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MS-291: Engineering Economy(3 Credit Hours)

Pervious Lecture!!!

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Single payment Factors

F/P FactorF= P(F/P, i%, n)

P/F FactorP= F(P/F, i%, n)

P is given

F = ?

n and i is given

F = given

n and i is given

P =?

Uniform Series Factors

P/A FactorP = A(P/A, i%, n)

A/P FactorA = P(A/P, i%, n)

F/A FactorF = A(F/A, i%, n)

A/F FactorA = F(A/F, i%, n)

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Athematic Gradient

PA = A(P/A, i%, n)

PG = G(P/G, i%, n)

FG = ?

F = PT(F/P, i%, n)

or

/ = 1 (1 + ) 1

Athematic Gradient

PA = A(P/A, i%, n)

PG = G(P/G, i%, n)

A = PT(A/P, i%, n)or

AT = AA + AG

= 1 (1 + ) 1AA = A (Annual Worth) &AG = G(A/G, i, n)

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Geometric Gradient

Pg = ?

= 1 1 +1 +for g i:

for g = i:= 1 +

Fg = ?

F = Pg(F/P, i%, n)Similarly

A = Pg(A/P, i%, n)

Engineering Economy

Chapter 3Combining Factorsand Spreadsheet

Functions

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This Chapter Objectives

1. Shifted uniform series

2. Shifted series and single cash flows

3. Shifted gradients

Example

0 62 3 4 51 7 8 9 10 11 12 13 Year

A = $50

P = ?

Use the P/F factor to find the present worth of each disbursement at year 0 and addthem.

Use the F/P factor to find the future worth of each disbursement in year 13, addthem, and then find the present worth of the total, using P/F= F( P/F, i ,13).

Use the F/A factor to find the future amount F/A =A( F/A, i ,10), and then computethe present worth, using P/F=F(P/F, i ,13).

Use the P/A factor to compute the present worth P3 =A( P/A , i ,10) (which will belocated in year 3, not year 0), and then find the present worth in year 0 by using the(P/F , i ,3) factor.

How can we get Present worth of this series ?

P3 = ? F

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Shifted Uniform Series Typically the last method is used for calculating the present worth

of a uniform series that does not begin at the end of period 1.

Note that a P value is always located 1 year or period prior to thebeginning of the first series amount. Why? Because the P/A factorwas derived with P in time period 0 and A beginning at the end ofperiod 1.

The most common mistake made in working problems of this typeis improper placement of P .

Remember: When using P/A or A/P factor, PA is always one year ahead of first A When using F/A or A/F factor, FA is in same year as last A The number of periods n in the P/A or F/A factor is equal to the number of uniform

series values

0 62 3 4 51 7 8 9 10 11 12 13 Year

A = $50

P3 = ?

0 62 3 4 51 7 8 9 10 11 12 13 Year

A = $50

F = ?

PA is always one year aheadof first A

FA is in same year as last A

The number of periods n in the P/A orF/A factor is equal to the number ofuniform series values

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Steps for applying factors toShifted Cash Flows

1. Draw a diagram of the positive and negative cash flows.2. Locate the present worth or future worth of each series on

the cash flow diagram.3. Determine n for each series by renumbering the cash flow

diagram.4. Draw another cash flow diagram representing the desired

equivalent cash flow. (Optional)5. Set up and solve the equations.

ExampleThe offshore design group at Bechtel just purchasedupgraded CAD software for $5000 now and annualpayments of $500 per year for 6 years starting 3 yearsfrom now for annual upgrades. What is the presentworth in year 0 of the payments if the interest rate is8% per year?

1. Draw a diagram of the positive and negative cash flows.Solution

0 62 3 4 51 7 8

P0 = $5000

A = $500

PT = ? 2. Locate the presentworth or future worth ofeach series on the cashflow diagram.PA = ?PA = ? i= 8% per year

3. Determine n for eachseries by renumbering thecash flow diagram.

0 62 3 4 51 nYear

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P' A = $500( P /A ,8%,6)PA = P' A ( P /F ,8%, 2)

P T = P 0 + P A=5000 + 500( P /A ,8%,6)( P / F ,8%,2)=5000 +500(4.6229)(0.8573)$6981.60

5. Set upand solvetheequations.

PA = $500( P /A ,8%,6) ( P /F ,8%, 2)

Class Practice 5 Minutes Time

A = $10,000

0 1 2 3 4 5 6

i = 10%

Calculate the present worth of the cash flow shown below at i = 10%

Actual year

10% Single Payments Uniform Series Factorsn Compoun

d Amount(F/P)

PresentWorth(P/F)

SinkingFund(A/F)

CompoundAmount(F/A)

CapitalRecovery(A/P)

PresentWorth(P/A)

1 1.1000 0.9091 1.00000 1.0000 1.10000 0.90915 1.6105 0.6209 0.16380 6.1051 0.26380 3.7908

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Class Practice 5 Minutes Time

A = $10,000

PT = ?

0 1 2 3 4 5 6

i = 10%

Calculate the present worth of the cash flow shown below at i = 10%

PT = A(P/A,10%, 5) (P/F,10%,1)

$34462

0 1 2 3 4 5

PA = ?Actual yearSeries year

(2) Use P/F factor with n = 1 to move PA back for PT in year 0

= 10,000(3.7908)(0.9091)

(1) Use P/A factor with n = 5 (for 5 arrows) to get PA in year 1Solution

A(P/A,10%, 5)----(P/F,10%, 1)----

Shifted Series and RandomSingle Amounts

For cash flows that include uniform series and randomly placedsingle amounts:

Uniform series procedures are applied to the series amounts

Single amount formulas are applied to the one-time cash flows

The resulting values are then combined per the problem statement

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Example

Find the present worth in year 0 for the cash flows shown using an interestrate of 10% per year.

0 1 2 3 4 5 6 7 8 9 10

A = $5000

i = 10%

Find the cash flows both positive and negatives

$2000

0 1 2 3 4 5 6 7 8 9 10

PT = ?

A = $5000

i = 10%

$2000

0 1 2 3 4 5 6 7 8

Solution:

Actual year

Series year

Locate the present worth/ future worth Determine the n by re-numbering the cash flows series Uniform series procedures are applied to the series amounts. Single amount

formulas are applied to the one-time cash flows The resulting values are then combined per the problem statement

Use P/A to get PA in year 2: PA = 5000(P/A,10%,8) = 5000(5.3349) = $26,675

Move PA back to year 0 using P/F: P0 = 26,675(P/F,10%,2) = 26,675(0.8264) = $22,044Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933Now, add P0 and P2000 to get PT: PT = 22,044 + 933 = $22,977

Example:

PA = ?

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Class Practice: 8 MinutesAn engineering company lease the mineral rights to a miningcompany on its land. The engineering company makes aproposal to the mining company that it pay $20,000 per year for20 years beginning 1 year from now, plus $10,000 six years fromnow and $15,000 sixteen years from now. If the mining companywants to pay off its lease immediately, how much should it paynow if the investment is to make 16% per year?

16% Single Payments Uniform Series Factorsn Compound

Amount (F/P)Present Worth(P/F)

CapitalRecovery (A/P)

Present Worth(P/A)

6 2.4364 0.4104 0.27139 3.68477 2.8262 0.3538 0.24761 4.038616 10.7480 0.0930 0.17641 5.668517 12.4677 0.0802 0.17395 5.748720 19.4608 0.0514 0.16867 5.9228

A =$20,000P = ?

0 1 2 3 4 5 6 7 16 17 18 19 20

$10,000 $15,000

P = 20,000(P/A ,16%,20)+

= $124,075

10,000( P /F ,16%,6) +15,000(P/F,16%,16)

P = $20,000(5.9288)+ $ 10,000( 0.4104) + $ 15,000(0.0930)

Solution

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Shifted Gradient Series

We already learnt how to get P (Presentvalue) or A ( Annuity or a Uniform series)from a Gradient Series

We will now discuss how to calculate P orA from Shifted Gradient Series agradient series not starting from year 1.

PA = A(P/A, i%, n)PG = G(P/G, i%, n)

P from Shifted Gradient Series

Shifted gradient begins at atime other than between periods1 and 2

Present worth PG is located 2periods before gradient starts

Must use multiple factors to find PTin actual year 0

P T =P A +P G=100(P/A , i ,8) + 50(P/G, i ,5)(P/F, i ,3)

Shifted Arithmetic gradient Series

PG = ?

PT = ?

PA = ?

PG = ?

P from Normal Arithmetic Gradient Series

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What will be the procedure for calculating P fromShifted Geometric Gradient Series?

Lets discuss it directly from a Numerical Example

Example:Shifted Geometric Gradient

Weirton Steel signed a 5-year contract to purchase water treatment chemicals from alocal distributor for $7000 per year. When the contract ends, the cost of the chemicals isexpected to increase by 12% per year for the next 8 years. If an initial investment instorage tanks is $35,000, determine the equivalent present worth in year 0 of all of thecash flows at i = 15% per year.

0 2 31 4 5

$7000

Year

$7840

6 7 8 9 10 11 12 13

$1733112% increaseper year

$35000

i =15% per year

GeometricGradient n

0 1 2 3 4 5 6 7 8

Pg = ?PT = ?

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P from Shifted Gradient Series

PT = 35,000 + A ( P /A ,15%, 4) + A1 ( P/A ,12%,15%,9) (P/F ,15%,4)

$83,232

PT = 35,000 + 7000 ( 2.8550) + 7000. .. . (0.5718)

A from Shifted Gradient SeriesShifted gradient Series

0 62 3 4 51 7 8

To calculate A for shifted Gradient Series (Arithmetic orGeometric), there are several possibilities.

The easiest way (and recommended also) is to get the P ofshifted Gradient Series first (procedure just explained in pervious slides) thenuse A/P factor to get A for the shifted gradient series

A, for above example will be: A = PT (A/PT, i%, n), where PT refers to the present value of the shifted gradient series that procedure is

already explained on pervious slide.

A (Annuity or Uniform Series)

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Important Points for P and Aof Shifted Gradient Series

Must use multiple factors to find P in actual year 0, for shiftedgradient series

The present worth (P) of an arithmetic gradient will always belocated two periods before the gradient starts.

To find the equivalent A series of a shifted gradient throughall the n periods, first find the present worth of the gradient atactual time 0, then apply the (A/P, i, n) factor.

F from gradient series can also be find by first calculating Pand then using F/P factor

Example: Shifted ArithmeticGradient

Solution:

John Deere expects the cost of a tractor part to increase by $5 per year beginning 4 years fromnow. If the cost in years 1-3 is $60, determine the present worth in year 0 of the cost through year10 at an interest rate of 12% per year.

i = 12%

First find P2 for G = $5 and base amount ($60) in actual year 2P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41

Next, move P2 back to year 0 P0 = P2(P/F,12%,2) = $295.29

Next, find PA for the $60 amounts of years 1 and 2 PA = 60(P/A,12%,2) = $101.41

Finally, add P0 and PA to get PT in year 0 PT = P0 + PA = $396.70

60 60 6065

7095G = 5

0 1 2 3 104 5 Actual years0 1 2 3 8 Gradient years

PT = ? P2 = ?

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Class Question.. 5 minutesFor the cash flows shown, find the future value in year 7 at i = 10% per year

F = ?

0 1 2 3 4 5 6 7

700 650

500 450550600

G = $-50Solution:

Actual years0 1 2 3 4 5 6 Gradient years

PG is located in gradient year 0 (actual year 1); base amount of $700 is in gradient years 1-6

F = PG(F/P,10%,6) = 2565(1.7716) = $4544

i = 10%PGPG = ?

PG = A(P/A,10%,6) G(P/G,10%,6)PG = 700(P/A,10%,6) 50(P/G,10%,6) = 700(4.3553) 50(9.6842) = $2565

PG= PG(F/P,10%,1) = 2565(0.9091) = $2331.84

F = PG(F/P,10%,7) = 2331.84(1.9487) = $4544

Set up theequationsonly

Method 1

Method 2

Using Single Amount factors (Correctbut not Standard methods)

Method 3

Method 4

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THANK YOU