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ycow [4]
3 years ago
11

You are a network engineer tasked with performing a site survey for a multiple-channel architecture (MCA) system in a three-stor

y building. Which characteristic must be considered while performing a site survey?
Computers and Technology
1 answer:
blagie [28]3 years ago
5 0

Answer:

20/20 vision, and quick internal math if you are talking about human characteristics. If you are talking about the sight survey itself and considering other aspects; You would want to make sure the site has an easily accessible area for the client, doing this by considering the following: easy access, building height (cell phone tower correspondents), multi-channel sight points, and wiring less visible than the access point.

Explanation:

You might be interested in
Consider two different implementations, M1 and M2, of the same instruction set. There are three classes of instructions (A, B, a
kkurt [141]

Answer:

(a) Calculate the average CPI for each machine, M1, and M2

Average CPI = Cycles per Instruction * Frequency of each  Instruction

                     = CPI * F

For Machine M1:

Cycles per Instruction/ Clocks per Instruction =  (60%)* 1 + (30%)*2 + (10%)*4

                                                                        = (0.60) * 1 + (0.30) * 2 + (0.10) * 4

                                                                          = 0.6 + 0.6 + 0.4

                                                                          = 1.6

For Machine M2:

Cycles per Instruction/ Clocks per Instruction=  (60%)*2 + (30%)*3 + (10%)*4

                                                                         = (0.60) *2 + (0.30) * 3 + (0.10) *4        

                                                                         = 1.2 + 0.9 + 0.4

                                                                          = 2.5

(b) Calculate the average MIPS ratings for each machine, M1 and M2.

MIPS rating can be found by using this formula

MIPS rating = Instruction Count / Execution Time * 10^6

                   = Instruction count  / IC X CPI * Clock cycle time * 10^6

                   = IC X Clock rate  / IC X CPI X 10^6

                   = Clock Rate/(CPI * 10^6)

MIPS rating for Machine M1:

The clock rate for M1 is 80 MHz and CPI calculated in (a) is 1.6 So:

MIPS rating = (80 * 10^6) / (1.6 * 10^6 )

                   =  80000000/ 1600000

                   =  50

MIPS rating for Machine M2:

The clock rate for M1 is 100 MHz and CPI calculated in (a) is 2.5 So:

MIPS rating = (100 * 10^6 ) / (2.5 * 10^6 )

                   =  100000000 / 2500000

                  =  40

(c) Which machine has a smaller MIPS rating?

  • M2 has a smaller MIPS rating which is 40.

Which individual instruction class CPI do you need to change, and by how much, to have this machine have the same or better performance as the machine with the higher MIPS rating.

Machine M1 has the higher MIPS rating than M2 and in order to make M2 perform better than M1, the CPI of instruction class A should be modified. Lets change the instruction class A CPI to 1 in place of 2.

Then:

Cycles per Instruction/ Clocks per Instruction= (60%)*1 + (30%)*3 + (10%)*4

                                                                       = (0.60 * 1) + (0.30 * 3) + (0.10 * 4)

                                                                       = 0.60 + 0.9 + 0.4

                                                                       = 1.9

Average MIPS rating = (100 * 10^6) / (1.9 * 10^6 )

                                   = 100000000 / 1900000

                                   =  52.6

Average MIPS rating is of M2 after changing instruction class A CPI is 52.6 which is better than the average MIPS rating of M1 which is 50.0

4 0
3 years ago
a town with 4 lakh inhabitants is to be supplied with water from a reservoir 6.4 km away from a town with 25 m available head ca
Komok [63]

Answer: B

Explanation:

4 0
3 years ago
The programmer must initialize variables when they are declared
lorasvet [3.4K]

Answer: When you declare a variable, you should also initialize it. Two types of variable initialization exist: explicit and implicit. Variables are explicitly initialized if they are assigned a value in the declaration statement. Implicit initialization occurs when variables are assigned a value during processing.

Explanation:

For example, in JavaScript

var PaintAmount = 50; -declare and initialize  

function setup() {

   creatCanvas(200, 200);

}  

function draw() {

   ellipse(PaintAmount, PaintAmount) -use the variable PaintAmount

}

or rather in Java,

package random;

public class something() {

Public static void Main(String []args) {

        string name;  // this is declaring the variable with the example type  

     

       string name = new string; //this initializes the declared variable

}

}

6 0
3 years ago
The process of using or controlling two or more windows at a time is known as. a threading .b multitasking. c hyperthreading.d s
mixas84 [53]
<span>The process of using or controlling two or more windows at a time is known as multitasking.</span>
5 0
3 years ago
Write programs for two MSP430 boards to do the following:
Rainbow [258]

Answer:

Explanation:

#include "msp430g2553.h"

#define TXLED BIT0

#define RXLED BIT6

#define TXD BIT2

#define RXD BIT1

const char string[] = { "Hello World\r\n" };

unsigned int i; //Counter

int main(void)

{

WDTCTL = WDTPW + WDTHOLD; // Stop WDT

DCOCTL = 0; // Select lowest DCOx and MODx settings

BCSCTL1 = CALBC1_1MHZ; // Set DCO

DCOCTL = CALDCO_1MHZ;

P2DIR |= 0xFF; // All P2.x outputs

P2OUT &= 0x00; // All P2.x reset

P1SEL |= RXD + TXD ; // P1.1 = RXD, P1.2=TXD

P1SEL2 |= RXD + TXD ; // P1.1 = RXD, P1.2=TXD

P1DIR |= RXLED + TXLED;

P1OUT &= 0x00;

UCA0CTL1 |= UCSSEL_2; // SMCLK

UCA0BR0 = 0x08; // 1MHz 115200

UCA0BR1 = 0x00; // 1MHz 115200

UCA0MCTL = UCBRS2 + UCBRS0; // Modulation UCBRSx = 5

UCA0CTL1 &= ~UCSWRST; // **Initialize USCI state machine**

UC0IE |= UCA0RXIE; // Enable USCI_A0 RX interrupt

__bis_SR_register(CPUOFF + GIE); // Enter LPM0 w/ int until Byte RXed

while (1)

{ }

}

#pragma vector=USCIAB0TX_VECTOR

__interrupt void USCI0TX_ISR(void)

{

P1OUT |= TXLED;

UCA0TXBUF = string[i++]; // TX next character

if (i == sizeof string - 1) // TX over?

UC0IE &= ~UCA0TXIE; // Disable USCI_A0 TX interrupt

P1OUT &= ~TXLED; }

#pragma vector=USCIAB0RX_VECTOR

__interrupt void USCI0RX_ISR(void)

{

P1OUT |= RXLED;

if (UCA0RXBUF == 'a') // 'a' received?

{

i = 0;

UC0IE |= UCA0TXIE; // Enable USCI_A0 TX interrupt

UCA0TXBUF = string[i++];

}

P1OUT &= ~RXLED;

}

7 0
4 years ago
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