Exam 1 Solution

Exam 1 Solution - University of Alabama at Birmingham CE...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: University of Alabama at Birmingham CE 395 Engineering Economics _ Exam 1 Summer 2005 Name i z Date True or False ( 1 point each) (circle the correct answer): I. ®0r F A typical life cycle of a project is divided into the acquisition phase and the operational phase. Typically the greatest commitment of costs occurs during the acquisition phase. 2. @or F The initial investment cost for a project is a nonrecurring cost. T or® A noncash cost is a cash flow. DJ 4. ®or F Overhead costs normally include all costs that are not direct costs. 5. T GI® The Balance Sheet explains the cash flow balance for each year 6. @or F The effective annual interest rate is greater than or equal to nominal annual interest. 7. T or® You should always take sunk costs into account when considering a future economic choice between alternative courses of action. 8. '1‘ or® Optimal volume (demand) is when total costs equal total revenues. 9. @or F Normally it is necessary to use an iterative process when calculating the internal rate of return on a cash flow. 10. T oi® The typical maturity date for a bond is less than 10 years from the date of initial offering. ll©or F Internal Rate of Return (IRR) analysis is probably the most widely used investment evaluation method used today. 12. T o® When determining the IRR you must always use present worth analysis. l3©or F In a conducting an IR analysis it is possible to find multiple answers for IR. 14. T or® The face value of a bond is the price a willing buyer will pay for a bond. 15®or F The income statement includes the impact of debt interest, depreciation, and taxes on the ultimate stated profitability of a company. 16.®or F Fixed costs are unchanged by the variable output level. 17. T o F Opportunity cost is the sum of fixed and variable costs in a financial analysis. l . 18.®0r F Traditional cost accounting typically spreads overhead costs over all productive work activities without specific detailed data allocating costs to their primary source. 19(1)» F The yield on a stock or bond investment includes capital appreciation and interest or dividend payments. 20. T 01® The Minimum Attractive Rate of Return (MARR) is published annually by the government to protect consumers from over optimistic promoters. N€_+ PmQ-f 7:: eevanqes —- Fter C55 _. Mint-Elli cost} at g a Comm-l". N9: [2,(lo,noo) - 40,000 —~Vc 1': 30., 000 $ V‘i: :: 50,4510 #S‘omao : fishnet-1L .._.——-— l¢,000 Vc. lunri‘ = New.) Fina 05+; :2 1490,0003 =344Mo ~$ . VanebleflnL-f : |~Z('3) : é/unql- #Unlfi : i2, 000 a" P“:- hE-+ ng‘H = gamma; a C¢>s+ P :: Nfioo- 49m) _.-(_ 0002M)~!0,ddo P: —-.omz*...aoozuz’+loow —IO,ooo P: —, o IOZNL—l-IDON #107000 (Epmfit d9 :0 —__—_ _.,02,04.H+ 100 n..._-- dd , N: 4,902 and-'5 b. Check 4:? 4.0 cm" 612;? I “.0204 - l- max‘mum AN?- 2. a. pm“ : ——.OIOZ(4‘102.3.+I00C4‘102> — Io,ooo 2: 1‘; 2553098. 04" TAP M520 i=7dlo ' P =7so(P/Al‘12vzo\ +- |?(P{1:,7znzo) P Pg 750 1+.o‘1}zi| + . |P(l+.o7)-'ZO P ________.. .o7(t+.oj)2° :73!“ 10.5% ) +. I p(.z€5>4) HMZP: 7,945.: P: Mimi/72. H. 750 P: mso(P/A/ i) 5) + 130(P/é. [A 5) C L = 279 P21544755) 4—50 (1.2%} "-7 $2,063 .07 -.. D _ 1.93331 £9 (my-i _ .3” GIL-- P__._ 15.0 (050,020514- ,03 "'37;— s (1.03? -_- 6,36% Jr W332 "5-" $2/020.29 7.:ng“? Goo ' .. lam- 75 F x / mewkj H Ti: _ I “1:501:51: lama/Ur = 550 ' =- I,Doo)ooo .2: A(F'/Al 1.17mf 360) “0001000: A ( 5, MM 9 +1) 9 A: $de.xs/mo la - Jam. 5- 129F=(/+TE “I =JZ£8 ,L‘rg': : 12.682) “ma = erhrg 4. ___: M) :_ filf/flw A, mag: 27o/aJnr 0'“ 3:23" 2:2) .—.. W '- h...- ‘A "'"' T’Ioao ‘6 (‘5 2:31 " P P=15(p/&, 2%,. 4°) +rooo(P/F12‘%4°J = 15(27.35-533+;m (.152?) 12:. 59943.23 17m. cos+ mam Mar G. o D .— z 3 4.. 5 (9 c ‘ ® :EZFDDflWB‘L/ “ 2900(P/61‘39 ...
View Full Document

Page1 / 5

Exam 1 Solution - University of Alabama at Birmingham CE...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online