InvestmentWorth Investment Worth

Given a minimum attractive rate-of-return, be able to evaluate the investment worth of a project using Net Present Worth Equivalent Annual Worth Internal Rate of Return External Rate of Return Capitalized Cost Method

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

NPW( MARR ) > 0Good Investment

EUAW( MARR ) > 0Good Investment

NPW( MARR ) > 0Good Investment EUAW( MARR ) > 0Good Investment IRR > MARRGood Investment

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 12% Is it worth it to the company to buy the machine?

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

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

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

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

Repeating our PW example, we have AW(12)= -16(A/P,12,5) (A/F,12,5) 3, ,000 16,000

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

Repeating our PW example, we have AW(12)= -16(A/P,12,5) (A/F,12,5) = -16(.2774) (.1574) = = -$808 3, ,000 16,000

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

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 = A t (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 = A t (1 + i*) n - t PW revenue (i*) = PW costs (i*) 

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

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

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

216% 16 yrs

F = P(F/P,i *,16) (F/P,i *,16) = F/P = 2.16 (1+i * ) 16 = 2.16

(1+i * ) 16 = ln(1+i * ) = ln(2.16) =.7701

(1+i * ) 16 = ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481

(1+i * ) 16 = ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481 (1+i * ) = e.0481 =

(1+i * ) 16 = ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481 (1+i * ) = e.0481 = i* =.0493 = 4.93%

We know i = 4.93%, is that significant growth?

We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period.

We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period. d ij j      d= 1.4%

NPW > 0 Good Investment

EUAW > 0 Good Investment

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

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

1,000 4,100 5,580 2,520 n 0123 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR).

1,000 4,100 5,580 2,520 n 0123 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR). PW R (i * ) = PW C (i * ) 4,100(1+i * ) ,520(1+i * ) -3 = 1, ,580(1+i * ) -2

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

Purpose: to get around a problem of multiple roots in IRR method Notation: A t = net cash flow of investment in period t A t, A t > 0 0, else -A t, A t < 0 0, else r t = reinvestment rate (+) cash flows (MARR) i’ = rate return (-) cash flows   R t = C t =

Method find i = ERR such that R t (1 + r t ) n - t = C t (1 + i ’ ) n - t Evaluation If i ’ = ERR > MARR Investment is Good 

Example MARR = 15% R t (1 +.15) n - t = C t (1 + i ’ ) n - t 4,100(1.15) 2 + 2,520 = 1,000(1 + i ’ ) 3 + 5,580(1 + i ’ ) 1 i ’ =.1505  ,000 4,100 5,580 2,520

Example MARR = 15% R t (1 +.15) n - t = C t (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  ,000 4,100 5,580 2,520

Example MARR = 15% R t (1 +.15) n - t = C t (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  ,000 4,100 5,580 2,520

Method 1 Let i = MARR SIR(i) = R t (1 + i) -t C t (1 + i) -t = PW (positive flows) - PW (negative flows)  

Relationships among MARR, IRR, and ERR If IRR < MARR, then IRR < ERR < MARR If IRR > MARR, then IRR > ERR > MARR If IRR = MARR, then IRR = ERR = MARR

Method #2 SIR(i) = A t (1 + i) -t C t (1 + i) -t SIR(i) = PW (all cash flows) PW (negative flows)  

Method #2 SIR(i) = A t (1 + i) -t C t (1 + i) -t SIR(i) = PW (all cash flows) PW (negative flows) Evaluation: Method 1: If SIR(t) > 1 Good Investment Method 2: If SIR(t) > 0 Good Investment  

Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) 16 =.818 <

Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) 16 =.818 <

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

Example m= 5 years n c o R t n 0 

Perpetuity (Capitalized Cost) Occasionally, donors sponsor perpetual awards or programs by a lump sum of money earning interest.Occasionally, donors sponsor perpetual awards or programs by a lump sum of money earning interest. The interest earned each period (A) equals the funds necessary to pay for the ongoing award or program.The interest earned each period (A) equals the funds necessary to pay for the ongoing award or program. The relationship is A = P( i ) This concept is also called capitalized cost (where CC = P).This concept is also called capitalized cost (where CC = P).

Perpetuity Example A donor has decided to establish a $10,000 per year scholarship. The first scholarship will be paid 5 years from today and will continue at the same time every year forever. The fund for the scholarship will be established in 8 equal payments every 6 months starting 6 months from now. Determine the amount of each of the equal initiating payments, if funds can earn interest at the rate of 6% per year with semi- annual compounding.

Perpetuity Problem Given: A = per year, every year after Year 5 n = 8 6 mo. intervals, 6 mo. i = 6%, cpd semi-annually Find Amount of Initiating payments (A i ):

Perpetuity Problem Given: A = per year, every year after Year 5 n = 8 6 mo. intervals, 6 mo. i = 6%, cpd semi-annually Find Amount of Initiating payments (A i ):

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.

,000 P c = 10,000 + A/i = 10, /.08 = 11,250 Capitalized Cost = $11.25 million

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.

Compute an annuity for the 1,000 every 5 years: ,000 1,000

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

,000 P c = 10, /.08 = 13, ,000 1,000

How does Bill Gates change a light bulb?

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

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

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