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Leokris [45]
3 years ago
9

You have a site (Site1) that has about 20 users. For the last few months, users at Site1 have been complaining about the perform

ance when accessing multiple files at the corporate office, particularly if the files are relatively large. They have no dedicated server to configure DFS replication. Therefore, what else can you do to improve performance when accessing these files?
Computers and Technology
1 answer:
SIZIF [17.4K]3 years ago
6 0
Upgrade Site1's (probably WAN) network connection.
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My friend Leo wants to have an emergency plan for his final exams on University of Southern Algorithmville. He has N subjects to
leonid [27]

Answer:

Greedy is an algorithmic paradigm that builds up a solution piece by piece, always choosing the next piece that offers the most obvious and immediate benefit. Greedy algorithms are used for optimization problems. An optimization problem can be solved using Greedy if the problem has the following property: At every step, we can make a choice that looks best at the moment, and we get the optimal solution of the complete problem.

If a Greedy Algorithm can solve a problem, then it generally becomes the best method to solve that problem as the Greedy algorithms are in general more efficient than other techniques like Dynamic Programming. But Greedy algorithms cannot always be applied. For example, the Fractional Knapsack problem (See this) can be solved using Greedy, but 0-1 Knapsack cannot be solved using Greedy.

The following are some standard algorithms that are Greedy algorithms.

1) Kruskal’s Minimum Spanning Tree (MST): In Kruskal’s algorithm, we create an MST by picking edges one by one. The Greedy Choice is to pick the smallest weight edge that doesn’t cause a cycle in the MST constructed so far.

2) Prim’s Minimum Spanning Tree: In Prim’s algorithm also, we create an MST by picking edges one by one. We maintain two sets: a set of the vertices already included in MST and the set of the vertices not yet included. The Greedy Choice is to pick the smallest weight edge that connects the two sets.

3) Dijkstra’s Shortest Path: Dijkstra’s algorithm is very similar to Prim’s algorithm. The shortest-path tree is built up, edge by edge. We maintain two sets: a set of the vertices already included in the tree and the set of the vertices not yet included. The Greedy Choice is to pick the edge that connects the two sets and is on the smallest weight path from source to the set that contains not yet included vertices.

4) Huffman Coding: Huffman Coding is a loss-less compression technique. It assigns variable-length bit codes to different characters. The Greedy Choice is to assign the least bit length code to the most frequent character. The greedy algorithms are sometimes also used to get an approximation for Hard optimization problems. For example, the Traveling Salesman Problem is an NP-Hard problem. A Greedy choice for this problem is to pick the nearest unvisited city from the current city at every step. These solutions don’t always produce the best optimal solution but can be used to get an approximately optimal solution.

6 0
3 years ago
"what is the common information listed under the ip section of pdu details as compared to the information listed under the osi m
frez [133]

Solution:

The common information listed under the ip section of pdu details as compared to the information listed under the osi model tab is Packet Tracer.

This simulation activity is intended to provide a foundation for understanding the TCP/IP protocol suite and the relationship to the OSI model. Simulation mode allows you to view the data contents being sent across the network at each layer. As data moves through the network, it is broken down into smaller pieces and identified so that the pieces can be put back together when they arrive at the destination. Each piece is assigned a specific name (protocol data unit [PDU]) and associated with a specific layer of the TCP/IP and OSI models. Packet Tracer simulation mode enables you to view each of the layers and the associated PDU. The following steps lead the user through the process of requesting a web page from a web server by using the web browser application available on a client PC.  Even though much of the information displayed will be discussed in more detail later, this is an opportunity to explore the functionality of Packet Tracer and be able to visualize the encapsulation process.

This is the required answer.

7 0
3 years ago
A detective agency is looking to bust an arms sale. According to available intelligence, the sale is likely to happen in one of
Paha777 [63]

Answer:

Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}

H(X)=5.4224 bits per symb

H(X|Y="not C")=0.54902 bits per symb

Explanation:

P(B)=2P(C)

P(A)=2P(B)

But

P(A)+P(B)+P(C)=1

4P(C)+2P(C)+P(C)=1

P(C)=1/7

Then

P(A)=4/7

P(B)=2/7

Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}

iii

If X={A,B,C}

and P(Xi)={4/7,2/7,1/7}

where  Id =logarithm to base  2

Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}

=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}

=5.4224 bits  per symb

if P(C)  =0

P(A)=2P(B)

P(B)=1/3

P(A)=2/3

H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)

=0.54902 bits per symb

4 0
3 years ago
Start
Crank

Answer:

N=5  key task decision start and end

Explanation:

8 0
3 years ago
Array, destination contains the number of 10 cities numbered 1 to 10 that the products were shipped to. Hence, 10 quantities of
Naddik [55]

Answer:

#include <stdlib.h>

#include <stdio.h>

void func1(int product[]){

int orders[6]={0};

for(int i=0;i<70;i++){

orders[product[i]]++;

}

printf("Total number of each type of products that were bought\n");

for(int i=1;i<=5;i++){

printf("Product %d = %d\n",i,orders[i]);

}

}

void func2(int product[],int quantity[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

total_cost+= price[product[i]]* quantity[i];

}

printf("The total cost of all 70 orders = %.2f\n",total_cost);

}

void func3(int product[],int quantity[],int destination[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

if(destination[i]==8){

total_cost+= price[product[i]]* quantity[i];

}

}

printf("The total cost of all products shipped to destination 8 = %.2f\n",total_cost);

}

void func4(int product[],int quantity[],float price[]){

int total_orders=0;

for(int i=0;i<70;i++){

if(price[product[i]]* quantity[i]>=50){

total_orders++;

}

}

printf("The total number of orders where each order is $50 or more = %d\n",total_orders);

}

void func5(int product[],int quantity[],int origination[],float price[]){

int total_orders=0;

for(int i=0;i<70;i++){

if(origination[i]==3 && price[product[i]]* quantity[i]>=50){

total_orders++;

}

}

printf("The total number of orders that originated from 3 where each order is $50 or more. = %d\n",total_orders);

}

void func6(int product[],int quantity[],int origination[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

if(origination[i]==3 && price[product[i]] * quantity[i]>=50){

total_cost += price[product[i]] * quantity[i];

}

}

printf("The total number of orders that originated from 3 where each order is $50 or more. = %.2f\n",total_cost);

}

void func7(int origination[],int destination[]){

int total_orders=0;

for(int i=0;i<70;i++){

if(origination[i]==3 && destination[i]==8){

total_orders++;

}

}

printf("The total number of orders that originated from 3 and shipped to 8. = %d\n",total_orders);

}

void func8(int product[],int quantity[],int origination[],int destination[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

if(origination[i]==3 && destination[i]==8){

total_cost += price[product[i]] * quantity[i];

}

}

printf("The total cost of orders that originated from 3 and shipped to 8. = %.2f\n",total_cost);

}

void func9(int destination[]){

int total_orders=0;

for(int i=0;i<70;i++){

if(destination[i]!=8){

total_orders++;

}

}

printf("The total number of orders that was shipped to all destinations except to 8. = %d\n",total_orders);

}

void func10(int product[],int quantity[],int destination[],float price[]){

float total_cost=0;

for(int i=0;i<70;i++){

if(destination[i]!=8){

total_cost += price[product[i]] * quantity[i];;

}

}

printf("The total cost of orders that was shipped to all destinations except to 8. = %.2f\n",total_cost);

}

int main(){

int product[70] = {4, 2, 4, 2, 4, 5, 5, 2, 2, 5, 5, 4, 3, 5, 4, 2, 5, 3, 1, 2, 2, 3, 3, 4, 5, 5, 4, 5, 3, 5, 5, 1, 4, 5, 1, 5, 3, 2, 4, 1, 2, 4, 5, 1, 5, 5, 5, 5, 5, 2, 5, 1, 4, 4, 4, 2, 3, 3, 3, 3, 4, 3, 5, 5, 3, 2, 3, 5, 3, 2};

int quantity[70] = {10, 9, 6, 4, 10, 4, 9, 6, 10, 7, 3, 4, 4, 9, 1, 8, 9, 1, 5, 8, 7, 2, 3, 4, 10, 5, 6, 2, 1, 7, 2, 8, 6, 9, 8, 8, 7, 7, 9, 10, 6, 7, 8, 2, 1, 7, 6, 3, 3, 1, 8, 4, 10, 7, 1, 10, 6, 9, 8, 2, 4, 6, 1, 8, 2, 6, 10, 2, 6, 2};

int origination[70] = {2, 7, 5, 5, 7, 2, 7, 2, 7, 7, 5, 2, 5, 5, 5, 2, 2, 7, 2, 7, 7, 2, 2, 2, 2, 5, 7, 5, 7, 7, 5, 5, 2, 2, 5, 7, 2, 5, 7, 2, 5, 7, 2, 5, 7, 2, 2, 7, 2, 7, 5, 2, 2, 2, 5, 7, 2, 5, 5, 5, 7, 7, 2, 5, 2, 7, 5, 2, 5, 7};

int destination[70] = {8, 7, 3, 10, 2, 6, 4, 5, 1, 3, 5, 9, 5, 8, 6, 4, 3, 7, 1, 2, 7, 2, 8, 2, 2, 1, 2, 6, 10, 2, 7, 7, 8, 6, 8, 8, 4, 8, 3, 10, 6, 9, 4, 9, 5, 1, 7, 3, 1, 7, 5, 5, 4, 9, 3, 10, 8, 1, 1, 1, 1, 8, 10, 3, 5, 2, 8, 7, 4, 10};

float price[6]={0,11.95,7.95,19.95,24.95,15.25};

func1(product);

func2(product,quantity,price);

func3(product,quantity,destination,price);

func4(product,quantity,price);

func5(product,quantity,origination,price);

func6(product,quantity,origination,price);

func7(origination,destination);

func8(product,quantity,origination,destination,price);

func9(destination);

func10(product,quantity,destination,price);

}

Explanation:

The program inputs order, products and cities products are shipped.

After a series of conditional requirements being met, will output the destination each product os going to and the number of products with its associated price.

3 0
4 years ago
Read 2 more answers
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