Scope 1. 2. 3. 4. 5. 6.
ENGR 301 Lecture 14 Evaluation of Projects
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ENGR 301 Lecture 14
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Project Cash Flows Loan Repayment
Customer
Project Cash flow Investment Company
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Return
Project
ENGR 301 Lecture 14
Identifying Project Cash Flow Year (n) 0
Cash inflows Cash out flows Net (Benefits) (Costs) Cash flows
1 . . .
Example Given: $650,000 Initial investment to improve product Before improvement; 30,000 kg / year of chemical product sold for $ 15 per kg (benefits) 3500 operating hr / year (cost) After improvement; increase in price $2 / kg (benefits) 4000 kg / year (benefits) reduction in operators $25 / hr (benefits) additional maintenance $53,000/year (cost) useful life 8 years Find : Net cash flow in each year over the 8 years S. El-Omari
ENGR 301 Lecture 14
Identifying Project Cash Flow • Revenues from price increase; 30,000 kg/year x $2 = $60,000 / year 4,000 kg/year additional x ($15/kg + $2/kg) = 68,000 Savings of 3,500 operating hr/year x $25 = $87,500 Total Benefits / year = 60,000 + 68,000 + 87,500 = $215,500 / year
• Maintenance Cost = $53,000 / year • Net $ / year = 215,000–53,000 = $162,500 / year
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ENGR 301 Lecture 14
Identifying Project Cash Flow
Loan Cash Flow Bank
Pay Back Period Net Present Worth (NPW) Future Worth (FW) Annual Worth (AW) Rate of Return (ROR) Internal Rate of Return (IRR)
ENGR 301 Lecture 14
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ENGR 301 Lecture 14
1
Identifying Project Cash Flow Year (n) 0
Cash inflows Cash out flows Net (Benefits) (Costs) Cash flows 0 $650,000 -$650,000
1
215,000
53,000
162,000
2 . . .
215,000
53,000
162,000
8
215,000
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Pay Back Period A $162,500 0 1
2
3
4
5
6
7
8 Years
Payback period = Initial cost / Uniform annual payment = 650,000 / 162,500 = 4 years
53,000
162,000
ENGR 301 Lecture 14
$650,000 S. El-Omari
ENGR 301 Lecture 14
Pay Back Period
Pay Back Period with Salvage Value
• Determine how long it takes to repay the initial loan. • Doesn't consider the time value of money. • Helps in selecting investments with pay back periods less than maximum acceptable pay back periods.
Example Automotive company has just bought a new machine at a cost of $105,000 to replace one that had a salvage value of $20,000. the projected annual after tax savings via improved efficiency, which will exceed the investment cost, are as follows. Find the pay back period
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ENGR 301 Lecture 14
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ENGR 301 Lecture 14
ENGR 301 Lecture 14
Pay Back Period with Salvage Value Period
Cash inflows
0
-105,000 + 20,000
Cumulative Cash flows -$85,000
1
15,000
-70,000
2
25,000
-45,000
3
35,000
-10,000
4
45,000
35,000
5
45,000
80,000
6
35,000
115,000
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ENGR 301 Lecture 14
2
Discounted Pay Back Period Cash flow 0
-$85,000
Cost of funds 15% 0
1
15,000
-85,000(.15)=-12,750
-82,750
2
25,000
-82,750(.15)=-12,413
-70,163
3
35,000
-70,163(.15)=-10,524
-45,687
4
45,000
-45,687(.15)=-6,853
-7,540
5
45,000
-7,540(.15)=-1131
36,329
6
35,000
36,329(.15)=5449
76,778
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Cumulative Cash flows -$85,000
ENGR 301 Lecture 14
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ENGR 301 Lecture 14
Present Worth Analysis
Net Present Worth
• Payback method was widely used until 1950s • Net Present Worth (NPW) or Present Equivalent (PE) is a method to account for the time value of money and is part of Discounted Cash Flow techniques (DCF).
• Application steps 1. Determine the interest rate that the company wishes to earn on its investment and is referred to the Minimum Attractive Rate of Return (MARR) and is subject to management decision. 2. Estimate the service life of the project. 3. Estimate the cash inflow for each period over the service life of the project. 4. Determine the net cash flow for each period 5. Find the NPW at each cash flow at the MARR
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ENGR 301 Lecture 14
ENGR 301 Lecture 14
Net Present Worth If PE(MARR) > 0, Accept the investment
Net Present Worth Year (n) 0
PE(MARR) = 0, Remain indifferent to the investment
PE(MARR) < 0, Reject the investment
1
215,000
53,000
162,000
2 . . . 8
215,000
53,000
162,000
215,000
53,000
162,000
MARR = 15% S. El-Omari
ENGR 301 Lecture 14
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Cash inflows Cash out flows Net (Benefits) (Costs) Cash flows 0 $650,000 -$650,000
Find: PE ENGR 301 Lecture 14
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Net Present Worth
Net Present Worth
PE(15%) outflow = $650,000 PE(15%) inflow = $162,500(P/A,15%,8) = $729,190. PE(15%) = $729,190 - $650,000 = $79,190 Since PE(15%) > 0, the project would be acceptable
Example: Uneven flows Tiger machine tool company is considering the acquisition of a new metal cutting machine. The required initial investment of $75,000 and the projected cash benefits over the 3-year’s project are as follows End of year Net cash flow 0 -$75,000 Find PE at 1 24,400 MARR = 15% 2 3
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ENGR 301 Lecture 14
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27,340 55,760
ENGR 301 Lecture 14
Net Present Worth 55,760 24,400
27,340
0 1
2
3
75,000
PE(15%) = -$75,000 + 24400(P/F,15%,1) + 27,340(P/F,15%,2)+55,760(P/F,15%,3) PE(15%) = $3553 S. El-Omari
ENGR 301 Lecture 14
$3,553
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