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2 years ago

In the circuit below I 1=20mA V1=10 R1=400 R2=2000 R3=1000. Use node analysis to find R2

Engineering

1 answer:

Hatshy [7]2 years ago

6 0

Answer:0.2A

Explanation:

First, write KCL

i1-i2-i3=0

Then, replace currents with viltages and resistors.

V2-10v/100-v2/200-v2/400=0

V2-40=0

V2=40v

I hope it was helpful

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Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in

Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

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Execution Time = ∑ $(\frac{Clock cyles}{Clock rate})$

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Clock cycles = ( $CPI_{FP}$ x No. FP instructions )+ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)

Clock cycles = ( 50 x $10^{6}$ x 1) + ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

= $\frac{512(10^{6}) }{2(10^{9}) }$

Initial execution Time = 256 x 10⁻³

For 16 processors ,

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Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

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$CPI_{FP improved}$ x No. FP instructions = ( $\frac{Clockcycles}{2}$ ) -[ ( $CPI_{INT}$ x No. INT instructions) + ( $CPI_{L/S}$ x No. L/S instructions ) + ( $CPI_{branch}$ x No. branch instructions)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = ( $\frac{512(10)^{6} }{2}$ ) - [ ( 110 x $10^{6}$ x 1) + ( 80 x $10^{6}$ x 4) + ( 16 x $10^{6}$ x 2)]

$CPI_{FP improved}$ x 50 x $10^{6}$ = - 206 x $10^{6}$

$CPI_{FP improved}$ = - 206 x $10^{6}$ / 50 x $10^{6}$

$CPI_{FP improved}$ = - 4.12 < 0

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