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2 years ago

based on the graph shown, for any given output level, the veritcal distance between the avc and the atc curves represents

1 answer:

4 0

A a result of the graph shown, for any given output level, the veritcal distance between the avc and the atc curves represents total fixed cost (TFC).

<h3>What is Vertical distance?</h3>

This is known as vertical separation and it is said to be the distance that exist between two vertical positions.

Note that vertical coordinates are seen for expressing vertical position: depth, and others and as such,

See full question below

The vertical distance between the ATC and AVC curve represents

a) marginal cost (MC)

b) average fixed cost (AFC)

c) total fixed cost (TFC)

d) total cost (TC) - total variable cost (TVC)

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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

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Answer:

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Explanation:

Processor clock rate = 2 GHz

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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 ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

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

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

$CPI_{FP improved}$ 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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3 years ago

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Answer:

0.0406 m/s

Explanation:

Given:

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Now,

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mass flow rate = ρAV

where,

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V = 0.0203 m/s

also,

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thus,

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Re = 611.39 < 2000

thus,

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