**Problem 3****Objective:** Use the Branch and Bound Algorithm to solve the following integer programming problem. Also, use the simplex algorithm to solve this problem.**Maximize:**\[ 2x - y - r \]**Subject to the constraints:**\[\begin{align*}x - y - 2r &\leq 3 \\x + y + r &\leq 4 \\r &\leq 6 \\x, y, r &\geq 0 \\x, y, r &\in \mathbb{Z}\end{align*}\]**Explanation:**- The objective function to maximize is $2x - y - r$.- There are three constraints involving the variables $x$, $y$, and $r$.- The variables $x$, $y$, and $r$ must all be non-negative integers.- $r$ must be less than or equal to 6.- The solution involves solving this problem using both the Branch and Bound method, which is suitable for integer programming, and the simplex method, which is typically used for linear programming problems.

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Description: **Problem 3****Objective:** Use the Branch and Bound Algorithm to solve the following integer programming problem. Also, use the simplex algorithm to solve this problem.**Maximize:**\[ 2x - y - r \]**Subject to the constraints:**\[\begin{align*}x -

Given :- Using the simplex algorithm, max 2x-y-rs.t. x-y-2r≤3 x+y+r≤4 r≤6 x,y,r≥0 x,y,r ∈ℤ STEP 1 :- Check whether the objective function is to be maximised and all b's are positive. The problem being of maximization type and all b's being ≥ 0 . STEP 2 :- Express the problem in the standard form. By introducing the slack variables s1 , s2 , s3 , the problem n standard form becomes max 2x-y-r +0s1 +0s2 +0s3s.t. x-y-2r +1s1 +0s2 +0s3 =3 x+y+r +0s1 +1s2 +0s3 =4 r +0s1 +0s2 +1s3 =6 x,y,r,s1 ,s2 , s3≥0 x,y,r ∈ℤSTEP 3 :- Find an initial basic feasible solution. There are three equations involving six unknowns and for obtaining a solution , we assign zero values to any two of the variables. We start with a basic solution for which we set x =0, y =0 , r =0. (This basic solution corresponds to the origin in the graphical method ). Substituting x =0, y =0 , r =0 in given equations, we get the basic solution s1 =3 , s2 =4, s3 =6 Since all s1 , s2 , s3 are positive , the basic solution is also feasible and non-degenerate. The basic feasible solution is x =0, y =0 , r =0 (non-basic) and s1 =3 , s2 =4, s3 =6 (basic) 2 -1 -1 0 0 0 CB XB B b a1 a2 a3 a4 a5 a6 Minimum ratio 0 x4 a4 3 1 -1 -2 1 0 0 31 =3 → 0 x5 a5 4 1 1 1 0 1 0 41=4 0 x6 a6 6 0 0 1 0 0 1 Zj - Cj = -2↑ 1 1 0 0 0 For Z1 -C1 = 1.0 +1.0 +0.0 -2 =-2 Similarly others can be calculated. As Cj is positive under some columns , the initial basic feasible solution is not optimal and we proceed to the next steps. Step 5 :- Identify the incoming and outgoing variables. The above table shows that x1 is the incoming variable as its incremental contribution is Cj =-2 is minimum and the column in which it appears is the key column . Dividing the elements under b-column by the corresponding elements of key-column, we find minimum positive ratio is 3. We, therefore, arbitrarily choose the row and column intersection known as the pivot element. The pivot element is 1.Step 6 :- Iterate towards the optimal solution. 2 -1 -1 0 0 0 CB XB B b a1 a2 a3 a4 a5 a6 Minimum ratio 2 x1 a1 3 1 -1 -2 1 0 0 0 x5 a5 1 0 2 3 -1 1 0 13→ 0 x6 a6 6 0 0 1 0 0 1 61 =6 Zj - Cj = 0 -1 -3↑ 2 0 0 For Z1 -C1 = 1.0 +1.0 +2.1-2 =0 Similarly others can be calculated. As Cj is positive under some columns , the initial basic feasible solution is not optimal and we proceed to the next steps. Step 6 :- Identify the incoming and outgoing variables. The above table shows that x3 is the incoming variable as its incremental contribution is Cj =-3 is minimum and the column in which it appears is the key column . Dividing the elements under b-column by the corresponding elements of key-column, we find minimum positive ratio is13 . We, therefore, arbitrarily choose the row and column intersection known as the pivot element. The pivot element is 3.Step 6 :- Iterate towards the optimal solution. 2 -1 -1 0 0 0 CB XB B b a1 a2 a3 a4 a5 a6 Minimum ratio 2 x1 a1 73 1 13 0 13 23 0 7313 =7 -1 x3 a3 13 0 23 1 -13 13 0 12→ 0 x6 a6 173 0 -23 0 13 -13 1 Zj - Cj = 0 -1↑ 0 1 1 0 For Z1 -C1 = 1.0 +1.0 +2.1-2 =0 Similarly others can be calculated. As Cj is positive under some columns , the initial basic feasible solution is not optimal and we proceed to the next steps. Step 6 :- Identify the incoming and outgoing variables. The above table shows that x2 is the incoming variable as its incremental contribution is Cj =-1 is minimum and the column in which it appears is the key column . Dividing the elements under b-column by the corresponding elements of key-column, we find minimum positive ratio is 12. We, therefore, arbitrarily choose the row and column intersection known as the pivot element. The pivot element is 23.This solution is unbounded because of no feasible solution.

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3. Use the Branch and Bound Algorithm to solve the following integer programming problem: Use the simplex algorithm to solve the following problem. тал 2r - у — r s.t. I – y – 2r < 3 I +y+r<4 r<6 I, Y, r > 0 I, Y,r € Z

3. Use the Branch and Bound Algorithm to solve the following integer programming problem: Use the simplex algorithm to solve the following problem. тал 2r - у — r s.t. I – y – 2r < 3 I +y+r<4 r<6 I, Y, r > 0 I, Y,r € Z

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Topic Video

Question

**Problem 3**

**Objective:** Use the Branch and Bound Algorithm to solve the following integer programming problem. Also, use the simplex algorithm to solve this problem.

**Maximize:**
\[
2x - y - r
\]

**Subject to the constraints:**
\[
\begin{align*}
x - y - 2r &\leq 3 \\
x + y + r &\leq 4 \\
r &\leq 6 \\
x, y, r &\geq 0 \\
x, y, r &\in \mathbb{Z}
\end{align*}
\]

**Explanation:**

- The objective function to maximize is $2x - y - r$.
- There are three constraints involving the variables $x$, $y$, and $r$.
- The variables $x$, $y$, and $r$ must all be non-negative integers.
- $r$ must be less than or equal to 6.
- The solution involves solving this problem using both the Branch and Bound method, which is suitable for integer programming, and the simplex method, which is typically used for linear programming problems.

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