entire scheduling cycle does the prioritized process run?
Computers and Technology
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

50% of the entire scheduling cycle

Explanation:

Number of processes in round-robin scheduler  = 6

The prioritized process run time is 100 ms while the other runs are 20 ms.

All processes are run during an entire scheduling cycle, therefore the total time for an entire scheduling cycle will be 100 ms + (20* 5) ms = 200ms.

Now we have total time 200 ms and in this time prioritized process runs for 100 ms.

Therefore the percentage of prioritized process run = \frac{100}{200}  *100 = 50%

Therefore the prioritized process runs 50% of the entire scheduling cycle.

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amid [387]

Answer:

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Explanation:

Hi there!

Please see below for the circuit diagram.

<u>1) Find the total resistance of the resistors in parallel</u>

Total resistance in parallel equation: \frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2}

Both the resistors measure 200 Ω. Plug these into the equation as R₁ and R₂:

\frac{1}{R_T} = \frac{1}{200} +\frac{1}{200}\\\frac{1}{R_T} = \frac{1}{100}\\R_T=100

Therefore, the total resistance of the resistors in parallel is 100 Ω.

<u>2) Find the total resistance of the circuit</u>

Now, to find the total resistance of the circuit, we must add the 100 Ω we just solved for and the 100 Ω for the other resistor placed in series:

100 Ω + 100 Ω = 200 Ω

Therefore, the total resistance of the circuit is 200 Ω.

I hope this helps!

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