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Anna11 [10]
3 years ago
13

A program consists of 100,000 instructions as follows:

Engineering
1 answer:
vesna_86 [32]3 years ago
6 0

Answer:

Execution time is 13.65ms

CPI is 5.46

MIPS is 73.3\ MIPs

Explanation:

Given:

s = 40MHz --- processor speed [Missing from the question]

\begin{array}{ccc}{Instruction\ Type} & {Instruction\ Count} & {Cycles\ per\ Instruction} & {Integer\ Arithmetic} & {45000} & {4} \ \\ {Data\ Transfer} & {32000} & {6} & {Floating\ point\ arithmetic} & {15000} & {10} & {Control\ Transfer} &{8000} & {3} \ \end{array}

Solving (a): The program execution time

First, we solve for (b)

Solving (b): The effective CPI

This is calculated as:

CPI = \frac{\sum IC * CI}{\sum IC}

Where: <em>IC = Instruction Count and CI = Cycles per Instruction</em>

So, the equation becomes:

CPI = \frac{45000 *4 + 32000 * 6 + 15000 * 10 + 8000 * 3}{45000+32000+15000+8000}

CPI = \frac{546000}{100000}

CPI = 5.46

Solving (c): MIPS

This is calculated as:

MIPS = Speed * \frac{1}{CPI} * \frac{1}{\sum IC}

MIPS = 40 * \frac{1}{5.46} * \frac{1}{100000}

MIPS = \frac{40 * 1 * 1}{5.46*100000}

MIPS = \frac{40}{546000}

MIPS = 0.00007326007

Convert to MIPs

MIPS = 73.3\ MIPs

Solving (a): Execution Time

This is calculated as:

Time = Instructions * CPI * \frac{1}{Speed}

Time = 100000 * 5.46* \frac{1}{40M}

Time = \frac{546000}{40M}

Time = \frac{546000}{40*1000000}

Time = \frac{546000}{40000000}

Time = 0.01365s

Time = 13.65ms

Execution time is 13.65ms

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Answer:

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Explanation:

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7 0
3 years ago
A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a velocity of 0.2 m/s .A jet of w
BabaBlast [244]

A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.

Answer:

Net power of 1.2 KW is being extracted

Explanation:

We are given;

Mass flow rate; m' = 0.1kg/s

Inlet temperature; T1 = 20°C = 293K

Inlet pressure; P1 = 1 atm = 10^(5) pa

Inlet velocity; v1 = 0.2 m/s

Exit Pressure; P2 = 1 atm = 10^(5) pa

Exit Temperature; T2 = 1 atm = 296K

Exit velocity; V2 = 20m/s

Change in elevation; h = Z2 - Z1 = 5m

We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.

Thus;

Q = -0.1W

From Bernoulli equation;

Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

Where;

∆Potential energy = mg(z2 - z1)

∆Kinetic energy = ½m(v2² - v1²)

∆Pressure energy = mc_p(T2 - T1)

Thus;

-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]

Where C_p is specific heat capacity of water = 4200 J/Kg.k

Plugging in the relevant values, we have;

-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))

-1.1W = 4.905 + 19.998 + 1260

-1.1W = 1284.903

W = -1284.903/1.1

W ≈ -1168 J/s ≈ -1.2 KW

The negative sign means that work is extracted from the system.

7 0
3 years ago
Calculate the osmotic pressure of seawater containing 3.5 wt % NaCl at 25 °C . If reverse osmosis is applied to treat seawater,
AlladinOne [14]

Answer:

Highest osmotic pressure that membrane may experience is

' =58.638 atm

Explanation:

Suppose sea-water taken is M= 1 kg

Density of water = 1000 kg/m3

Therefore Volume of water= Mass,M/Density of water

V= 1 kg/(1000 kg/m3)

V= 10-3 m3= 1 Litre

Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water

Mass present of NaCl= m= 0.035*1000 g

m= 35 g

Since molecular weight of NaCl= 58.44 g/mol =M.W.

Thus its Number of moles of Nacl= m/M.W

nNaCl= 35g/58.44 gmol-1

= 0.5989 mol

ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres

C= 0.5989mol/ 1L

=0.5989 M

Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2

And osmotic pressure  = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)

Putting in equation 1 ,we get  = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K

=29.319 atm

Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)

Thus New concentration of NaCl C'= 2*C

C'=2*0.5989 M

=1.1978 M

and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure  ' also doubles from  ,

Thus,Highest osmotic pressure that membrane may experience is,  '=2*  

=2*29.319 atm

' =58.638 atm

3 0
3 years ago
A part has been tested to have Sut = 530 MPa, f = 0.9, and a fully corrected Se = 210 MPa. The design requirements call for the
kobusy [5.1K]

Answer:

126984 cycles

Explanation:

Given data :

Sut = 530 MPa

f = 0.9

fully corrected Se = 210 MPa

using Miner's method attached below is the detailed solution of the given problem

when loaded with ± 225 MPa the number of cycles before it fails will be

≈  126984

8 0
3 years ago
You doubled the voltage frequency in an RL series AC circuit, the inductive resistance would?
Ivenika [448]

Answer:

Remain the same.

Explanation:

If you doubled the voltage frequency in an RL series AC circuit, the inductive resistance would <u>remain the same</u>.

6 0
3 years ago
Read 2 more answers
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