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mars1129 [50]
3 years ago
7

A disadvantage of a bus network is that _____. a. failure in one workstation necessarily affects other workstations on the netwo

rk b. data flows in only one direction from one computer to the next c. devices cannot be attached or detached without disturbing the rest of the network d. performance can decline as more users and devices are added
Computers and Technology
1 answer:
andrew11 [14]3 years ago
6 0

A disadvantage of a bus network is that <u>performance can decline as more users and devices are added</u>

<u></u>

Explanation:

  • A bus network is an arrangement in a local area network (LAN) in which each node is connected to a main cable or link called the bus.
  • The advantages of a bus network are: it is easy to install. it is cheap to install, as it doesn't require much cable.

The disadvantages of a bus network are:

  1. If the main cable fails or gets damaged the whole network will fail.
  2. As more workstations are connected the performance of the network will become slower because of data collisions.
  3. Every workstation on the network "sees" all of the data on the network – this is a security risk.
  • Bus topology is used for small workgroup local area networks whose computers are connected using a thinnet cable. Trunk cables connecting hubs or switches of departmental LANs to form a larger LAN.
  • Bus topology uses one main cable to which all nodes are directly connected. The main cable acts as a backbone for the network. One of the computers in the network typically acts as the computer server. The first advantage of bus topology is that it is easy to connect a computer or peripheral device.

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The rules of a programming language constitute its syntax.
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Look at the logic program below. Set a breakpoint after each commented instruction. You can also single step through the program
Goshia [24]

Using programming knowledge, it is possible to use programming logic, we can differentiate and find the terms

<h3>Writing the code and understanding the programming logic:</h3>

<em>.data</em>

<em>sourceword:</em>

<em>.word 0xAB</em>

<em>wordmask:</em>

<em>.word 0xCF</em>

<em>operand:</em>

<em>.long 0xAA</em>

<em>.text</em>

<em>.globl _start</em>

<em>_start:</em>

<em>movw sourceword, %ax # ax=0x00AB ( move the word value of sourceword =0x00AB to ax register.)</em>

<em>movw %ax, %bx # bx=0x00AB ( copy the value of ax to bx register.)</em>

<em>xorw %cx, %cx # cx=0000 ( cx=cx XOR cx, that is clear cx register.)</em>

<em />

<em>andw wordmask, %ax # ax=0x008B ( ax=ax AND wordmask= 0x00AB AND 0x00CF= 0x008B)</em>

<em>orw wordmask, %bx # bx=0x00EF (bx=bx OR wordmask= 0x00AB OR 0x00CF= 0x00EF.)</em>

<em>xorw wordmask, %cx # cx=0x00CF (cx=cx XOR wordmask= 0x0000 XOR 0x00CF=0x00CF.)</em>

<em />

<em>movl operand, %eax # eax=0x000000AA. ( copy the 32-bit value of operand=0x000000AA to eax.)</em>

<em>movl %eax, %ebx # ebx=0x000000AA.( copy the value of eax to ebx.)</em>

<em>movl %eax, %ecx # ecx=0x000000AA. ( copy the value of eax to ecx.)</em>

<em />

<em>shll $3,%eax # eax=0x00000550. ( logical shift left of eax=0x000000AA by 3 bits.)</em>

<em />

<em># 0x000000AA=00000000000000000000000010101010</em>

<em />

<em># =>00000000000000000000010101010000=0x00000550.</em>

<em />

<em>rorl $4,%ebx # ebx=0xA000000A. (Rotate right ebx=0x000000AA by 4 bits.)</em>

<em />

<em># 0x000000AA=00000000000000000000000010101010</em>

<em />

<em>3 =>10100000000000000000000000001010= 0xA000000A.</em>

<em />

<em>sarl %ecx # here the count value to shift the bits not mentioned.</em>

<em />

<em># It is the arithmetic shift right instruction. the shifted left bits filled with MSB bit.</em>

See more about logic program at brainly.com/question/14970713

#SPJ1

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