If You Can Only Answer A Couple Of My Questions Still

If You Can Only Answer A Couple Of The Questions I Am Still Willing To

If you can only answer a couple of the questions I am still willing to pay you $5 for #1 and $3 for the other five (#2-#6). Show all work (step by step) Reference text is Chapters 1-5 of Engineering Economy 15th Edition Sullivan, Wicks, Koelling - Older editions work fine.

Paper For Above instruction

This analytical essay addresses selected financial and economic problems based on engineering economy principles derived from chapters 1-5 of Sullivan, Wicks, and Koelling's "Engineering Economy" (15th edition). The scope includes profit maximization through sales volume analysis, inflation rate computation, present value calculations, retirement savings planning, loan amortization, and cash flow equivalence, demonstrating application of core economic analysis tools in engineering contexts.

Analysis of Selected Engineering Economic Problems

Problem 1: Revenue and Profit Maximization Analysis

The company produces widgets with fixed costs of $45,000 per month and variable costs of $45 per widget. The selling price per widget is p = $0.4D, where D represents the demand. The goal is to determine the sales volume at which revenue and profit are maximized, as well as identifying the breakeven points.

Assuming the demand D equates to the sales volume (Q), the revenue is R = p × Q = (0.4 × Q) × Q = 0.4Q^2, which suggests a quadratic relationship between price, demand, and revenue. Close examination indicates a need to clarify the demand-price relationship; assuming a linear demand function is typical, but given the information, the focus is on revenue maximization first.

Revenue is maximized when the incremental revenue of additional units equals zero, which implies considering the revenue function's derivative concerning Q. However, without explicit demand constraints, approaching the problem through profit analysis provides clearer insights.

The profit function Π is: Π = Total Revenue – Total Costs = pQ – (Fixed Costs + Variable Costs per unit × Q) = pQ – (45,000 + 45Q).

Maximize Π with respect to Q by taking the derivative: dΠ/dQ = p – 45.

Set dΠ/dQ = 0 for maximum profit: p – 45 = 0 ⇒ p = 45.

Recall p = 0.4D; to find the sales volume at maximum profit, substitute p = 45: 45 = 0.4D ⇒ D = 45 / 0.4 = 112.5.

This demand level corresponds to a sales volume Q = D = 112.5 units. The maximum revenue at this point is R = p × Q = 45 × 112.5 = $5,062.50.

For breakeven analysis, set profit to zero: 0 = pQ – (45,000 + 45Q). Rearranged: pQ – 45Q = 45,000 ⇒ (p – 45)Q = 45,000. At breakeven, p = variable cost per unit, which is $45, so: (45 – 45)Q = 45,000; 0 × Q = 45,000, which is impossible; thus, the breakeven volume occurs at p = 45 when total revenue exactly matches total costs: R = 45Q = 45,000 + 45Q, implying 45Q – 45Q = 45,000, which can't hold; therefore, the breakeven volume occurs when the revenue just covers fixed costs plus variable costs:

Total Revenue: pQ = fixed costs + variable costs ⇒ pQ = 45,000 + 45Q ⇒ pQ – 45Q = 45,000 ⇒ (p – 45)Q = 45,000.

At breakeven, p must be greater or equal to 45 to produce positive volume; if p > 45, then:

Q = 45,000 / (p – 45).

But since p = 0.4D and demand depends on price, the actual calculation requires specific demand-price functions or constraints not provided here. Therefore, assumptions are necessary for precise numeric values, but the fundamental approach involves setting revenue equal to total costs for breakeven point calculation.

Problem 2: Inflation Rate Calculation

The subway fare increased from 10 cents in 1952 to $2.25 today, over a period of 60 years. To find the average yearly inflation rate, we use the compound interest formula for growth:

\[ P_{\text{final}} = P_{\text{initial}} \times (1 + r)^n \]

Where:

- \( P_{\text{final}} = 2.25 \) dollars,

- \( P_{\text{initial}} = 0.10 \) dollars,

- \( n = 60 \) years,

- \( r \) is the annual inflation rate.

Rearranged to solve for \( r \):

\[ r = \left(\frac{P_{\text{final}}}{P_{\text{initial}}}\right)^{1/n} - 1 \]

Substituting the values:

\[ r = \left(\frac{2.25}{0.10}\right)^{1/60} - 1 = (22.5)^{1/60} - 1 \]

Calculating:

\[ \ln(r + 1) = \frac{1}{60} \times \ln(22.5) \]

Using calculator:

\[ \ln(22.5) \approx 3.113 \]

\[ \ln(r + 1) \approx \frac{3.113}{60} \approx 0.0519 \]

Exponentiating:

\[ r + 1 = e^{0.0519} \approx 1.0533 \]

Therefore,

\[ r \approx 0.0533 \text{ or } 5.33\% \]

This indicates an average annual inflation rate of approximately 5.33% over the past 60 years.

Problem 3: Present Worth of Future Receipt

Given a future receipt of $22,000 five years from now and a nominal interest rate of 8% compounded quarterly, the present worth (PW) is calculated using the formula:

\[ PW = \frac{F}{(1 + i/n)^{nt}} \]

Where:

- \( F = 22,000 \),

- \( t = 5 \) years,

- \( i = 8\% = 0.08 \),

- \( n = 4 \) (quarterly compounding).

Calculating the periodic rate:

\[ i/n = 0.08/4 = 0.02 \]

Total number of periods:

\[ nt = 4 \times 5 = 20 \]

Now, substitute:

\[ PW = \frac{22,000}{(1 + 0.02)^{20}} = \frac{22,000}{(1.02)^{20}} \]

Calculating denominator:

\[ (1.02)^{20} \approx e^{20 \times \ln(1.02)} \approx e^{20 \times 0.0198} = e^{0.396} \approx 1.486 \]

Finally:

\[ PW \approx \frac{22,000}{1.486} \approx 14,808.89 \]

The present worth of the future $22,000 receipt is approximately $14,809.

Problem 4: Retirement Savings Planning

A 30-year-old student wants to save $1,000,000 by age 65, so in 35 years. Assuming an annual interest rate of 5%, the annual savings needed are calculated using the future value of an ordinary annuity:

\[ FV = P \times \frac{(1 + r)^n - 1}{r} \]

Rearranged to find annual payment \( P \):

\[ P = \frac{FV \times r}{(1 + r)^n - 1} \]

Where:

- \( FV = 1,000,000 \),

- \( r=0.05 \),

- \( n=35 \).

Calculating denominator:

\[ (1.05)^{35} - 1 \approx e^{35 \times 0.04879} - 1 \approx e^{1.70765} - 1 \approx 5.514 - 1 = 4.514 \]

Calculating numerator:

\[ 1,000,000 \times 0.05 = 50,000 \]

Thus, annual savings needed:

\[ P = \frac{50,000}{4.514} \approx 11,078.65 \]

The student must save approximately $11,079 annually to reach his goal of $1 million by age 65.

Problem 5: Car Loan Payment Calculation

A loan of $20,000 is to be repaid over 4 years with monthly payments. The APR is 6% compounded monthly, so the monthly interest rate is 0.06/12 = 0.005. The total number of payments is 4 years × 12 months/year = 48 months.

The monthly payment \( M \) is calculated using the amortization formula:

\[ M = P \times \frac{i (1 + i)^n}{(1 + i)^n - 1} \]

Where:

- \( P = 20,000 \),

- \( i = 0.005 \),

- \( n = 48 \).

Calculations:

\[ (1 + i)^n = (1.005)^{48} \approx e^{48 \times 0.0049875} \approx e^{0.239} \approx 1.27 \]

Now, compute payment:

\[ M = 20,000 \times \frac{0.005 \times 1.27}{1.27 - 1} = 20,000 \times \frac{0.00635}{0.27} \approx 20,000 \times 0.02352 \approx 470.40 \]

The monthly payment is approximately $470.40.

Problem 6: Cash Flows and Uniform Annual Equivalent

The cash flows at year-end are: $7,000, $15,000, $23,000, $31,000, and $39,000. With an interest rate of 10% per year, the goal is to determine the uniform annual equivalent (UAE).

The approach involves calculating the present worth of all cash flows and then converting this present worth into an equivalent annual amount.

Present worth (PW):

\[ PW = \sum_{i=1}^n \frac{C_i}{(1 + i)^i} \]

Calculations for each cash flow:

• Year 1: \( \frac{7,000}{(1.10)^1} \approx 7,000 / 1.10 \approx 6,364 \)
• Year 2: \( 15,000 / (1.10)^2 \approx 15,000 / 1.21 \approx 12,397 \)
• Year 3: \( 23,000 / (1.10)^3 \approx 23,000 / 1.331 \approx 17,278 \)
• Year 4: \( 31,000 / (1.10)^4 \approx 31,000 / 1.4641 \approx 21,157 \)
• Year 5: \( 39,000 / (1.10)^5 \approx 39,000 / 1.6105 \approx 24,209 \)

Sum of PVs: approximately \( 6,364 + 12,397 + 17,278 + 21,157 + 24,209 \approx 81,405 \).

The UAE is then calculated as the annuity payment over five years that has this present worth at 10%. Using the capital recovery factor:

\[ \text{UAE} = PW \times \frac{i}{1 - (1 + i)^{-n}} \]

Calculate denominator:

\[ 1 - (1.10)^{-5} \approx 1 - \frac{1}{1.6105} \approx 1 - 0.6209 = 0.3791 \]

Therefore:

\[ \text{UAE} \approx 81,405 \times \frac{0.10}{0.3791} \approx 81,405 \times 0.2638 \approx 21,491 \]

Thus, the uniform annual equivalent of the cash flows, at 10% interest, is approximately $21,491 per year.

References

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