Let's solve this dynamic programming resource allocation problem **step-by-step**. ### **Step 1: Define the Problem** - **Projects:** 4 (let's call them P1, P2, P3, P4) - **Resources:** 4 (to allocate among projects) - **Profit Matrix:** Entry `(i, j)` = profit from assigning `j` resources to project `i`. We need to assign a total of 4 resources among the 4 projects to **maximize the total profit**. --- ### **Step 2: Dynamic Programming Table Setup** Let \( f(i, j) \) be the maximum profit by assigning `j` resources to the **first `i` projects**. - \( f(0, j) = 0 \) (no profit if there are no projects) - \( f(i, 0) = 0 \) (no profit if no resources) Our goal: **Find \( f(4, 4) \)** --- ### **Step 3: Recursive Relation** For each project \( i \) and resources \( j \): \[ f(i, j) = \max_{k = 0}^{j} \left\{ f(i-1, j-k) + \text{profit}[i][k] \right\} \] Where: - \( k \) = number of resources allocated to project \( i \) - \( \text{profit}[i][k] \) = profit for \( k \) resources to project \( i \) --- ### **Step 4: Build the Table** #### **Profit Matrix (from question):** | | 1 | 2 | 3 | 4 | |---------|---|---|---|---| | **P1** | 3 | 7 | 10| 12| | **P2** | 1 | 2 | 6 | 9 | | **P3** | 2 | 4 | 8 | 9 | | **P4** | 4 | 2 | 7 | 10| **Note:** Usually, profit[project][allocated resources], with 0-based indexing, so adjust accordingly. --- #### **Step 4a: Initialize Table** Let’s initialize \( f(0, j) = 0 \) for all \( j \). --- #### **Step 4b: Fill Table Iteratively** Let’s compute for each project: --- #### **Project 1 (i = 1):** Allocate all resources to P1: - \( f(1, 0) = 0 \) - \( f(1, 1) = 3 \) - \( f(1, 2) = 7 \) - \( f(1, 3) = 10 \) - \( f(1, 4) = 12 \) --- #### **Project 2 (i = 2):** We need to consider the allocation between P1 and P2. **For \( f(2, j) \):** \[ f(2, j) = \max_{k=0}^{j} \left\{ f(1, j-k) + \text{profit}[2][k] \right\} \] where profit[2][k] = profit for P2 with k resources. - profit[2][1]=1, [2]=2, [3]=6, [4]=9 \[ \begin{align*} f(2, 0) &= f(1,0) + \text{profit}[2][0] = 0 + 0 = 0 \\ f(2, 1) &= \max \{f(1,1)+\text{profit}[2][0],\ f(1,0)+\text{profit}[2][1]\} \\ &= \max \{3+0,\ 0+1\} = 3 \\ f(2, 2) &= \max \{f(1,2)+\text{profit}[2][0],\ f(1,1)+\text{profit}[2][1],\ f(1,0)+\text{profit}[2][2]\} \\ &= \max \{7+0,\ 3+1,\ 0+2\} = 7 \\ f(2, 3) &= \max \{f(1,3)+\text{profit}[2][0],\ f(1,2)+\text{profit}[2][1],\ f(1,1)+\text{profit}[2][2],\ f(1,0)+\text{profit}[2][3]\} \\ &= \max \{10+0,\ 7+1,\ 3+2,\ 0+6\} = \max\{10, 8, 5, 6\} = 10 \\ f(2, 4) &= \max \{f(1,4)+\text{profit}[2][0],\ f(1,3)+\text{profit}[2][1],\ f(1,2)+\text{profit}[2][2],\ f(1,1)+\text{profit}[2][3],\ f(1,0)+\text{profit}[2][4]\} \\ &= \max \{12+0, 10+1, 7+2, 3+6, 0+9\} = \max\{12, 11, 9, 9, 9\} = 12 \\ \end{align*} \] --- #### **Project 3 (i = 3):** profit[3][1]=2, [2]=4, [3]=8, [4]=9 \[ \begin{align*} f(3, 0) &= 0 \\ f(3, 1) &= \max \{f(2,1)+0,\ f(2,0)+2\} = \max\{3,2\}=3 \\ f(3, 2) &= \max \{f(2,2)+0,\ f(2,1)+2,\ f(2,0)+4\} = \max\{7,5,4\}=7 \\ f(3, 3) &= \max \{f(2,3)+0,\ f(2,2)+2,\ f(2,1)+4,\ f(2,0)+8\} = \max\{10,9,7,8\}=10 \\ f(3, 4) &= \max \{f(2,4)+0,\ f(2,3)+2,\ f(2,2)+4,\ f(2,1)+8,\ f(2,0)+9\} = \max\{12,12,11,11,9\}=12 \\ \end{align*} \] --- #### **Project 4 (i = 4):** profit[4][1]=4, [2]=2, [3]=7, [4]=10 \[ \begin{align*} f(4, 0) &= 0 \\ f(4, 1) &= \max \{f(3,1)+0,\ f(3,0)+4\} = \max\{3,4\}=4 \\ f(4, 2) &= \max \{f(3,2)+0,\ f(3,1)+4,\ f(3,0)+2\} = \max\{7,7,2\}=7 \\ f(4, 3) &= \max \{f(3,3)+0,\ f(3,2)+4,\ f(3,1)+2,\ f(3,0)+7\} = \max\{10,11,5,7\}=11 \\ f(4, 4) &= \max \{f(3,4)+0,\ f(3,3)+4,\ f(3,2)+2,\ f(3,1)+7,\ f(3,0)+10\} = \max\{12,14,9,10,10\}=14 \\ \end{align*} \] --- ### **Step 5: Final Answer** \[ \boxed{14} \] The **maximum possible profit is 14**. --- ### **Step 6: Find the Allocation** Let’s **trace back** for \( f(4,4) = 14 \): - \( f(4,4) = f(3,3) + 4 = 10 + 4 \), so Project 4 gets **1 resource**. - \( f(3,3) = f(2,3) + 0 = 10 + 0 \), so Project 3 gets **0 resources**. - \( f(2,3) = f(1,3) + 0 = 10 + 0 \), so Project 2 gets **0 resources**. - All 4 resources go to Project 1 (since \( f(1,3) \) is used, but we need the initial allocation): But we see this doesn't add up, let's check allocations more carefully. At \( f(4,4)=14 \), this came from \( f(3,3)+4 \), so Project 4 gets 1, the remaining 3 resources are allocated among Projects 1-3. At \( f(3,3)=10 \), this came from \( f(2,3)+0 \), so Project 3 gets 0, the remaining 3 resources go to Projects 1-2. At \( f(2,3)=10 \), this came from \( f(1,3)+0 \), so Project 2 gets 0, thus Project 1 gets all 3 resources. But the profits don't add up to 14: - P1 with 3 resources: profit = 10 - P2 with 0 resources: profit = 0 - P3 with 0 resources: profit = 0 - P4 with 1 resource: profit = 4 Total: 10 + 0 + 0 + 4 = 14 Wait, this matches! So the optimal allocation is: --- ### **Step 7: Optimal Allocation** - **Project 1:** 3 resources (profit 10) - **Project 2:** 0 resources (profit 0) - **Project 3:** 0 resources (profit 0) - **Project 4:** 1 resource (profit 4) **Total Profit = 10 + 0 + 0 + 4 = 14** --- ## **Final Answer** ### **Maximum Profit:** \(\boxed{14}\) ### **Optimal Allocation:** - **Project 1:** 3 resources - **Project 2:** 0 resources - **Project 3:** 0 resources - **Project 4:** 1 resource --- **If you want all possible optimal allocations (since there may be ties), you can check other paths, but this is one optimal allocation.**