...the reporting time. INDIAN INSTITUTE OF MANAGEMENT KOZHIKODE POST GRADUATE PROGRAMME OFFICE End-Term Examinations (PGP 17, Term-III, 2013-15) Seating Plan Course Venue C2 C3 C4 FM-II OM-II CC1 CC2 C1 A1 A2 Section-A 01-15 16-27 28-38 39-62, D/205,219 Section-B 63-77 78-87 88-97 98-125 D/180,238 Roll Numbers Section-C Section-D 126-139 140-149 150-160 161-188 D/324 189-209 210-224 225-238 239-250 357-359 Section-E Section-F 251-270 271-284 285-298 299-304 360-361 FPM 2-5 305-324 325-338 339-352 353-356 362-364 FPM 8-11 Venue CC2 C3 EM MM-II IE C4 A1 A2 C1 C2 CC1 Section-A 01-33 34-48 49-62 D/205, 219 Section-B 63-95 96-111 112-125 D/180 Roll Numbers Section-C Section-D Section-E Section-F 126-146 147-167 168-188 D/111 189-210 212-232 233-250 357-359 251-271 272-304 360-361 FPM 2-5 305-325 326-356 362-364 FPM 8-11 Venue CC1 C4 C3 HRM C2 C1 A2 A1 CC2 Section-A 01-30 31-41 42-51 52-62 D/205,219 Section-B 63-91 92-101 102-111 112-125 D/180 Roll Numbers Section-C Section-D 126-153 154-163 164-175 176-188 D/324 189-203 204-218 219-232 233-250 357-359 Section-E Section-F 251-264 265-278 279-292 293-304 360-361 FPM 2-5 305-318 319-332 333-346 347-356 362-364 FPM 8-11 Course Venue A1 A2 C1 Section-A 01-15 16-30 31-44 45-62 D/219...
Words: 1153 - Pages: 5
...IMSE561/EMGT525 - TOTAL QUALITY MANAGEMENT AND SIX SIGMA HOMEWORK-6 Name: Shyam V.H Karumanchi UM ID 0833-3824 Question 1: Describe the design procedure for robust design REFERENCES: 1. Chapter 20 Quality engineering lecture notes Products and services should be designed to be inherently defect free and of high quality. Robustness The product or process performs its intended function well within user profiles and insensitive to the variation including: Variation in production Differences in materials Differences in users or operators aging of the product or process, and the product or process accomplishes this without major cost impact. General procedure for robust design 1. Problem Formulation: This step consists of identifying the main function, developing the P-diagram, defining the ideal function and S/N ratio, and planning the experiments. The experiments involve changing the control, noise and signal factors systematically using orthogonal arrays. 2. Data Collection/Simulation: The experiments may be conducted in hardware or through simulation. It is not necessary to have a full-scale model of the product for the purpose of experimentation. It is sufficient and more desirable to have an essential model of the product that adequately captures the design concept. Thus, the experiments can be done more economically. 3. Factor Effects Analysis: The effects of the control factors are calculated in this step and the results are analyzed to select optimum setting of the...
Words: 1847 - Pages: 8
...2014 年秋季中国精算师资格考试公告 中精协发〔2014〕19 号 中国精算师资格考试是中国保险监督管理委员会主办的国家级 职业资格考试,委托中国精算师协会组织实施。现将 2014 年秋季中 国精算师资格考试的有关事项公告如下: 一、开考科目 本次开考中国精算师资格考试准精算师部分 A1~A8 科目; 精算 师部分 F3、F4、F5、F8 及 F10 科目。具体开考科目如下: 科目代码 A1 A2 A3 A4 A5 A6 A7 A8 数学 金融数学 精算模型 经济学 寿险精算 非寿险精算 会计与财务 精算管理 科目 科目代码 F3 F4 F5 F8 F10 科目 个人寿险与年金精算实务 员工福利计划 非寿险实务 投资学 健康保险 二、考试时间 日期 时间 上午 9:00-12:00 10 月 18 日 下午 2:00-5:00 10 月 19 日 上午 9:00-12:00 A1 数学 A2 金融数学 A4 经济学 考试科目 1 下午 2:00-5:00 上午 8:30-12:30 10 月 20 日 下午 2:00-5:00 上午 8:30-12:30 10 月 21 日 下午 2:00-5:00 上午 8:30-12:30 10 月 22 日 下午 2:00-5:00 上午 8:30-12:30 10 月 23 日 下午 2:00-5:00 10 月 24 日 上午 8:30-12:30 A7 会计与财务 F5 非寿险实务 A3 精算模型 F3 个人寿险与年金精算实务 A5 寿险精算 F8 投资学 A6 非寿险精算 F10 健康保险 A8 精算管理 F4 员工福利计划 三、考试地点 考区 北京 天津 上海 武汉 广州 成都 合肥 西安 考试中心 中央财经大学 南开大学 复旦大学 武汉大学 中山大学 西南财经大学 中国科学技术大学 西安交通大学 联系电话 010-62288158/62288156/13381412909 13072232967 021-55665146/65642341/13818889826 027-68752134/18971480365 020-84114060/18620907703 028-87352399 0551-63607241/13215698056 029–82656851/13772069342 2 重庆 南京 大连 济南 长沙 厦门 哈尔滨 重庆大学 南京大学 大连理工大学 山东大学 湖南大学 厦门大学 黑龙江大学 023-65678620/13883302157 025-83593881 0411-84706101-605/13998500066 0531-88364894/13806409823 0731-88684770/13975804736 0592-2181450/18959218870 0451-86604652/13796002792 备注:加拿大滑铁卢大学考试中心考试安排另行通知 四、考试报名 考试报名包括网上报名和报名确认两个阶段。 (一)网上报名 网上报名时间:8 月 1 日至 9 月 10 日。 报名网址:中国精算师协会网站(www.e-caa.org.cn)。 (二)报名确认 为保证考生报名信息提交成功, 考生须向所报考考区考试中心确 认网上报名信息并缴费。 1、报名确认时间与方式 报名确认分为现场确认和邮寄确认两种方式。 现场确认时间:9 月 11 日至 9 月 15 日 邮寄确认时间:9 月 6 日至 9 月 10 日 考生可在报名确认时间内直接到向所报考考区考试中心现场确...
Words: 738 - Pages: 3
........................................... 9 2.2 History .................................................................................................................................. 10 2.3 Share Price Performance ...................................................................................................... 12 2.4 Products and Services ........................................................................................................... 14 2.5 Toll Roads ............................................................................................................................ 15 2.5.1 CityLink ......................................................................................................................... 15 2.5.2 M2 Hills ......................................................................................................................... 15 2.5.3 Lane Cove Tunnel .......................................................................................................... 15 2.5.4 M1 Eastern Distributor .................................................................................................. 15 2.5.5 M7 Westlink .................................................................................................................. 16 2.5.6 M5 Motorway ................................................................................................................ 16 2.5.7 Pocahontas 895...
Words: 12462 - Pages: 50
...| | | NET PROFIT | | | III. INTENSITY OF DAMAGE DUE TO INSECTS- Pest | Autumn-Winter(Y/N) | Intensity Of Damage(Low/Med/High) | Spring-Summer | Intensity Of Damage(Low/Med/High | P1 | | | | | P2 | | | | | P3 | | | | | P4 | | | | | OTHERS | | | | | NOTE- 1. P1-FSB, P2- Whitefly, P3-Jassid, P4-Mite 2. Low- 0-10%, Med- 10-40%,High >40% IV. DO YOU USE INSECTICIDE- YES/NO V. INSECTICIDE USAGE PATTERN- Insect | Order of importance | General period Occurrence | Molecule used | Brand | PriceRs./acre | Doseper acre | Time of appl. | Control duration | Criterias considered | Application behaviorN/H | Satisfaction(Y/N) | P1 | | | M1 | | | | | | | | | | | | M2...
Words: 640 - Pages: 3
...On Tuesday, June 13, Blythe City Council approved the recommendation set forth by Chief Jeffrey Wade of the Blythe Police Department, for the acquisition of a 2000 Dodge Peacekeeper armored rescue truck, from the Glendale California Police Department. Through the 1033 program, a federal government policy regulated by the Secretary of Defense, underfunded law enforcement agencies that are unable to obtain supplies and equipment to fulfill their duties are inclined to receive a transfer of excess military equipage at no cost to the agency. Valued at $65,100, staff report made available at the council meeting said, due to budget concerns, purchasing a new Peacekeeper was fiscally impossible. Having already received issued patrol rifles through this program, Chief Wade said, Sgt. Joshua Coe of Blythe PD, worked tirelessly to make sure the City was able to receive at a time where the nation is persistently seeing an uprise in violence. Orating this report at last week’s City Council meeting, Wade said, “According to the FBI, violent crime and “active shooter” situations are on the rise all across the United States. The question is not “if,”we will have an active shooter situation in Blythe, but rather “when?” Wade said the vehicle will be used to rescue injured citizens and officers when an active threat is present, which will also provide officers with a safe cover. Before procuring this vehicle, the City never had an armored rescue truck, although, Coe said, Blythe, and the...
Words: 620 - Pages: 3
...d = ax + by + c + a/2 + b d = 0 + a/2 + b d = a/2+b ×2 d = a + 2b 2 By substituting the value fora and b we will able to find the value of d. If d > 0, then we will select the N (North) pixel If d < 0, then we will select NE (Northeast) pixel Then there are two possibilities to go the line through M1 or M2 Equation for the above two points; M1 and M2 M1: d1 – Equation for the M1 midpoint d1 = a(x+1/2) + b(y+2) + c d1 = ax + by + c + a/2 +2 b d1 = 0 + a/2 + 2b ×2; d1= a+4 b 3 3 – 2;d b = d1 – d d1 = b + d 4 The equation for the M1 is d1 = b+d M2 : d2 – Equation for the M2 midpoint d1 = a(x+3/2) + b(y+2) + c d1 = ax + by + c +3 a/2 +2 b d1 = 0+ 3a/2 + 2b ×2; d2 = 3a +4b 5 5 – 2; a + b = d2 – d d2 = a + b + d 6 The equation of the M2 midpoint is d1 = a + b + d As mentioned above the algorithm chooses between N and NE pixels on the sign of the decision variable calculated in previous method. * Algorithm for the ellipse The equation for the ellipse is : x2/a2 + y2/b2 = 1 x2/a2 + y2/b2 – 1 = 0 x2 b2 + y2 a2 - a2 b2 = 0 Center of the ellipse (x,y) = (0,0) a = rx , b = ry The first point is considered from the origin (x,y) = (0,b) The ellipse has 4 -fold symmetry, so that we have to consider ¼ of the ellipse to implement the algorithm. When we consider about the first region of the ellipse, m<0 In region 2 of the ellipse, m>0 ...
Words: 822 - Pages: 4
...Answer all questions: QUESTION 1 (20 Marks) (a) (i) continuous (a2) (ii) discrete (a2) (iii) continuous (a2) (b) (i) (m2, a1) (ii) arrange the data in ascending order 18, 18, 24, 27, 35, 36, 45, 47, 51, 61, 63, 81, 82, 87, 90 (a2) Median is 47 (a1) (iii) x | | (A) | x2 (B) | 24 | 27 | 729 | 576 | 47 | -4 | 16 | 2209 | 90 | 39 | 1521 | 8100 | 61 | 10 | 100 | 3721 | 51 | 0 | 0 | 2601 | 87 | 36 | 1296 | 7569 | 36 | -15 | 225 | 1296 | 35 | -16 | 256 | 1225 | 18 | -33 | 1089 | 324 | 81 | 30 | 900 | 6561 | 82 | 31 | 961 | 6724 | 18 | -33 | 1089 | 324 | 45 | -6 | 36 | 2025 | 63 | 12 | 144 | 3969 | 27 | -24 | 576 | 729 | 765 | - | 8938 | 47953 | (column A 2 marks, m1, a1 ) OR (column B 2 marks, m1, a1) (iv) range = 90 – 18 = 72 (a1) (v) (m2, a1) QUESTION 2 (15 Marks) (a) Index (Feb) = (m1, a1) Index (March) = (a1) Index (April) = (a1) Index (May) = (a1) (b) (i) Size | 2011 | 2012 | | | | | | Price P | Quantity | Price pn | Quantity | | | | | Large | 0.60 | 500 | 0.75 | 600 | 300 | 360 | 375 | 450 | Medium | 0.55 | 700 | 0.60 | 1000 | 385 | 550 | 420 | 600 | Small | 0.40 | 300 | 0.45 | 500 | 120 | 200 | 135 | 225 | | | | Total | 805 | 1110 | 930 | 1275 | (last 4 columns, 1 mark each) (a1) (a1) (ii) Size | Weighting | Po | Pn | Index Relative | Large | 25 | 0.60 | 0.75 | 125 | ...
Words: 523 - Pages: 3
...Engineer-to-Engineer Note EE-218 a Technical notes on using Analog Devices DSPs, processors and development tools Contact our technical support at dsp.support@analog.com and at dsptools.support@analog.com Or visit our on-line resources http://www.analog.com/ee-notes and http://www.analog.com/processors Writing Efficient Floating-Point FFTs for ADSP-TS201 TigerSHARC® Processors Contributed by Boris Lerner Rev 2 – March 4, 2004 Introduction So, you want to write efficient code for the ADSP-TS201 TigerSHARC® processor? Or, maybe, you have come across the optimized example floating-point FFT for this processor and would like to understand how it works and what the author had in mind when writing it. This application note tries to answer both questions by going through that FFT example and all its levels of optimization in detail. This example can be followed in developing other algorithms and code optimized for the ADSPTS201S processor. Generally, most algorithms have several levels of optimization, all of which are discussed in detail in this note. The first and most straightforward level of optimization is paralleling of instructions, as the processor architecture will allow. This is simple and boring. The second level of optimization is loop unrolling and software pipelining to achieve maximum parallelism and to avoid pipeline stalls. Although more complex than the simple parallelism of level one, this can be done in prescribed steps without good understanding...
Words: 9075 - Pages: 37
...ACOUSTIC VISION – Acoustic Perception Based On Real Time Video Acquisition for Navigation Assistance Supreeth K Rao#, Arpitha Prasad B*, Anushree R Shetty&, Chinmai$ , Rajeshwari Hegde@ #,*,&,, Department of Telecommunication Engineering, @ Guide and faculty BMS College of Engineering, Bangalore, India # supreethkrao@gmail.com arpithaprasad@gmail.com & anushree.shetty12@gmail.com $ cpchinmai@gmail.com * Abstract— A smart navigation system based on an object detection mechanism has been designed to detect the presence of obstacles that immediately impede the path, by means of real time video processing. This paper is discussed keeping in mind the navigation of the visually impaired. A video camera feeds images of the surroundings to a Da-Vinci Digital Media Processor, DM642, which works on the video, frame by frame. The processor carries out image processing techniques whose result contains information about the object in terms of image pixels. The algorithm aims to select that object, among all others, that poses maximum threat to the navigation. A database containing a total of three sounds is constructed. Hence, each image translates to a beep, where every beep informs the navigator of the obstacles directly in front of him. This paper implements a more efficient algorithm compared to its predecessor, NAVI. Keywords— Navigation, Edge Detection, Flood Function, Object Detection, DM642, Acoustic Transformation I. INTRODUCTION Assistance...
Words: 2605 - Pages: 11
...CHAPTER 0 Contents Preface v vii Problems Solved in Student Solutions Manual 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Matrices, Vectors, and Vector Calculus Newtonian Mechanics—Single Particle Oscillations 79 127 1 29 Nonlinear Oscillations and Chaos Gravitation 149 Some Methods in The Calculus of Variations 165 181 Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics Central-Force Motion 233 277 333 Dynamics of a System of Particles Motion in a Noninertial Reference Frame Dynamics of Rigid Bodies Coupled Oscillations 397 435 461 353 Continuous Systems; Waves Special Theory of Relativity iii iv CONTENTS CHAPTER 0 Preface This Instructor’s Manual contains the solutions to all the end-of-chapter problems (but not the appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T. Thornton and Jerry B. Marion. It is intended for use only by instructors using Classical Dynamics as a textbook, and it is not available to students in any form. A Student Solutions Manual containing solutions to about 25% of the end-of-chapter problems is available for sale to students. The problem numbers of those solutions in the Student Solutions Manual are listed on the next page. As a result of surveys received from users, I continue to add more worked out examples in the text and add additional problems. There are now 509 problems, a significant number over the 4th edition. The instructor will find a large...
Words: 97501 - Pages: 391
...FACULTY OF BUSINESS AND MANAGEMENT MATHEMATICS FOR MANAGEMENT Question 1: A) 7 A + 10 B = 73 4 53 3 15 2 4 + 101 5 40 1 33 2 2 = 10 35 4221 21 735 14 28 + 10 50 400 10 3030 20 20 = 24 85 8221 31 3765 34 48 B) I. Matrix inverse method: 2x1 + 3x2 + 4x3 = 29 X1 + x2 + 2x3 = 13 3x1 + 2x2 + x3 = 16 1) Matrix form ,A: A =211 312 413121 122 223331 232 133 XX1 X2 X3 b291316 Ax = b ; x = A-1b A-1 =1A - A djoint A 2) Determine ( Row 2 as sample) A = (1) (-1)2+13421 + (1) (-1)2+2 2431 + 2(1)2+32332 = (-1) [(3)(1) – (4)(2)] + (1) [(2)(1)-(4)(3)] + (-2) [(2)(2)-(3)(3)] = (-1)(3-8) + (1)(2-12) + (-2)(4-9) = 5 – 10 + 10 A=5 3) Minor A A11 = 1221 = 1(1) – (2)(2) = 1 – 4 = -3 A12 = 1231 = 1(1) – (2)(3) = 1 – 6 = -5 A13 = 1232 = 1(1) – (1)(3) = 2 – 3 = -1 A21 = 3421 = 3(1) – (4)(2) = 3 – 8 = -5 A22 = 2431 = 2(1) – (4)(3) = 2 – 12 = -10 A23 = 2332 = 2(2) –(3)(3) = 4 – 9 = -5 A31 = 3412 = 3(2) – (4)(1) = 6 – 4 = 2 A33 = 2311 = 2(1) – (3)(1) = 2 – 3 = -1 Minor A = -3-5-1-5-10-520-1 4) Cofactor A = +-3--5+-1--5+-10--5+2-0+-1 = -35-15-10520-1 5) Ad-joint A = [Cofactor A] T = -35-15-10520-1T = -3525-100-15-1 6) Inverse A, A-1 A-1 = 1A – Ad-Joint A = 15-3525-100-15-1 = -3/512/51-20-1/55-1/5 7) Find X1...
Words: 1468 - Pages: 6
...trebled, what is the area of the new square? [ 441 cm2 ] 8. The product of two numbers is 1. One number is 1 2/5. What is the other number? [ 5/7 ] 9. The sum of two whole numbers is 100. One number is divisible by 7 and the other by 11. Find the number divisible by 7. [ 56 ] 10. The sum of the edges of a cube is 240 cm. Find the total surface area. [ 2400 cm2 ] 11. Which fraction is the smallest among the fractions in the set {11/12, 13/16, 17/20,19/20}? [ 13/16 ] 30-Second Questions 1. If 4/5 of a number is 3/10, what is 2/3 of the number? [ 1/4 ] 2. A residential compound has parallel frontage and back side whose measures are 93 m and 74 m. If the distance of those sides is 72 m, what is the area of the compound? [ 6012 m2 ] 5 3 3. How much is 3 greater than 5 ? [ 118 ] 4. What percent of 230 is 25% of 184? [ 20% ] 5. From Manila, it takes 5 1/3 hours to reach Baguio via the newly constructed expressway. To reach Baguio at 4:30 pm, what time should you leave Manila? [ 11:10 am ] 6. Bernadette...
Words: 1354 - Pages: 6
...Area = (base)(height) ; Perimeter = a+b+c. 2 3 2 Area of equilateral triangle = a . 4 4 Sphere : Surface Area = 4 π r2 ; Volume = π r3. 3 2 3 Cube : Surface Area = 6a ; Volume = a . 1 Cone : Curved Surface Area = π rl ; Volume = π r2 h 3 π r l + π r2 Total surface area = . Cuboid : Total surface area = 2 (ab + bh + lh); Volume = lbh. Cylinder : Curved surface area = 2 π rh; Volume = π r2 h Total surface area (open) = 2 π rh; Total surface area (closed) = 2 π rh+2 π r2 . SOME BASIC ALGEBRAIC FORMULAE: 1.(a + b)2 = a2 + 2ab+ b2 . 2. (a - b)2 = a2 - 2ab+ b2 . 3.(a + b)3 = a3 + b3 + 3ab(a + b). 4. (a - b)3 = a3 - b3 - 3ab(a - b). 2 2 2 2 5.(a + b + c) = a + b + c +2ab+2bc +2ca. 6.(a + b + c)3 = a3 + b3 + c3+3a2b+3a2c + 3b2c +3b2a +3c2a +3c2a+6abc. 7.a2 - b2 = (a + b)(a – b ) . 8.a3 – b3 = (a – b) (a2 + ab + b2 ). 9.a3 + b3 = (a + b) (a2 - ab + b2 ). 10.(a + b)2 + (a - b)2 = 4ab. 11.(a + b)2 - (a - b)2 = 2(a2 + b2 ). 12.If a + b +c =0, then a3 + b3 + c3 = 3 abc . INDICES AND SURDS m n mn (ab)m = a m b m am 1. am an = am + n 2. = a m − n . 3. (a ) = a . 4. . an m am −m = 1 a 5. = . 6. a 0 = 1, a ≠ 0 . 7. a . 8. a x = a y ⇒ x = y m am b b 9. a x = b x ⇒ a = b 10. a ± 2 b = x ± y , where x + y = a and xy = b. S B SATHYANARAYANA M. Sc., M.I.E ., M Phil . 9481477536 2 LOGARITHMS a x = m ⇒ log m = x (a > 0 and a ≠ 1) a 1. loga mn = logm + logn. m 2. loga = logm – logn. n 3. loga mn = n logm. log...
Words: 2819 - Pages: 12
...European Journal of Operational Research 224 (2013) 507–519 Contents lists available at SciVerse ScienceDirect European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor Production, Manufacturing and Logistics Pricing decisions for complementary products with firms’ different market powers Jie Wei a,⇑, Jing Zhao b, Yongjian Li c a General Courses Department, Military Transportation University, Tianjin 300161, PR China School of Science, Tianjin Polytechnic University, Tianjin 300160, PR China c Business School, Nankai University, Tianjin 300071, PR China b a r t i c l e i n f o Article history: Received 25 July 2011 Accepted 5 September 2012 Available online 11 September 2012 Keywords: Pricing Complementary products Market power Stackelberg game a b s t r a c t This article reports the results of a study that explores the pricing problems with regard to two complementary products in a supply chain with two manufacturers and one common retailer. The authors establish five pricing models under decentralized decision cases, including the MS-Bertrand, MS-Stackelberg, RS-Bertrand, RS-Stackelberg, and NG models, with consideration of different market power structures among channel members. By applying a game-theoretical approach, corresponding analytic solutions are obtained. Then, by comparing the maximum profits and optimal pricing decisions obtained in different decision cases, interesting and valuable managerial...
Words: 11966 - Pages: 48