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4 introduction to reactor design

Introduction to Optimum Design phần 4 pdf

Introduction to Optimum Design phần 4 pdf

Kỹ thuật lập trình

... identied to be replaced with x2 in the basic set Basicỉ x1 x2 x3 x4 x5 b Ratio: bi /aiq x3 1 0 16 16 x4 x5 28 14 14 24 Cost -40 0 = 14 ă smallest 1 = 16 1/ 14 1/ 24 0 1 0 f-0 -600 = 24 Second ... Optimum Design 251 6 .45 Minimize subject to f = -x1 - 4x2 x1 + x2 Ê 16 x1 + 2x2 Ê 28 24 2x1 + x2 x1, x2 6 .46 Minimize f = x1 - x2 subject to 4x1 + 3x2 Ê 12 x1 + 2x2 Ê 4 2x1 + x2 x1, x2 6 .47 Maximize ... subject to x1 - x2 x1 + x2 2x1 + x2 Ê x1, x2 6.19 248 Maximize z = 2x1 + 5x2 - 4. 5x3 + 1.5x4 subject to 5x1 + 3x2 + 1.5x3 Ê 1.8x1 - 6x2 + 4x3 + x4 -3.6x1 + 8.2x2 + 7.5x3 + 5x4 = 15 xi 0; i = to...
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Chapter 4 Introduction to Portfolio Theory

Chapter 4 Introduction to Portfolio Theory

Chuyên ngành kinh tế

... varied from -0 .4 to 1 .4 in increments of 0.1 and, since xB = − xA , the weight on asset is then varies from 1 .4 to -0 .4 This gives us 18 portfolios with weights (xA , xB ) = (−0 .4, 1 .4) , (−0.3, 1.3), ... investments in the two stocks The investor s problem is to decide how much wealth to put in asset A and how much to put in asset B Let xA denote the share of wealth invested in stock A and xB denote ... (0. 542 )(0.175) + (0 .45 8)(0.055) = 0.110, the variance of the tangency portfolio is σ2 = T ³ xT A ´2 ³ σ + xT A B ´2 σ + 2xT xT σ AB B A B = (0. 542 )2 (0.067) + (0 .45 8)2 (0.013) + 2(0. 542 )(0 .45 8)...
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Chapter 15  introduction to the design of electric machinery

Chapter 15 introduction to the design of electric machinery

Điện - Điện tử

... the machine resistance, but increases conductor mass from 0.9 14 to 1. 64 kg and stator mass from 4. 12 to 6 .40   kg The increase in stator mass is in order to accommodate larger slots for the larger ... either mass or loss; 616 Introduction to the Design of Electric Machinery Pareto-Optimal Front 500 45 0 40 0 Loss, W 350 300 250 200 150 100 50 10 12 14 16 18 20 22 24 26 Mass, kg Figure 15.10-2.  ... peak line -to- line voltage is essentially equal to the maximum value allowed for the dc voltage of 40 0  V It is also interesting to observe that the stator tooth, stator backiron, and rotor tangential...
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Data Mining Classification: Basic Concepts, Decision Trees, and Model Evaluation Lecture Notes for Chapter 4 Introduction to Data Mining pptx

Data Mining Classification: Basic Concepts, Decision Trees, and Model Evaluation Lecture Notes for Chapter 4 Introduction to Data Mining pptx

Cơ sở dữ liệu

... > Yes 3 3 2 3 3 No 4 4 Gini © Tan,Steinbach, Kumar 0 .42 0 0 .40 0 0.375 0. 343 0 .41 7 Introduction to Data Mining 0 .40 0 0.300 0. 343 0.375 0 .40 0 0 .42 0 37 Alternative Splitting Criteria ... P(C2) = 4/ 6 Error = – max (2/6, 4/ 6) = – 4/ 6 = 1/3 Introduction to Data Mining 43 Comparison among Splitting Criteria For a 2-class problem: © Tan,Steinbach, Kumar Introduction to Data Mining 44 Misclassification ... (5/6)2 = 0.278 P(C2) = 4/ 6 Gini = – (2/6)2 – (4/ 6)2 = 0 .44 4 Introduction to Data Mining 32 Splitting Based on GINI Used in CART, SLIQ, SPRINT When a node p is split into k partitions (children),...
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introduction to the design & analysis of algorithms

introduction to the design & analysis of algorithms

Đại cương

... Exercises 12.3 12 .4 Algorithms for Solving Nonlinear Equations 44 1 44 3 45 3 45 7 Exercises 12 .4 459 46 0 46 4 46 4 46 7 Summary 46 8 Epilogue 47 1 Bisection Method Method of False Position Newton’s Method ... 338 342 344 345 346 347 351 357 359 361 371 372 378 380 383 3 84 387 388 389 390 390 391 393 3 94 395 397 399 40 1 40 2 40 6 40 9 xiv Contents 11 .4 Challenges of Numerical Algorithms Exercises 11 .4 Summary ... 138 142 144 144 146 148 x Contents 4. 4 Decrease-by-a-Constant-Factor Algorithms Binary Search Fake-Coin Problem Russian Peasant Multiplication Josephus Problem 150 150 152 153 1 54 Exercises 4. 4...
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Introduction to digital design using digilent FPGA boards

Introduction to digital design using digilent FPGA boards

Điện - Điện tử

... 5 14 3 84 – 3 ,45 6 1 24 – 7 84 192 – 8,320 108 – 376 240 – 3,688 180 – 512 3 84 – 6, 144 176 – 8 04 3 84 – 16,2 24 88 – 1,108 64 – 11, 648 Block RAM (bits) 36 - 288 360 – 2,016 642 – 5,556 16,3 84 – 57, 344 ... 0.08192 12 6103.52 0.163 84 13 3051.76 0.32768 14 1525.88 0.65536 15 762. 94 1.31072 16 381 .47 2.62 144 17 190.73 5. 242 88 18 95.37 10 .48 576 19 47 .68 20.97152 20 23. 84 41. 943 04 21 11.92 83.88608 22 ... statement and in Section 7 .4 we design a quad 4- to- 1 multiplexer Prerequisite knowledge: Example – 2 -to- 1 Multiplexer 7.1 Designing a 4- to- 1 MUX Using 2 -to- 1 Modules A 4- to- 1 multiplexer has the...
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Introduction to Optimum Design phần 1 pdf

Introduction to Optimum Design phần 1 pdf

Kỹ thuật lập trình

... Optimization Problem 41 4 41 4 41 5 41 8 42 0 43 4 43 5 43 6 43 6 44 0 44 2 44 5 44 6 44 7 44 8 44 8 44 8 44 9 45 0 45 0 45 1 45 3 45 4 45 4 45 5 45 5 45 7 46 2 46 5 46 6 46 6 46 7 Contents xvii 14. 2 14. 3 Gradient Evaluation ... 383 383 3 84 3 84 385 388 388 389 393 40 0 40 0 40 3 40 4 40 6 40 6 40 7 40 7 40 9 41 0 41 1 Contents 41 3 12.1 xvi Introduction to Optimum Design with MATLAB 41 3 41 3 Introduction to Optimization Toolbox 12.1.1 ... for Chapter 14 Chapter 15 47 3 47 8 47 8 47 8 47 9 48 0 48 0 48 1 48 3 48 3 48 4 49 0 49 1 49 2 49 2 49 3 49 3 49 7 503 505 508 508 Discrete Variable Optimum Design Concepts and Methods 513 15.1 5 14 Basic Concepts...
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Introduction to Optimum Design phần 2 pdf

Introduction to Optimum Design phần 2 pdf

Kỹ thuật lập trình

... point (6 ,4) , P is P = ¥ 40 0 + ¥ 600 = 48 00 To plot the P = 48 00 contour, we plot the function 40 0x1 + 600x2 = 48 00 This contour is shown in Fig 3 -4 Step 6: Identification of Optimum Solution To locate ... pij x j ; i = to n (4. 16) j =1 Also, we can rewrite Eq (4. 14) as n Ê n ˆ  xi Á  pij x j ˜ ¯ i =1 Ë j =1 F (x ) = (4. 17) n  xi yi i =1 (4. 18) Substituting Eq (4. 16) into Eq (4. 17), F (x ) = ... cv3=[0 5]; const4=contour(x1,x2,g4,cv3,‘k’); clabel(const4) text(.25,20,‘g4’) const5=contour(x1,x2,g5,cv3,‘k’); clabel(const5) text(19,.5,‘g5’) text(1.5,7,’Feasible Region’) fv=[ 240 0, 48 00, 7200,...
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Introduction to Optimum Design phần 3 pptx

Introduction to Optimum Design phần 3 pptx

Kỹ thuật lập trình

... unit 4. 97 4. 98 Exercise 4. 45 4. 99 Exercise 4. 46 4. 100 Exercise 4. 47 4. 101 Exercise 4. 48 4. 102 Exercise 4. 49 4. 103 Exercise 4. 50 4. 1 04 Exercise 4. 51 4. 105 Exercise 4. 52 4. 106 Exercise 4. 53 4. 107 ... Exercise 4. 54 4.108 Exercise 4. 55 4. 109 Exercise 4. 56 4. 110 Exercise 4. 57 4. 111 Exercise 4. 58 4. 112 Exercise 4. 59 172 Exercise 4. 44 4.113 Exercise 4. 60 4. 1 14 Exercise 4. 61 INTRODUCTION TO OPTIMUM DESIGN ... DESIGN 4. 115 Exercise 4. 62 4. 116 Exercise 4. 63 4. 117 Exercise 4. 64 4.118 Exercise 4. 65 4. 119 Exercise 4. 66 4. 120 Exercise 4. 67 4. 121 Exercise 4. 68 4. 122 Exercise 4. 69 4. 123 Exercise 4. 70 4. 1 24 Exercise...
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Introduction to Optimum Design phần 5 pptx

Introduction to Optimum Design phần 5 pptx

Kỹ thuật lập trình

... [0 .46 646 4] 1.500000 [0 .48 1689] 1.809017 [0.868376] 0.809017 16 1.38 543 8 [0 .45 48 24] 1.386031 [0 .45 4823] 1.386398 [0 .45 4823] 1.386991 [0 .45 48 24] 0.001553 17 1.386031 [0 .45 4823] 1.386398 [0 .45 4823] ... II of the method 8 .41 Exercise 8.21 8 .42 Exercise 8.22 8 .43 Exercise 8.23 8 .44 Exercise 8. 24 8 .45 Exercise 8.25 8 .46 Exercise 8.26 8 .47 Exercise 8.27 8 .48 Exercise 8.28 8 .49 Exercise 8.29 8.50 ... = (2, 4, 10) c = f = (2x1 + 2x2, 4x2 + 2x1 + 2x3, 4x3 + 2x2); c(0) = (12, 40 , 48 ) c ( ) = 40 48 = 63.6 > e (continue) d(0) = -c(0) = (-12, -40 , -48 ) Calculate a0 by golden section search to minimize...
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Introduction to Optimum Design phần 6 potx

Introduction to Optimum Design phần 6 potx

Kỹ thuật lập trình

... 11, 41 6.3, g1 = 0.25 94, g2 = 0 .40 22, g3 = -46 .70, g4 = -0.618 (j) V0 ,1 = max{0; 0.25 94, 0 .40 22, - 46 .70, - 0.618} = 0 .40 22 (k) F ,2 = f0 ,2 + RV0 ,2 = 11, 41 6.3 + ( 24, 280)(0 .40 22) = 21,182 (l) ... point ( -4. 5, -4. 5), we obtain 380 INTRODUCTION TO OPTIMUM DESIGN g1 = 1 ( -4. 5) + ( -4. 5) - 1.0 = (active) 18 36 (d) g2 = [-( -4. 5) + 60( -4. 5)] = -2.655 < (inactive) 100 (e) g3 = -4. 5 - 1.0 = -1 .45 < ... - 6) = (- 14, 18) g1 (4, 4) = -1 < (inactive), a (1) = g1 = ấ1 , ậ3 (d) (e) ấ 1 , ậ 12 (f) g3 (4, 4) = -4 < (inactive), a (3) = g3 = (-1, 0) (g) g4 (4, 4) = -4 < (inactive), a ( ) = g4 = (0, -...
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Introduction to Optimum Design phần 7 pot

Introduction to Optimum Design phần 7 pot

Kỹ thuật lập trình

... -28 .45 40 - 3.300 ˚ Since K(x) is already known in Example 14. 1, we use Eq ( 14. 20) to calculate dU/dx as È 16.9090 37. 645 0 383 .42 00 ˘ dU Í 0 .47 11 5.0 247 ˙ = 0. 244 3 ˙ dx Í Í -0. 244 3 -0 .47 11 - 5.0 247 ˙ ... 1.00000E+00 1 .42 246 E-01 8.92216E-03 6 .47 793E-03 3.2 144 8E-02 7.68889E-02 8.80280E-02 8.87076E-02 6.66881E-02 7.90 647 E-02 6.86892E-02 4. 5 048 2E-02 1. 942 56E-02 4. 93063E-03 2.69 244 E-05 9.76924E-08 2.08000E-01 ... 2.19 948 E-02 3 .44 115E- 04 9 .40 469E-08 6 .40 000E+03 2.27873E+01 1.31993E+00 1.03 643 E-01 1.67 349 E-03 4. 30513E-07 7.20000E+03 7.87500E+02 7.13063E+02 6.99734E+02 7.03880E+02 7.03 947 E+02 1.0000E+01 9. 943 8E+00...
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Introduction to Optimum Design phần 8 pot

Introduction to Optimum Design phần 8 pot

Kỹ thuật lập trình

... 16100 15000 13200 12100 1110 1030 953 895 837 719 6 64 15.20 15.10 15.00 15.00 14. 90 14. 60 14. 60 1300 1200 1090 1010 940 41 1 375 156 144 132 123 1 14 67.5 61.9 3.830 3.810 3.780 3.750 3.730 2.580 2.560 ... feasible set S} 546 INTRODUCTION TO OPTIMUM DESIGN FIGURE 17 -4 Illustration of Pareto optimal set and utopia point in the criterion space The feasible points in the design space map onto a set of ... 88.30 82 .40 76.50 72.10 67.60 61.80 57.00 36. 74 36.52 36.26 36.08 35.90 36.69 36 .49 0. 945 0.885 0. 840 0.800 0.760 0.830 0.765 16.655 16.595 16.550 16.510 16 .47 0 12.180 12.115 1.680 1.570 1 .44 0 1.350...
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Introduction to Optimum Design phần 9 ppsx

Introduction to Optimum Design phần 9 ppsx

Kỹ thuật lập trình

... = to 11) are given as follows: (0.1975, 4) , (0.1 947 , 2), (0.1735, 1), (0.16, 0.5), (0.0 844 , 0.25), (0.0627, 0.1667), (0. 045 6, 0.125), (0.0 342 , 0.1), (0.0323, 0.0833), (0.0235, 0.07 14) , (0.0 246 , ... = 24) using the single payment compound amount factor of Eq (A.1) will be S 24 = spcaf (0.0075, 24) (1000) 24 = (1 + 0.0075) (1000) = (1.19 641 )(1000) = $1196 .41 and at the end of years (n = 48 ) ... 89, 715 .43 + 16, 768 .46 + 31 34. 14 + 585.79 - 27.37 = $1, 043 , 43 3.7 PWB = 900, 000 + 40 , 000 uspwf (0.15, 60) - 60, 000 sppwf (0.15, 20) + 900, 000 uspwf (0.15, 20) - 60, 000 sppwf (0.15, 40 ) +...
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Introduction to Optimum Design phần 10 ppsx

Introduction to Optimum Design phần 10 ppsx

Kỹ thuật lập trình

... correction algorithm, 44 0 44 2 constraint correction with minimum increase in cost, 44 6 44 7 cost reduction algorithm, 43 6 44 0 example—constraint correction at constant cost, 44 3 44 4, 44 4 44 5 example—constraint ... constant cost, 44 3 44 4, 44 4 44 5 constraint correction at specified increase in cost, 44 5 constraint correction step, 44 1 44 2 constraint correction with minimum increase in cost, 44 6 44 7 constraint ... algorithm, 44 0 44 2 cost reduction algorithm, 43 6 44 0 example—constraint correction at constant cost, 44 3 44 4, 44 4 44 5 example—constraint correction at specified increase in cost, 44 5 695 Algorithms,...
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Introduction to e design

Introduction to e design

Cơ khí - Chế tạo máy

... time Natural frequency 18.9 ksi 7.1 0 .43 8813 in.3 13.2 1515 Hz 10.5 ksi 14. 2 0. 548 8 in.3 13.2 1 840 Hz 44 .4 100 25.1 21.5 30 Chapter Figure 1.21: Physical prototypes of engine parts 1.2 and 1.3, ... employed in design modification to assign qualitative weighting factors to product performance and design changes e -Design employes a systematic and quantitative approach to design modifications (for ... marketing ideas In addition, the prototypes can be given to potential customers for feedback, thus bringing customers into the design loop early in product development 1 .4. 2 CNC Machining The machining...
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Phát triển ứng dụng web – web programming  chương 4   introduction to HTML5

Phát triển ứng dụng web – web programming chương 4 introduction to HTML5

Quản trị mạng

... classid="clsid:D27CDB6E-AE6D-11cf-96B8 -44 4553 540 000" codebase="http://download.macromedia.com/pub/shockwave/cabs/flash/swflash.cab #version=9,0,16,0" width="320" height= "40 0"> ... browser/ereader, so you will not be able to listen to this song.)
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an introduction to chemical engineering kinetics and reactor design

an introduction to chemical engineering kinetics and reactor design

Hóa học - Dầu khí

... in Reactor Design and Ideal Reactor Models 245 Selectivity and Optimization Considerations in the Design of Isothermal Reactors 317 10 Temperature and Energy Effects in Chemical Reactors 349 ... foundation for the practice of reactor design The text is designed as a teaching instrument It can be used to introduce the novice to chemical kinetics and reactor design and to guide him until he understands ... exposed to the concepts involved and has learned to use them in attacking reactor design problems I believe that the subject of computer-aided design should be deferred to graduate courses in reactor...
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DESIGN OF MACHINERYAN INTRODUCTION TO THE SYNTHESIS AND ANALYSIS OF MECHANISMS AND MACHINES phần 4 pptx

DESIGN OF MACHINERYAN INTRODUCTION TO THE SYNTHESIS AND ANALYSIS OF MECHANISMS AND MACHINES phần 4 pptx

Điện - Điện tử

... made in regard to graphical velocity analysis in Section 6.2 (p 244 ) apply as well to graphical acceleration analysis Historically, graphical methods were the only practical way to solve these ... asymptotes Centrodes of double-crank linkages will be closed curves Program FOURBARwill calculate and draw the fixed and moving centrodes for any linkage input to it Input the datafiles F06-l4.4br, ... centrode of link Read the diskfile F06-17bAbr into program FOURBARto animate that linkage with its coupler curve or centrodes Note in Figure 6- 14 (p 2 64) that choosing any location of instant center...
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