Fn = P (1+i)n SPCAF = single-payment compound-amount factor
Outline Modul 2
Engineering Economy Factors Single-Payment Factors
Uniform-Series Present-Worth Factor Capital-Recovery Factor
Sinking-Fund Factor
Gradient Formulas Module 2: Factors
SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.
2-2 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.
Engineering Economy Factors Single-Payment Factors F P
P = present value, usually used to represent initial
? investment
1
2 3 n-1 n F = future value, accumulated value at some future time
SPCAF = single-payment compound-amount factor n
A = a series of consecutive, equal, end-of-period F n = P (1+i)
Notation: (F/P, i, n)
amount of payment I = interest, usually compounded
From the opposite side, if F is given:
i = interest rate per interest period, %
F n = number of (interest) periods P
?
G = arithmetic gradient, uniform arithmetic change in magnitude of payment / receipts
1
2 3 n-1 n g = geometric gradient, constant rate of change in
1 SPPWF = single-payment present worth factor magnitude of payment / receipts, %
P F
Notation: (P/F, i, n)
n
2-3 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D. 2 - 4 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.
1 i Uniform-Series Present-Worth Factor 2 - 5 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
? P
3 A A A A A A A
1 2 n-1 n
2 - 7 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
? Sinking-Fund Factor
What A is needed to equal P at t=0? Notation: (A/P, i, n)
P A CRF = capital recovery factor
1 n n i i i
1
1
3 A A A A A A A
1 2 n-1 n
P
Capital-Recovery Factor 2 - 6 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
Notation: (P/A, i, n)
1
1
A P
n n i i i
i i
(F/A , i% , n) Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
(A/F , i% , n) Uniform-Series Compound-Amount USCAF
(A/P , i% , n) Sinking-Fund SFF
(P/A , i% , n) Capital-Recovery CRF
(F/P , i% , n) Uniform-Series Present-Worth USPWF
Factor Name Standard Notation Single-Payment Present-Worth SPPWF (P/F , i% , n) Single-Payment Compound-Amount SPCAF
Standard Factor Notations 2 - 8
Notation: (A/F, i, n) Notation: (F/A, i, n)
USCAF = uniform-series compound-amount factor
1
A F n
1 2 n-1 n
F A
1 n i i
1
F
1 n n n i i i i F A
1
1
1
1
SPPWF = single-payment present-worth factor
1 1 3 2 1
1
1
1
1
1
1
1
1
n n i A i A i A i A i A P
1
1
1
1 1 3 2 1
1
1
1
1
1
1
1
n n i i i i i A P
1 USPWF = uniform-series present-worth factor
3 A A A A A A A
1 SFF = sinking-fund factor
3 A A A A A A A
2 - 12 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
1 A
1 A
1 A
1 A
1 A
1 A
1
1 2 n-1 n
3
3G
2G G (n-3)G (n-2)G
(n-1)G + Arithmetic Gradient (3)
Equivalent with: A = A 1 + A
1 2 n-1 n
2
1 2 n-1 n
3 A
1 A
1 A
1 A
1 A
1 A
1 A
1
2 n-1 n
3 A 2 A 2 A 2 A 2 A 2 A 2 A 2 =
1 2 n-1 n
3 A
Equivalent with:
Extreme Values 2 - 9 factor n =
Equivalent with:
∞
; i known n = ; i known
( F / P, i % , n ) ∞
1
( P / F, i % , n )
1
( P / A , i % , n ) 1 /i ( A / P, i % , n ) i
NA
( A / F, i % , n )
NA
( F / A , i % , n ) ∞
Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik Arithmetic Gradient (1)
2 - 10 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik In some cases, periodic payments do not occur in an equal series, but they maybe in a constant increase amount
1 2 n-1 n
2 - 11 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
3
1 2 n-1 n
3 A
1 A
1 A
1 A
1 A
1 A
1 A
1
3G
2G G (n-3)G (n-2)G
(n-1)G Arithmetic Gradient (2)
- 1
A = payment at the end of the first year 1 G = gradient, annual change n = number of period n
A = equivalent equal annual payment A = A 1 + A 2 1
1 i 1 n
P G n n
1
1 n i i i
A G 2
1 i
n
1
1 i i
UGPWF = uniform-gradient-present-worth factor UGSF = uniform-gradient-series factor n 1
1 i 1
F G n i i
UGFWF = uniform-gradient-future-worth factor 2 - 13 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D. 2 - 14 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.
Negative Arithmetic Gradient (1) Arithmetic Gradient (2)
In some cases, periodic payments do not occur in an equal series, but they Equivalent with: maybe in a constant decrease amountA A A A A A A
1
1
1
1
1
1
1
1
2 3 n-1 n
- (n-1)G (n-2)G (n-3)G
1
2 3 n-1 n
3G
2G G
1
2 3 n-1 n 2 - 15 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D. 2 - 16 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.
- – A
- 1
F P n i 1 1
n P i F 1
n n i i i A P 1 1 1
1 1 1 P A 1 n n i i
i i A F n 1 1
1 F A 1 n i i
Using Factors 2 - 19
n n n i n i i i i G P 1 1 1 1 1
1 1 1 n i n i G A
Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik Interest Table
2 - 20 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
For the shake of simplicity and easier calculation all values based on interest formulas have been tabularized
For any values that are not available in the interest table, interpolation or extrapolation can be applied:
n Single Payment Equal Payment Series Uniform Gradient Series
(F/P, i, n) (P/F, i, n) (F/A, i, n) (A/F, i n) (P/A, i n) (A/P, i, n) (A/G, i n)
1 1.128 0.8869 1.000 1.0000 0.8869 1.1275 0.0000 2 1.271 0.7866 2.128 0.4700 1.6736 0.5975 0.4700 3 1.433 0.6977 3.399 0.2942 2.3712 0.4217 0.9202 4 1.616 0.6188 4.832 0.2070 2.9900 0.3345 1.3505 5 1.822 0.5488 6.448 0.1551 3.5388 0.2826 1.7615 10 3.320 0.3012 18.197 0.0550 5.4810 0.1825 3.5332
20 11.023 0.0907 78.625 0.0127 7.1318 0.1402 5.8480
12% interest factors for continuous compounding 1.352
Finding Given Factor Equation Formula P F ( P / F , i % , n ) P = F (P/F, i%, n) F P ( F / P , i % , n ) F = P (F/P, i%, n) P A ( P / A , i % , n ) P = A (P/A, i%, n) A P ( A / P , i % , n ) A = P (A/P, i%, n) A F ( A / F , i % , n ) A = F (A/F, i%, n) F A ( F / A , i % , n ) F = A (F/A, i%, n) A G ( A / G , i % , n ) A = G (A/G, i%, n) P G ( P / G , i % , n ) P = G (P/G, i%, n)
g i g growth-free rate
Arithmetic Gradient (3) 2 - 17 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
1 A
Equivalent with: A = A
1
2
1 2 n-1 n
3 A
1
2 n-1 n
3 A 2 A 2 A 2 A 2 A 2 A 2 A 2 =
1 2 n-1 n
3 A A A A A A A A
1 A
1 A
1 A
1 A
1 Geometric Gradient 2 - 18 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
1
1
1
1
1 F 1 (1+g) n-2 F 1 (1+g) 3 F 1 (1+g) 2 F 1 (1+g)
3 F
1 2 n-1 n
1 1 GGSF = geometric-gradient-series factor F 1 (1+g) n-1
1
In many situations, periodic payments decreases or increases not by a constant amount, but at a constant percentage1
1
n n g g g g F P
Example: find value of (F/P, i, n) for i = 12% and n = 2.5 period n = 2 , (F/P, 12, 2) = 1.271; n = 3 , (F/P, 12, 3) = 1.433 n = 2.5, (F/P, 12, 2,5) = 1.271 + 0.5 (1.433-1.271) = Spreadsheet Formulas Exercises (1)
1.
How much money should you be willing to pay now for a guaranteed $600 per year for 9 years starting next year, at a rate of return of 16% per year?
Factors in Engineering Economy Excel Functions
2. Determine the value of the A/P factor for an interest rate of 7.3% and n of 10 years, that is, (A/P, 7.3%, 10).
P = Present Value = PV (i%,n,A,F)
3. HP has completed a study indicating that $50,000 in reduced F = Future Value = FV (i%,n,A,P) maintenance this year on one processing line resulted from
A = Equal, Periodic Value = PMT (i%,n,P,F) improved integrated circuit (IC) fabrication technology based on rapidly changing designs. n = Number of Periods = NPER (i%,A,P,F)
a. If HP considers these types of savings worth 20% per year, find the equivalent value of this result after 5 years. i = Compound Interest Rate = RATE (n,A,P,F)
b. If the $50,000 maintenance savings occur now, find its equivalent value 3 years earlier with interest at 20% per year.
2 - 21 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.
22 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.
Exercises (2) Exercises (3) 4.
5. How long will it take for $1000 to double if the interest rate is 5% per Three contiguous counties in Florida have agreed to pool tax year? resources already designated for county-maintained bridge
6. Engineers at SeaWorld, a division of Busch Garden, Inc., have completed refurbishment. At a recent meeting, the county engineers an innovation on an existing water sports ride to make it more exciting. estimated that a total $500,000 will be deposited at the end
The modification costs only $8,000 and is expected to last 6 years with a of the next year into an account for the repair of old and
$1,300 salvage value for the solenoid mechanisms. The maintenance cost safety-questionable bridges throughout the three-county is expected to be high at $1,700 the first year, increasing by 11% per year thereafter. Determine the equivalent present worth of the modification area. Further, they estimate that the deposits will increase by and maintenance cost. The interest rate is 8% per year.
$100,000 per year for only 9 years thereafter, then cease.
7. If Laurel can make an investment in a friend’s business of $3,000 now in Determine the equivalent: order to receive $5,000 five years from now, determine the rate of a. Present worth if county funds earn interest at a rate of 5% per year. return. If Laurel can receive 7% per year interest on a certificate of
b. Annual series amounts if county funds earn interest at a rate of 5%
deposit, which investment should be made? per year.
23 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D.
24 SI-4251 Ekonomi Teknik Muhamad Abduh, Ph.D. Exercises (4) Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
25
4. Annual series over the first 10 years.
Exercises (6) Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
12. Chemical engineers at a Coleman Industries plants in the Midwest have determined that a small amount of a newly available chemical additive will increase the water repellency of Coleman’s tent fabric by 20%. The plant superintendent has arranged to purchase the additive through a 5- year contract at $7,000 per year, starting 1 year from now. He expects the annual price to increase by 12% per year thereafter for the next 8 years. Additionally, an initial investment of $35,000 was made now to prepare a site suitable for the contractor to deliver the additive. Use i = 15% to determine the equivalent total present worth for all cash flows.
11. Gerri, an engineer at Fujitsu, Inc., has tracked the average inspection cost on a robotics manufacturing line for 8 years. Cost averages were steady at $100 per completed unit for the first 4 years, but have increased consistently by $50 per unit for each of the last 4 years. Gerri plans to analyze the gradient increase using the P/G factor. Where is the present worth located for gradient? What is the general relation used to calculate total present worth in year 0?
26
Exercises (5) Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
5. Annual series over the last 12 years.
3. Annual series over all 22 years.
Homework #2 Muhamad Abduh, Ph.D. SI-4251 Ekonomi Teknik
2. Future worth in year 22.
lease the mineral rights to a mining company. The primary objective is to obtain long term income to finance ongoing projects 6 and 16 years from the present time. The engineering company makes a proposal to the mining company that it pay $20,000 per year for 20 years beginning 3 years from now, plus $10,000 six years from now and $15,000 sixteen years from now. Determine the equivalent values listed below at 16% per year: 1. Total present worth in year 0.
10. An engineering company in Wyoming that owns 50 hectares of valuable land has decided to
recalibrated for each of 6 years starting 3 years after purchase, calculate the 8-year equivalent uniform series at 16% per year.
9. Recalibration of sensitive measuring devices costs $8,000 per year. If the machine will be
annual payment of $500 per year for 6 years starting 3 years from now for annual upgrades. What is the present worth of the payments if the interest rate is 8% per year?
8. An engineering technology group just purchased new CAD software for $5,000 now and
13. Assume that you are planning to invest money at 7% per year as shown
1
2
3
4
5
6
7
8 9 10 11 12 $2000 $2500
$3000 $3500 $4000 $5000
$4000 $3000 $2000 $1000 $1000 $1000
by the increasing gradient of the following figure. Further, you expect to withdraw according to the decreasing gradient shown. Find the net present worth and equivalent annual series for the entire cash flow sequence and interpret the results.
28
Redo all exercises (#1 to #13) in this session using a computer (spreadsheet application).
27