chapter9 multiple compartment model

Solving Crossword Puzzles Using ExtendedPotts ModelKazuki Jimbo1,Hiroya Takamura2,and Manabu Okumura21Tokyo Institute of Technology,Department of Computer Science2Tokyo Institute of Technology,Precision and Intelligence LaboratoryAbstract.Solving crossword puzzles by computers is a challenging taskin artificial intelligence.It requires logical inference and association aswell as vocabulary and common sense knowledge.For this task,wepresent an extension of the Potts model.This model can incorporatevarious clues for solving puzzles and require less computational cost com-pared with other existing models.1IntroductionThe crossword puzzle is one of the most famous puzzles in the world.Solving crossword puzzles automatically by computers is a challenging task in artificial intelligence and is more difficult than solving other logical puzzles.One has to use his or her linguistic knowledge,world knowledge,and logical and association inference when solving crossword puzzles,while one does not have to use them when solving sudoku or other logical puzzles.There are strict rules on sudoku or other logical puzzles and one can verify that his or her solution is correct by checking whether the solution follows the rules completely.On the other hand,rules on crossword puzzles are not strict because of the diversity of human language expressions.One cannot verify that his or her solution is correct by simply checking whether the solution follows the rules.In this writing,we introduce an extended Potts model for solving Japanese crossword ing this model,one can incorporate various clues for solv-ing crossword puzzles.Moreover,this model requires less computational cost compared with other existing models.2Related WorkKeim et al.constructed Proverb,a system which solves American-style cross-word puzzles in English automatically[2].Proverb divides the process of solving crossword puzzles into two steps:generating a list of candidate answers for each clue using many language resources and algorithms,andfinding the best solu-tion combining the candidate answers for the clues.In thefirst step,they used a database which contains more than350,000clue-answer pairs from5,133puz-zles.In American-style crossword puzzles,this database of clue-answer pairs is H.Hattori et al.(Eds.):JSAI2008,LNAI5447,pp.39–47,2009.c Springer-Verlag Berlin Heidelberg200940K.Jimbo,H.Takamura,and M.Okumuravery useful because clues in American-style crossword puzzles are usually short and the same clue-answer pair can be found many times in different puzzles.In the second step,they used Shazeer et al.’s algorithm[1],whichfinds the optimal answer from the list of candidate answers.This algorithm recursively updates the weight of each candidate so that the expected number of correct answers will be maximized.Additionally,itfinds all feasible combinations of candidate answers,and then calculates the score of each feasible combination using the weights of candidate answers.The combination with the maximum score should be the solution.This algorithm is computationally complex because it needs tofind all fea-sible combinations of candidate answers.Moreover,this algorithm outputs no solutions if there is no feasible combination of candidate answers.On the other hand,the extended Potts model,which we propose in this writing,is computa-tionally less complex.Additionally,our model outputs a solution even if there are no feasible combinations.Sato attempted to automatically solve crossword puzzles in Japanese[3].His system also divides the solving process into two steps as in Proverb.Clues of Japanese crossword puzzles tend to be longer and hence more difficult than those of English ones.In order to tackle the difficulty,he classified the clues according to how their answers would be found.The algorithm used in his second step is Shazeer’s,the same one as in Proverb.Ernandes et al.constructed a crossword solver named WebCrow[4].The sys-tem uses the World Wide Web as a knowledge base.3The Proposed Model for Solving Crossword3.1Solving Crossword PuzzlesThe process of automatically solving crossword puzzles by computers is divided into two steps:making a list of candidate answers for each clue andfinding the correct combination of answers from the lists.Thefirst step is to create a list of candidate answers for each clue.This step is to answer some quizzes.Each crossword puzzle contains one grid and some clues.A clue is a hint which represents one word in the puzzle grid.Clues are often so ambiguous that one clue can represent more than one word.That is why most algorithms create a list of candidate answers for each clue,instead of a unique answer.In this step,the system searches linguistic resources or World Wide Web,retrieves candidate answers,and calculates their confidence scores. These candidate answers are inputs to the next step.Thisfirst step is similar to the question answering task.The second step is tofind the correct combination of answers from the lists. The combination should satisfy the following constraint:if two words cross each other,they should have the same character at the crossing point.Violation of this constraint means that at least one of the two words is incorrect.Solving Crossword Puzzles Using Extended Potts Model41 3.2Candidate Answer DetectionIn this section,we explain how to detect the candidate answers and their scores from clues of crossword puzzles.We introduce two types of methods:the method exploiting dictionaries as in Sato[3]and the method exploiting the World Wide Web as in Ernandes[4].We use multiple methods to generate candidate answers.First,the system ex-tracts content words from each clue.We refer to those content words as keywords. Next,the system applies multiple methods independently to each keyword and extracts candidate answers.Finally,the system merges all the extracted candi-dates into one list.We describe the details of these methods below.Dictionary-based method.First,we explain methods using dictionaries.Japanese-Japanese dictionary:This method looks up each keyword in a Japanese-Japanese dictionary(Iwanami Japanese dictionary),and returns the words found in the glosses(i.e.,definition sentences)as candidates.The score is the number of the words found.Japanese-Japanese dictionary(inverted):This method searches a Japanese-Japanese dictionary(Iwanami Japanese dictionary)for the entry words whose glosses contain the keyword.The candidate answers are those entry words.The score is the number of the keywords in the glosses.Japanese-Japanese dictionary(filling-in):Thefilling-in method attacksfilling-in clues,which contain a blank,such as“bon-ni kaerazu(It’s no use crying over spilt).”Only in this method,instead of extracting keywords in preprocess-ing,we take the clue as a pattern including blanks.Then,the method searches a Japanese-Japanese dictionary(Iwanami Japanese dictionary)for expressions matching the clue pattern.The candidate answers are the found expressions. The score is the number of the matched expressions in the dictionary. Thesaurus:This method returns the words categorized as in the same semantic group as the keyword,which is given by a thesaurus(Bunrui Goihyo,Word List by Semantic Principles).The score is the depth of the most specific ancestor shared by the keyword and the candidate words.Thesaurus(antonym):This methodfinds antonyms of keywords from a the-saurus(Kadokawa thesaurus).The score is the number of the found descriptions on the antonym information in the thesaurus.Web-based method.Second,we explain methods using the World Wide Web. Dictionary-based method has an advantage that we can obtain reliable answers from high-quality linguistic resources,but also has a disadvantage that we cannot obtain an answer if the knowledge is not contained in the resources.We expect that we can obtain answers for more various questions by using huge amount of information on the World Wide Web.42K.Jimbo,H.Takamura,and M.OkumuraSearch by keyword:For each clue,we create a query for web search(Yahoo! JAPAN)by listing keywords.The words in the snippets of the top50results retuned by the search engine with the query are used as candidates.The score is the number of the words found in the snippets.Filling-in pattern match:This method is a variation of‘Japanese-Japanese dic-tionary(filling-in)’method above.This method uses the World Wide Web in-stead of the dictionary.For eachfilling-in clue,we create a query of web search(Yahoo!JAPAN) consisting of the expression around the blank.Then,the method searches the snippets of the top50results for expressions matching the clue pattern.The score is the number of the found expressions in the50snippets.Candidate merging.After obtaining the lists of candidate answers by the above methods,the system merges these lists into one.If more than one method generate the same candidate,the score for the candidate the merged list is the sum of the scores given by each method.This merged list contains words of various parts of speech.However,most of the words used as answers of crossword puzzles are nouns.Therefore,the candidate words with other parts of speech than noun are excluded from the list.3.3Finding the Best CombinationGenerally,more than one answer candidates will be found for each clue.The correct solution of the puzzle may consist of words without the highest scores in the candidate lists.In other words,the highest-scored words may not be in the best combination.Then,it is important how tofind the best combination from the candidate lists.We use the extended Potts model for this purpose,which is described below.Originally,the Potts model is used for describing behaviors of spins on crystal trellis in the domain of statistical dynamics.This model is used for describing states of nodes on the problem concerned with network[6].For graph(or network)G(V,E),c i∈{1,...,n}represent states of nodes v i∈V The weight between v i and v j is represented by w ij.L is the set of indices for the observed nodes,a i∈{1,...,n}is the state of each observed variable indexed by i.Let H(c)denote an energy function,which indicates a state of the whole network:H(c)=−βij w ijδ(c i,c j)+αi∈L−δ(c i,a i),(1)whereβis a constant called the inverse-temperature,andαis a positive constant representing a weight on labeled data.Functionδreturns1if two arguments are equal to each other,0otherwise.The state is penalized if c i(i∈L)is different from a ing H(c),the probability distribution of the network is represented asP(c)=exp(−H(c))Z,(2)Solving Crossword Puzzles Using Extended Potts Model 43where Z is a normalization factor.One can estimate the state of this network at ˆc which maximizes P (ˆc ).However,it is computationally difficult to exactly estimate the state of this network.A mean-field approximation method described by Nishimori [5]can be applied in order to avoid this difficulty.In the method,P (c )is replaced by factorized function ρ(c )= i ρi (c i ),where ρi (c i )corresponds to the probability that the state of node v i is c i .Then we can obtain the function with the smallest value of the variational free energy:F (c )= cP (c )H (c )− c −P (c )log P (c )(3)=−αi c i ρi (c i )δ(c i ,a i )(4)−βij c i ,c j ρi (c i )ρj (c j )w ij δ(c i ,c j )(5)− ic i −ρi (c i )log ρi (c i ).(6)By minimizing F (c )under the condition that ∀i, c i ρi (c i )=1,we obtainthe following fixed point equation for i ∈L :ρi (c )=exp(αδ(c,a i )+β j w ij ρj (c ))n exp(αδ(n,a i )+β j w ij ρj (n )).(7)The fixed point equation for i ∈L can be obtained by removing αδ(c,a i )from above.This fixed point equation is solved by an iterative computation.We convert the crossword puzzle into a problem of finding a state of the network modeled by a variant of the Potts model.As pointed out by Shazeer et al.[1],crossword puzzles can be described by a graph as follows:Node:Blank for answer of a clueState:Selected answer of a clueEdge:Crossing of down and across blanks (nodes)However,the original Potts model cannot be applied straightforwardly for solving crossword puzzles in two reasons.First,in the Potts model,the value of energy function is low and the probability of combination is high,if the states of nodes connected with an edge are the same.However,in crossword puzzles,crossing blanks does not mean that the blanks have the same answer,while the character at the crossing point has to be shared by the blanks.Second,in the task of solving crossword puzzles,a list containing weighted multiple candidate words is given for each clue.Now,we define the extended Potts model,which is applicable to the task of solving crossword puzzles.First,we define the energy function H cross (c )asH cross (c )=−βij M ij (c i ,c j )+α i−S i (c i )(8)44K.Jimbo,H.Takamura,and M.Okumurawhile M ij is a function which indicates whether the words in two nodes v i and v j satisfy the constraint,and S i is the normalized score of node v i in the candidate list.When the states of v i and v j are c i and c j respectively,M ij (c i ,c j )=1if the two nodes have the same character at the crossing point,or the two nodes do not cross,and M ij (c i ,c j )=0otherwise.The score S i satisfies ∀i, c S i (c )=1.Then,we derive the probability of state vector P cross (c )and the free energy F cross (c )in the task of solving crossword puzzles from the energy function H cross as follows:P cross (c )=exp(−H cross (c ))Z,(9)F cross (c )=c P cross (c )H cross (c )(10)− c −P cross (c )log P cross (c )(11)=−α i c i ρi (c i )S i (c i )(12)−β ij c i ,c j ρi (c i )ρj (c j )M ij (c i ,c j )(13)− ic i −ρi (c i )log ρi (c i ),(14)while Z is the normalization factor and ρi (c i )is the probability that the state of node v i is c i .Note that ∀i, c i ρi (c i )=1.Minimizing F cross (c )on the condition that ∀i, c i ρi (c i )=1,we obtain the equationρi (c )=exp(αS i (c )+β j c j M ij (c,c j )ρj (c j )) n exp(αS i (n )+β j c jM ij (n,c j )ρj (c j )).(15)We can minimize F cross (c )by updating ρi recursively using this equation.In our experiments,we iteratively update ρi until the value of the energy function H cross (c )converges.After the convergence of H cross (c ),we select the word c i which maximizes value of ρi (c i )for each clue v i and output all c i as the final solution.In this writing,we call this model the extended Potts model.The constraints in the puzzle are embedded in this model as penalty factors.That is why this model outputs a solution even if there are no feasible combinations.In this writing,we compare two initial values of ρi :score-based initial value ρi (c i )=S i (c i )and random initial value satisfying ρi (c i )(0≤ρi (c i )≤1)andc i ρi (c i )=1.Annealing process can be applied to the Potts model.Annealing process is a process which increases the inverse temperature βby Δβ>0after every convergence of ρand executes the iterative updating again.In the extendedSolving Crossword Puzzles Using Extended Potts Model45 Potts model,largeβemphasizes the constraints.However,the scores of words tend to be neglected in this case.On the other hand,largeαemphasizes the scores of words.However,the constraints tend to be neglected in this case.Annealing processfirst computes the state of the network emphasizing the scores of words by settingβsmall.After that,the process setsβslightly larger and computes the state again.These two computations are iterated by turns until convergence.4ExperimentsWe conducted some experiments in order to test the performance on real cross-word puzzles.To measure the performance of our method,we used the word accuracy,which indicates the ratio of the correct words to all words for clues. This measure is also used by[1]and[3].4.1Data SetWe tested the performance using15puzzles from http://cross.matrix.jp/. The size of all these puzzles is7x7letters.The maximum,minimum,and average number of clues in each puzzle are22,18,and20.5,respectively.4.2ExperimentsWe tried search for the optimal solution using candidate answers that we ob-tained by the methods in section3.2.In this section,we used candidate answers from all the methods including Web-based ones,unless otherwise noted. Variation of parameters.We conducted the experiments with various param-eter values.We adjusted the parametersαandβin range of0–100000,executed the iterations without annealing,and calculated the word accuracy.The val-ues of the parameters which give the best performance are(α=1000,β=2) with the score-based initialization(14.9%correct),and(α=1000,β=2)and (α=500,β=5)with the random initialization(13.0%correct).Table1shows the variation of the word accuracy forfixedα=1000and variousβ.The word accuracy with the score-based initialization is higher than that with the random initialization in most cases.Next,we conducted the same experiment over the candidates from only dictionary-based methods.The word accuracy without the Web-based methods is much lower than that with the Web-based methods.Table1.The word accuracy(%)forfixedα=1000and variousββInitial value01251020304050Score12.313.014.914.011.713.39.18.810.1Random11.911.411.412.311.713.011.47.89.446K.Jimbo,H.Takamura,and M.OkumuraTable2.The word accuracy(%)with annealingΔβWithoutβInitial value0.010.050.10.5125annealing2Score14.014.015.314.014.014.0—14.92Random10.712.011.711.412.710.1—11.410Score13.613.013.314.014.313.614.311.710Random11.411.412.713.012.310.712.711.7Annealing.We compared results with and without annealing,inα=1000, which gives high performance in the above experiment.We tested both the score-based initialization and the random initialization.The incrementΔβin one annealing step is set in range of0.01–5.We executed them withβfrom0to 2and from0to10.Table2shows the result.In most cases,annealing increases the word accuracy.However,how to decide the parameter values is also important when we employ annealing,because the word accuracy also depends on the parameter value.Comparison with related work.We conducted another experiment for com-paring Shazeer et al.’s method[1]and our method.The computer used in this experiment has2.66GHz x2dual core CPU and2GB memory.We compared only execution time.Shazeer et al.’s method takes more than3days for one puzzle,while our method takes15.47seconds in average for one puzzle with10times annealing. This result shows that our method using the extended Potts model is quite effective to shorten the execution time.Additionally,the execution time tends to be long if the number of candidates increases.5ConclusionWe proposed to use the extended Potts model for automatically solving cross-word puzzles.Our method is much faster than existing methods.However,the maximum word accuracy was around15%.This is because the method for can-didate answer detection is too rough.Consequently,the inaccurate result made a bad effect on the solution of optimization problems.Therefore,our future work will include making the candidate answer detection more accurate.For this work, question answering technology can be applied.Moreover,the extended Potts model has some parameters.How to decide the values of these parameters is also an open problem.References1.Shazeer,N.M.,Littman,M.L.,Keim,G.A.:Solving Crossword Puzzles as Probabilis-tic Constraint Satisfaction.In:Proceedings of the sixteenth national conference on Artificial intelligence and the eleventh Innovative applications of artificial intelligence conference innovative applications of artificial intelligence,pp.156–162(1999)Solving Crossword Puzzles Using Extended Potts Model47 2.Keim,G.A.,Shazeer,N.,Littman,M.L.,Agarwal,S.,Cheves,C.M.,Fitzgerald,J.,Grosland,J.,Jiang,F.,Pollard,S.,Weinmeister,K.:Proverb:The Probabilistic Cruciverbalist.In:Proceedings of the Sixteenth National Conference on Artificial Intelligence,pp.710–717(1999)3.Sato,S.:Solving Japanese Crossword Puzzles.IPSJ SIG Notes,NL-147-11,69–76(2002)(in Japanese)4.Ernandes,M.,Angelini,G.,Goli,M.:WebCrow:a WEB-based system for CROss-Word solving.In:Proceedings of the Twentieth National Conference of Artificial Intelligence,pp.1412–1417(2005)5.Nishimori,H.:Statistical Physics of Spin Glasses and Information Processing.Ox-ford University Press,Oxford(2001)6.Wu,F.-Y.:The Potts model.Reviews of Modern Physics54(1),235–268(1982)。

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Aquarius with Therapeutic Plasma Exchange (TPE): Practical GuideChoosing an appropriate filtero Use a Membrane Plasma Separation (MPS) filter or equivalent: MPS03 Low Volume), MPS05 (Low Volume or Adults) or MPS07 (Adult).o Low Volume : clinician's choice: either MPS03 or MPS05, considering overall extracorporeal volume and total plasma exchange amount.o Adult: MPS05 and MPS07 are interchangeable but for plasma exchange rates of more than 1 litre / hour an MPS07 commonly maintains Trans Membrane Pressure (TMP) <50mmHg.o Maximum plasma exchange rates rise with each incremental size of MPS filter, MPS type filter allows removal of molecules up to 980,000 Dalton Molecular Weight** IMPORTANT**: ONLY an MPS filter is used for TPE treatment. Incorrect use of a MPS filter for any other type of therapy such as Continuous Veno-Venous Haemofiltration (CVVH) could lead to significant patient harm.TPE filters are clearly labelled for Plasma Separation Only and should be stored separately, away from Haemofilters. NIKKISO RECOMMENDS 2 X TRAINED USERS MUST ALWASYS VERIFY AND SUITABLY DOCUMENT THE CORRECT CHOICE OFFILTER HAS BEEN SELECTED FOR THE CORRECT THERAPYEssentials1.o Choose 'TPE' on Aquarius therapy selection screeno Lining of the Aquarius is covered in the on-screen graphics, Help and Zoom Graphic m enus.o MPS filters are identified by the cardboard box packaging, a red warning label and identifying collar ‘for plasma separation only’. It is suggested that the collar is left in place after assembly and during treatment.o Attach 1l Sodium Chloride 0.9% to replacement scale hook as priming solution.o TPE may be anticoagulated with Heparin or Regional Citrate Anticoagulation.Recirculationo Clamp pre-dilution line (all plasma is replaced post-dilution).o Input programme. Training sessions typically specify: Set Total Plasma first, then titrate Plasma Rate within MPS filter TMP limit to give an appropriate time.o At the end of Recirculation time, remove 1l Sodium Chloride 0.9% bag on replacement scale and attach prescribed plasma replacement solution (For example: Albumin / Fresh Frozen Plasma / Plasma Expander / Sodium Chloride 0.9%). o If replacement fluid is in bottled presentation, air inlets avoid multiple balance a larms.o Practical tip: If more than one container is hung on the replacement scale hook, a close observational watch for emptying of containers should be maintained. To switch from one container to the next, a temporary pause of the treatment pumps allows the transfer of replacement fluid flow.Programmingo There are several methods available to physicians responsible for prescribing an appropriate volume of plasma to exchange during each treatment.o Individualised patient prescription considers that removal during a single plasma exchange is limited to components from the intravascular compartment. Multiple plasma volumes may be removed over additional treatments, typically on alternate days2o Plasma Volume = (0.065 x W) X (1- HCT)3o Plasma Volume = W x 70 x [1-(HCT x 0.91)]4o Plasma Volume = 30-40ml/kg5o Plasma Volume = between 2-5 litres in adultso W = Body weight (kg), HCT = Haematocrit (%), Mean Blood Volume = 70-80ml/kg.o Aquarius "Time" programming box may be left as 0:00 or used as an additional goal reminder.o"Plasma Rate" Plasma Exchange rate (ml/Hr) and "Total Plasma" Plasma Exchange volume (ml) are prescribed at the clinician's choice. “Plasma Rate” is titrated to give the "Total Plasma" over the chosen time p eriod.o It is usual to set entire Total Plasma Volume as one programme: For example: "Total Plasma" first, then titrate "Plasma Rate” to give an appropriate time.o Practical tips "Container weight" – the weight of the empty plasma replacement container(s) influences Aquarius "Bags Change in:" advisory message, and will stop the treatment pumps at the chosen weight.o As a 'rule of thumb' for glass bottles, container weight is approximately half of volume; e.g. typical 500ml Albumin bottle weighs 280-300g, allowing for overage, 300-350g is a common choice6.o For bagged fluids: (For example: Albumin / Fresh Frozen Plasma / Plasma Expander / Sodium Chloride 0.9%), a 100g 'counterweight' may be hung on the replacement fluid scale. This counterweight should not be connected to theAqualine.Aquarius with Therapeutic Plasma Exchange (TPE): Practical GuideConnectiono Single and Double connection choices can be used.o Adults: Start blood flow at about 80ml/min, gradual increase as tolerated; up to approximately 100-150 ml/min3 based on the Physician prescription.o Independent start of treatment after 2-5 minutes allows fibres to acclimatise to blood c ontact.Treatmento Transmembrane Pressure (TMP) must be maintained below 80mmHg3 to protect the integrity of the membrane fibres.o Aquarius will alarm "High TMP" at 100mmHg1.o Practical tip: Typical troubleshoot for rising TMP: decrease "Plasma Rate" first (i.e. extend duration of exchange), secondary option to decrease blood flow.Troubleshooting Essentialso Patient preparation.o Anticoagulation.o Your training may cover: Removal of air from Aqualine, Filtration fraction, Treatment sequence and local troubleshooting advice.Replacement Fluido Considerations for Paediatric and low blood flow therapies.o Practical Tip: If Aquarius displays “Change bag now”, and there is still replacement solution in the container, user may decrease the "container weight" by 10g and restart treatment.o Practical tip: be ready to stop treatment as soon as the container is empty.Blood Leako Aquarius sensitivity 4ml blood in 1l waste plasma.o Alarm and blood observed in filtrate - stop treatment, do not wash back, disconnect.o Alarm and NO visible blood seen in filtrate - typical troubleshooting may include checks on: Blood Leak Detection ( BLD) chamber, mirror, % in More screen, consideration of exchange rate and blood flow rate with TMP, considerations; length of treatment left, sample of line filtrate for whole erythrocytes or whole haemoglobin. If in doubt, best advice is to stop, do not wash back, disconnect, recommence.TPE with Regional Citrate Anticoagulationo Serum Calcium/Phosphate levels may be affected by certain presentations of Albumin which include Citrate, typical clinical choices include careful monitoring and supplementation7.o Anticoagulant volumes may be considered.End TreatmentFollow the on-screen instructions for reinfusion, safe disconnection, and line set removal of the Aqualine before switching Aquarius off.References1)Nikkiso Europe GMBH. IFU Aquarius System SW 6.02.16, Rev. 5.1 (02/2019). Section 5.11.5 TPE (Therapeutic PlasmaExchange. P157.2)Ismail,N. Roxana, N. and Hakim, R. Chapter 11: Plasmapheresis. In: Daudirgas, J. Blake, P. and Ing, T. 2004.Handbook of Dialysis., 4th Edition. Philadelphia: Lippincott Williams. ISBN:0316173819. P236.3)MPS Filter Packaging insert. Micropes Filter for Plasma Separation Instructions for use. Bellco, Via Camurana,41037 Mirandola (MO), Italy. IB0510133000-REV05.4)UK Blood Transfusion & Tissue Transplantation Services: Transfusion Handbook: Therapeutic Plasma Exchange.link: /?Publication=HTM&Section=9&pageid=1137#lnk01. P129, section 11.1.Accessed 19042017.5)Kaplan, A. 2008. CORE CURRICULUM IN NEPHROLOGY: Therapeutic Plasma Exchange: Core Curriculum 2008.University of Connecticut Health Center, John Dempsey Hospital, and the UConn Dialysis Center, Farmington,Connecticut. /article/S0272-6386(08)00707-5/pdf . Accessed 19042017.6)CRRT Aquarius Filtration/PICU/Protocols/picuSpecific/Clinical%20guidelines%20CVVH.pdf Accessed 19042017, P23-25.7)Kissling et al. A new prescription model for regional citrate anticoagulation in therapeutic plasma exchange. BMCNephrology (2017) 18:81. P8 para 4 lines 3-28.https:///pmc/articles/PMC5333425/pdf/12882_2017_Article_494.pdf。

人教版全国全部高考专题英语高考真卷1.阅读理解第1题.If you hand my son an orange, he'll ask, "Does it have nuts in it?" You might think, What's wrong with that kid? Of course not!" But I think,"A smart boy."You see, my son Gus has food allergies to peanuts and tree nuts, and even the smallest amount of accidental ingestion can touch off anaphylaxis—a severe immune response that affects multiple systems and can throw the body into shock.The hardest thing about raising a child with food allergies is that you have to be perfect. You always have to ask that question about nuts. You need to read labels every time, even for foods you've bought a zillion times before. And to be honest, that makes me anxious a lot of the time.Once, I brought home an Italian combo sandwich for my son without asking about nuts. After a single bite, my son said, "My tongue feels funny." Oh no. Those are words you never want to hear if your child has food allergies. I pulled apart his sandwich and found pistachios（开心果）, one of the nuts he is most allergic to. What had I done? We immediately injected the contents of an EpiPen into his outer thigh（大腿）and raced to the Emergency Room.My son's first serious reaction happened when he was 3 years old, on a vacation to Puerto Rico. I still thank God he was OK, because that day could have turned out different if we were deep in the rainforest instead of right in achain hotel in San Juan. His face swelled up after eating trail mix with nuts, and he had difficulty breathing.The one thing that calms me, to be honest, is Gus. Kids with food allergies have a superpower: they are mature far beyond their years. Gus is 11 now and is able to speak up for himself. When he orders his meal, whether it's at a sit-down restaurant or a fast food counter, he says, "I am allergic to all nuts. Can you make sure there are no nuts in anything I order? Thanks."（1）How does the author start her story?A: By stating her son's unique hobby.B: By asking the readers a question.C: By creating an imaginary scene.D: By comparing nuts with fruits.（2）What does the underlined phrase "touch off" in paragraph 2 probably mean?A: Describe.B: Recover.C: Affect.D: Cause.（3）Why did Gus say his tongue felt funny?A: He hated to eat pistachios.B: He had some allergic food.C: He wanted to amuse his mother.D: He found his sandwich half-cooked.（4）What can we learn about Gus's disease from the text?A: It can be deadly when it begins to attack.B: It's about three years since it was first found.C: It kept Gus away from fast food.D: It was due to his parents' ignorance in his babyhood.【答案】CDBA【解答】（1）C 细节理解题。

107/dewe-3020All-In-OneDEWE-3020Most portable all-in-one instrument8 isolated DAQP or 16 differential MDAQ analog inputs Simple expansion to 16 isolated DAQP modules3 PCI slots for A/D and other cards (1394, ARINC, 1553…)15.4” TFT wide-screen display with touch-screenAdditional interfaces and sensorsMeasurements are not limited to just classic analog and digital signals. Please find further detailed information to expand your system in the chapter “Components”. Needed to complete the systemDEWE-ORION “A/D Boards” offer simultaneous sampled analog inputs, synchronous digital I/Os, high-performance counters and high-speed CAN interfaces. DAQP or MDAQ signal amplifiers and software are needed as well.Options to expand the systemAdd further “Interface Cards” like ARINC-429, 1553, PCM telemetry, FireWire ® and analog output or special “Sensors” like synchronized video, industrial encoders (RIE-360) or GPS.DEWESoft DAQP/MDAQ A/D card DEWE-30-8-EXP VIDEO FlexRayChoose from two modelsAdd your choice of signal conditioning,A/D board(s) and software to complete these systems1981108DEWE-3020DEWE-3020Most flexible model, prepared for DAQP isolated analog input amplifier modules. DAQP conditioners offer highest band-width, great accuracy, different input ranges and integrated filters. Besides the single channel modularity – a module eas-ily can be changed by the user at any time – the main advan-tage of these modules is the high galvanic isolation which ensures safe measurements, high quality results and make them almost indestructible. See chapter “Signal Conditioning” for details.DEWE-3021Version for sensor input via differential MDAQ analog input amplifiers. MDAQ modules are available in cost efficient and space saving 8-channel blocks. See chapter “Signal Conditioning” for details.In case a 16 channel DEWE-ORION series A/D board is installed inside the DEWE-3020, the system can be expanded to 16 isolated DAQP modules easily. Just add a DEWE-30-8-EXP expansion unit by a convenient single cable connection.DEWE-3020 with DEWE-30-8-EXP (16 DAQP modules in total)DEWE-3021 with 4 encoder inputsBNC configurationDSUB-9 configurationMixed layoutDEWE-3021DEWE-3020Adding more counters or digital inputsEspecially many automotive applications require multiple counter inputs, e.g. for connecting wheel pulse transducers. By choosing a suitable DEWE-ORION series A/D board up to 10 counters per card are available. The connectors for attach-ing the sensors are placed on the sidepanel near the analog inputs.DEWE-3020 with openedbattery compartment3020-DC-UPSDEWE-UPS-300-DC3020-CSMK1109EPAD2-TH81103020-CB8-BNC and 3020-CB8-B connection blocksDEWE-31-16PS-BATBattery power supply DEWE-POWERBOX-12DC Power distribution boxDEWE-30-8-EXP8 additional amplifier slotsOptional battery charger (range: 90 to 260 V AC )15.4“ touch-screenDVD-BURNERfor data backup to DVD or CDThe carrying bag is included as standard accessoryInternal 256 GB Solid State Disk for maximum reliabilityDEWE-3020Channel ExpansionSignal conditioning for slow signals is added by connecting EPAD2 series modulesto the systems EPAD interface.A N A L O GDEWE-30-16DEWE-30-16Analog cable : Additional A/D boards are installed into the basic instrument and external signal conditioning, e.g. DAQ modules in a DEWE-30 chassis, is connected by means of an analogsignal cable.E T HE RN ETDEWE-3021DEWE-NET : Several instruments are con-nected via Ethernet. Each unit requires an ORION-SYNC option. For short dis-tances a sync cable is used if the units are far from each other a sync interfacelike DEWE-CLOCK is used.P C IDEWE-50-32PCI expansion : A PCI-HOST card is installed into the basic instrument and external signal conditioning, e.g. DAQ modules in a DEWE-50 chassis, is con-nected by means of a PCI cable.For expanding the number of dynamic channels there some choices:DEWETRONGmbH•Parkring4•A-8074Graz-Grambach•Tel(0043)31630700•Fax(0043)316307090•******************。

多水平模型(英文原著) chap8Chapter 8Multilevel cross classifications8.1 Random cross classificationsIn previous chapters we have considered only data where the units have a purely hierarchical or nested structure. In many cases, however, a unit may be classified along more than one dimension. An example is students classified both by the school they attend and by the neighbourhood where they live. We can represent this diagramatically as follows for three schools and four neighbourhoods with between one and six students per school/neighbourhood cell. The cross classification is at level 2 with students at level 1.School 1 School 2 School 3 Neighbourhood 1 x x x x x x xNeighbourhood 2 x x x x x x x x x xNeighbourhood 3 x x x x x x xNeighbourhood 4 x x x x x x xFigure 8.1 A random cross classification at level 2Another example is in a repeated measures study where children are measured by different raters at different occasions. If each child has its own set of raters not shared with other children then the cross classification is at level 1, occasions by raters, nested within children at level 2. This can be represented diagramatically as follows for three children with up to 7 measurement occasions and up to three raters per child.We see that the cross classification takes place entirely within the level 2 units. We note that, by definition, a level 1 cross classification has only one unit per cell. We can, however, also view such a cross classification as a special case of a level 2 cross classification with, at most, a single level 1 unit per cell. It seems appropriate to view such cases as level 1 cross classifications only where the substantive context determines that there is at most one unit per cell (see section 8.6).Child 1 Child 2 Child 3Occasion: 1 2 3 4 5 6 7 1 2 3 4 6 1 4 7Rater 1 x x x x xRater 2 x x x x xRater 3 x x x x xRater 4 x x x x xRater 5 x x xRater 6 xFigure 8.2 A random cross classification at level 1.If now the same set of raters is involved with all the children the crossing is at level 2 as can be seen in the following diagram with three raters and three children and up to five occasions.Child 1 Child 2 Child 3Occasion: 1 2 3 4 1 2 1 2 3 4 5Rater 1 x x x xxx xRater 2xRater 3 x x xFigure 8.3Figure 8.3 is formally the same structure as Figure 8.1 with the level 1 variance being that between occasions.These basic cross classifications occur commonly when a simple hierarchical structure breaks down in practice. Consider, for example, a repeated measures design which follows a sample of students over time, say once a year, within a set of classes for a single school. We assume first that each class group is taken by the same teacher. The hierarchical structure is then a three level one with occasions grouped within students who are grouped within classes. If we had several schools then schools would constitute the level 4 units. Suppose, however, that students change classes during the course of the study. For three students, three classes and up to three occasions we might have the following pattern in Figure 8.4..Student 1 Student 2 Student 3 Occasion: 1 2 3 1 2 1 2 3 Class/teacher 1 x x x xClass/teacher 2 xClass/teacher 3 x x xFigure 8.4 Students changing classes/teachers.Formally this is the same structure as Figure 8.3, that is a cross classification at level 2 for classes by students. Such designs will occur also in panel or longitudinal studies of individuals who move from one locality to another, or workers who change their place of employment. If we now include schools these will be classified as level 3 units, but if students also change schools during the course of the study then we obtain a level 3 cross classification of students by schools with classes nested at level 2 within schools and occasions as the level 1 units. The students have moved from being crossed with classes to being crossed with schools. Note that since students are crossed at level 3 with schools they are also automatically crossed with any units nested within schools and we do not need separately to specify the crossing of classes with students.Suppose now that, instead of the same teachers taking the classes throughout the study, the classes are taken by a completely new set of teachers every year and where new groupings of students are formed each year too. Such a structure with four different teachers at two occasions for three students is given in Figure 8.5.Student 1 Student 2 Student 3Occasion: 1 2 1 2 1 2Teacher 1 1 x xTeacher 2 xTeacher 3 2 xxTeacher 4xThis is now a cross classification of teachers by students at level 2 with occasion as the level 1 unit. We note that most of the cells are empty and that there is at most one level 1 unit per cell so that no independent between occasion variance can be estimated as pointed out above. In fact we can also view this as a level 1 cross classification of teachers by students, with missing data, and occasion can be modelled in the fixed part, for example using a polynomial function of age. Raudenbush (1993) gives an example of such a design, and provides details of an EM estimation procedure for 2-level 2-way cross classifications with worked examples.We can have a design which is a mixture of those given by Figure 8.4 and Figure 8.5 where some teachers are retained and some are new at each occasion. In this case we would have a cross classification of teachers by students at level 2 where some of the teachers only had observations at one occasion. More generally, we can have an unbalanced design where each teacher is present at a variable number of occasions. Other examples of such designs occur in panel studies of households where, over time, some households split up and form new households. The total set of all households is crossed with individual at level 2 with occasion at level 1. The households which remain intact for more than one occasion provide the information for estimating level 1 variation.Occasion 2Teacher 1 Teacher 2 Teacher 3Teacher 1 x x x x x x x xOccasionTeacher 2 x x x x x x1Teacher 3 x x x x x x x xWith two occasions where we have the same teachers or intact groups we can formulate an alternative cross classification design which may be more appropriate in some cases. Instead of cross classifying students by teachers we consider cross classifying the set of all teachers at the first occasion by the same set at the second occasion, as follows.We have 22 students who are nested within the cross classification of teachers at each occasion. The difference between this design and that in Figure 8.4 is analogous to the difference between a two-occasion longitudinal design where a second occasion measurement is regressed on a first occasion measurement and the two-occasion repeated measures design where a measurement is related to age or time. In Figure 8.6 we are concerned with the contribution from each occasion to the variation in, say, a measurement made at occasion 2. In Figure 8.4 on the other hand, although we could fit a separate between teacher variance for each occasion, the response variable is essentially the same one measured at each occasion. Designs such as that of Figure 8.6 are useful where, for example, measurements are made on the same set of studentsand schools at the start and end of schooling, as in school effectiveness studies, and where students can move between schools. In such cases we may also wish to introduce a ‘weight’ to reflect the time spent in each school, and we shall discuss this below.We now set out the structure of these basic models and then go on to consider extensions and special cases of interest.8.2 A basic cross classified modelGoldstein (1987a) sets out the general structure of a model with both hierarchical and cross classified structures and Rasbash and Goldstein (1994) provide further elaborations. We consider first the simple model of Figure 8.1 with variance components at level 2 and a single variance term at level 1.We shall refer to the two classifications at level 2 using the subscripts j j 12, and in general parentheses will group classifications at the same level. We write the model asy Xu u e i j j i j j j j i j j ()()()12121212=+++β(8.1)The covariance structure at level 2 can be written in the following formc o v ()c o v ()v a r ()c o v ()()()()()()()()/////y y y y y y y i j j i j j ui j j i j j u i j j i j j i j j u u 1212112122121212122222====+σσσσ(8.2)Note that if there is no more than one unit per cell, then model (8.1) is still valid and can be used to specify a level 1 cross classification as defined in Section 8.1.Thus the level 2 variance is the sum of the separate classification variances, the covariance for two level 1 units in the same classification is equal to the variance for that classification and the covariance for two level 1 units which do not share either classification is zero. If we have a model where random coefficients are included for either or both classifications, then analogous structures are obtained. We can also add further ways of classification with obvious extensions to the covariance structure.Appendix 8.1 shows how cross classified models can be specified and estimated efficiently using a purely hierarchical formulation and we can summarise the procedure using the simple model of 8.1. We specify one of the classifications, most efficiently the one with the larger number of units, as a standard hierarchical level 2 classification. For the other classification we define a dummy (0,1) variable for each unit which is one if the observation belongs to that unit and zero if not. Then we specify that each of these dummy variables has a coefficient random at level 3 and in addition constrain the resulting set of level 3 variances to be equal. The variance estimate obtained is that required for this classification and the level 2 variance for the other classification is the one we require for that.If we have a third classification at level 2 then we can obtain the third variance by defining a similar set of dummy variables with coefficients varying at level 4 and variances constrained to be equal. This procedure generalises straightforwardly to sets of several random coefficients for each classification, with dummy variables defined as the products of the basic (0,1) dummy variables used in the variance components case and with corresponding variances and covariances constrained to be equal within classifications. In general a p-way cross classification at any level can be modelled by inserting sets of random variables at the next p-1 higher levels. Thus in a 2-level model with two crossed classifications at level 1 we would obtaina three level model with the original level 2 at level 3 and the level 1 cross classifications occupying levels 1 and 2.8.3 Examination results for a cross classification of schoolsThe data consist of scores on school leaving examinations obtained by 3435 students who attended 19 secondary schools cross classified by 148 primary schools in Fife, Scotland (Paterson, 1991). Before their transfer to secondary school at the age of 12 each student obtained a score on a verbal reasoning test, measured about the population mean of 100 and with a population standard deviation of 15. The model is as followsy xu u e i j j i j j j j i j j ()()()12121212011=++++ββ(8.3)and the results are given in Table 8.1. Random coefficients for verbal reasoning were also fitted but the coefficients are estimated as zero.Ignoring the verbal reasoning score we see that the between-primary school variance is estimated to be more than three times that between secondary schools. The principal reason for this is that the secondary schools are on average far larger than primary schools, so that within a secondary school, primary school differences are averaged. Such an effect will often be observed where one classification has far fewer units than another, for example where a small number of schools is crossed with a large number of small neighbourhoods or a small number of teachers is crossed with a large number of students at level 1 within schools. In such circumstances we need to be careful about our interpretation of the relative sizes of the variances.Table 8.1 Analysis of Examination Scores by Secondary by Primary school attended. The subscript 1 refers to primary and 2 to secondary school.ParameterEstimate (s.e.)Estimate (s.e.)Estimate (s.e.)Fixed A B C Intercept 5.50 5.98 5.99 Verbal Reasoning - 0.16 (0.003)0.16 (0.003)Randomσu ()1021.12 (0.20) 0.27 (0.06) - σu ()2020.35 (0.16) 0.011 (0.021) 0.28 (0.06) σe 28.1 (0.2)4.25 (0.10)4.26 (0.10)When the verbal reasoning score is added to the fixed part of the model the between secondary school variance becomes very small, the between primary school variance is also considerably reduced and the level 1 variance also. The third analysis shows the effect of removing the cross classification by primary school. The between secondary school variance is now only a little smaller than in analysis A without verbal reasoning score. Using analysis C alone, which is typically the case with school effectiveness studies which control for initial achievement, we would conclude that there were important differences between the progress made in secondaryschools. From analysis B, however, we see that most of this is explained by the primary schools attended. Of course, the verbal reasoning score is only one measure of initial achievement, but these results illustrate that adjusting for achievement at a single previous time may not be adequate.8.4 Computational considerationsAnalysis A in Table 8.1 took about 40 seconds per iteration on a 66 Mhz 486 PC using the ML3 software, approximately ten times longer than analysis C. This relative slowness is due to the size of the single level 3 unit which contains all the 3435 level 1 units. For very much larger problems the computing considerations will become of greater concern, so that some procedure for speeding up the computations would be useful.In the present analysis there are 120 cells of the cross classification which contain only one student. If we eliminate these from the analysis we obtain two disjoint subsets containing 14 and 5 secondary schools. There are a further 24 cells containing two students and if these are removed we obtain six disjoint subsets the largest of which contains eight secondary schools. Table 8.2 shows the estimates from the resulting analyses.The only substantial difference is in the between secondary school variance which is anyway poorly estimated. The first analysis took about 15 seconds and the second about six seconds. Such computational advantages in some cases may well outweigh a slight loss in precision.Table 8.2 Examination scores for Secondary by Primary school classification omitting small cells.Parameter Estimate (s.e.) Estimate (s.e.) Fixed ≤1 student ≤2students Intercept 6.00 6.00 Verbal reasoning 0.16 (0.003)0.16 (0.003)Randomσu ()1020.27 (0.06) 0.25 (0.06) σu ()2020.004 (0.021) 0.028 (0.030) σe 24.28 (0.11)4.29 (0.11)8.5 Interactions in cross classificationsConsider the following extension of equation (8.1)y X u u u e i j j i j j j j j j i j j ()()()()1212121212=++++β(8.4)We have now added an ‘interaction’ term to the model which was previously an additive onefor the two variances. The usual specification for such a random interaction term is that it hassimple variance σu ()122across all the level 2 cells (Searle et al, 1992). To fit such a model we would define each cell of the cross classification as a level 2 unit with a between cell variance σu ()122, a single level 3 unit with a variance σu 12 and a single level 4 unit with a variance σu 22. Theadequacy of such a model can be tested against an additive model using a likelihood ratio testcriterion. For the example in Table 8.1 this interaction term is estimated as zero. While this indicates that the cross classification is adequate, because the between secondary school variance is so small we would not expect to be able to detect such an interaction.Extensions to this model are possible by adding random coefficients for the interaction component, just as random coefficients can be added to the additive components. For example, the gender difference between students may vary across both primary and secondary schools in the example of section 8.3 and we can fit an extra variance and covariance term for this to both the additive effects and the interaction.8.6 Level 1 cross classificationsSome interesting models occur when units are basically cross classified at level 1. By definition we have a design with only one unit per cell, as shown for example inFigure 8.2 and we can also have a level 2 cross classification which is formally equivalent to a level 1 cross classification where there is just one unit per cell as in Figure 8.5. This case should be distinguished from the case where a level 2 cross classification happens to produce no more than 1 level 1 unit in a cell as a result of sampling, so that the confounding occurs by chance rather than by design.A 2-level variance components model with a cross classification at level 1 can be written asy X u e e e i i j i i j j i j i j i i j()()()12121212=++++β(8.5)where for level 1 we use a straightforward extension of the notation for a level 2 cross classification. The term e i i j (,)12 is analogous to the interaction term in (8.4). To specify this model we would define the u j as random at level 4, the e e i j i j 12, as random at levels 3 and 2, each with a single unit and the interaction term random across the cells of the cross classification at level 1, within the original level 2 units.Suppose now that we were able to extend the design by replicating measurements for each cell of the level 1 cross classification. Then (8.5) would refer to a 3-level model with replications as level 1 units, and which could be written as follows where the subscript h denotes replicationsy X u e e e h i i j h i i j ji ji jh i i j()()()12121212=++++β(8.6)Since (8.5) is just model (8.6) with one unit per cell, we could interpret the ‘interaction’ variance in (8.5) as an estimate of the extent to which the additive variances of thecross classification fail to account for the total level 1 variance.So called ‘generalisability theory’ models (Cronbach and Webb, 1975) can be formulated as level 1 cross classifications. The basic model is one where a test or other instrument consisting of a set of items, for example ratings or questions, is administered to a sample of individuals. The individuals are therefore cross classified by the items at level 1 and may be further nested within schools etc. at higher levels. In educational test settings the item responses are often binary so that we would apply the methods of chapter 7 to the present procedures in a straightforward way. Since each individual can only respond once to each item this an example of a genuine level 1 cross classification.8.7 Cross-unit membership modelsIn some circumstances units can be members of more than one higher level unit at the same time. An example is friendship patterns where at any time individuals can be members of more than one friendship group. Another example is where children belong to more than one ‘extended’ fa mily which includes aunts and uncles as well as parents. In an educational systemstudents may attend more than one institution. In all such cases we shall assume that for each higher level unit to which a lower level unit belongs there is a known weight, summing to 1.0 for each lower level unit, which represents, for example, the amount of time spent in that unit. We may also have data where, although there is no cross-unit membership, there is some uncertainty about which higher level unit some lower level units belong to. For example, in a survey of students information about their neighbourhood of residence may only be available for a few students for larger geographical units. For these cases it may be possible to assign a weight for each of the constituent neighbourhoods which is in effect a probability of belonging to each based upon available information. Such a structure can be analysed formally as a cross-unit membership model with most students having a single weight of 1.0 and the remainder zero.Consider the 2-level variance components model (8.1) with each level 1 unit belonging to at most two level 2 units where the j j 12, subscripts now refer to the same type of unit. y X w u w u e w w i j j i j j i j j i j j i j j i j i j ()()()12121122121212121=++++=β(8.7)The overall contribution at level 2 is therefore the weighted sum over the level 2 units to which each level 1 unit belongs. This leads to the following covariance structurev a r ()()c o v ()()c o v ()()()()()(y w w y y w w w w y y w w i jj i j i j u e i jj i jj i j i j i j i j u i jj i jj i j i j u12121212112212122212222211222222=++=+=''''''σσσσ This has the structure of a standard 2-level cross classified model with the additionalconstraint σσσu u u 12222== and where the explanatory indicator variables Z Z 12, described in Figure 8.1.1 in appendix 8.1 have the value 1 replaced by the relevant weights for each level 1 unit. As with the standard cross classification this model can be extended to include random coefficients and general p-unit membership (see Appendix 8.1).8.8 Multivariate cross classified modelsFor multivariate models the responses may have different structures. Thus in a bivariate model one response may have a 2-level hierarchical structure and the other may have a cross classification at level 2. Suppose, for example that we measure the height and the mathematics attainment of a sample of students from a sample of schools. The mathematics attainment is assessed by a different set of teachers in each school and the heights are measured by a single anthropometrist. For the mathematics scores there is a level 1 cross classification of students within each school whereas for height there is a 2-level hierarchy with students nested within schools. Height and mathematics attainment will be correlated at both the student and the school level and we can write a model for this structure as followsy X u e e X u e u u e e i f m a t h e m a t i c s i f h e i g h t h i ij h i ij j i j i j h i j j i jj j u i j i j e h h h()()()()c o v ()c o v (),,1212121111111111222221212121212111=++++++====-δβδβσσδδδ(8.8)where all other covariances are zero. This will therefore be specified as a 4-level model with the bivariate structure as level 1 and level 2 units being individual students. There will be a single level 3 unit with the coefficients of the dummy variables for teachers having variances random at this level, with level 4 being that of the school.Finally, we have already mentioned that cross classified models can have a discrete response and the models of chapter 7 can be fitted. We can also fit, for example, time series models as discussed in chapter 6 and in general cross classified structures can incorporate all the types of models which can be fitted for purely hierarchical structures.Appendix 8.1Random cross classified data structuresWe illustrate the procedure using a 2-level model with crossing at level 2.The 2-level cross classified model, using the notation in Appendix 2.1, can be writteny X z u z u e i jj i jj h i j h j h q h i j h j h q i jj ()()()121211122121112212=+∑+∑+==β(8.1.1)Parentheses group the ways of classification at each level. We have two sets of explanatory variables, type 1 and type 2, for the random components defined by the columns of Z n p q n p q 11122(),()⨯⨯ Z 2 where p p 12, are respectively the number of categories of each classification.Zz Zz z z i f j m f o r m t h t y p e l e v e l u n i t o t h e r w i s e z z i f j m f o r m t h t y p e l e v e l u n i t o t h e r w i s e h i j h i j h i j h i m h i j h i m 11221112221212120220====-==-{},{},, , ,These variables are dummy variables where for each level 2 unit of type 1 we have q 1 random coefficients with covariance matrix Ω()12 and likewise for the type 2 units. To simplify the exposition we restrict ourselves to the variance component case where we haveΩΩ()()()()()()()(),(~~)()()12122222221112212222212===++σσσσE Y Y V Z I Z Z I Z T p T p T(8.1.2)Consider figure 8.1 in chapter 8 where schools are ordered within neighbourhoods. The explanatory variables will have the following structure for the first 8 students It is clear that the second term in (8.1.2) can be written asZ IZ J J p T T112211221()()()()σσ=where J is a (n x 1) vector of ones. The third term is of the general form Z Z T333Ω, namely a level 3 contribution where in this case there is only a single level 3 unit and with no covariances between the random coefficients of the Z h 2 and with the variance terms constrained to be equal to a single value, σ()222.i,j1i,j2Z11Z12Z13Z14Z21Z22Z231,1 1,1 1 0 0 0 1 0 02,1 2,1 1 0 0 0 1 0 03,1 3,1 1 0 0 0 1 0 04,1 4,1 1 0 0 0 1 0 05,1 1,2 1 0 0 0 0 1 06,1 2,2 1 0 0 0 0 1 07,1 1,3 1 0 0 0 0 0 11,2 2,1 0 1 0 0 1 0 0 Figure 8.1.1 Explanatory variables for level 2 cross classification of Figure 8.1More generally we can specify a level 2 cross classified variance components model by modelling one of the classifications as a standard hierarchical component and the second as a set of dummy explanatory variables, one for each category, with the random coefficients uncorrelated and with variances constrained to be equal. If this second (type 2) classification has further explanatory variables with random coefficients as in (8.1.1) then we form extended dummy variable ‘interactions’ as the product of the basic dummy variables and the further explanatory variables with random coefficients, so that these coefficients have variances and covariances within the same type 2 level 2 unit but not across units. In addition the corresponding variances and covariances are constrained to be equal.To extend this to further ways of classification we add levels. Thus, for a three way cross classification at level 2 we choose one classification, typically that with the largest number of categories, to model in standard hierarchical fashion at level 2, the second to model with coefficients random at level 3 as above and the third to model in a similar fashion with coefficients random at level 4. The same principle applies to cross classifications at level 1 nested within level 2 units. The level 1 cross classification is modelled as a 2-level hierarchy with the original level 2 units becoming level 3 units. We can also allow simultaneous crossing at more than one level. Thus for example, if there is a 2-way cross classification at level 1 and a 3-way cross classification at level 2, we will require five levels, the first two describing the level 1 cross classification and the next three describing the level 2 cross classification.Chapter 8 discusses the level 2 cross unit membership model where level 1 units can belong to more than one level 2 unit with predetermined weights. Because the structure imposed above level 2 replicates that at level 2 we need only in fact specify a single level 2 unit with explanatoryas diagonal of order variable design matrix Z containing dummy weight vectors and Ωu2.equal to the number of level 2 units, and elements equal to σu。

投资学课后答案APT资料讲解Chapter 10 Arbitrage Pricing Theory and Multifactor Models of Risk and Return Multiple Choice Questions1. ___________ a relationship between expected return and risk.A. APT stipulatesB. CAPM stipulatesC. Both CAPM and APT stipulateD. Neither CAPM nor APT stipulateE. No pricing model has found2. Consider the multifactor APT with two factors. Stock A has an expected return of 17.6%, a beta of 1.45 on factor 1 and a beta of .86 on factor 2. The risk premium on the factor 1 portfolio is3.2%. The risk-free rate of return is 5%. What is the risk-premium on factor 2 if no arbitrage opportunities exit?A. 9.26%B. 3%C. 4%D. 7.75%E. 9.75%3. In a multi-factor APT model, the coefficients on the macro factors are often called ______.A. systemic riskB. factor sensitivitiesC. idiosyncratic riskD. factor betasE. both factor sensitivities and factor betas4. In a multi-factor APT model, the coefficients on the macro factors are often called ______.A. systemic riskB. firm-specific riskC. idiosyncratic riskD. factor betasE. unique risk5. In a multi-factor APT model, the coefficients on the macro factors are often called ______.A. systemic riskB. firm-specific riskC. idiosyncratic riskD. factor loadingsE. unique risk6. Which pricing model provides no guidance concerning the determination of the risk premium on factor portfolios?A. The CAPMB. The multifactor APTC. Both the CAPM and the multifactor APTD. Neither the CAPM nor the multifactor APTE. No pricing model currently exists that provides guidance concerning the determination of the risk premium on any portfolio7. An arbitrage opportunity exists if an investor can construct a __________ investment portfolio that will yield a sure profit.A. small positiveB. small negativeC. zeroD. large positiveE. large negative8. The APT was developed in 1976 by ____________.A. LintnerB. Modigliani and MillerC. RossD. SharpeE. Fama9. A _________ portfolio is a well-diversified portfolio constructed to have a beta of 1 on one of the factors and a beta of 0 on any other factor.A. factorB. marketC. indexD. factor and marketE. factor, market, and index10. The exploitation of security mispricing in such a way that risk-free economic profits may be earned is called___________.A. arbitrageB. capital asset pricingC. factoringD. fundamental analysisE. technical analysis11. In developing the APT, Ross assumed that uncertainty in asset returns was a result ofA. a common macroeconomic factor.B. firm-specific factors.C. pricing error.D. neither common macroeconomic factors nor firm-specific factors.E. both common macroeconomic factors and firm-specific factors.12. The ____________ provides an unequivocal statement on the expected return-beta relationship for all assets, whereas the _____________ implies that this relationship holds for all but perhaps a small number of securities.A. APT; CAPMB. APT; OPMC. CAPM; APTD. CAPM; OPME. APT and OPM; CAPM13. Consider a single factor APT. Portfolio A has a beta of 1.0 and an expected return of 16%. Portfolio B has a beta of 0.8 and an expected return of 12%. The risk-free rate of return is 6%. If you wanted to take advantage of an arbitrage opportunity, you should take a short position in portfolio __________ and a long position in portfolio _______.A. A; AB. A; BC. B; AD. B; BE. A; the riskless asset14. Consider the single factor APT. Portfolio A has a beta of 0.2 and an expected return of 13%. Portfolio B has a beta of 0.4 and an expected return of 15%. The risk-free rate of return is 10%. If you wanted to take advantage of an arbitrage opportunity, you should take a short position in portfolio _________ and a long position in portfolio _________.A. A; AB. A; BC. B; AD. B; BE. No arbitrage opportunity exists.15. Consider the one-factor APT. The variance of returns on the factor portfolio is 6%. The beta of a well-diversified portfolio on the factor is 1.1. The variance of returns on thewell-diversified portfolio is approximately __________.A. 3.6%B. 6.0%C. 7.3%D. 10.1%E. 8.6%16. Consider the one-factor APT. The standard deviation of returns on a well-diversified portfolio is 18%. The standard deviation on the factor portfolio is 16%. The beta of thewell-diversified portfolio is approximately __________.A. 0.80B. 1.13C. 1.2517. Consider the single-factor APT. Stocks A and B have expected returns of 15% and 18%, respectively. The risk-free rate of return is 6%. Stock B has a beta of 1.0. If arbitrage opportunities are ruled out, stock A has a beta of __________.A. 0.67B. 1.00C. 1.30D. 1.69E. 0.7518. Consider the multifactor APT with two factors. Stock A has an expected return of 16.4%, a beta of 1.4 on factor 1 and a beta of .8 on factor 2. The risk premium on the factor 1 portfolio is 3%. The risk-free rate of return is 6%. What is the risk-premium on factor 2 if no arbitrage opportunities exit?A. 2%B. 3%C. 4%D. 7.75%E. 6.89%19. Consider the multifactor model APT with two factors. Portfolio A has a beta of 0.75 on factor 1 and a beta of 1.25 on factor 2. The risk premiums on the factor 1 and factor 2 portfolios are 1% and 7%, respectively. The risk-free rate of return is 7%. The expected return on portfolio A is __________ if no arbitrage opportunities exist.A. 13.5%B. 15.0%C. 16.5%D. 23.0%E. 18.7%20. Consider the multifactor APT with two factors. The risk premiums on the factor 1 and factor 2 portfolios are 5% and 6%, respectively. Stock A has a beta of 1.2 on factor 1, and a beta of 0.7 on factor 2. The expected return on stock A is 17%. If no arbitrage opportunities exist, the risk-free rate of return is ___________.A. 6.0%B. 6.5%C. 6.8%D. 7.4%E. 7.7%21. Consider a one-factor economy. Portfolio A has a beta of 1.0 on the factor and portfolio B has a beta of 2.0 on the factor. The expected returns on portfolios A and B are 11% and 17%, respectively. Assume that the risk-free rate is 6% and that arbitrage opportunities exist. Suppose you invested $100,000 in the risk-free asset, $100,000 in portfolio B, and sold short $200,000 of portfolio A. Your expected profit from this strategy would be ______________.A. ?$1,000B. $0C. $1,00022. Consider the one-factor APT. Assume that two portfolios, A and B, are well diversified. The betas of portfolios A and B are 1.0 and 1.5, respectively. The expected returns on portfolios A and B are 19% and 24%, respectively. Assuming no arbitrage opportunities exist, therisk-free rate of return must be ____________.A. 4.0%B. 9.0%C. 14.0%D. 16.5%E. 8.2%23. Consider the multifactor APT. The risk premiums on the factor 1 and factor 2 portfolios are 5% and 3%, respectively. The risk-free rate of return is 10%. Stock A has an expected return of 19% and a beta on factor 1 of 0.8. Stock A has a beta on factor 2 of ________.A. 1.33B. 1.50C. 1.67D. 2.00E. 1.7324. Consider the single factor APT. Portfolios A and B have expected returns of 14% and 18%, respectively. The risk-free rate of return is 7%. Portfolio A has a beta of 0.7. If arbitrage opportunities are ruled out, portfolio B must have a beta of__________.A. 0.45B. 1.00C. 1.10D. 1.22E. 1.33There are three stocks, A, B, and C. You can either invest in these stocks or short sell them. There are three possible states of nature for economic growth in the upcoming year; economic growth may be strong, moderate, or weak. The returns for the upcoming year on stocks A, B, and C for each of these states of nature are given below:25. If you invested in an equally weighted portfolio of stocks A and B, your portfolio return would be ___________ if economic growth were moderate.A. 3.0%D. 16.0%E. 17.0%26. If you invested in an equally weighted portfolio of stocks A and C, your portfolio return would be ____________ if economic growth was strong.A. 17.0%B. 22.5%C. 30.0%D. 30.5%E. 25.6%27. If you invested in an equally weighted portfolio of stocks B and C, your portfolio return would be _____________ if economic growth was weak.A. ?2.5%B. 0.5%C. 3.0%D. 11.0%E. 9.0%28. If you wanted to take advantage of a risk-free arbitrage opportunity, you should take a short position in _________ and a long position in an equally weighted portfolio of _______.A. A; B and CB. B; A and CC. C; A and BD. A and B; CE. No arbitrage opportunity exists.Consider the multifactor APT. There are two independent economic factors, F1and F2. The risk-free rate of return is 6%. The following information is available about two well-diversified portfolios:29. Assuming no arbitrage opportunities exist, the risk premium on the factor F1portfolio should be __________.A. 3%B. 4%C. 5%D. 6%E. 2%30. Assuming no arbitrage opportunities exist, the risk premium on the factor F2 portfolio should be ___________.A. 3%B. 4%C. 5%D. 6%E. 2%31. A zero-investment portfolio with a positive expected return arises when _________.A. an investor has downside risk onlyB. the law of prices is not violatedC. the opportunity set is not tangent to the capital allocation lineD. a risk-free arbitrage opportunity existsE. a risk-free arbitrage opportunity does not exist32. An investor will take as large a position as possible when an equilibrium price relationship is violated. This is an example of _________.A. a dominance argumentB. the mean-variance efficiency frontierC. a risk-free arbitrageD. the capital asset pricing modelE. the SML33. The APT differs from the CAPM because the APT _________.A. places more emphasis on market riskB. minimizes the importance of diversificationC. recognizes multiple unsystematic risk factorsD. recognizes multiple systematic risk factorsE. places more emphasis on systematic risk34. The feature of the APT that offers the greatest potential advantage over the CAPM is the ______________.A. use of several factors instead of a single market index to explain the risk-return relationshipB. identification of anticipated changes in production, inflation, and term structure as key factors in explaining the risk-return relationshipC. superior measurement of the risk-free rate of return over historical time periodsD. variability of coefficients of sensitivity to the APT factors for a given asset over timeE. superior measurement of the risk-free rate of return over historical time periods and variability of coefficients of sensitivity to the APT factors for a given asset over time35. In terms of the risk/return relationship in the APTA. only factor risk commands a risk premium in market equilibrium.B. only systematic risk is related to expected returns.C. only nonsystematic risk is related to expected returns.D. only factor risk commands a risk premium in market equilibrium and only systematic risk is related to expected returns.E. only factor risk commands a risk premium in market equilibrium and only nonsystematic risk is related to expected returns.36. The following factors might affect stock returns:A. the business cycle.B. interest rate fluctuations.C. inflation rates.D. the business cycle, interest rate fluctuations, and inflation rates.E. the relationship between past FRED spreads.37. Advantage(s) of the APT is(are)A. that the model provides specific guidance concerning the determination of the risk premiums on the factor portfolios.B. that the model does not require a specific benchmark market portfolio.C. that risk need not be considered.D. that the model provides specific guidance concerning the determination of the risk premiums on the factor portfolios and that the model does not require a specific benchmark market portfolio.E. that the model does not require a specific benchmark market portfolio and that risk need not be considered.38. Portfolio A has expected return of 10% and standard deviation of 19%. Portfolio B has expected return of 12% and standard deviation of 17%. Rational investors willA. borrow at the risk free rate and buy A.B. sell A short and buy B.C. sell B short and buy A.D. borrow at the risk free rate and buy B.E. lend at the risk free rate and buy B.39. An important difference between CAPM and APT isA. CAPM depends on risk-return dominance; APT depends on a no arbitrage condition.B. CAPM assumes many small changes are required to bring the market back to equilibrium; APT assumes a few large changes are required to bring the market back to equilibrium.C. implications for prices derived from CAPM arguments are stronger than prices derived from APT arguments.D. CAPM depends on risk-return dominance; APT depends on a no arbitrage condition, CAPM assumes many small changes are required to bring the market back to equilibrium; APT assumes a few large changes are required to bring the market back to equilibrium, implications for prices derived from CAPM arguments are stronger than prices derived from APT arguments.E. CAPM depends on risk-return dominance; APT depends on a no arbitrage condition and assumes many small changes are required to bring the market back to equilibrium.40. A professional who searches for mispriced securities in specific areas such as merger-target stocks, rather than one who seeks strict (risk-free) arbitrage opportunities is engaged inA. pure arbitrage.B. risk arbitrage.C. option arbitrage.D. equilibrium arbitrage.E. covered interest arbitrage.41. In the context of the Arbitrage Pricing Theory, as a well-diversified portfolio becomes larger its nonsystematic risk approachesA. one.B. infinity.C. zero.D. negative one.E. None of these is correct.42. A well-diversified portfolio is defined asA. one that is diversified over a large enough number of securities that the nonsystematic variance is essentially zero.B. one that contains securities from at least three different industry sectors.C. a portfolio whose factor beta equals 1.0.D. a portfolio that is equally weighted.E. a portfolio that is equally weighted and contains securities from at least three different industry sectors.43. The APT requires a benchmark portfolioA. that is equal to the true market portfolio.B. that contains all securities in proportion to their market values.C. that need not be well-diversified.D. that is well-diversified and lies on the SML.E. that is unobservable.44. Imposing the no-arbitrage condition on a single-factor security market implies which of the following statements?I) the expected return-beta relationship is maintained for all but a small number ofwell-diversified portfolios.II) the expected return-beta relationship is maintained for all well-diversified portfolios.III) the expected return-beta relationship is maintained for all but a small number of individual securities.IV) the expected return-beta relationship is maintained for all individual securities.A. I and III are correct.B. I and IV are correct.C. II and III are correct.D. II and IV are correct.E. Only I is correct.45. Consider a well-diversified portfolio, A, in a two-factor economy. The risk-free rate is 6%, the risk premium on the first factor portfolio is 4% and the risk premium on the second factor portfolio is 3%. If portfolio A has a beta of 1.2 on the first factor and .8 on the second factor, what is its expected return?A. 7.0%B. 8.0%C. 9.2%D. 13.0%E. 13.2%46. The term "arbitrage" refers toA. buying low and selling high.B. short selling high and buying low.C. earning risk-free economic profits.D. negotiating for favorable brokerage fees.E. hedging your portfolio through the use of options.47. To take advantage of an arbitrage opportunity, an investor wouldI) construct a zero investment portfolio that will yield a sure profit.II) construct a zero beta investment portfolio that will yield a sure profit.III) make simultaneous trades in two markets without any net investment.IV) short sell the asset in the low-priced market and buy it in the high-priced market.A. I and IVB. I and IIIC. II and IIID. I, III, and IVE. II, III, and IV48. The factor F in the APT model representsA. firm-specific risk.B. the sensitivity of the firm to that factor.C. a factor that affects all security returns.D. the deviation from its expected value of a factor that affects all security returns.E. a random amount of return attributable to firm events.49. In the APT model, what is the nonsystematic standard deviation of an equally-weighted portfolio that has an average value of σ(e i) equal to 25% and 50 securities?A. 12.5%B. 625%C. 0.5%D. 3.54%E. 14.59%50. In the APT model, what is the nonsystematic standard deviation of an equally-weighted portfolio that has an average value of σ(e i) equal to 20% and 20 securities?A. 12.5%B. 625%C. 4.47%D. 3.54%E. 14.59%51. In the APT model, what is the nonsystematic standard deviation of an equally-weighted portfolio that has an average value of σ(e i) equal to 20% and 40 securities?A. 12.5%B. 625%C. 0.5%D. 3.54%E. 3.16%52. In the APT model, what is the nonsystematic standard deviation of an equally-weighted portfolio that has an average value of σ(e i) equal to 18% and 250 securities?A. 1.14%B. 625%C. 0.5%D. 3.54%E. 3.16%53. Which of the following is true about the security market line (SML) derived from the APT?A. The SML has a downward slope.B. The SML for the APT shows expected return in relation to portfolio standard deviation.C. The SML for the APT has an intercept equal to the expected return on the market portfolio.D. The benchmark portfolio for the SML may be any well-diversified portfolio.E. The SML is not relevant for the APT.54. Which of the following is false about the security market line (SML) derived from the APT?A. The SML has a downward slope.B. The SML for the APT shows expected return in relation to portfolio standard deviation.C. The SML for the APT has an intercept equal to the expected return on the market portfolio.D. The benchmark portfolio for the SML may be any well-diversified portfolio.E. The SML has a downward slope, the SML for the APT shows expected return in relation to portfolio standard deviation, and the SML for the APT has an intercept equal to the expected return on the market portfolio are all false.55. If arbitrage opportunities are to be ruled out, each well-diversified portfolio's expected excess return must beA. inversely proportional to the risk-free rate.B. inversely proportional to its standard deviation.C. proportional to its weight in the market portfolio.D. proportional to its standard deviation.E. proportional to its beta coefficient.56. Suppose you are working with two factor portfolios, Portfolio 1 and Portfolio 2. The portfolios have expected returns of 15% and 6%, respectively. Based on this information, what would be the expected return on well-diversified portfolio A, if Ahas a beta of 0.80 on the first factor and 0.50 on the second factor? The risk-free rate is 3%.A. 15.2%B. 14.1%C. 13.3%D. 10.7%E. 8.4%57. Which of the following is (are) true regarding the APT?I) The Security Market Line does not apply to the APT.II) More than one factor can be important in determining returns.III) Almost all individual securities satisfy the APT relationship.IV) It doesn't rely on the market portfolio that contains all assets.A. II, III, and IVB. II and IVC. II and IIID. I, II, and IVE. I, II, III, and IV58. In a factor model, the return on a stock in a particular period will be related toA. factor risk.B. non-factor risk.C. standard deviation of returns.D. both factor risk and non-factor risk.E. There is no relationship between factor risk, risk premiums, and returns.59. Which of the following factors did Chen, Roll and Ross not include in their multifactor model?A. Change in industrial productionB. Change in expected inflationC. Change in unanticipated inflationD. Excess return of long-term government bonds over T-billsE. Neither the change in industrial production, change in expected inflation, change in unanticipated inflation, nor excess return of long-term government bonds over T-bills were included in their model.60. Which of the following factors did Chen, Roll and Ross include in their multifactor model?A. Change in industrial wasteB. Change in expected inflationC. Change in unanticipated inflationD. Change in expected inflation and Change in unanticipated inflationE. All of these factors were included in their model61. Which of the following factors were used by Fama and French in their multi-factor model?A. Return on the market index.B. Excess return of small stocks over large stocks.C. Excess return of high book-to-market stocks over low book-to-market stocks.D. All of these factors were included in their model.E. None of these factors were included in their model.62. Consider the single-factor APT. Stocks A and B have expected returns of 12% and 14%, respectively. The risk-free rate of return is 5%. Stock B has a beta of 1.2. If arbitrage opportunities are ruled out, stock A has a beta of __________.A. 0.67B. 0.93C. 1.30D. 1.69E. 1.2763. Consider the one-factor APT. The standard deviation of returns on a well-diversified portfolio is 19%. The standard deviation on the factor portfolio is 12%. The beta of thewell-diversified portfolio is approximately __________.A. 1.58B. 1.13C. 1.25D. 0.76E. 1.4264. Black argues that past risk premiums on firm-characteristic variables, such as those described by Fama and French, are problematic because ________.A. they may result from data snoopingB. they are sources of systematic riskC. they can be explained by security characteristic linesD. they are more appropriate for a single-factor modelE. they are macroeconomic factors65. Multifactor models seek to improve the performance of the single-index model byA. modeling the systematic component of firm returns in greater detail.B. incorporating firm-specific components into the pricing model.C. allowing for multiple economic factors to have differential effects.D. modeling the systematic component of firm returns in greater detail, incorporatingfirm-specific components into the pricing model, and allowing for multiple economic factors to have differential effects.E. none of these statements are true.66. Multifactor models such as the one constructed by Chen, Roll, and Ross, can better describe assets' returns byA. expanding beyond one factor to represent sources of systematic risk.B. using variables that are easier to forecast ex ante.C. calculating beta coefficients by an alternative method.D. using only stocks with relatively stable returns.E. ignoring firm-specific risk.67. Consider the multifactor model APT with three factors. Portfolio A has a beta of 0.8 on factor 1, a beta of 1.1 on factor 2, and a beta of 1.25 on factor 3. The risk premiums on the factor 1, factor 2, and factor 3 are 3%, 5% and 2%, respectively. The risk-free rate of return is 3%. The expected return on portfolio A is __________ if no arbitrage opportunities exist.A. 13.5%B. 13.4%C. 16.5%D. 23.0%E. 11.6%68. Consider the multifactor APT. The risk premiums on the factor 1 and factor 2 portfolios are 6% and 4%, respectively. The risk-free rate of return is 4%. Stock A has an expected return of 16% and a beta on factor 1 of 1.3. Stock A has a beta on factor 2 of ________.A. 1.33B. 1.05C. 1.67D. 2.00E. .95。