Chapter 6 discounted cash flow valuation
... = 769 .31 – Year CF: N; -60 0 PV; I/Y; CPT FV = 8 46 . 95 – Total FV = 769 .31 + 8 46 . 95 = 1 ,61 6. 26 • Second way – use value at year 2: – N; -1, 248 .05 PV; I/Y; CPT FV = 1 ,61 6. 26 6C -6 Multiple Cash Flows ... I/Y; -4, 000 PV; CPT FV = 4 ,66 5 .60 – Year CF: N; I/Y; -4, 000 PV; CPT FV = 4, 320 – Year CF: value = 4, 000 – Total value in years = 8,817.98 + 4 ,66 5 .60 + 4, 320 + 4, 000 = 21,803.58 • Value at year 4: ... PV = 140 ,105 • Total Price – Closing costs = 04( 140 ,105) = 5 ,60 4 – Down payment = 20,000 – 5 ,60 4 = 14, 3 96 – Total Price = 140 ,105 + 14, 3 96 = 1 54, 501 6C-23 Annuities on the Spreadsheet - Example...
Ngày tải lên: 07/03/2015, 12:22
... j!)j (dB) 0.2 28 0 .4 22 0.8 16 1 .6 3.0 4. 0 4 .6 10.7 7.5 7.3 7.0 6. 0 0.9 10 20 40 À9.3 À28 À 36 À 54 G( j!)H( j!) (deg) À91 À92 À95 À100 À115 À138 À 162 À180 À217 À 244 À259 À 262 À 266 Plot the Bode ... Example 6. 4 (see also Appendix 1, examp64a.m and examp64b.m), when the controller gain set to K1 1:0, the open-loop transfer function is G(s)H(s) s(s2 2s 4) (6: 66) Equation (6. 66) represents ... 1) (6: 54) //SYS21/D:/B&H3B2/ACE/REVISES(08-08-01)/ACEC 06. 3D ± 166 ± [ 145 ±197/53] 9.8.2001 2:31PM 166 Advanced control engineering Controller Plant K1 s(s + 2s + 4) R(s) + C(s) – Fig 6. 20 Closed-loop...
Ngày tải lên: 09/08/2014, 06:23
... mốn húâi CHÛÚNG 6: LÂ M THÏË NÂ O ÀÏÍ KIÏË M HÂ N G TRIÏå U ÀƯ DOANH THU VÂ LÚÅ I NHÅ N 179 www.toitaigioi.com http://sach.tgm.vn Adam Khoo om Chiïën thåt tùng tó lïå mua hâng 4: Bấn cho khấch ... chiïëc xe cố àưång cú 340 mậ lûåc, hậy nối: “Tûúãng tûúång nêëy àïìu ngoấi àêìu nhòn bẩn hổ nghe tiïëng nưí giôn tan ca àưång cú” .v n g Chiïën thåt tùng tó lïå mua hâng 6: Xêy dûång uy tđn m ... bẩn vúái nhûäng sẫn phêím ca àưëi th :/ 14 Sûã dng hònh ẫnh hóåc video vïì trûúác vâ sau sûã dng sẫn phêím 15 Sûã dng bùng hònh giúái thiïåu sẫn phêím 16 Nhùỉc àïën tïn ngûúâi giúái thiïåu 17...
Ngày tải lên: 03/09/2013, 10:09
Tài liệu An Introduction to Intelligent and Autonomous Control-Chapter 6: Nested Hierarchical Control pdf
Ngày tải lên: 14/12/2013, 12:15
Tài liệu Linux Device Drivers-Chapter 6 : Flow of Time pptx
... command 45 12 944 9 8883 head 45 12 945 3 45 12 945 3 60 1 X 45 12 945 3 60 1 X 45 12 945 3 60 1 X 45 12 945 3 60 1 X 45 12 945 4 1 45 12 945 4 60 1 X 45 12 945 4 60 1 X 45 12 945 4 60 1 X 45 12 945 4 60 1 X 45 12 945 4 60 1 X 45 12 945 4 60 1 X 45 12 945 4 ... command 45 0 848 45 1 8783 cc1 45 0 848 46 1 8783 cc1 45 0 848 47 1 8783 cc1 45 0 848 48 1 8783 cc1 45 0 848 49 1 87 84 as 45 0 848 50 1 8758 cc1 45 0 848 51 1 8789 cpp 45 0 848 52 1 8758 cc1 45 0 848 53 1 8758 cc1 45 0 848 54 1 ... time 45 472377 delta interrupt pid cpu command 89 04 head 45 472378 1 0 swapper 45 472379 1 0 swapper 45 472380 1 0 swapper 45 472383 0 swapper 45 472383 60 1 X 45 472383 60 1 X 45 472383 60 1 X 45 472383 60 1...
Ngày tải lên: 24/12/2013, 01:17
ccna explorationg 4.0 - chapter 6 teleworker services
... www.bkacad.com 43 Task is configuring IPSec: Học viện mạng Bach Khoa - Website: www.bkacad.com 44 Học viện mạng Bach Khoa - Website: www.bkacad.com 45 IPsec Security Protocols • Activity 6. 3.7 Học ... DOCSIS specifies the channel widths (bandwidths of each channel) as 200 kHz, 40 0 kHz, 800 kHz, 1 .6 MHz, 3.2 MHz, and 6. 4 MHz DOCSIS also specifies modulation techniques (the way to use the RF signal ... Website: www.bkacad.com 46 IPsec Security Protocols Học viện mạng Bach Khoa - Website: www.bkacad.com 47 IPsec Security Protocols Học viện mạng Bach Khoa - Website: www.bkacad.com 48 IPsec Security...
Ngày tải lên: 06/07/2014, 09:29
THE MAN WHO LAUGHS VICTOR HUGO PART 2 BOOK 4 CHAPTER 6 pot
... lieutenant has no dislike to domestic quarrels, because he always has the pickings" (22nd July 17 04) As to the lieutenant of police, he was a redoubtable person, multiple and vague The best personification...
Ngày tải lên: 07/07/2014, 09:20
Control Engineering - A guide for beginners - Chapter 6 doc
... unsuitable controller setting Fig 69 : Switched auxiliary process variable correction 6. 5 Coarse/fine control Two control loops in series are used to maintain some parameter of a mass flow or energy flow ... controller The master controller, set for setpoint response, is usually a PI or PID controller 108 JUMO, FAS 525, Edition 02. 04 Improved control quality through special controls For cascade control, ... two different values of process gain here Fig 66 : Power switching JUMO, FAS 525, Edition 02. 04 103 Improved control quality through special controls 6. 3 Switched disturbance correction The effect...
Ngày tải lên: 07/08/2014, 15:20
Analysis and Control of Linear Systems - Chapter 6 ppsx
... ⎞ ⎟ = rank ⎜ ⎜ ⎝ H A⎠ ⎝ ⎞ ⎟=2 q⎟ ⎠ rank ⎜ [6. 45 ] 0) and we can [6. 46 ] In section 6. 2.2 we saw that ( A, B ) is controllable, so that hypotheses [6. 36] are verified The positive semi-defined solution ... and W = w ⎝ ⎠ ⎠ ⎝ [6. 65] where v and w are positive coefficients We can verify that ( A, J ) is controllable: ⎛ v ⎞ ⎟=2 ⎝ v − v⎠ rank ( J AJ ) = rank ⎜ [6. 66] In section 6. 3 .6 we saw that (C , ... 1 64 Analysis and Control of Linear Systems Figure 6. 2 Stabilization by pole placement 6. 3 Reconstruction of state and observers 6. 3.1 General principles The disadvantage of state feedback controls,...
Ngày tải lên: 09/08/2014, 06:23
BIOLOGICAL AND BIOTECHNOLOGICAL CONTROL OF INSECT PESTS - CHAPTER 6 potx
... Compound Chromosomes 6. 4. 4 Male-Linked Translocations 6. 4. 5 Hybrid Sterility 6. 5 Field Trials 6. 5.1 Lucilia cuprina, the Sheep Blowfly 6. 5.2 Mosquitoes 6. 6 Operational Programmes 6. 6.1 New World Screwworm, ... Nature 2 36, 44 56 45 7, 1972 Laven, H.E., E Jost, H Meyer, and R Selinger Semisterility for Insect Control Sterility Principle for Insect Control or Eradication IAEA-SM-138/ 16, 41 5 42 4, 1971 Lindquist, ... Post-Production Processes 6. 2.3 Field Monitoring 6. 3 Quantitative and Qualitative Approaches 6. 4 Mechanisms 6. 4. 1 Dominant Lethality 6. 4. 2 Inherited Partial Sterility 6. 4. 3 Autosomal Translocations...
Ngày tải lên: 11/08/2014, 04:20
AIR POLLUTION CONTROL TECHNOLOGY HANDBOOK - CHAPTER 6 pps
... sophisticated as would be implied by the system of Figure 6. 1 © 2002 by CRC Press LLC 9588ch 06 frame Page 73 Wednesday, September 5, 2001 9: 46 PM 6. 6 TYPICAL AIR SAMPLING TRAIN A typical air pollution ... on July 31, 1987 © 2002 by CRC Press LLC 9588ch 06 frame Page 74 Wednesday, September 5, 2001 9: 46 PM FIGURE 6. 2 Two-stage particulate sampler 6. 8 CONTINUOUS AIR QUALITY MONITORS Continuous emissions ... CRC Press LLC 9588ch 06 frame Page 76 Wednesday, September 5, 2001 9: 46 PM FIGURE 6. 3 Generalized automatic continuous air pollution monitor © 2002 by CRC Press LLC 9588ch 06 frame Page 77 Wednesday,...
Ngày tải lên: 11/08/2014, 06:22
Biological Risk Engineering Handbook: Infection Control and Decontamination - Chapter 6 pot
... 6. 14. 3 Medium-Efficiency Filters 6. 14. 4 High-Efficiency Extended Surface Filters 6. 14. 5 Gas and Volatile Organic Compound Removal Filters 6. 14 .6 Acoustical Lining 6. 15 Ducts 6. 16 Duct ... Devices 6. 22 Humidification and Dehumidification Equipment 6. 23 Self-Contained Units 6. 24 Controls 6. 25 Boilers 6. 26 Cooling Towers 6. 27 Water Chillers Resources In order to understand biological ... (7 34) 7 64 - 240 7; www.herb.lsa.umich.edu (specimen-based information on fungi; information on fungal ecology) University of Minnesota, Department of Environmental Health & Safety; (61 2) 62 6-58 04; ...
Ngày tải lên: 11/08/2014, 09:21
Environmental Pollution Control Microbiology - Chapter 6 pdf
... particles A typical nematode is shown in Figure 6- 6 Nematodes range in size from 1,000 to 2,000 um, making them easy to observe * •" ,»«*«! Figure 6- 6 PHOTOMICROGRAPH OF A TYPICAL NEMATODE under ... 23 days to reach its maximum population, 15 ,60 0/ml, when grown at 20°C in the concentrated nutrients The Colpidium used 245 mg/L oxygen, about 0.0 16 ug/cell Once the Colpidium reached their maximum ... in days and dropped to 10/ml by Day 10 The bacteria population reached 6. 9 x 1 06/ ml after one day and was down to 0.7 x 1 06/ ml by Day 10 Their study gave some additional data The growth of Aerobacter...
Ngày tải lên: 11/08/2014, 13:21
Handbook Of Pollution Control And Waste Minimization - Chapter 6 doc
... a minimum and 65 % as a maximum by weight of the packaging waste will be recovered Moreover, within this general target, and with the same time limit, between 25% as a minimum and 45 % as a maximum ... flooding, subsidence, landslides, or avalanches on the site Therefore the water control and leachate management must control water from precipitations entering into the landfill body, prevent surface ... incineration plants Copyright 2002 by Marcel Dekker, Inc All Rights Reserved 30 mg/m3 20 mg/m3 60 mg/m3 mg/m3 200 mg/m3 40 0 mg/m3 Until January 2007, the emission limit value for NOx does not apply to plants...
Ngày tải lên: 11/08/2014, 13:22
ADVANCED ONSITE WASTEWATER SYSTEMS TECHNOLOGIES - CHAPTER 6 pptx
... Systems” (EPA 832-B-03-001) Since ongoing management (operation © 20 06 by Taylor & Francis Group, LLC Chapter six: Management framework 169 and maintenance) of traditional onsite systems (septic systems) ... the house, there are no “sewage alarms” to worry about, there is no odor © 20 06 by Taylor & Francis Group, LLC 1 74 Advanced onsite wastewater systems technologies from the sewage system, and the ... discharge monitoring reports are submitted based on a schedule set in the permit © 20 06 by Taylor & Francis Group, LLC 1 76 Advanced onsite wastewater systems technologies If and when needed to meet higher...
Ngày tải lên: 11/08/2014, 17:21
Mechanisms and Mechanical Devices Sourcebook - Chapter 6
... Sclater Chapter 5/3/01 12: 24 PM Page 1 74 FLAT SPRINGS IN MECHANISMS Constant force is approached because of the length ... maximum axial movement 1 74 Increasing support area as the load increases on both upper and lower platens is provided by a circular spring Sclater Chapter 5/3/01 12: 24 PM Page 175 These mechanisms ... simply, efficiently, and at low cost by flatspring arrangement shown here 175 Sclater Chapter 5/3/01 12: 24 PM Page 1 76 POP-UP SPRINGS GET NEW BACKBONE An addition to the family of retractable coil...
Ngày tải lên: 22/10/2012, 14:26
Bài giảng marketing quốc tế CHAPTER 6
... vọng Giá=CP đơn vò/(1-%LN kỳ vọng/giá) • VD: Đònh giá bán sản phẩm A, biết: – Chi phí đơn vò: 24 USD/ sản phẩm – Lợi nhuận kỳ vọng so với giá bán: 20% ĐỊNH GIÁ THEO CHI PHÍ (tt) • Ưu điểm: * ... bán • * Chiết khấu hỗ trợ chức • * Chiết khấu theo mùa Khách hàng nghó thấy sản phẩm giảm giá? 14 THAY ĐỔI GIÁ (tt) • Tăng giá: Khi: * Cung không đủ cầu * Tăng giá không làm giảm thò phần, làm ... bớt tính sản phẩm 15 * Giảm hay loại bỏ bớt dòch vụ kèm Khách hàng nghó thấy sản phẩm tăng giá? 16 ...
Ngày tải lên: 25/10/2012, 08:58
Máy xúc lất HuynDai HL760 - Chapter 6
... of Replace fuse 6 12 Y VW 24 LOr LOr 26 CN -6 11 12 25 34 35 R B 36 45 CS- 36 46 ILL & HEAD LAMP SW 47 12 11 48 59 60 6 9 12 11 12 LOr R R B R CLUSTER 17 ILLUMINATION CN- 56 CN -4 R MONITOR 18 ILLUMINATION ... clean) disconnection in wiring harness or poor contact between CS-12 (4) and CN -4( 43) WIPER RY LO 3 CR- 26 B 26 40 41 42 43 44 45 59 A31 B02 B WASH SIG B+ R O L C WB L Y Lo BW R WIPWE RLY Hi Lo II ... CN-58A (B 36) and CN -6 (49 ) NO YES between CN -6 Voltage : 26~ 30V Remedy Repair or replace (after clean) Defective alternator Replace NO CONTROLLER(MCU) B 36 49 ALTERNATOR " " TERMINAL CN-13 CN -6 CN-58A...
Ngày tải lên: 27/10/2012, 08:17
Lập trình hệ điều hành Chapter 6.
... Lập trình hướng đối tượng CHƯƠNG VI Hà Văn Sang Khoa HTTT, Academy Of Finance, Hanoi 11/01/12 16: 38 Khuôn hình hàm Định nghĩa Ví dụ 1: xây dựng hàm tìm max hai số thực -Xây dựng hàm tính max...
Ngày tải lên: 01/11/2012, 16:37
Giáo trình Xử lý ảnh -Chapter 6
... N H (n1 , n2 ) = ∑ ∑ h(k1 , k )e − j 2π k1 = k = Công thức 6. 4 biến đổi Fourier rời rạc DFT Tham khảo giảng ĐH ( n1k1 + n2 k ) N⋅ (6. 4) Biến đổi Fourier 1D Biến đổi Fourier 1-D cho tín hiệu thời ... f(kT) tính theo công thức: N −1 F (n) = ∑ f (kT )e − j 2π N ⋅nk (6. 5) k =0 Công thức viết lại dạng N −1 − F (n) = ∑ f (k ) ¦ WN nk (6. 6) n =0 f(k) = f(kT) WN = e- j2 /N WN gọi hạt nhân phép biến ... ∞ ∑ ∞ h( k1 , k )e − j (ω1k1 +ω2 k ) (6. 1) ∑ k1 = −∞ k = −∞ Nếu h(k1,k2) có k1 ≥ 0, k2 ≥ xác định miền hữu hạn N × N thì: H (ω , ω ) = ∑ ∑ h(k , k )e ω ω (6. 2) N −1 N −1 Tham khảo giảng ĐH k1...
Ngày tải lên: 14/11/2012, 14:36