Allocation problems 2E
|
|
- Melinda Greene
- 5 years ago
- Views:
Transcription
1 Allocation problems 2E 1 The initial cost matrix is shown below (with the numbers representing minutes): Task C Task D Task E Worker L Worker M Worker N The following linear programming problem can be formulated to minimise the total time taken. 1 if worker i does task j 0 otherwise andj C, D, E where i L, M, N To minimise the total time (in minutes) taken P. Minimise P37 x 15 x 12x LC LD LE 25 x 13 x 16x MC MD ME 32 x 41 x 35x NC ND NE Each worker can be assigned to at most one task: x x x 1 or x 1 LC LD LE xmc xmd xme 1 or x 1 xnc xnd xne 1 or x 1 Each task must be done by just one worker: x x x 1 or x 1 LC MC NC xld xmd xnd 1 or x 1 xle xme xne 1 or x 1 Lj Mj Nj ic id ie Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 1
2 2 The initial cost matrix is shown below (with the numbers representing hundreds of pounds): Task S Task T Task U Task V Worker C Worker D Worker E Worker F The following linear programming problem can be formulated to minimise the total training cost. 1 if worker i is trained in task j 0 otherwise and j S, T, U, V where i C, D, E, F To minimise the total training cost (in hundreds of pounds) P. Minimise P36 x 34 x 32 x 35x CS CT CU CV 37 x 32 x 34 x 33x DS DT DU DV 42 x 35 x 37 x 36x ES ET EU EV 39 x 34 x 35 x 35x FS FT FU FV Each worker is to be trained in exactly one task: x x x x 1 or x 1 CS CT CU CV xds xdt xdu xdv 1 or x 1 xes xet xeu xev 1 or x 1 xfs xft xfu xfv 1 or x 1 Each task must have one worker trained to carry it out: x x x x 1 or x 1 CS DS ES FS xct xdt xet xft 1 or x 1 xcu xdu xeu xfu 1 or x 1 xcv xdv xev xfv 1 or x 1 Ej Fj is it iu iv Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 2
3 3 The initial cost matrix is shown below (with the figures representing the number of names and addresses): A B C D Replacing the entry with a large value of 1000 gives: A B C D The following linear programming problem can be formulated to maximise the number of names and addresses collected. 1 if worker i is assigned to site j 0 otherwise where i A, B, C, D and j 1, 2, 3 To maximise the number of names and addresses collected P. Maximise P11x A1 15x A2 1000x A3 14 x 18 x 17x B1 B2 B3 16 x 13 x 23x C1 C2 C3 15 x 14 x 22x D1 D2 D3 Each worker must be assigned to just one site: x x x or x 1 A1 A2 A3 1 B1 B2 B3 1 C1 C2 C3 1 D1 D2 D3 1 x x x or x 1 x x x or x 1 x x x or x 1 Each site must be assigned to just one worker: x x x x or x 1 1 A1 B1 C1 D1 1 A2 B2 C2 D2 1 A3 B3 C3 D3 1 Aj Bj x x x x or x 2 1 x x x x or x 3 1 i i i Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 3
4 4 a The initial cost matrix is shown below (with the numbers representing 100s): A B C D As the table above is an n by n matrix (with n = 4), we do not need to add any additional dummy rows or columns. To maximise the profit, subtract every number from the largest value in the table. b The largest value in the table is 14. Subtracting every value from 14 the table becomes: A B C D The smallest numbers in rows 1, 2, 3 and 4 are 2, 0, 2 and 1. We reduce rows first by subtracting these numbers from each element in the row. The table becomes: A B C D The smallest numbers in columns 1, 2, 3 and 4 are 0, 2, 0 and 1. We reduce columns by subtracting these numbers from each element in the column. Therefore, the reduced cost matrix is: A B C D We can cover all the zeros with four lines, so we have found our optimal solution of: A W (12) B Z (13) C Y (12) D X (11) Maximum profit = Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 4
5 4 c Using the initial cost matrix below the following linear programming problem can be formulated to maximise the amount of profit made. A if worker i does task j 0 otherwise and j W, X, Y, Z where i A, B, C, D To maximise the total profit (in 100s) P. Maximise P12 x 8 x 11 x 9x 14 x 10 x 9 x 13x BW BX BY BZ 11 x 9 x 12 x 10x CW CX CY CZ 13 x 11 x 10 x 12x Each worker can be assigned to at most one task: x x x x 1 or x 1 xbw xbx xby xbz 1 or x 1 xcw xcx xcy xcz 1 or x 1 x B C D x x x 1 or x 1 Each task must be done by just one worker: x x x x 1 or x 1 AW BW CW DW xax xbx xcx xdx 1 or x 1 xay xby xcy xdy 1 or x 1 xaz xbz xcz xdz 1 or x 1 Aj Bj iw ix iy iz Alternatively, the linear programming problem on the next page shows the problem formulated as a minimisation problem using the modified cost matrix found in part b. Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 5
6 4 c (continued) Modified cost matrix from part b: A B C D if worker i does task j 0 otherwise and j W, X, Y, Z where i A, B, C, D Objective: To minimise the objective function P. Minimise P2 x 6 x 3 x 5x 4 x 5 x x BX BY BZ 3 x 5 x 2 x 4x CW CX CY CZ x 3 x 4 x 2x Each worker can be assigned to at most one task: x x x x 1 or x 1 xbw xbx xby xbz 1 or x 1 xcw xcx xcy xcz 1 or x 1 x x x x 1 or x 1 Each task must be done by just one worker: x x x x 1 or x 1 AW BW CW DW xax xbx xcx xdx 1 or x 1 xay xby xcy xdy 1 or x 1 xaz xbz xcz xdz 1 or x 1 Aj Bj iw ix iy iz Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 6
Review Exercise 2. 1 a Chemical A 5x+ Chemical B 2x+ 2y12 [ x+ Chemical C [ 4 12]
Review Exercise a Chemical A 5x+ y 0 Chemical B x+ y [ x+ y 6] b Chemical C 6 [ ] x+ y x+ y x, y 0 c T = x+ y d ( x, y) = (, ) T = Pearson Education Ltd 08. Copying permitted for purchasing institution
More information7.2. Matrix Multiplication. Introduction. Prerequisites. Learning Outcomes
Matrix Multiplication 7.2 Introduction When we wish to multiply matrices together we have to ensure that the operation is possible - and this is not always so. Also, unlike number arithmetic and algebra,
More informationParts Manual. EPIC II Critical Care Bed REF 2031
EPIC II Critical Care Bed REF 2031 Parts Manual For parts or technical assistance call: USA: 1-800-327-0770 2013/05 B.0 2031-109-006 REV B www.stryker.com Table of Contents English Product Labels... 4
More informationProof by induction ME 8
Proof by induction ME 8 n Let f ( n) 9, where n. f () 9 8, which is divisible by 8. f ( n) is divisible by 8 when n =. Assume that for n =, f ( ) 9 is divisible by 8 for. f ( ) 9 9.9 9(9 ) f ( ) f ( )
More informationThe table below shows the improvement indices (the routes already in use are marked with a cross): 7 8 5
Review exercise 1 1 a Applying the north-west corner method gives: W1 W2 W3 Supply F1 2 2 4 F2 3 3 F3 4 4 8 Demand 2 9 4 15 Cost 2 7 2 8 3 2 4 6 4 3 72 b Set S(F1) = 0 and so solve the remaining equations.
More informationSummer 2017 Vol. 4 Issue 4 TLP Remembers the Man Who Taught Dallas to Dance. Darrell Cleveland started his professional modern dancing career as a
T L P S 2017 V. 4 I 4 TLP R M W T D D B C C DALLAS - I J, M Vw C, Aj D P D C, w w w ww D. A w C, w ;,,,. D C D B D T 1990, w, S A S. I, C. A M Vw 2015, MVC D P J M w 13. H D C I B E T D B D T,. w w C C
More informationProblem Max. Possible Points Total
MA 262 Exam 1 Fall 2011 Instructor: Raphael Hora Name: Max Possible Student ID#: 1234567890 1. No books or notes are allowed. 2. You CAN NOT USE calculators or any electronic devices. 3. Show all work
More informationTHE TRANSLATION PLANES OF ORDER 49 AND THEIR AUTOMORPHISM GROUPS
MATHEMATICS OF COMPUTATION Volume 67, Number 223, July 1998, Pages 1207 1224 S 0025-5718(98)00961-2 THE TRANSLATION PLANES OF ORDER 49 AND THEIR AUTOMORPHISM GROUPS C. CHARNES AND U. DEMPWOLFF Abstract.
More informationTaylor Series 6B. lim s x. 1 a We can evaluate the limit directly since there are no singularities: b Again, there are no singularities, so:
Taylor Series 6B a We can evaluate the it directly since there are no singularities: 7+ 7+ 7 5 5 5 b Again, there are no singularities, so: + + c Here we should divide through by in the numerator and denominator
More informationMathematics (JAN12MD0201) General Certificate of Education Advanced Level Examination January Unit Decision TOTAL.
Centre Number Candidate Number For Examiner s Use Surname Other Names Candidate Signature Examiner s Initials Mathematics Unit Decision 2 Wednesday 25 January 2012 General Certificate of Education Advanced
More informationPearson Edexcel GCE Decision Mathematics D2. Advanced/Advanced Subsidiary
Pearson Edexcel GCE Decision Mathematics D2 Advanced/Advanced Subsidiary Friday 23 June 2017 Morning Time: 1 hour 30 minutes Paper Reference 6690/01 You must have: D2 Answer Book Candidates may use any
More information( ) ( ) 2 1 ( ) Conic Sections 1 2E. 1 a. 1 dy. At (16, 8), d y 2 1 Tangent is: dx. Tangent at,8. is 1
Conic Sections E a y y so At (6, ), d y y ( 6) y 6 y+ 6 y 6+ y+ 6 d y y At, 6 When, y, angent at, is y 6 y 6+ 6+ y 6+ y 6 y y so,, d y At y ( ) y ( ) y y+ y + y+ e 6+ y 6 y 7 7 y 7 so d d y 7 At ( 7, 7),
More informationQ Scheme Marks AOs Pearson. Notes. Deduces that 21a 168 = 0 and solves to find a = 8 A1* 2.2a
Further Maths Core Pure (AS/Year 1) Unit Test : Matrices Q Scheme Marks AOs Pearson Finds det M 3 p p 4 p 4 p 6 1 Completes the square to show p 4 p 6 p M1.a Concludes that (p + ) + > 0 for all values
More informationDraft Version 1 Mark scheme Further Maths Core Pure (AS/Year 1) Unit Test 5: Algebra and Functions. Q Scheme Marks AOs. Notes
1 b Uses α + β = to write 4p = 6 a TBC Solves to find 3 p = Uses c αβ = to write a 30 3p = k Solves to find k = 40 9 (4) (4 marks) Education Ltd 018. Copying permitted for purchasing institution only.
More informationFinds the value of θ: θ = ( ). Accept awrt θ = 77.5 ( ). A1 ft 1.1b
1a States that a = 4. 6 + a = 0 may be seen. B1 1.1b 4th States that b = 5. 4 + 9 + b = 0 may be seen. B1 1.1b () Understand Newton s first law and the concept of equilibrium. 1b States that R = i 9j (N).
More informationPredicate Logic. 1 Predicate Logic Symbolization
1 Predicate Logic Symbolization innovation of predicate logic: analysis of simple statements into two parts: the subject and the predicate. E.g. 1: John is a giant. subject = John predicate =... is a giant
More informationDecision Mathematics D2 Advanced/Advanced Subsidiary. Thursday 6 June 2013 Morning Time: 1 hour 30 minutes
Paper Reference(s) 6690/01R Edexcel GE Decision Mathematics D2 Advanced/Advanced Subsidiary Thursday 6 June 2013 Morning Time: 1 hour 30 minutes Materials required for examination Nil Items included with
More informationMark scheme Mechanics Year 1 (AS) Unit Test 7: Kinematics 1 (constant acceleration)
1a Figure 1 General shape of the graph is correct. i.e. horizontal line, followed by negative gradient, followed by a positive gradient. Vertical axis labelled correctly. Horizontal axis labelled correctly.
More informationAdvanced Microeconomics Note 1: Preference and choice
Advanced Microeconomics Note 1: Preference and choice Xiang Han (SUFE) Fall 2017 Advanced microeconomics Note 1: Preference and choice Fall 2017 1 / 17 Introduction Individual decision making Suppose that
More informationHypothesis testing Mixed exercise 4
Hypothesis testing Mixed exercise 4 1 a Let the random variable X denote the number of vehicles passing the point in a 1-minute period. 39 Use the Poisson distribution model X Po, i.e. X Po(.5).5 e.5 P(
More informationQ Scheme Marks AOs Pearson ( ) 2. Notes. Deduces that 21a 168 = 0 and solves to find a = 8 A1* 2.2a
Further Maths Core Pure (AS/Year 1) Unit Test : Matrices Q Scheme Marks AOs Pearson Finds det M = 3 p p+ 4 = p + 4 p+ 6 1 ( )( ) ( )( ) ( ) Completes the square to show p + 4p+ 6= p+ + M1.a Concludes that
More information1a States correct answer: 5.3 (m s 1 ) B1 2.2a 4th Understand the difference between a scalar and a vector. Notes
1a States correct answer: 5.3 (m s 1 ) B1.a 4th Understand the difference between a scalar and a vector. 1b States correct answer: 4.8 (m s 1 ) B1.a 4th Understand the difference between a scalar and a
More informationReview For the Final: Problem 1 Find the general solutions of the following DEs. a) x 2 y xy y 2 = 0 solution: = 0 : homogeneous equation.
Review For the Final: Problem 1 Find the general solutions of the following DEs. a) x 2 y xy y 2 = 0 solution: y y x y2 = 0 : homogeneous equation. x2 v = y dy, y = vx, and x v + x dv dx = v + v2. dx =
More informationHomework 10 (due December 2, 2009)
Homework (due December, 9) Problem. Let X and Y be independent binomial random variables with parameters (n, p) and (n, p) respectively. Prove that X + Y is a binomial random variable with parameters (n
More informationConstant acceleration, Mixed Exercise 9
Constant acceleration, Mixed Exercise 9 a 45 000 45 km h = m s 3600 =.5 m s 3 min = 80 s b s= ( a+ bh ) = (60 + 80).5 = 5 a The distance from A to B is 5 m. b s= ( a+ bh ) 5 570 = (3 + 3 + T ) 5 ( T +
More informationLinear Algebra. Hoffman & Kunze. 2nd edition. Answers and Solutions to Problems and Exercises Typos, comments and etc...
Linear Algebra Hoffman & Kunze 2nd edition Answers and Solutions to Problems and Exercises Typos, comments and etc Gregory R Grant University of Pennsylvania email: ggrant@upennedu Julyl 2017 2 Note This
More informationMATH 2413 TEST ON CHAPTER 4 ANSWER ALL QUESTIONS. TIME 1.5 HRS
MATH 1 TEST ON CHAPTER ANSWER ALL QUESTIONS. TIME 1. HRS M1c Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Find the general solution of the differential
More informationDecision Mathematics D1
Pearson Edexcel GCE Decision Mathematics D1 Advanced/Advanced Subsidiary Friday 16 June 2017 Afternoon Time: 1 hour 30 minutes Paper Reference 6689/01 You must have: D1 Answer Book Candidates may use any
More informationDISCRIMINANT EXAM QUESTIONS
DISCRIMINANT EXAM QUESTIONS Question 1 (**) Show by using the discriminant that the graph of the curve with equation y = x 4x + 10, does not cross the x axis. proof Question (**) Show that the quadratic
More informationQ Scheme Marks AOs. 1a States or uses I = F t M1 1.2 TBC. Notes
Q Scheme Marks AOs Pearson 1a States or uses I = F t M1 1.2 TBC I = 5 0.4 = 2 N s Answer must include units. 1b 1c Starts with F = m a and v = u + at Substitutes to get Ft = m(v u) Cue ball begins at rest
More informationQ Scheme Marks AOs. Notes. Ignore any extra columns with 0 probability. Otherwise 1 for each. If 4, 5 or 6 missing B0B0.
1a k(16 9) + k(25 9) + k(36 9) (or 7k + 16k + 27k). M1 2.1 4th = 1 M1 Þ k = 1 50 (answer given). * Model simple random variables as probability (3) 1b x 4 5 6 P(X = x) 7 50 16 50 27 50 Note: decimal values
More informationAnswers: Year 4 Textbook 1 Pages 4 13
Answers: Year Textbook Pages Page. + = 00. + = 00. 00 0 =. + 7 = 00. 00 = 77. + = 00 7. 00 =. 00 =. + 7 = 00 0. + = 00. 00 = 7. 00 =. p. 7p. p. p 7. 7. 7p. p 0.... p. 7p Page. + = 700. + 7 = 00. 00 = 7.
More informationChapter Four: Linear Programming: Modeling Examples PROBLEM SUMMARY 4-1
Instant download and all chapters Solutions Manual Introduction to Management Science 11th Edition Bernard W. Taylor III https://testbankdata.com/download/solutions-manual-introduction-managementscience-11th-edition-bernard-w-taylor-iii/
More informationCHAPTER X. SIMULTANEOUS EQUATIONS.
CHAPTER X. SIMULTANEOUS EQUATIONS. 140. A SINGLE equation which contains two or more unknown quantities can be satisfied by an indefinite number of values of the unknown quantities. For we can give any
More informationThe normal distribution Mixed exercise 3
The normal distribution Mixed exercise 3 ~ N(78, 4 ) a Using the normal CD function, P( 85).459....4 (4 d.p.) b Using the normal CD function, P( 8).6946... The probability that three men, selected at random,
More informationConformal Mapping, Möbius Transformations. Slides-13
, Möbius Transformations Slides-13 Let γ : [a, b] C be a smooth curve in a domain D. Let f be a function defined at all points z on γ. Let C denotes the image of γ under the transformation w = f (z).
More informationCS 520: Introduction to Artificial Intelligence. Review
CS 520: Introduction to Artificial Intelligence Prof. Louis Steinberg : First-order Logic Proof Algorithms 1 Review Representing a Domain in FOL A domain is a portion of the world about which we wish to
More informationb UVW is a right-angled triangle, therefore VW is the diameter of the circle. Centre of circle = Midpoint of VW = (8 2) + ( 2 6) = 100
Circles 6F a U(, 8), V(7, 7) and W(, ) UV = ( x x ) ( y y ) = (7 ) (7 8) = 8 VW = ( 7) ( 7) = 64 UW = ( ) ( 8) = 8 Use Pythagoras' theorem to show UV UW = VW 8 8 = 64 = VW Therefore, UVW is a right-angled
More informationDecision Mathematics D1
Pearson Edexcel International Advanced Level Decision Mathematics D1 Advanced/Advanced Subsidiary Monday 1 February 2016 Afternoon Time: 1 hour 30 minutes Paper Reference WDM01/01 You must have: D1 Answer
More informationElastic collisions in one dimension 4D
Elastic collisions in one dimension 4D 1 a First collision (between A and B) 5+ 1 1= u+ v u+ v= 11 (1) 1 v u e= = 5 1 v u= () Subtracting equation () from equation (1) gives: 3u=9 u=3 Substituting into
More informationComplex symmetry Signals and Systems Fall 2015
18-90 Signals and Systems Fall 015 Complex symmetry 1. Complex symmetry This section deals with the complex symmetry property. As an example I will use the DTFT for a aperiodic discrete-time signal. The
More informationDecision Mathematics D1 (6689) Practice paper A mark scheme
Decision Mathematics D1 (6689) Practice paper A mark scheme 1. e.g. C 2 = A 5 = E 4 cs C = 2 A = 5 E = 4 M1 F 1 = B 3 = D 6 cs F = 1 B = 3 D = 6 M1 A =1, B = 3, C = 2, D = 6, E = 4, F = 1 (5) (5 marks)
More informationPartial Differential Equations
Part II Partial Differential Equations Year 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2015 Paper 4, Section II 29E Partial Differential Equations 72 (a) Show that the Cauchy problem for u(x,
More information1. Definition of a Polynomial
1. Definition of a Polynomial What is a polynomial? A polynomial P(x) is an algebraic expression of the form Degree P(x) = a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 3 x 3 + a 2 x 2 + a 1 x + a 0 Leading
More informationCircles, Mixed Exercise 6
Circles, Mixed Exercise 6 a QR is the diameter of the circle so the centre, C, is the midpoint of QR ( 5) 0 Midpoint = +, + = (, 6) C(, 6) b Radius = of diameter = of QR = of ( x x ) + ( y y ) = of ( 5
More informationUNIVERSITY OF SOUTHAMPTON. A foreign language dictionary (paper version) is permitted provided it contains no notes, additions or annotations.
UNIVERSITY OF SOUTHAMPTON MATH055W SEMESTER EXAMINATION 03/4 MATHEMATICS FOR ELECTRONIC & ELECTRICAL ENGINEERING Duration: 0 min Solutions Only University approved calculators may be used. A foreign language
More informationLinear Functions (I)
4. MAPPING BY ELEMENTARY FUNCTIONS There are many practical situations where we try to simplify a problem by transforming it. We investigate how various regions in the plane are transformed by elementary
More informationThe gradient of the radius from the centre of the circle ( 1, 6) to (2, 3) is: ( 6)
Circles 6E a (x + ) + (y + 6) = r, (, ) Substitute x = and y = into the equation (x + ) + (y + 6) = r + + + 6 = r ( ) ( ) 9 + 8 = r r = 90 = 0 b The line has equation x + y = 0 y = x + y = x + The gradient
More information( ) ( ) or ( ) ( ) Review Exercise 1. 3 a 80 Use. 1 a. bc = b c 8 = 2 = 4. b 8. Use = 16 = First find 8 = 1+ = 21 8 = =
Review Eercise a Use m m a a, so a a a Use c c 6 5 ( a ) 5 a First find Use a 5 m n m n m a m ( a ) or ( a) 5 5 65 m n m a n m a m a a n m or m n (Use a a a ) cancelling y 6 ecause n n ( 5) ( 5)( 5) (
More informationLesson 3: Linear differential equations of the first order Solve each of the following differential equations by two methods.
Lesson 3: Linear differential equations of the first der Solve each of the following differential equations by two methods. Exercise 3.1. Solution. Method 1. It is clear that y + y = 3 e dx = e x is an
More informationBertie3 Exercises. 1 Problem Set #3, PD Exercises
Bertie3 Exercises 1 Problem Set #3, PD Exercises 1. 1 ( x)qx P 2 ( x)w x P ( x)(qx W x) 2. 1 ( x)(ax (Bx Cx)) P 2 Bd P Ad 3. 1 ( x)jxx P 2 ( y) Ry P ( y)(jyy Ry) 4. 1 ( x)(lx (Mx Bx)) P 2 ( y)(my Jy) P
More informationCalculus II Practice Test Problems for Chapter 7 Page 1 of 6
Calculus II Practice Test Problems for Chapter 7 Page of 6 This is a set of practice test problems for Chapter 7. This is in no way an inclusive set of problems there can be other types of problems on
More informationFourier Analysis Overview (0B)
CTFS: Continuous Time Fourier Series CTFT: Continuous Time Fourier Transform DTFS: Fourier Series DTFT: Fourier Transform DFT: Discrete Fourier Transform Copyright (c) 2009-2016 Young W. Lim. Permission
More informationAssignment 11 Assigned Mon Sept 27
Assignment Assigned Mon Sept 7 Section 7., Problem 4. x sin x dx = x cos x + x cos x dx ( = x cos x + x sin x ) sin x dx u = x dv = sin x dx du = x dx v = cos x u = x dv = cos x dx du = dx v = sin x =
More informationHow to construct international inputoutput
How to construct international inputoutput tables (with the smallest effort) Satoshi Inomata Institute of Developing Economies JETRO OVERVIEW (1) Basic picture of an international input-output table (IIOT)
More informationDraft Version 1 Mark scheme Further Maths Core Pure (AS/Year 1) Unit Test 1: Complex numbers 1
1 w z k k States or implies that 4 i TBC Uses the definition of argument to write 4 k π tan 1 k 4 Makes an attempt to solve for k, for example 4 + k = k is seen. M1.a Finds k = 6 (4 marks) Pearson Education
More informationLECTURE 16 GAUSS QUADRATURE In general for Newton-Cotes (equispaced interpolation points/ data points/ integration points/ nodes).
CE 025 - Lecture 6 LECTURE 6 GAUSS QUADRATURE In general for ewton-cotes (equispaced interpolation points/ data points/ integration points/ nodes). x E x S fx dx hw' o f o + w' f + + w' f + E 84 f 0 f
More informationQ Scheme Marks AOs Pearson Progression Step and Progress descriptor. and sin or x 6 16x 6 or x o.e
1a A 45 seen or implied in later working. B1 1.1b 5th Makes an attempt to use the sine rule, for example, writing sin10 sin 45 8x3 4x1 States or implies that sin10 3 and sin 45 A1 1. Solve problems involving
More informationi t 1 v d B & N 1 K 1 P M 1 R & H 1 S o et al., o 1 9 T i w a s b W ( 1 F 9 w s t t p d o c o p T s a s t b ro
Pgm PII: S0042-6989(96)00205-2 VR, V 37, N 6, 705 720,1997 01997 E S L Pd G B 0042-698997$1700+ 000 P M L I B L P S M S E C S W A U O S R E R O 1 f 1 A 1 f fm J 1 A bqu g vy bd z gu u vd mvg z d,, -dg
More informationMöbius transformations and its applications
Möbius transformation and its applications Every Möbius transformation is the composition of translations, dilations and the inversion. Proof. Let w = S(z) = az + b, ad bc 0 be a Möbius cz + d transformation.
More informationDecision Mathematics D1
Pearson Edexcel GCE Decision Mathematics D1 Advanced/Advanced Subsidiary Friday 17 June 2016 Afternoon Time: 1 hour 30 minutes Paper Reference 6689/01 You must have: D1 Answer Book Candidates may use any
More informationEUROPEAN JOURNAL OF PHYSICAL AND REHABILITATION MEDICINE. An abridged version of the Cochrane review of exercise therapy for chronic fatigue syndrome
EUROPEAN JOURNAL OF PHYSICAL AND REHABILITATION MEDICINE EDIZIONI MINERVA MEDICA T PDF. A. T j. A C L LARUN J ODGAARD-JENSEN J R PRICE Kj G BRURBERG E J P R M S 6 [E ] EUROPEAN JOURNAL OF PHYSICAL AND
More informationTrigonometry and modelling 7E
Trigonometry and modelling 7E sinq +cosq º sinq cosa + cosq sina Comparing sin : cos Comparing cos : sin Divide the equations: sin tan cos Square and add the equations: cos sin (cos sin ) since cos sin
More informationCS 138b Computer Algorithm Homework #14. Ling Li, February 23, Construct the level graph
14.1 Construct the level graph Let s modify BFS slightly to construct the level graph L G = (V, E ) of G = (V, E, c) out of a source s V. We will use a queue Q and an array l containing levels of vertices.
More informationProblem Set 1. This week. Please read all of Chapter 1 in the Strauss text.
Math 425, Spring 2015 Jerry L. Kazdan Problem Set 1 Due: Thurs. Jan. 22 in class. [Late papers will be accepted until 1:00 PM Friday.] This is rust remover. It is essentially Homework Set 0 with a few
More informationLINEAR SYSTEMS AND MATRICES
CHAPTER 3 LINEAR SYSTEMS AND MATRICES SECTION 3. INTRODUCTION TO LINEAR SYSTEMS This initial section takes account of the fact that some students remember only hazily the method of elimination for and
More informationMathematics I. Exercises with solutions. 1 Linear Algebra. Vectors and Matrices Let , C = , B = A = Determine the following matrices:
Mathematics I Exercises with solutions Linear Algebra Vectors and Matrices.. Let A = 5, B = Determine the following matrices: 4 5, C = a) A + B; b) A B; c) AB; d) BA; e) (AB)C; f) A(BC) Solution: 4 5 a)
More informationSOME VARIANTS OF LAGRANGE S FOUR SQUARES THEOREM
Acta Arith. 183(018), no. 4, 339 36. SOME VARIANTS OF LAGRANGE S FOUR SQUARES THEOREM YU-CHEN SUN AND ZHI-WEI SUN Abstract. Lagrange s four squares theorem is a classical theorem in number theory. Recently,
More informationHousing Market Monitor
M O O D Y È S A N A L Y T I C S H o u s i n g M a r k e t M o n i t o r I N C O R P O R A T I N G D A T A A S O F N O V E M B E R İ Ī Ĭ Ĭ E x e c u t i v e S u m m a r y E x e c u t i v e S u m m a r y
More informationMark scheme. 65 A1 1.1b. Pure Mathematics Year 1 (AS) Unit Test 5: Vectors. Pearson Progression Step and Progress descriptor. Q Scheme Marks AOs
Pure Mathematics Year (AS) Unit Test : Vectors Makes an attempt to use Pythagoras theorem to find a. For example, 4 7 seen. 6 A.b 4th Find the unit vector in the direction of a given vector Displays the
More informationMath 21B - Homework Set 8
Math B - Homework Set 8 Section 8.:. t cos t dt Let u t, du t dt and v sin t, dv cos t dt Let u t, du dt and v cos t, dv sin t dt t cos t dt u v v du t sin t t sin t dt [ t sin t u v ] v du [ ] t sin t
More information5 Flows and cuts in digraphs
5 Flows and cuts in digraphs Recall that a digraph or network is a pair G = (V, E) where V is a set and E is a multiset of ordered pairs of elements of V, which we refer to as arcs. Note that two vertices
More informationHomework For each of the following matrices, find the minimal polynomial and determine whether the matrix is diagonalizable.
Math 5327 Fall 2018 Homework 7 1. For each of the following matrices, find the minimal polynomial and determine whether the matrix is diagonalizable. 3 1 0 (a) A = 1 2 0 1 1 0 x 3 1 0 Solution: 1 x 2 0
More informationA combinatorial algorithm minimizing submodular functions in strongly polynomial time
A combinatorial algorithm minimizing submodular functions in strongly polynomial time Alexander Schrijver 1 Abstract We give a strongly polynomial-time algorithm minimizing a submodular function f given
More informationMatrices and Linear transformations
Matrices and Linear transformations We have been thinking of matrices in connection with solutions to linear systems of equations like Ax = b. It is time to broaden our horizons a bit and start thinking
More information0.1 Problems to solve
0.1 Problems to solve Homework Set No. NEEP 547 Due September 0, 013 DLH Nonlinear Eqs. reducible to first order: 1. 5pts) Find the general solution to the differential equation: y = [ 1 + y ) ] 3/. 5pts)
More informationElastic collisions in one dimension Mixed Exercise 4
Elastic collisions in one dimension Mixed Exercise 1 u w A (m) B (m) A (m) B (m) Using conseration of linear momentum for the system ( ): mu m= mw u = w (1) 1 w e= = 3 u ( ) u+ = 3 w () Adding equations
More information9 Mixed Exercise. vector equation is. 4 a
9 Mixed Exercise a AB r i j k j k c OA AB 7 i j 7 k A7,, and B,,8 8 AB 6 A vector equation is 7 r x 7 y z (i j k) j k a x y z a a 7, Pearson Education Ltd 7. Copying permitted for purchasing institution
More informationMark Scheme (Results) June GCE Decision D2 (6690) Paper 1
Mark (Results) June 20 GCE Decision D2 (6690) Paper Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including academic,
More information550 = cleaners. Label the managers 1 55 and the cleaners Use random numbers to select 5 managers and 45 cleaners.
Review Exercise 1 1 a A census observes every member of a population. A disadvantage of a census is it would be time-consuming to get opinions from all the employees. OR It would be difficult/time-consuming
More information2 a Substituting (1,4,7) and the coefficients. c The normal vector of the plane
Exam-style practice: Paper 1 1 a First we find y and d x. y t t ( = + y = t t + 1 1 y t t t t = + + 1. x= t+ t = 1+ t. We now find the limits in terms of t. x= t = x=.75= t+ t t t +.75= 1± 1+ 15 t = We
More informationIIOCTAHOBJIEHILN. 3AKOHOITATEJIbHOEc oe pahiie riejtfl Br,rHc Kofr on[q.c ru ]s. llpe4ce4arenr 3 arono4are,rbuoro Co6paurax B.B.
KTTJbc p Jf Bc Kf n[q.c u CTBJL 20.12.2018 ] {nx6nnc 170 J KnpKTbx p6 nb C1pu 9g 6unc
More informationPhys 170 Lecture 4 1. Physics 170 Lecture 4. Solving for 3 Force Magnitudes in Known Directions
Phys 170 Lecture 4 1 Physics 170 Lecture 4 Solving for 3 Force Magnitudes in Known Directions Phys 170 Lecture 4 2 Mastering Engineering Issues There is definitely a problem with the campusebookstore.com
More informationTest 2 - Answer Key Version A
MATH 8 Student s Printed Name: Instructor: CUID: Section: Fall 27 8., 8.2,. -.4 Instructions: You are not permitted to use a calculator on any portion of this test. You are not allowed to use any textbook,
More informationSOLUTION FOR HOMEWORK 11, ACTS 4306
SOLUTION FOR HOMEWORK, ACTS 36 Welcome to your th homework. This is a collection of transformation, Central Limit Theorem (CLT), and other topics.. Solution: By definition of Z, Var(Z) = Var(3X Y.5). We
More informationFURTHER MATHEMATICS AS/FM/CP AS LEVEL CORE PURE
Surname Other Names Candidate Signature Centre Number Candidate Number Examiner Comments Total Marks FURTHER MATHEMATICS AS LEVEL CORE PURE CM Bronze Set B (Edexcel Version) Time allowed: 1 hour and 40
More informationQuality of tests Mixed exercise 8
Quality of tests Mixed exercise 8 1 a H :p=.35 H 1 :p>.35 Assume H, so that X B(15,.35) Significance level 5%, so require c such that P( Xc)
More informationMathematics Extension 1
BAULKHAM HILLS HIGH SCHOOL TRIAL 04 YEAR TASK 4 Mathematics Etension General Instructions Reading time 5 minutes Working time 0 minutes Write using black or blue pen Board-approved calculators may be used
More information2x + 5 = 17 2x = 17 5
1. (i) 9 1 B1 (ii) 19 1 B1 (iii) 7 1 B1. 17 5 = 1 1 = x + 5 = 17 x = 17 5 6 3 M1 17 (= 8.5) or 17 5 (= 1) M1 for correct order of operations 5 then Alternative M1 for forming the equation x + 5 = 17 M1
More informationORIE Pricing and Market Design
1 / 15 - Pricing and Market Design Module 1: Capacity-based Revenue Management (Two-stage capacity allocation, and Littlewood s rule) Instructor: Sid Banerjee, ORIE RM in the Airline Industry 2 / 15 Courtesy:
More informationPhysicsAndMathsTutor.com. Mark Scheme (Results) Summer Pearson Edexcel GCE Decision Mathematics /01
Mark (Results) Summer 2016 Pearson Edexcel GCE Decision Mathematics 2 6690/01 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications are awarded by Pearson, the UK s largest awarding body. We
More informationProblem Set # 1 Solution, 18.06
Problem Set # 1 Solution, 1.06 For grading: Each problem worths 10 points, and there is points of extra credit in problem. The total maximum is 100. 1. (10pts) In Lecture 1, Prof. Strang drew the cone
More informationMark scheme. Statistics Year 1 (AS) Unit Test 5: Statistical Hypothesis Testing. Pearson Progression Step and Progress descriptor. Q Scheme Marks AOs
1a Two from: Each bolt is either faulty or not faulty. The probability of a bolt being faulty (or not) may be assumed constant. Whether one bolt is faulty (or not) may be assumed to be independent (or
More informationTrigonometric Functions Mixed Exercise
Trigonometric Functions Mied Eercise tan = cot, -80 90 Þ tan = tan Þ tan = Þ tan = ± Calculator value for tan = + is 54.7 ( d.p.) 4 a i cosecq = cotq, 0
More informationYear 2011 VCE. Mathematical Methods CAS. Trial Examination 1
Year 0 VCE Mathematical Methods CAS Trial Examination KILBAHA MULTIMEDIA PUBLISHING PO BOX 7 KEW VIC 30 AUSTRALIA TEL: (03) 908 5376 FAX: (03) 987 4334 kilbaha@gmail.com http://kilbaha.com.au IMPORTANT
More informationMark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
Mark scheme Pure Mathematics Year 1 (AS) Unit Test : Coordinate in the (x, y) plane Q Scheme Marks AOs Pearson 1a Use of the gradient formula to begin attempt to find k. k 1 ( ) or 1 (k 4) ( k 1) (i.e.
More informationo V fc rt* rh '.L i.p -Z :. -* & , -. \, _ * / r s* / / ' J / X - -
-. ' ' " / ' * * ' w, ~ n I: ».r< A < ' : l? S p f - f ( r ^ < - i r r. : '. M.s H m **.' * U -i\ i 3 -. y$. S 3. -r^ o V fc rt* rh '.L i.p -Z :. -* & --------- c it a- ; '.(Jy 1/ } / ^ I f! _ * ----*>C\
More information4 Differential Equations
Advanced Calculus Chapter 4 Differential Equations 65 4 Differential Equations 4.1 Terminology Let U R n, and let y : U R. A differential equation in y is an equation involving y and its (partial) derivatives.
More informationAlgorithms: COMP3121/3821/9101/9801
NEW SOUTH WALES Algorithms: COMP32/382/90/980 Aleks Ignjatović School of Computer Science and Engineering University of New South Wales LECTURE 7: MAXIMUM FLOW COMP32/382/90/980 / 24 Flow Networks A flow
More informationMathematics of Physics and Engineering II: Homework answers You are encouraged to disagree with everything that follows. P R and i j k 2 1 1
Mathematics of Physics and Engineering II: Homework answers You are encouraged to disagree with everything that follows Homework () P Q = OQ OP =,,,, =,,, P R =,,, P S = a,, a () The vertex of the angle
More information