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Nuetrik [128]
3 years ago
9

You have two LANs connected via a router. Network 1 has a DHCP server with a single Ethernet interface. Network 2 does not have

a DHCP server. You want the h​osts on Network 2 to be able to get IP addresses from the DHCP server on Network 1. What should you configure?
Computers and Technology
2 answers:
earnstyle [38]3 years ago
5 0

Answer:

You will Configure a LAN with DHCP and VLANs.

Explanation:

Routers generally support clients on both physical Local Area Networks ork (LANs) and Virtual Local Area Networks (VLANs). The routers can use the Dynamic Host Configuration Protocol (DHCP) to enable automatic assignment of IP configurations for nodes on these networks i.e the network 2.

The Dynamic Host Configuration Protocol (DHCP), uses a client/server model for address allocation. It means it automatically assigns an IP address to each client.

Virtual Local Area Networks (VLANs) enable networks to be segmented or broken and formed into logical groups of users, regardless of the user's physical location or Local Area Network connection.

slava [35]3 years ago
4 0

Answer:

The best thing to configure is the DHCP Server.

Explanation:

DHCP stand for Dynamic Hosting Configuration Protocol which is a sever used to automatically assign IP addresses, default gateways, domain names, etc to various users connected on a network. From the question above, since Network 1 has already configured DHCP server which provides IP addressing to its hosts, and Network 1 and Network 2 are connected to the same router directly, then then there is need to configure  DHCP server for Network 2 users for IP addressing . DHCP server is basically flexible and robust enough to do hots IP configurations which it was build for other than the router which may likely result into delay in packet delivery and communication with connected networks.

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It's formula and functions

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The following method public String removeFromString(String old, String frag) removes all occurences of the string frag from the
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private static String removeFromString(String old, String frag)

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int i = old.indexOf(frag);

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Here the index of first occurrence is obtained outside the “while loop” and if this loop runs until index value is >-1. It extracts the rest of the characters other than “frag” from the index and all the characters for which the given set of characters are removed.

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2 years ago
The discipline of building hardware architectures, operating systems, and specialized algorithms for running a program on a clus
kondaur [170]

The discipline of building hardware architectures, operating systems, and specialized algorithms for running a program on a cluster of processors is known as <u>parallel computing.</u>

<u></u>

<h3>What is Parallel Computing?</h3>

Parallel computing refers to the process of breaking down larger problems into smaller, independent, often similar parts that can be executed simultaneously by multiple processors communicating via shared memory, the results of which are combined upon completion as part of an overall algorithm. The primary goal of parallel computing is to increase available computation power for faster application processing and problem solving.

<h3>Types of parallel computing</h3>

There are generally four types of parallel computing, available from both proprietary and open source parallel computing vendors:

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Learn more about parallel computing

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5 0
1 year ago
Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. Computer B has a CPI of 2.5 and can be run at a
kirill115 [55]

Answer:

Check the explanation

Explanation:

CPI means Clock cycle per Instruction

given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M

Execution time is given by following Formula.

Execution Time(CPU time) = CPI*Instruction Count * clock time = \frac{CPI*Instruction Count}{ClockRate}

a)

for system A CPU time is 1.3 * 100, 000 600 106

= 216.67 micro sec.

b)

for system B CPU time is =\frac{2.5*100,000}{750*10^6}

= 333.33 micro sec

c) Since the system B is slower than system A, So the system A executes the given program in less time

Hence take CPU execution time of system B as CPU time of System A.

therefore

216.67 micro = =\frac{2.5*Instruction}{750*10^6}

Instructions = 216.67*750/2.5

= 65001

hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A

3 0
3 years ago
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