1. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Process of Developing Project Cash Flows Lecture No.33 Chapter 10 Contemporary Engineering Economics Copyright © 2016

2. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Chapter Opening Story  Eclipse has spent $1.4 billion to develop its six-seat 550 twin engine jet. o Base price: $2,895,000 o Fuel efficiency: 59 gallons per hour o Flying range: 430 miles o Operating cost: $648 per hour o Demand: unknown  At Issue: How would you determine the cash flows from this jet investment?

3. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Key Elements of Investment Decision Step 1 Identify investment opportunities Step 2 Estimate project cash flows Step 3 Measure the investment worth Step 4 Select the best project Step 5 Implement the project Step 6 Post-audit the project

4. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Types of Cash Flow Elements in Project Analysis

5. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 10.1: When Projects Require Only Operating and Investing Activities  Given: Financial Data o Investment: $125,000 o Project life: 5 years o Salvage value: $50,000 o Annual labor savings: $100,000 o Annual manufacturing costs o Labor: $20,000 o Materials:$12,000 o Overhead:$8,000 o Depreciation method: 7-year MACRS o Income tax rate: 40% o MARR: 15%  Find: Determine the project cash flows

6. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution

7. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Return on Invested Capital n = 0n =1n = 2n = 3n = 4n = 5 Beginning Balance −$125,000 − $116,380 − $100,279 − $83,231 − $63,974 Return on Investment (27.62%) − $34,525 − $32,144 − $27,697 − $22,988 − $17,670 Payment − $125,000 $43,145$48,245$44,745$42,245$81,619 Project Balance − $125,000 − $116,380 − $100,279 − $83,231 − $63,974 ≈0 The firm earns a 27.62% return on funds that remain internally invested in the project.

8. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved When Projects Require Working-Capital Investments What is Working Capital? The amount carried in cash, accounts receivable, and inventory that is available to meet day-to-day operating needs. How to treat working capital investments: just like a capital expenditure except that no depreciation is allowed. Working Capital Equations Accounting definition: WC = Current Asset – Current Liabilities  WC =  CA -  CL where  WC = changes in working capital  CA = changes in current assets  CL = changes in current liabilities If  WC > 0, working capital requirement. With the net change being positive, the firm has a net requirement of working capital that has to be financed during the year. Therefore, the WC requirement appears as uses of cash in the cash flow statement. If  WC < 0, working capital release. If this amount were negative, there would have been a cash inflow from working capital release, which could add to the sources of cash.

9. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved When Projects Require Working-Capital Investments What is working capital?  Definition: The amount carried in cash, accounts receivable, and inventory that is available to meet day-to-day operating needs.  How to treat working capital investments  Just like a capital expenditure except that no depreciation is allowed Working Capital Equations o Accounting definition WC = Current Asset − Current Liabilities Δ WC = Δ CA − Δ CL where Δ WC = changes in working capital Δ CA = changes in current assets Δ CL = changes in current liabilities o If Δ WC > 0, working capital requirement. o If Δ WC < 0, working capital release.

10. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 10.2: Working Capital Requirements  Given: Elements of Working Capital  Find: Working capital requirement

11. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution

12. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 10.3: Cash Flow Statement with Working Capital

13. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Changes in Profitability  Changes in Profitability o NPW without the Working Capital Requirement PW(15%) = $43,152 o NPW with the Working Capital Requirement PW(15%) = $31,420  Difference: $11,732 (lost earnings due to funds tied up in working capital)

14. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved When Projects Results in Negative Taxable Income Negative taxable income (project loss) means you can reduce your taxable income from regular business operation by the amount of loss, which results in a tax savings. Regular Business ProjectCombine d Operation Taxable income Income taxes (35%) $100M $35M (10M) ? $90M $31.5M Tax Savings = $35M − $31.5M = $3.5M Or (10M)(0.35) = −$3.5M Tax savings

15. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 10.5: Project Cash Flows for a Cost-Only Project  Project Nature: Installing a cooling-fan at Alcoa Aluminum’s McCook plant to reduce the work-in- process inventory buildup  Given: Financial facts o Required investment: $536,000 o Service life: 16 years o Salvage value: 0 o Reduction of WIP (working-capital release): $2,121,000 o Depreciation Method: 7-year MACRS o Annual electricity cost: $86,000 o Income tax rate:40% o MARR: 20%  Find: Develop the project cash flow

16. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Cash Flow Statement

17. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Analysis o PW(20%) = $991,008 o i* = 4.24% and 291.56% o A nonsimple and mixed investment o RIC = 241.87% >20% o Good investment

18. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved When Projects Are Financed with Borrowed Funds Key issue: Interest payment is a tax- deductible expense. What needs to be done Once a loan repayment schedule is known, separate the interest payments from the annual installments. What about principal payments? As the amount of borrowing is NOT viewed as income to the borrower, the repayments of principal are NOT viewed as expenses either—NO tax effect.

19. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Example 10.4: Loan Repayment Schedule End of Year Beginning Balance Interest Payment Principal Payment Ending Balance 1$62,500$6,250$10,237$52,263 252,2635,22611,26141,002 3 4,10012,38728,615 4 2,86113,62614,989 5 1,49914,9880 o Amount financed: $62,500, or 50% of total capital expenditure o Financing rate: 10% per year o Annual installment: $16,487 or, A = $62,500(A/P, 10%, 5) $16,487

20. Contemporary Engineering Economics, 6 th edition Park Copyright © 2016 by Pearson Education, Inc. All Rights Reserved Solution  Effects of debt financing on profitability o MARR = 15%, debt interest rate = 10% o NPW without debt financing (100% equity) PW(15%) = $31,420 o NPW with debt financing (50% debt) PW(15%) = $44,439 o The debt financing increases the present worth by $13,019. This result is largely caused by the firm’s being able to borrow the funds at a cheaper rate (10%) than its MARR of 15%.