Presentation is loading. Please wait. # TM 661 Engineering Economics for Managers

## Presentation on theme: "TM 661 Engineering Economics for Managers"— Presentation transcript:

TM 661 Engineering Economics for Managers
Investment Worth

Investment Worth MARR Suppose a company can earn 12% / annum in U. S. Treasury bills No way would they ever invest in a project earning < 12% Def: The Investment Worth of all projects are measured at the Minimum Attractive Rate of Return (MARR) of a company.

MARR MARR is company specific MARR based on
utilities - MARR = % mutuals - MARR = % new venture - MARR = % MARR based on firms cost of capital Price Index Treasury bills

Investment Worth Alternatives
NPW(MARR) > 0 Good Investment

Investment Worth Alternatives
NPW(MARR) > 0 Good Investment EUAW(MARR) > 0 Good Investment

Investment Worth Alternatives
NPW(MARR) > 0 Good Investment EUAW(MARR) > 0 Good Investment IRR > MARR Good Investment

Present Worth Example: Suppose you buy and sell a piece of equipment. Purchase Price \$16,000 Sell Price (5 years) \$ 4,000 Annual Maintenance \$ 3,000 Net Profit Contribution \$ 6,000 MARR % Is it worth it to the company to buy the machine?

Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5) 4,000 4,000 6,000
16,000 6,000 3,000 5 4,000 3,000 5 4,000 16,000 NPW = (P/A,12,5) + 4(P/F,12,5)

Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)
16,000 6,000 3,000 5 4,000 3,000 5 4,000 16,000 NPW = (P/A,12,5) + 4(P/F,12,5) = (3.6048) + 4(.5674)

Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)
16,000 6,000 3,000 5 4,000 3,000 5 4,000 16,000 NPW = (P/A,12,5) + 4(P/F,12,5) = (3.6048) + 4(.5674) = = -\$2,916

Annual Worth  Annual Worth (AW or EUAW) AW(i) = PW(i) (A/P, i%, n)
= [ At (P/F, i%, t)](A/P, i%, n) AW(i) = Annual Worth of Investment AW(i) > 0 **OK Investment**

Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)
3,000 5 4,000 16,000 Repeating our PW example, we have AW(12) = -16(A/P,12,5) (A/F,12,5)

Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)
3,000 5 4,000 16,000 Repeating our PW example, we have AW(12) = -16(A/P,12,5) (A/F,12,5) = -16(.2774) (.1574)

Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)
3,000 5 4,000 16,000 Repeating our PW example, we have AW(12) = -16(A/P,12,5) (A/F,12,5) = -16(.2774) (.1574) = -.808 = -\$808

Alternately AW(12) = PW(12) (A/P, 12%, 5) = -2.92 (.2774)
3,000 5 4,000 16,000 AW(12) = PW(12) (A/P, 12%, 5) = (.2774) = - \$810 < 0 NO GOOD

Internal Rate of Return
IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment**

Internal Rate of Return
IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t

Internal Rate of Return
IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t PWrevenue(i*) = PWcosts(i*)

Internal Rate of Return
Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) 3,000 5 4,000 16,000

Internal Rate of Return
Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) 3,000 5 4,000 16,000

Internal Rate of Return
Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) i* = 5 1/4 % i* < MARR 3,000 5 4,000 16,000

Public School Funding

Public School Funding 216% 16 yrs

School Funding F = P(F/P,i*,16) (F/P,i*,16) = F/P = 2.16
3 16 100 216 F = P(F/P,i*,16) (F/P,i*,16) = F/P = 2.16 (1+i*)16 = 2.16

School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 1 2 3 16
100 216 (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701

School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701
3 16 100 216 (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481

School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701
3 16 100 216 (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481 (1+i*) = e.0481 =

School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701
3 16 100 216 (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 ln(1+i*) = .0481 (1+i*) = e.0481 = i* = = 4.93%

School Funding We know i = 4.93%, is that significant growth? 1 2 3 16
100 216 We know i = 4.93%, is that significant growth?

School Funding We know i = 4.93%, is that significant growth?
1 2 3 16 100 216 We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period.

School Funding We know i = 4.93%, is that significant growth?
1 2 3 16 100 216 We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period. d i j 1 0493 0350 . d= 1.4%

School Funding We know that d, the real increase in school funding
1 2 3 16 100 216 We know that d, the real increase in school funding after we discount for the effects of inflation is 1.4%. So schools have experienced a real increase in funding?

School Funding We know that d, the real increase in school funding
1 2 3 16 100 216 We know that d, the real increase in school funding after we discount for the effects of inflation is 1.4%. So schools have experienced a real increase in funding? Rapid City growth rate 3% / yr.

Summary NPW > Good Investment

Summary NPW > Good Investment EUAW > Good Investment

Summary NPW > 0 Good Investment EUAW > 0 Good Investment
IRR > MARR Good Investment

Summary NPW > 0 Good Investment EUAW > 0 Good Investment
IRR > MARR Good Investment Note: If NPW > EUAW > 0 IRR > MARR

Internal Rate of Return
IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = At(1 + i*)n - t PWrevenue(i*) = PWcosts(i*)

Internal Rate of Return
Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) 3,000 5 4,000 16,000

Internal Rate of Return
Example PW(i) = (P/A, i, 5) + 4(P/F, i, 5) i* = 5 1/4 % i* < MARR 3,000 5 4,000 16,000

Spreadsheet Example

Spreadsheet Example

Spreadsheet Example

Spreadsheet Example

Spreadsheet Example

IRR Problems Consider the following cash flow diagram. We
1,000 4,100 5,580 2,520 n 1 2 3 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR).

IRR Problems Consider the following cash flow diagram. We
1,000 4,100 5,580 2,520 n 1 2 3 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR). PWR(i*) = PWC(i*) 4,100(1+i*)-1 + 2,520(1+i*)-3 = 1, ,580(1+i*)-2

IRR Problems NPV vs. Interest (\$5) \$0 \$5 \$10 \$15 \$20 \$25 0% 10% 20%
30% 40% 50% 60% Interest Rate Net Present Value

External Rate of Return
Purpose: to get around a problem of multiple roots in IRR method Notation: At = net cash flow of investment in period t At , At > 0 0 , else -At , At < 0 0 , else rt = reinvestment rate (+) cash flows (MARR) i’ = rate return (-) cash flows Rt = Ct =

External Rate of Return
Method find i = ERR such that Rt (1 + rt) n - t = Ct (1 + i’) n - t Evaluation If i’ = ERR > MARR Investment is Good

External Rate of Return
Example MARR = 15% Rt ( ) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505 1 2 3 1,000 4,100 5,580 2,520

External Rate of Return
Example MARR = 15% Rt ( ) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505 ERR > MARR 1 2 3 1,000 4,100 5,580 2,520

External Rate of Return
Example MARR = 15% Rt ( ) n - t = Ct (1 + i’) n - t 4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1 i’ = .1505 ERR > MARR Good Investment 1 2 3 1,000 4,100 5,580 2,520

Critical Thinking IRR < MARR a. IRR < MARR < ERR
b. IRR < ERR < MARR c. ERR < IRR < MARR

Critical Thinking IRR > MARR a. IRR > MARR > ERR
b. IRR > ERR > MARR c. ERR > IRR > MARR

Savings Investment Ratio
Method 1 Let i = MARR SIR(i) = Rt (1 + i)-t Ct (1 + i)-t = PW (positive flows) - PW (negative flows)

Savings Investment Ratio
Method #2 SIR(i) = At (1 + i) -t Ct (1 + i) -t SIR(i) = PW (all cash flows) PW (negative flows)

Savings Investment Ratio
Method #2 SIR(i) = At (1 + i) -t Ct (1 + i) -t SIR(i) = PW (all cash flows) PW (negitive flows) Evaluation: Method 1: If SIR(t) > 1 Good Investment Method 2: If SIR(t) > 0 Good Investment

Savings Investment Ratio
16 1 2 3 4 5 7 Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) = < 1.0

Savings Investment Ratio
16 1 2 3 4 5 7 Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) = < 1.0

Payback Period   Method Find smallest value m such that where
Co = initial investment m = payback period investment m t = 1 R t C o

Payback Period R t n å Example c n o 16 1 2 3 4 5 7 m = 5 years

Capitalized Costs Example: \$10,000 into an account @ 20% / year
A = P(A/P, i, ) P = A / i A 10,000

Class Problem A flood control project has a construction cost of \$10 million, an annual maintenance cost of \$100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep.

Class Problem Pc = 10,000 + A/i = 10,000 + 100/.08 = 11,250
10,000 A flood control project has a construction cost of \$10 million, an annual maintenance cost of \$100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep. Pc = 10,000 + A/i = 10, /.08 = 11,250 Capitalized Cost = \$11.25 million

Class Problem 2 Suppose that the flood control project has major repairs of \$1 million scheduled every 5 years. We now wish to re-compute the capitalized cost.

Class Problem 2 Compute an annuity for the 1,000 every 5 years:
10,000 1,000 Compute an annuity for the 1,000 every 5 years:

Class Problem 2 Compute an annuity for the 1,000 every 5 years:
10,000 1,000 Compute an annuity for the 1,000 every 5 years: A = ,000(A/F,8,5) = ,000(.1705) = 270.5

Class Problem 2 10,000 1,000 10,000 Pc = 10, /.08 = 13,381

How Many to Change a Bulb?
How does Bill Gates change a light bulb?

How Many to Change a Bulb?
How does Bill Gates change a light bulb? He doesn’t, he declares darkness a new industry standard!!!

How Many to Change a Bulb?
How many Industrial Engineers does it take to change a light bulb?

How Many to Change a Bulb?
How many Industrial Engineers does it take to change a light bulb? None, IE’s only change dark bulbs!!!!!

Download ppt "TM 661 Engineering Economics for Managers"

Similar presentations

Ads by Google