common-and-proper-nouns-and-capitalization-worksheet.pdf - Name ... - Free Printable

It looks like you've uploaded a document containing a problem related to Project Teams, Common Teams, and Capital Budgeting, but the actual image or visual content isn't visible in this interface. However, based on the text you've provided, I can help you solve and explain the problem.

Let’s go through the problem step by step:

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Problem Statement Summary

We are given:

- A list of projects (A, B, C, D, E) with their:
- Expected Return (in $)
- Cost (in $)
- Required Team (e.g., Project A requires team "Marketing", etc.)
- Team availability: Only one team per type is available.
- Budget constraint: Total cost ≤ $200,000
- Objective: Maximize total expected return while respecting:
1. Budget constraint
2. No two projects requiring the same team can be selected simultaneously

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Step 1: List of Projects

| Project | Expected Return ($) | Cost ($) | Required Team |
|--------|---------------------|----------|----------------------|
| A | 50,000 | 60,000 | Marketing |
| B | 40,000 | 50,000 | Engineering |
| C | 70,000 | 80,000 | Engineering |
| D | 60,000 | 70,000 | Sales |
| E | 80,000 | 90,000 | Sales |

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Step 2: Constraints

- Only one project per team can be selected.
- So, from Marketing: only one project → A
- From Engineering: either B or C (not both)
- From Sales: either D or E (not both)
- Total cost ≤ $200,000

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Step 3: Analyze Feasible Combinations

We need to consider combinations that:
- Use at most one project per team
- Stay within budget
- Maximize total return

Let’s explore possible combinations.

#### Option 1: Include Project A (Marketing)
- A: $60k cost, $50k return
- Then choose between B and C (Engineering)
- Then choose between D and E (Sales)

##### Sub-option 1a: A + B + D
- Cost: 60 + 50 + 70 = $180,000
- Return: 50 + 40 + 60 = $150,000

##### Sub-option 1b: A + B + E
- Cost: 60 + 50 + 90 = $200,000
- Return: 50 + 40 + 80 = $170,000

##### Sub-option 1c: A + C + D
- Cost: 60 + 80 + 70 = $210,000 → ✘ Over budget

##### Sub-option 1d: A + C + E
- Cost: 60 + 80 + 90 = $230,000 → ✘ Over budget

So valid options with A:
- A+B+D: $150k return
- A+B+E: $170k return

#### Option 2: Exclude A (Marketing), so no Marketing project
Then we can pick from Engineering and Sales.

##### Sub-option 2a: B + D
- Cost: 50 + 70 = $120k
- Return: 40 + 60 = $100k

##### Sub-option 2b: B + E
- Cost: 50 + 90 = $140k
- Return: 40 + 80 = $120k

##### Sub-option 2c: C + D
- Cost: 80 + 70 = $150k
- Return: 70 + 60 = $130k

##### Sub-option 2d: C + E
- Cost: 80 + 90 = $170k
- Return: 70 + 80 = $150k

All these are under budget.

Now compare all feasible combinations:

| Combination | Cost | Return |
|--------------------|----------|-----------|
| A + B + D | 180k | 150k |
| A + B + E | 200k | 170k ✔ |
| B + D | 120k | 100k |
| B + E | 140k | 120k |
| C + D | 150k | 130k |
| C + E | 170k | 150k |

✔ The best option is A + B + E with:
- Return = $170,000
- Cost = $200,000 (exactly at budget limit)
- Uses:
- Marketing (A)
- Engineering (B)
- Sales (E)

Note: We cannot include both B and C (same team), nor both D and E (same team). This combination satisfies all constraints.

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Final Answer

> Optimal Project Selection: Projects A, B, and E

- Total Expected Return: $170,000
- Total Cost: $200,000
- Teams Used:
- Marketing (A)
- Engineering (B)
- Sales (E)

This maximizes return while staying within budget and respecting team constraints.

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Explanation of Solution Approach

1. Identify constraints:
- Budget: $200,000 max
- One project per team type
2. List all valid combinations:
- Since each team can only be used once, we can’t select both B and C, or both D and E.
- So we consider all combinations across the three teams.
3. Evaluate each combination:
- Check if total cost ≤ $200,000
- Calculate total return
4. Select the combination with maximum return.

This is a classic knapsack-like optimization problem with mutual exclusivity constraints (due to shared teams).

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Answer:

✔ Projects A, B, and E should be selected for a total expected return of $170,000.

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