Act like a helpful tutor and exlain me :

Of course! Here is a step-by-step solution to the problem.

### **1. Analysis of Subproblems**

Here's a breakdown of the reason for branching or fathoming a

Question

Act like a helpful tutor and exlain me :

Of course! Here is a step-by-step solution to the problem.

### **1. Analysis of Subproblems**

Here's a breakdown of the reason for branching or fathoming at each step of the branch-and-bound process. The process aims to find the best integer solution, and we keep track of the best one found so far, called the **incumbent**, with its objective value as the **Lower Bound (LB)**. Initially, $LB = -\infty$.

* **Subproblem 1 (t=1):** This is the initial LP relaxation.
    * **Action: Branched.**
    * **Reason:** The solution is $(x_1, x_2) = (\frac{15}{4}, \frac{9}{4})$ or $(3.75, 2.25)$, which is not integer-feasible. Since the solution is non-integer, we must branch on a fractional variable to continue the search. The algorithm branches on $x_1$, creating two new subproblems with the constraints $x_1 \le 3$ and $x_1 \ge 4$. The objective value $z=41.25$ serves as the initial overall upper bound.

* **Subproblem 2 (t=2):** This node adds the constraint $x_1 \ge 4$ to Subproblem 1.
    * **Action: Branched.**
    * **Reason:** The solution is $(x_1, x_2) = (4, \frac{9}{5})$ or $(4, 1.8)$. This is not integer-feasible because $x_2$ is a frac

VIPSolutions · Accepted Answer

Absolutely! Here’s a step-by-step explanation of the branch-and-bound process, closely following your structure and logic, but with clear, tutor-style guidance and the same content.

---

## **1. Analysis of Subproblems**

Let’s walk through each subproblem in the branch-and-bound process. Remember, our goal is to find the best integer solution (the **incumbent**), keeping track of its objective value as the **Lower Bound (LB)**. Initially, $LB = -\infty$.

---

### **Subproblem 1 (t=1):** *Initial LP Relaxation*

- **Action:** Branched
- **Why?** The solution to the relaxed LP is $(x_1, x_2) = (3.75, 2.25)$, which is not integer-feasible. Since we need integer values, we must branch on a fractional variable. Here, we branch on $x_1$, creating two new cases: $x_1 \le 3$ and $x_1 \ge 4$. The LP relaxation’s optimal value ($z=41.25$) is our initial overall upper bound.

---

### **Subproblem 2 (t=2):** *Add $x_1 \ge 4$*

- **Action:** Branched
- **Why?** The solution is now $(x_1, x_2) = (4, 1.8)$, which is still not integer-feasible (since $x_2$ is fractional). So, we branch on $x_2$, giving us two more subproblems: $x_2 \ge 2$ and $x_2 \le 1$.

---

### **Subproblem 4 (t=3):** *Add $x_2 \ge 2$ to previous*

- **Action:** Fathomed (Pruned)
- **Why?** Adding $x_1 \ge 4$ and $x_2 \ge 2$ makes the problem **infeasible** (no solution satisfies all constraints). Since this branch can't produce a solution, we prune it.

---

### **Subproblem 5 (t=4):** *Add $x_2 \le 1$ to previous*

- **Action:** Branched
- **Why?** Now the solution is $(x_1, x_2) \approx (4.44, 1)$. $x_1$ is not an integer, so we branch on $x_1$, making two subproblems: $x_1 \ge 5$ and $x_1 \le 4$.

---

### **Subproblem 7 (t=5):** *Add $x_1 \le 4$ to previous*

- **Action:** Fathomed (Pruned)
- **Why?** The solution here is $(x_1, x_2) = (4, 1)$, which **is** integer-feasible, with $z=37$. This is our first integer solution, so we update $LB = 37$. We can fathom this node.

---

### **Subproblem 6 (t=6):** *Add $x_1 \ge 5$ to previous*

- **Action:** Fathomed (Pruned)
- **Why?** The solution is $(x_1, x_2) = (5, 0)$, another integer-feasible solution, with $z=40$. Since $z=40 > LB=37$, we update our incumbent and set $LB=40$. Node is fathomed.

---

### **Subproblem 3 (t=7):** *Add $x_1 \le 3$ to initial*

- **Action:** Fathomed (Pruned)
- **Why?** This subproblem’s optimal value is $z=39$, but we already have a better integer solution ($LB=40$). Because this branch cannot improve upon our best solution, we prune it (**fathoming by bound**).

---

## **2. LP Model for Subproblem 5**

Let’s write the full LP for Subproblem 5. This subproblem has all constraints inherited from its path in the tree:

- **Original constraints:** $x_1 + x_2 \le 6$, $9x_1 + 5x_2 \le 45$
- **From Subproblem 2:** $x_1 \ge 4$
- **From Subproblem 5:** $x_2 \le 1$

So, the LP for Subproblem 5 is:

**Maximize:** $z = 8x_1 + 5x_2$  
**Subject to:**
$$
\begin{align*}
x_1 + x_2 &\leq 6 \
9x_1 + 5x_2 &\leq 45 \
x_1 &\geq 4 \
x_2 &\leq 1 \
x_1, x_2 &\geq 0
\end{align*}
$$

---

## **3. Optimal Solution for the IP Problem**

After fully exploring the branch-and-bound tree:

1. At **t=5**, we found integer solution $(4, 1)$ with $z=37$. So, $LB=37$.
2. At **t=6**, we found a better integer solution $(5, 0)$ with $z=40$. Update $LB=40$.
3. At **t=7**, the last subproblem has $z=39 < LB=40$, so it’s pruned.

The process ends. The best integer-feasible solution we found is the **optimal solution**:

$$
\boxed{
x_1 = 5, \quad x_2 = 0, \quad z = 40
}
$$

---

Let me know if you’d like any further details or clarifications!

VIPSolutions · Answer

Absolutely! Let's go through this step-by-step, in a clear and friendly way, just like a Chegg tutor would. I'll explain each part carefully and keep it easy to follow.

---

## 1. Breaking Down the Subproblems

In branch-and-bound, we're trying to find the best **integer** solution to our problem. We start with a relaxed version (allowing fractions), and as we go along, we branch on variables that are fractional, adding constraints to force them to be integers.

At the beginning, our **best known solution** (called the **incumbent**) has an **objective value** which we'll track as the **Lower Bound** (LB). Initially, LB is set to negative infinity because we haven't found any feasible solutions yet.

### **Step 1: The initial problem (t=1)**

- We solve the LP relaxation (ignoring integer constraints). The solution is (x1, x2) = (3.75, 2.25).
- This solution isn't integer because both variables are fractional.
- Since it's not feasible as an integer solution, we **branch** on one of the fractional variables, say x1.
- We create two subproblems: one with x1 ≤ 3 and another with x1 ≥ 4.
- The LP relaxation's optimal value here is 41.25, which is our current **upper bound** on the best possible solution.

---

### **Step 2: Branching on x1 (t=2)**

- Now, we look at the subproblem with x1 ≥ 4.
- The LP solution here gives us (x1, x2) = (4, 1.8), which again isn't integer because x2 is fractional.
- So, we branch again, this time on x2, creating two more subproblems:
  - One with x2 ≥ 2
  - One with x2 ≤ 1

---

### **Step 3: Pruning infeasible branch (x2 ≥ 2)**

- The subproblem with x2 ≥ 2 (adding to x1 ≥ 4) turns out to have **no feasible solution**.
- We call this **infeasible**, so we **prune** this branch, meaning we don't explore it further.

---

### **Step 4: Branching on x2 ≤ 1 (t=4)**

- The other branch with x2 ≤ 1 gives us a solution around (x1 ≈ 4.44, x2 = 1).
- Here, x1 is still fractional, so we branch again on x1:
  - x1 ≥ 5
  - x1 ≤ 4

---

### **Step 5: Finding integer solutions**

- For the branch with x1 ≤ 4 (t=5), the solution is (x1=4, x2=1), which **is** an integer solution.
- The objective value here is 37.
- Since this is our **first feasible integer solution**, we update our **LB** to 37.
- We **fathom** this branch because we've found a feasible solution, so no need to explore it further.

- For the branch with x1 ≥ 5 (t=6), the solution is (x1=5, x2=0), which is also integer-feasible.
- The objective value is 40, which is **better** than our current LB of 37.
- We update LB to 40, meaning we've found a better solution now.

---

### **Step 6: Pruning the last branch (x1 ≤ 3)**

- The remaining subproblem with x1 ≤ 3 has an optimal value of 39.
- Since 39 is **less than** our current LB of 40, we **fathom** this branch because it can't produce a better solution.

---

## 2. Writing the LP Model for the Important Subproblem (Subproblem 5)

Let's write down the LP for the subproblem where:

- We keep all previous constraints
- Add the new constraints from the branch

Starting with the original constraints:

- x1 + x2 ≤ 6
- 9x1 + 5x2 ≤ 45

From the previous branches, for Subproblem 5, we also include:

- x1 ≥ 4 (from earlier branch)
- x2 ≤ 1 (from the current branch)

Putting it all together, the LP for this subproblem is:

**Maximize**: 8x1 + 5x2

**Subject to:**

- x1 + x2 ≤ 6

- 9x1 + 5x2 ≤ 45

- x1 ≥ 4

- x2 ≤ 1

- x1 ≥ 0, x2 ≥ 0

---

## 3. The Final Answer: What's the Best Integer Solution?

After exploring all branches:

- The best feasible integer solution we found was (x1=5, x2=0) with an objective value of 40.
- We update our **incumbent** to this solution.
- The process stops because all other branches are either infeasible or have bounds worse than 40.

**Therefore, the optimal solution to the integer problem is:**

- x1 = 5
- x2 = 0
- with an **optimal objective value** of 40

---

**Hope that clears things up!** Feel free to ask if you'd like me to go over any step again or clarify anything further.

VIPSolutions · Answer

Absolutely! Let's walk through this step-by-step, in a clear and friendly way, just like a Chegg tutor would. I'll explain each part carefully and keep it easy to follow.

---

## 1. Breaking Down the Subproblems

In branch-and-bound, we're trying to find the best **integer** solution to our problem. We start with a relaxed version (allowing fractions), and as we go along, we branch on variables that are fractional, adding constraints to force them to be integers.

At the beginning, our **best known solution** (called the **incumbent**) has an **objective value** which we'll track as the **Lower Bound** (LB). Initially, LB is set to negative infinity because we haven't found any feasible solutions yet.

### **Step 1: The initial problem (t=1)**

- We solve the LP relaxation (ignoring integer constraints). The solution is *(x₁, x₂) = (3.75, 2.25)*.
- This solution isn't integer because both variables are fractional.
- Since it's not feasible as an integer solution, we **branch** on one of the fractional variables, say **x₁**.
- We create two subproblems: one with **x₁ ≤ 3** and another with **x₁ ≥ 4**.
- The LP relaxation's optimal value here is **41.25**, which is our current **upper bound** on the best possible solution.

---

### **Step 2: Branching on x₁ (t=2)**

- Now, we look at the subproblem with **x₁ ≥ 4**.
- The LP solution here gives us *(x₁, x₂) = (4, 1.8)*, which again isn't integer because **x₂** is fractional.
- So, we branch again, this time on **x₂**, creating two more subproblems:
  - **x₂ ≥ 2**
  - **x₂ ≤ 1**

---

### **Step 3: Pruning infeasible branch (x₂ ≥ 2)**

- The subproblem with **x₂ ≥ 2** (adding to x₁ ≥ 4) turns out to have **no feasible solution**.
- We call this **infeasible**, so we **prune** this branch, meaning we don't explore it further.

---

### **Step 4: Branching on x₂ ≤ 1 (t=4)**

- The other branch with **x₂ ≤ 1** gives us an approximate solution *(x₁ ≈ 4.44, x₂=1)*.
- Here, **x₁** is still fractional, so we **branch again** on **x₁**:
  - **x₁ ≥ 5**
  - **x₁ ≤ 4**

---

### **Step 5: Finding integer solutions**

- For the branch **x₁ ≤ 4** (t=5), the solution is *(x₁=4, x₂=1)*, which **is** an integer solution.
- The **objective value** at this point is **37**.
- Since this is our **first feasible integer solution**, we **update** our **LB** to **37**.
- We **fathom** this branch because we've found a feasible solution, so no need to explore it further.

- For the branch **x₁ ≥ 5** (t=6), the solution is *(x₁=5, x₂=0)*, which is also integer-feasible.
- The **objective value** here is **40**, which is **better** than our current LB of 37.
- We **update** LB to **40**, meaning we've now found our **best solution** so far.

---

### **Step 6: Pruning the last branch (x₁ ≤ 3)**

- The remaining subproblem with **x₁ ≤ 3** has an optimal value of **39**.
- Since **39** is **less than** our current LB of **40**, we **fathom** this branch because it can't improve our current best solution.

---

## 2. Writing the LP Model for the Key Subproblem (Subproblem 5)

Let's write down the LP for the subproblem where:

- All previous constraints are included
- Additional constraints from the branch are added

Starting with the original constraints:

- **x₁ + x₂ ≤ 6**
- **9x₁ + 5x₂ ≤ 45**

From the previous branches, for **Subproblem 5**, we include:

- **x₁ ≥ 4** (from earlier branch)
- **x₂ ≤ 1** (from current branch)

Putting it all together, the LP for this subproblem is:

**Maximize:** 8x₁ + 5x₂

**Subject to:**

- x₁ + x₂ ≤ 6

- 9x₁ + 5x₂ ≤ 45

- x₁ ≥ 4

- x₂ ≤ 1

- x₁ ≥ 0, x₂ ≥ 0

---

## 3. The Final Answer: What's the Best Integer Solution?

After exploring all branches, the **best feasible integer solution** found is:

- **x₁ = 5**

- **x₂ = 0**

- with an **objective value** of **40**.

This is the **optimal solution** to the integer program.

---

**Hope that helps!** Feel free to ask if you'd like me to clarify any part further.

VIPSolutions · Answer

Certainly! Here's the explanation with equations in rendered form, presented clearly and in a friendly, tutor-like style:

---

## 1. Breaking Down the Subproblems

In branch-and-bound, our goal is to find the **best integer solution** to the problem. We start with the LP relaxation (ignoring integrality constraints), then branch on variables with fractional values to enforce integrality.

We keep track of the **incumbent** (best integer solution found so far) and its **objective value**, which we call the **Lower Bound (LB)**. Initially, set LB = $-\infty$.

---

### **Step 1: The initial problem (t=1)**

- Solve the LP relaxation:

$$
\text{Maximize } z = 8x_1 + 5x_2
$$
Subject to:
$$
x_1 + x_2 \leq 6
$$
$$
9x_1 + 5x_2 \leq 45
$$
$$
x_1, x_2 \geq 0
$$

- The LP solution (relaxation) gives:
$$
x_1 = 3.75, \quad x_2 = 2.25
$$
- Since these are **not** integers, we **branch** on $x_1$, creating two subproblems:
  - $x_1 \leq 3$
  - $x_1 \geq 4$

- The LP’s optimal value at this node is:
$$
z = 41.25
$$
which provides an **upper bound** for the branch.

---

### **Step 2: Branching on $x_1 \geq 4$ (t=2)**

- The LP solution here is:
$$
x_1 = 4, \quad x_2 = 1.8
$$
- Not integer because $x_2$ is fractional, so we **branch** on $x_2$:

- $x_2 \geq 2$
  - $x_2 \leq 1$

---

### **Step 3: Infeasible branch with $x_2 \geq 2$**

- Adding $x_2 \geq 2$ to the constraints with $x_1 \geq 4$, the problem becomes **infeasible**:

$$
\text{Infeasible} \implies \text{prune this branch}
$$

---

### **Step 4: Branching on $x_2 \leq 1$ (t=4)**

- The LP solution here is approximately:

$$
x_1 \approx 4.44, \quad x_2 = 1
$$

- Since $x_1$ is fractional, we **branch** on $x_1$:

- $x_1 \geq 5$
  - $x_1 \leq 4$

---

### **Step 5: Integer solutions found**

- **Branch with $x_1 \leq 4$ (t=5):**

- Solution: $\boxed{x_1=4, \quad x_2=1}$
  - Objective value:

$$
    z = 8 \times 4 + 5 \times 1 = 32 + 5 = 37
    $$

- This is a feasible integer solution, so **update**:

$$
    \text{LB} = 37
    $$

- **Fathom** this branch, as we've found a feasible solution.

- **Branch with $x_1 \geq 5$ (t=6):**

- Solution: $\boxed{x_1=5, \quad x_2=0}$
  - Objective:

$$
    z = 8 \times 5 + 5 \times 0 = 40 + 0 = 40
    $$

- Since 40 > current LB (37), **update**:

$$
    \text{LB} = 40
    $$

- **Fathom** this branch as well.

---

### **Step 6: Remaining branch $x_1 \leq 3$**

- The LP solution yields an optimal value:

$$
z = 39
$$

- But since our **best integer solution** is with $z=40$, and $39 < 40$, we **fathom** this branch by **bound**—no need to explore further.

---

## 2. The LP Model for Subproblem 5

The subproblem is constructed by adding the constraints from the relevant branches:

$$
\begin{cases}
x_1 + x_2 \leq 6 \
9x_1 + 5x_2 \leq 45 \
x_1 \geq 4 \
x_2 \leq 1 \
x_1, x_2 \geq 0
\end{cases}
$$

And the objective:

$$
\boxed{
\text{Maximize } z = 8x_1 + 5x_2
}
$$

---

## 3. Final Answer: The Optimal Integer Solution

After all the exploration, the **best integer-feasible solution** found is:

$$
\boxed{
x_1 = 5, \quad x_2 = 0, \quad z = 40
}
$$

This is the **optimal solution** for the integer problem.

---

I hope this clear, equation-rich explanation helps! Feel free to ask if you'd like me to clarify any step further.

VIPSolutions · Answer

Certainly! Here's the final summary in three lines:

The optimal integer solution is **x1 = 5, x2 = 0** with an objective value of **40**. This solution was found after exploring the branch-and-bound tree and updating the lower bound. All other branches were pruned because they were infeasible or had worse objective values.

✨ VIPSolutions

Question:

Final Answer:

Question:

Final Answer:

Question:

Final Answer:

Question:

Final Answer:

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

Question:

Final Answer:

Question:

Final Answer:

Question:

Final Answer:

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers