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3 years ago

In an all-integer linear program,

Computers and Technology

1 answer:

xenn [34]3 years ago

8 0

Answer: E

All decision variables must be integer

Explanation:

The integer-linear program is a program in which the objective function and any constraints are all linear ie all of the variables are restricted to be integers.

A linear program is Np complete. Also every decision variable appear in any constraints must also appear in the objective function, possibly with zero coefficient if needed.

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What is the best anime of all time?

Blizzard [7]

(My opinion) Naruto! (then Boruto is on it's way to becoming good)

but other people might say my hero academia or any of the dragon ball saga

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3 years ago

____ gets its name from the notion that it ignores the traditional A, B, and C class designations for IP addresses and can there

Sedbober [7]

Classless Inter-Domain Routing gets its name from the notion that it ignores the traditional A, B, and C class designations for IP addresses and can therefore set the network-host ID boundary wherever it wants to, in a way that simplifies routing across the resulting IP address spaces.

Explanation:

A router is a networking device that helps in connecting multiple networks. Classless Inter-Domain Routing (CIDR) is used for creating IP addresses and IP routing. CIDR was introduced in the year 1993 by “The Internet Engineering Task Force”. The classful network design was replaced by CIDR in the Internet.

The IP addresses are responsible for sending the particular information packets to specific computers. Classless inter-domain routing helps in improving the allocation of IP addresses.

7 0

3 years ago

Read 2 more answers

Write a method called classSplit that accepts a file scanner as input. The data in the file represents a series of names and gra

Lelu [443]

Answer:

public static void classSplit(Scanner file){

int studentsOf21 = 0, studentsOf22 = 0,studentsOf23 = 0, studentsOf24 = 0;

while (file.hasNext() == true){

String studentName = file.next();

int year = file.nextInt();

switch (year){

case 2021:

studentsOf21++;

break;

case 2022:

studentsOf22++;

break;

case 2023:

studentsOf23++;

break;

case 2024:

studentsOf24++;

break;

}

int totalGraduate = studentsOf21+studentsOf22+studentsOf23+studentsOf24;

System.out.printf("students in class of 2021: %.2f%n", students2021*100.0/totalGraduate);

System.out.printf("students in class of 2022: %.2f%n", students2022*100.0/totalGraduate);

System.out.printf("students in class of 2023: %.2f%n", students2023*100.0/totalGraduate);

System.out.printf("students in class of 2024: %.2f%n", students2024*100.0/totalGraduate);

}

Explanation:

The classSplit method of the java class accepts a scanner object, split the object by iterating over it to get the text (for names of students) and integer numbers (for graduation year). It does not need a return statement to ask it prints out the percentage of graduates from 2021 to 2024.

4 0

3 years ago

Suppose we compute a depth-first search tree rooted at u and obtain a tree t that includes all nodes of g.

Temka [501]

G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

An algorithm for navigating or examining tree or graph data structures is called depth-first search. The algorithm moves as far as it can along each branch before turning around, starting at the root node.

The breadth-first search strategy can be used to look for a node in a tree data structure that has a specific property. Before moving on to the nodes at the next depth level, it begins at the root of the tree and investigates every node there.

First, we reveal that G exists a tree when both BFS-tree and DFS-tree are exact.

If G and T are not exact, then there should exist a border e(u, v) in G, that does not belong to T.

In such a case:

- in the DFS tree, one of u or v, should be a prototype of the other.

- in the BFS tree, u and v can differ by only one level.

Since, both DFS-tree and BFS-tree are the very tree T,

it follows that one of u and v should be a prototype of the other and they can discuss by only one party.

This means that the border joining them must be in T.

So, there can not be any limits in G which are not in T.

In the two-part of evidence:

Since G is a tree, per node has a special path from the root. So, both BFS and DFS have the exact tree, and the tree is the exact as G.

The complete question is:

We have a connected graph G = (V, E), and a specific vertex u ∈ V.

Suppose we compute a depth-first search tree rooted at u, and obtain a tree T that includes all nodes of G.

Suppose we then compute a breadth-first search tree rooted at u, and obtain the same tree T.

Prove that G = T. (In other words, if T is both a depth-first search tree and a breadth-first search tree rooted at u, then G cannot contain any edges that do not belong to T.)

To learn more about DFS and BFS, refer to:

brainly.com/question/13014003

#SPJ4

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2 years ago

Write a SELECT statement that displays each product purchased, the category of the product, the number of each product purchased

Blababa [14]

Answer:

y6ou dont have mind

Explanation:

6 0

3 years ago