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mash [69]
3 years ago
11

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

structions and 16 x 10^6 branch instructions. The CPI for each type of instruction is 1, 1, 4 and 2, respectively. Assume that the processor has a 2 GHz clock rate.By how much must we improve the CPI of FP instructions if we want the program to run two times faster?
Engineering
1 answer:
Svetradugi [14.3K]3 years ago
3 0

Answer:

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

Explanation:

Processor clock rate = 2 GHz

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

Clock cycles can be determined using following formula

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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4-6. A vertical cylindrical storage vessel is 10 m high and 2 m in diameter. The vessel contains liquid cyclohexane currently at a liquid level of 8 m. The vessel is vented to the atmosphere. A 1-cm-diameter hole develops in the bottom of the vessel. Calculate and plot the discharge rate of cyclohexane as a function of time until the vessel is completely empty. Show that the time to empty is the same as the time calculated using Equation 4-21.

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3 years ago
A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and
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Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg

The quality at state 4 is determined from the condition  s_{4} =s_{3} and the entropies of the components at the condenser pressure taken from table:

 q_{4} =\frac{s_{4}-s_{liq50}  }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935

The enthalpy at state 4 then is:  

h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg\\

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:  

h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg

The enthalpy at state 2 is determined from an energy balance on the pump:

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg

The enthalpy at state 2 is then determined from an energy balance on the pump:  

h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1}  )

    =299.79 kJ/kg  

The thermal efficiency in this case then is:  

=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345

8 0
3 years ago
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german
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5 0
3 years ago
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Question 54 (1 point)
Rudiy27
The answer should be Oc 15
6 0
3 years ago
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A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum
antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

  • Density = 61 veh/km.
  • Speed = 59 km/hr.
  • Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

q_{max}=\frac{V_f \times K_i}{4}

<u>Where:</u>

  • V_f is the free flow speed.
  • K_i is the Jam density.

In order to calculate the free flow speed, we would use this formula:

V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr

Substituting the parameters into the model, we have:

q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}

Max flow = 3,599 veh/hr.

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