Engineering Economy

Sheet1

PP3.1	Example Using Excel PMT Function							Sample Table for Using NPV
											i	0.05
N	5		A = P(A/P, 10%, 5)					N	Profits	Expenses	First Cost	Total
i	10%		A = 15000*.2638	=	$ 3,957.00			0			-10000	-10000
P	15000							1	34000	-23000		11000
"(A/P,10%,5)	0.2638		=-PMT(B4,B3,B5,)		$3,956.96			2	34000	-23000		11000
A	?	=	$3,956.96					3	34000	-23000		11000
								4	12000	-23000		-11000
								5	34000	-23000		11000
		Example using Factors						6	34000	-46000		-12000
PP3.2		Given			Find P			7	34000	-23000		11000
								8	34000	-23000		11000
first cost	-1200000	P	Already P		$ (1,200,000.00)			9	34000	-23000		11000
savings	350000	A	* (P/A,10%,15)	7.606	$ 2,662,100.00			10	34000	-34000		0
salvage	200000	F	* (P/F,10%,15)	0.2394	$ 47,880.00			11	34000	-23000		11000		$56,593.99
i	0.1			Present Worth	$ 1,509,980.00			12	36000	-23000		13000		$46,593.99
N	15											$56,593.99	NPV	Yrs 1 - N
												$46,593.99	Present Worth (includes zero yr)
		Example using Excel PW function
PP3.2
			PW
first cost	-1200000		Present Worth of 1st cost		$ (1,200,000.00)
savings	350000		Present Worth of savings		$2,662,127.83		$2,662,127.83
salvage	200000		Present Worth of salvage		$47,878.41		$47,878.41
i	0.1	PW total	Combined Present Worth		$ 1,510,006.24
N	15
HW iii							What is the monthly payment for a 3 year new car loan, when the nominal annual interest is 6%.
	Ann	Monthly					After the down payment and other up-front charges, the amount borrowed is $44,000. Show your calculations and results using two methods:
n =	3	36
i =	0.06	0.50%		$1,338.57			a.       Use Engineering Economy factors and the compound interest tables found in Appendix C.
loan =	44000			=-PMT(C34,C33,B35)			b.      Using spreadsheet. Set up table with known values and use Excel payment function (PMT).
							c.      What is the monthly payment if the loan interest rate is the same, but loan is for 4 years?
Find ?	A
Given ?	P
			A = P(A/P, 0.5%, 36)
HW iii			from page 593
	Find A given P		A = P(A/P, 0.5%, 36)
n			0.0304
i			Using Factor
loan = P
HW viii	R&D project for new project already spent $200,000.
	Spending another $100,000 for patent will complete project.
	If company goes forward net returns will be $20,000 per year for 10 years.
	Company's interest rate is 10%
what is the $200,000
what is the $100,000
what is the $20,000
N =
I =
Find ?
Given ?
								Cash Flow
First Cost	-100000		Revenues	$122,891.34		0		-100000
Profits	20000		$122,891.34			1		20000
N	10		=-PV(B71,B70,20000)			2		20000
I	0.1					3		20000
		PW	$22,891.34			4		20000
						5		20000
						6		20000
						7		20000
						8		20000
						9		20000
						10		20000
						PW of Revenues =		$122,891.34
						Present Worth of		$22,891.34
						project going forward
Example: Problem setup similar to Homework problem ix.
Interest =	5%
First cost =	-4000
Life (yrs) =	5
Salvage =	1000
Annual Revenue =	1000
Repeat at end of 5 years and 10 years.
What is Present Worth for entire
15 years of Service
What is Present Worth for entire
15 years of Service if there was no
salvage value
Years	Cost $	Salvage	Revenue	Cash Flow
0	-4000			-4000
1			1000	1000
2			1000	1000
3			1000	1000
4			1000	1000
5	-4000	1000	1000	-2000
6			1000	1000
7			1000	1000
8			1000	1000
9			1000	1000
10	-4000	1000	1000	-2000
11			1000	1000
12			1000	1000
13			1000	1000
14			1000	1000
15		1000	1000	2000
				$6,668.36
				$2,668.36
					Interest =	5%
					First cost =	-4000
					Life (yrs) =	5
					Salvage =	1000
Years	Cost $	Salvage	Revenue	Cash Flow	Annual Revenue =	1000
0	-4000			-4000	Repeat at end of 5 years and 10 years.
1			1000	1000	What is Present Worth for entire
2			1000	1000	15 years of Service
3			1000	1000	What is Present Worth for entire
4			1000	1000	15 years of Service if there was no
5	-4000	1000	1000	-2000	salvage value
6			1000	1000
7			1000	1000
8			1000	1000
9			1000	1000
10	-4000	1000	1000	-2000
11			1000	1000
12			1000	1000
13			1000	1000
14			1000	1000
15		1000	1000	2000
			NPV Yrs 1-15	$6,668.36	=NPV(G118,E124:E138)
			Add in Zero Yr	$2,668.36	=+E140+E123