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dbp11e_tif_ch06

Course: IE IE 224, Spring 2010
School: Lehigh
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Processing, Database 11e (Kroenke) Chapter 6: Transforming Data Models into Database Designs 1) The first step in the database design process is to create tables and columns from entities and attributes. Answer: TRUE Diff: 2 Page Ref: 193 Fig 6-1 2) When creating a relational database design from E-R diagrams, first create a relation for each relationship. Answer: FALSE Diff: 3 Page Ref: 193 Fig 6-1 3) Each entity...

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Processing, Database 11e (Kroenke) Chapter 6: Transforming Data Models into Database Designs 1) The first step in the database design process is to create tables and columns from entities and attributes. Answer: TRUE Diff: 2 Page Ref: 193 Fig 6-1 2) When creating a relational database design from E-R diagrams, first create a relation for each relationship. Answer: FALSE Diff: 3 Page Ref: 193 Fig 6-1 3) Each entity in the extended E-R model is represented as a table in the relational database design. Answer: TRUE Diff: 1 Page Ref: 193 Fig 6-1 4) An entity needs to be examined according to normalization criteria before creating a table from it in the relational database design. Answer: FALSE Diff: 2 Page Ref: 193, 197 Fig 6-1 5) When creating a table in the relational database design from an entity in the extended E-R model, the attributes of the entity become the rows of the table. Answer: FALSE Diff: 1 Page Ref: 193 Fig 6-1 6) By default, the identifier of the entity becomes the foreign key of the corresponding table. Answer: FALSE Diff: 1 Page Ref: 194-195 7) The ideal primary key is short, numeric, and fixed. Answer: TRUE Diff: 2 Page Ref: 194 8) A surrogate key is appropriate when the primary key of a table contains a lengthy text field. Answer: TRUE Diff: 1 Page Ref: 194 9) One of the important properties of an attribute is whether or not it is required. Answer: TRUE Diff: 2 Page Ref: 195 10) Whether or not an attribute is required is determined during the database modeling phase. Answer: FALSE Diff: 3 Page Ref: 195 1 Copyright 2010 Pearson Education, Inc. 11) A null value is an attribute value that has been set to zero. Answer: FALSE Diff: 3 Page Ref: 195 12) Data types are consistent across all DBMS products. Answer: FALSE Diff: 1 Page Ref: 196 13) A default value is the value the user enters into the row the first time the user enters data. Answer: FALSE Diff: 1 Page Ref: 197 14) A data constraint is a limitation on data values. Answer: TRUE Diff: 1 Page Ref: 197 15) The last step in creating a table is to verify table normalization. Answer: TRUE Diff: 1 Page Ref: 197 Fig 6-1 16) A foreign key is a key that does not belong in any table. Answer: FALSE Diff: 2 Page Ref: 198 17) In a relational database design, all relationships are expressed by creating a foreign key. Answer: TRUE Diff: 1 Page Ref: 198 18) When the key of one table is placed into a second table to represent a relationship, the key is called a "relational key" in the second table. Answer: FALSE Diff: 2 Page Ref: 198 19) To represent a 1:1 binary relationship in a relational database design, the key of one table is placed into the second table. Answer: TRUE Diff: 2 Page Ref: 198 20) When placing a foreign key for a 1:1 relationship, the key of either table can be used as the foreign key in the other table. Answer: TRUE Diff: 1 Page Ref: 198 21) In a 1:1 relationship, the foreign key is defined as an alternate key to make the DBMS enforce uniqueness. Answer: TRUE Diff: 2 Page Ref: 199 2 Copyright 2010 Pearson Education, Inc. 22) In a 1:N relationship, the term "parent" refers to the table on the "many" side of the relationship. Answer: FALSE Diff: 1 Page Ref: 199 23) In representing a 1:N relationship in a relational database design, the key of the table representing the parent entity is placed as a foreign key in the table representing the child entity. Answer: TRUE Diff: 2 Page Ref: 199 24) In representing a 1:N relationship in a relational database design, the key of the table representing the entity on the "many" side is placed as a foreign key in the table representing the entity on the "one" side of the relationship. Answer: FALSE Diff: 3 Page Ref: 199 25) To represent a 1:N relationship in a relational database design, an intersection table is created. Answer: FALSE Diff: 2 Page Ref: 199-201 26) To represent an M:N relationship in a relational database design, a table is created to represent the relationship itself. Answer: TRUE Diff: 2 Page Ref: 199-201 27) To represent an M:N relationship in a relational database design, an intersection table is created. Answer: TRUE Diff: 1 Page Ref: 200 28) The key of an intersection table is always the combination of the keys of both parents. Answer: TRUE Diff: 2 Page Ref: 200 29) An intersection table is always ID-dependent on both of its parent tables. Answer: TRUE Diff: 2 Page Ref: 200-201 30) An intersection table can have additional attributes besides the keys of its parent tables. Answer: FALSE Diff: 2 Page Ref: 200-201 31) Like all ID-dependent relationships, the parents of an association table are required. Answer: TRUE Diff: 2 Page Ref: 200-201 3 Copyright 2010 Pearson Education, Inc. 32) Association tables sometimes connect more than two entities. Answer: TRUE Diff: 2 Page Ref: 201-203 33) An ID-dependent table can be used to represent multivalued attributes. Answer: TRUE Diff: 1 Page Ref: 204 34) All identifying relationships are 1:N. Answer: TRUE Diff: 3 Page Ref: 204 35) The design transformation for all IS-A relationships can be summarized by the phrase "place the key of the parent table in the child table." Answer: FALSE Diff: 3 Page Ref: 207 36) When transforming supertype/subtype entities into a relational database design, an entity is created for the supertype only. Answer: FALSE Diff: 1 Page Ref: 208 37) When transforming supertype/subtype entities into a relational database design, the key of the supertype table is placed into the subtype table typically as the key. Answer: TRUE Diff: 2 Page Ref: 208 38) To represent an IS-A relationship in a relational database design, the IS-A relationship must be converted into a HAS-A relationship. Answer: FALSE Diff: 2 Page Ref: 208 39) When transforming supertype/subtype entities into a relational database design, all of the attributes for the supertype table are placed into the subtype relations. Answer: FALSE Diff: 3 Page Ref: 208 40) When transforming an extended E-R model into a relational database design, recursive relationships are treated fundamentally the same as other HAS-A relationships. Answer: TRUE Diff: 2 Page Ref: 208 41) Recursive M:N relationships are represented with an intersection table that shows pairs of related rows from a single table. Answer: TRUE Diff: 3 Page Ref: 210 4 Copyright 2010 Pearson Education, Inc. 42) When the parent entity is required (M) in a relationship, every row of the child table must have a valid, non-null value of the foreign key. Answer: TRUE Diff: 2 Page Ref: 216 43) Propagating a change made to a parent's primary key to the child's foreign key is known as cascading changes. Answer: FALSE Diff: 2 Page Ref: 216 44) Deleting the children when the parent is deleted is known as cascading deletes. Answer: TRUE Diff: 1 Page Ref: 216 45) Cascading deletions are generally not used with relationships between strong entities. Answer: TRUE Diff: 2 Page Ref: 216 46) Cascading deletions are generally not used with relationships for weak child entities. Answer: FALSE Diff: 2 Page Ref: 195216 47) If the parent is required, then a new child row must be created with a valid foreign key value. Answer: TRUE Diff: 2 Page Ref: 216 48) When the parent entity has a surrogate key, the enforcement actions are the same for both parent and child. Answer: FALSE Diff: 3 Page Ref: 217 49) When the child entity is required (M) in a relationship, there needs to be at least one child row for each parent row at all times. Answer: TRUE Diff: 2 Page Ref: 217 50) It is easy to enforce the referential integrity actions for M-M relationships. Answer: FALSE Diff: 1 Page Ref: 219 51) Which of the following is not a step in the database design process? A) Create tables and columns from entities and attributes B) Select primary keys C) Represent relationships D) Create constraints and triggers E) All of the above are steps in the database design process. Answer: D Diff: 2 Page Ref: 193 Fig 6-1 5 Copyright 2010 Pearson Education, Inc. 52) The first step in transforming an extended E-R model into a relational database design is to ________. A) create a table for each relationship B) evaluate the entities against the normalization criteria C) create a table for each entity D) remove any recursive relationships E) document referential integrity constraints Answer: C Diff: 1 Page Ref: 193 Fig 6-1 53) Each attribute of an entity becomes a ________ of a table. A) column B) primary key C) foreign key D) alternate key E) B or D Answer: A Diff: 1 Page Ref: 193 Fig 6-1 54) The identifier of the entity becomes the ________ of the corresponding table. A) primary key B) foreign key C) supertype D) subtype E) either A or B Answer: A Diff: 1 Page Ref: 194 55) The ideal primary key is ________. A) short B) numeric C) fixed D) A and B E) A, B and C Answer: E Diff: 2 Page Ref: 194 56) A surrogate key should be considered when ________. A) a relationship is M:N B) a composite key is required C) the key contains a lengthy text field D) the key contains a number E) an index needs to be created Answer: C Diff: 3 Page Ref: 194 6 Copyright 2010 Pearson Education, Inc. 57) Which of the following is not true about surrogate keys? A) They are identifiers that are supplied by the system, not the users. B) They have no meaning to the users. C) They are nonunique within a table. D) They can be problematic when combining databases. E) The DBMS will not allow their values to be changed. Answer: C Diff: 2 Page Ref: 194 58) One of the important properties of an attribute is whether or not it is ________. A) found in more than one entity B) required C) character or numeric D) subject to normalization E) subject to denormalization Answer: B Diff: 2 Page Ref: 195-197 59) In a relational database design, all relationships are expressed by ________. A) creating a primary key B) creating a foreign key C) creating a supertype D) creating a subtype E) creating a line between entities Answer: B Diff: 1 Page Ref: 198 60) When representing a 1:1 relationship in a relational database design ________. A) the key of each table must be placed as foreign keys into the other B) the key of either table may be placed as a foreign key into the other C) the key of both tables must be the same D) the intersection table gets the key from both relations E) B and C Answer: B Diff: 1 Page Ref: 198-199 61) To represent a one-to-many relationship in a relational database design ________. A) the key of the child is placed as a foreign key into the parent B) the key of the parent is placed as a foreign key into the child C) an intersection table must be created D) the key of the table on the "many" side is placed in the table on the "one" side E) the keys of both tables are joined into a composite key Answer: B Diff: 2 Page Ref: 199 7 Copyright 2010 Pearson Education, Inc. 62) When representing a one-to-many relationship in a relational database design ________. A) the parent is always on the one side of the "one-to-many" relationship B) the child is always on the one side of the "one-to-many" relationship C) either parent or child can be on the one side of the "one-to-many" relationship, and the choice is arbitrary D) either parent or child can be on the one side of the "one-to-many" relationship, and special criteria indicate which table should be on the one side. E) None of the above is correct. Answer: A Diff: 2 Page Ref: 199 63) Many-to-many relationships are represented by ________. A) two tables with an M:N relationship B) two tables with a 1:N relationship C) an intersection table which has M:N relationships with the two tables D) an intersection table which has 1:N relationships with the two tables E) two intersection tables which each have 1:N relationships with the two tables Answer: D Diff: 2 Page Ref: 199-201 64) In many-to-many relationships in a relational database design ________. A) the key of the child is placed as a foreign key into the parent B) the key of the parent is placed as a foreign key into the child C) the keys of both tables are placed in a third table D) the keys of both tables are joined into a composite key E) C and D Answer: E Diff: 2 Page Ref: 199-201 65) In many-to-many relationships in a relational database design ________. A) the intersection table is ID-dependent on one of the parents B) the intersection table is ID-dependent on both of the parents C) the minimum cardinality from the intersection table to the parents is always M D) A and B E) B and C Answer: E Diff: 3 Page Ref: 199-201 66) In relational database design, ID-dependent entities are used to ________. A) represent N:M relationships B) handle associative relationships C) handle multivalued attributes D) handle archetype/instance relationships E) All of the above. Answer: E Diff: 2 Page Ref: 201 Fig 6-11 8 Copyright 2010 Pearson Education, Inc. 67) When transforming an E-R data model into a relational database design, the key of the parent entity should be placed as part of the primary key into the child entity ________. A) when the child entity is ID-dependent B) when the child entity is non-ID-dependent C) when the child entity has a 1:1 relationship with the parent entity D) when the child entity has a 1:N relationship with the parent entity E) when the child entity has a recursive relationship with the parent entity Answer: A Diff: 2 Page Ref: 201, 207 68) When transforming an ID-dependent E-R data model relationship into a relational database design and the parent entity has a surrogate primary key and the child entity has a data identifier, then the primary key of the child table should be ________. A) the parent's surrogate key + the child's data key B) the parent's surrogate key + a surrogate key in the child C) either A or B will work, and neither way is preferable D) either A or B will work, and method A is preferable E) either A or B will work, but method B is preferable Answer: E Diff: 3 Page Ref: 201, 206 69) When transforming an ID-dependent E-R data model relationship into a relational database design and the child entity is designed to use a surrogate key, then ________. A) the parent entity must also use a surrogate key B) the relationship remains an ID-dependent relationship C) the relationship changes to a non-ID-dependent relationship D) A and B E) A and C Answer: C Diff: 2 Page Ref: 206 70) Which of the following is not true about representing subtypes in a relational database design? A) One table is created for the supertype and one for each subtype. B) All of the attributes of the supertype are added to the subtype relations. C) The key of the supertype is made the key the of subtypes. D) A subtype and its supertype are representations of the same underlying table. E) An instance of the supertype may be related to one instance each of several subtypes. Answer: B Diff: 2 Page Ref: 208 9 Copyright 2010 Pearson Education, Inc. 71) In a supertype-subtype structure, discriminator attributes ________. A) are easily represented in a relational design B) cannot be represented in a relational design C) require application logic to determine which subtypes correspond to a specific supertype instance D) A and C E) B and C Answer: E Diff: 3 Page Ref: 208 72) Which of the following is not true of recursive relationships? A) When the recursive relationship is M:N, an intersection table is created. B) The rows of a single table can play two different roles. C) The techniques for representing the tables are the same as for non-recursive relationships except the rows are in the same table. D) Recursive relationships can be 1:1, 1:N, or M:N relationships. E) Even when the relationship is 1:N, a new table must be defined to represent the relationship. Answer: E Diff: 2 Page Ref: 208-210 73) Which of the following is not true of ternary relationships? A) Can be treated as a collection of binary relationships B) Are represented in the same way as binary relationships C) Must involve at least one M:N relationship D) Involve three entities of different logical types E) Often require documentation of special considerations as business rules Answer: C Diff: 3 Page Ref: 210-212 74) The binary constraint MUST NOT indicates that ________. A) a specific binary relationship must not be included in a ternary relationship B) a table includes values that must not occur in a binary relationship C) a table includes values that must not occur in a ternary relationship D) a binary relationship includes value combinations that must not occur in a binary relationship E) a binary relationship includes value combinations that must not occur in a ternary relationship Answer: E Diff: 2 Page Ref: 211-212 75) The binary constraint MUST COVER indicates that ________. A) a specific binary relationship must be included in a ternary relationship B) a table includes values that must occur in a binary relationship C) a table includes values that must occur in a ternary relationship D) a binary relationship includes value combinations that must occur in a binary relationship E) a binary relationship includes a set of value combinations that must all occur in a ternary relationship Answer: E Diff: 2 Page Ref: 211-212 10 Copyright 2010 Pearson Education, Inc. 76) To which of the following actions are referential integrity constraints not applied? A) create B) insert C) modify D) delete E) referential integrity constraints are applied to all of the listed actions Answer: A Diff: 2 Page Ref: 214-216 Fig 6-28 77) A referential integrity constraint policy that insures that foreign key values in a table are correctly maintained when there is a change to the primary key value in the parent table is called ________. A) incremental updates B) incremental deletes C) controlled key adjustments D) cascading updates E) cascading deletes Answer: D Diff: 1 Page Ref: 216 78) A referential integrity constraint policy that insures that all rows containing a particular foreign key value in a table are eliminated from the table when the row containing the corresponding primary key value in a parent table is eliminated from the database is called ________. A) incremental updates B) incremental deletes C) controlled key adjustments D) cascading updates E) cascading deletes Answer: E Diff: 1 Page Ref: 216 79) A referential integrity constraint policy that guarantees that a row in a parent table always has a required entry in a child table ________. A) is known as a minimum cardinality enforcement action B) needs to be documented by the database development team C) is enforced in most DBMS products D) A and B E) A, B and C Answer: D Diff: 3 Page Ref: 216-217 Fig 6-28 11 Copyright 2010 Pearson Education, Inc. 80) When transforming an ID-dependent E-R data model relationship into a relational database design, the referential integrity constraints should allow ________. A) parent updates to cascade B) child updates to cascade C) child deletes to cascade D) A and B E) A, B and C Answer: A Diff: 3 Page Ref: 216 Fig 6-28 81) When transforming an extended E-R model into a relational database design, first create a(n) ________ for each entity. Answer: table Diff: 1 Page Ref: 193 Fig 6-1 82) The final step in the database design process of creating tables is to ________. Answer: verify normalization Diff: 3 Page Ref: 193 Fig 6-1 83) By default, the identifier of the entity becomes the ________ of the corresponding table. Answer: primary key Diff: 1 Page Ref: 193 84) The ideal primary key is ________, ________ and ________. Answer: short; numeric; fixed Diff: 2 Page Ref: 194 85) A(n) ________ key is a unique, system-supplied identifier used as the primary key of a table. Answer: surrogate Diff: 1 Page Ref: 194 86) The values of a(n) ________ have no meaning to the users. Answer: surrogate key Diff: 2 Page Ref: 194 87) A surrogate key should be considered when the key contains a(n) ________. Answer: lengthy text field Diff: 3 Page Ref: 194 88) One of the important properties of an attribute is whether or not it can have a ________. Answer: NULL value Diff: 2 Page Ref: 195-196 89) In a relational database design, all relationships are expressed by creating a(n) ________. Answer: foreign key Diff: 1 Page Ref: 198 12 Copyright 2010 Pearson Education, Inc. 90) When the primary key of one table is stored in another table, it is called a(n) ________. Answer: foreign key Diff: 1 Page Ref: 198 91) When representing a 1:1 relationship in a relational database design the key of ________ may be placed as a foreign key into the other table. Answer: either table Diff: 2 Page Ref: 198-199 92) In 1:N relationships, the table on the "one" side is called the ________. Answer: parent Diff: 1 Page Ref: 199 93) In 1:N relationships, the table on the "many" side is called the ________. Answer: child Diff: 2 Page Ref: 199 94) In 1:N relationships, the foreign key is placed inside the ________ table. Answer: child Diff: 2 Page Ref: 199 95) To represent a one-to-many relationship in a relational database design the key of ________ is placed as a foreign key into the other table. Answer: the parent table Diff: 2 Page Ref: 199 96) In representing an M:N relationship in a relational database design, a(n) ________ table is created. Answer: intersection Diff: 2 Page Ref: 200 97) To represent an M:N relationship in a relational database design, in essence it is reduced to two ________ relationships. Answer: 1:N Diff: 3 Page Ref: 199-201 98) In many-to-many relationships in a relational database design the primary keys of both tables are joined into a(n) ________ in the intersection table. Answer: composite primary key Diff: 2 Page Ref: 200 99) Like all ________ relationships, the parents of an association table are required. Answer: ID-dependent Diff: 2 Page Ref: 201 100) ________ sometimes connect more than two entities. Answer: Association tables Diff: 2 Page Ref: 203 13 Copyright 2010 Pearson Education, Inc. 101) An ________ table can be used to represent multivalued attributes. Answer: ID-dependent Diff: 1 Page Ref: 204 102) All identifying relationships are ________. Answer: 1:N Diff: 3 Page Ref: 204 103) When transforming an ID-dependent E-R data model relationship into a relational database design and the child entity is designed to use a surrogate key, then the relationship ________. Answer: changes to a weak but not ID-dependent relationship Diff: 2 Page Ref: 206 104) When creating a table for an ID-dependent entity, both the key of the ________ and the key of the entity itself must appear in the table. Answer: parent Diff: 3 Page Ref: 204 105) The key of the subtypes will be the key of the ________. Answer: supertype Diff: 1 Page Ref: 208 106) ________ attributes cannot be represented in relational designs. Answer: Discriminator Diff: 2 Page Ref: 208 107) ________ are treated fundamentally the same as other HAS-A relationships. Answer: Recursive relationships Diff: 3 Page Ref: 208 108) Referential integrity constraints should ________ adding a new row to a child table when the foreign key does not match a primary key value in the parent table. Answer: disallow Diff: 2 Page Ref: 216 Fig 6-28(a) 109) A referential integrity constraint policy that insures that foreign key values in a table are correctly maintained when there is a change to the primary key value in the parent table is called ________. Answer: cascading updates Diff: 1 Page Ref: 216 110) A referential integrity constraint policy that insures that all rows containing a particular foreign key value in a table are eliminated from the table when the row containing the corresponding primary key value in a parent table is eliminated from the database is called ________. Answer: cascading deletes Diff: 1 Page Ref: 216 14 Copyright 2010 Pearson Education, Inc. 111) Briefly describe the process of converting an extended E-R model into a relational database design. Answer: To convert an extended E-R model into a relational database design, first tables and columns must be defined for each entity attribute. The attributes of the entity are represented as the columns of the table. Primary keys are then selected for each table, after which the relationships between the entities are represented through the creation of foreign keys. Next, constraints are specified. Finally the normalization is verified. Diff: 2 Page Ref: 193-198 Fig 6-1 112) Explain the pragmatic reason for using surrogate keys. Answer: Primary keys are commonly included in indexes, and are used to identify records to be retrieved by users. The ideal primary key is short, numeric and fixed. When the primary key contains a lengthy text field, this creates a large amount of duplicated data that must be frequently manipulated. For these reasons, it is often practical to use a surrogate key that is generated by the system and is relatively small and easy to manipulate. Diff: 1 Page Ref: 194 113) How are one-to-one recursive relationships addressed using the relational model? Answer: One-to-one recursive relationships are addressed just the same as one-to-one nonrecursive relationships. The only difference is that both of the related entity instances are in the same entity class. The key of either instance is placed in the other instance as a foreign key. In the case of a recursive relationship, this means that a new attribute is added to the entity class with a recursive relationship. For each instance, this new attribute will contain the value of the key attribute of the instance that is related. Diff: 2 Page Ref: 198-199 114) Explain the representation of a one-to-many strong entity relationship in a relational database design. Answer: One-to-many relationships are represented by placing the primary key of the table on the one side of the relationship into the table on the many side of the relationship as a foreign key. The term "parent" refers to the table on the one side of a 1:N relationship, and the term "child" refers to the table on the many side of the 1:N relationship. Therefore, the rule for representing a one-to-many relationship can be summarized as "Place the key of the parent table in the child table as a foreign key." Diff: 2 Page Ref: 199 115) Explain the representation of a many-to-many strong entity relationship in a relational database design. Answer: Many-to-many relationships cannot be directly represented in a relational database design. Therefore, many-to-many relationships are essentially broken into two one-to-many relationships by creating an intersection table that represents the relationship itself. The intersection table takes its key as a combination of the keys of the two original, or parent, entities. Each of the parent entities has a one-to-many relationship with the intersection table that is represented by placing the keys of the parents into the intersection table. Diff: 2 Page Ref: 199-201 15 Copyright 2010 Pearson Education, Inc. 116) What is an association relationship and how does it differ from an N:M relationship? Answer: An association relationship is every similar to an N:M relationship except that the intersection table has attributes of its own. This means that in addition to the foreign key fields linking to the two strong entities, there is at least one additional field in what would otherwise be called the intersection table but is now an association table. For example, intersection table ENROLLMENT for STUDENT and CLASS showing student enrollment in each class would normally have two columns: StudentID and ClassID. However, we can turn this intersection table into an association table by adding the column Grade, which records each student's grade in each class. Diff: 2 Page Ref: 201-204 117) What are MUST NOT and MUST COVER constraints? Answer: Both MUST NOT and MUST COVER constraints occur when working with three tables in a ternary relationship. When a MUST NOT constraint occurs, it indicates that one binary relationship of two of the tables contains combinations of values that are not allowed to occur in the ternary relationship. When a MUST COVER constraint occurs, it indicates that one binary relationship of two of the tables contains combinations of values that must all appear in the ternary relationship. Diff: 3 Page Ref: 210-212 118) Write the schema to represent the entities below, including tables, the proper placement of the foreign key, and referential integrity constraint. Answer: STUDENT (StudentID, StuName, StuMajor, StuPhone, AdvisorID) ADVISOR (AdvisorID, AdvName, AdvOffice, AdvPhone) AdvisorID in STUDENT must exist in AdvisorID in ADVISOR. Diff: 2 Page Ref: 198-201 119) What are the four sets of minimum cardinalities that can be present in a 1:N binary relationships? Answer: If we use the labels parent and child to represent, respectively, the one and many side of a 1:N relationship, we can then consider whether the parent and child must participate in the relationship. If one must participate, it is mandatory (M). If one does not have to participate it is optional (O). The four possible sets of minimum cardinalities are therefore: Parent optional and Child optional (O-O) Parent mandatory and Child optional (M-O) Parent optional and Child mandatory (O-M) Parent mandatory and Child mandatory (M-M) Diff: 2 Page Ref: 214 16 Copyright 2010 Pearson Education, Inc.
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HIT2308 Software Development PracticesTutorialTutorial 8 in week 9Task 1Consider the Automatic Submission System described in tutorial 6 again. SCENARIO Dr Burns has to write a review of her unit "Introduction to Programming" at the end of semester. A
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Software Development Practices S2 2009HIT2308Tutorial 9 in Week 101. Describe, relate and contrast the following terms, giving examples where possible from your experience: Error The process of making mistake in the code Defect The error that appears o
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Software Development Practices S2 2009HIT2308Tutorial 10 in Week 11Task 1. Apply the BlackBox Equivalence partitioning test selection technique to the brief description of question 1 in the quick quiz. You will need to supply as many tests as there are
Swinburne - ENG - HET489
HET489 Tutorials Week 21.Answer: k=4.3 2.Answer: k=8, T=0.25
Swinburne - ENG - HET489
Tutorial questions for week 3 1.2.Answers: 1. One of the possible answer is: Gc ( s ) = 4 s + 0.5 s+32. One of the possible answer is: Gc ( s ) = s + 0.05 s + 0.01
Swinburne - ENG - HET489
Tutorials week 4 1.answer: K=5000, Kh=0.01982.
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Tutorial question for week 5 1.2.answer:40.7766 s + 1 0.16308s + 1
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Tutorial questions of week 6 1. Find a state space representation if the output is the current through the resistor.2.3. For each system below, write the state equations and the output equation for the phase variable representation.Answers: 1. vc - 1
Swinburne - ENG - HET489
Tutorial questions of week 7 1. e - t cos 2t Answer: - t - e cos 2t - 2e - t sin 2t 2.Figure 12-49Answer: K = [ 20055 8]
Swinburne - ENG - HET489
Tutorial questions of week 8: Part 1: 1. Obtain the z transform of x(t ) = 2. (B-2-7)1 (1 - e - aT ) , where aais a constant.3. Obtain the z transform of Y ( s) =5 , where the sampling time is 0.2 sec. s ( s + 1)( s + 5)4. Obtain the z transform of
Swinburne - ENG - HET489
Tutorial questions for week 9: Part A: 1.B-3-6: Obtain the z transform of1 - e -Ts 1 X ( s) = s (s + a) 22.B-3-7: Consider the DE:y (k + 1) + 0.5 y (k ) = x(k ) ,Where y (0) = 0 . Obtain the response y (k ) when the input x(k ) is a unit-step sequence
Swinburne - ENG - HET489
Tutorial questions for week 10 Part A: 1. Apply the w-transform to the following characteristic equations of discrete-data control systems, and determine the conditions of stability (stable, marginally stable, or unstable) using the Routh-Hurwitz criterio
Swinburne - ENG - HET489
Tutorial questions for week 11 Part A: The system below shows a closed loop speed control systemDesign a lead compensator to give a step response with a damping ratio of 0.7 with a natural frequency of 6 r/s.One of the answers: 45.5 z - 0.9 z - 0.66Par
Swinburne - ENG - HET489
Tutorial questions for week 12 Part A:1.Obtain a state-space representation ofz -1 + 2 z -2 G( z) = 1 + 4 z -1 + 3 z - 2 .Answer:2.B-5-6: Obtain the state equation and output equation for the system shown below.Answer:Part B: B-6-11Answer:
Swinburne - ENG - HET489
INTRODUCTION TO CONTROL SYSTEM DESIGNThe purpose of the control system design is to make the system performance to meet the specifications.HET489 Robotic ControlPerformance specifications (Ch 5)The time response of a control system consists of two par
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Lag compensation The lag compensator has a phase lag frequency response.1Its transfer function is:1 1 s+ Eo ( s ) R4 C1 Ts + 1 R1C1 T = = Kc = Kc 1 Ei ( s ) R3C2 s + 1 Ts + 1 s+ R2C2 T s+ ^ s + z1 Gc ( s ) = K c s + p1whereRC ^ T = R1C1 , T = R2 C2 ,
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DESIGN IN TIME DOMAIN (ROOT LOCUS METHOD)The root locus method is a graphical method for determining the locations of all closed loop poles from knowledge of the locations of the open loop poles and zeros as some parameter (usually the gain) is varied fr
Swinburne - ENG - HET489
REVIEW OF CONTROL SYSTEM ANALYSIS IN FREQUENCY DOMAIN (Ch 8) One advantage of the frequency response approach is that we can use the data obtained from measurements on the physical system without deriving its mathematic model.Bode Diagram A Bode diagram
Swinburne - ENG - HET489
CONTROL SYSTEM MODELLING, ANALYSIS AND DESIGN IN STATE SPACEThe classical technique is based on converting a system's differential equation to a transfer function, which relates the output to the input. The disadvantage of the classical approach is its l
Swinburne - ENG - HET489
Introduction to discrete-time control systemAdvantages of using computer based control system Flexibility of program Reliability of the system Simplified and reduced design time (Maximum productivity) Hardware reusability Maximum productivityAdvantages
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z-plane analysisImpulse samplingx(t)t x(k) x(t) x(k)k x(0) x(1) x(2)1x * (t ) = x(0) (t ) + x(1) (t - 1) + x(2) (t - 2) + . L[ (t )] = 1; L[ (t - nT )] = e - nTs X * ( s ) = x(0) + x(1)e -Ts + x(2)e - 2Ts + .X (s)*is called the starred Laplace tr
Swinburne - ENG - HET489
Discrete EquivalentsContinuous-control designs are well established and one can take advantage of a good continuous design by finding a discrete equivalent to the continuous compensator. This method of design is called emulation.The task is to find a di
Swinburne - ENG - HET489
z domain AnalysisMapping between the s plane and the z planez = eTs s = + j z = eT ( + j ) = eT e jT1Periodic strips in the s-plane are mapped into the unit circle in the z-plane. Signals in complementary strips will be folded into primary strip after
Swinburne - ENG - HET489
Transient and steady-state response analysisTransient response:Relation between Characteristic-Equation Roots and Transient Response1Roots on the positive real axis inside the unit circle of the z-plane give rise to responses that decay exponentially
Swinburne - ENG - HET489
Design of Discrete-time control system in z domain Analysis of discrete time control systems using the Root locusThe characteristic equation is:1 + F ( z) = 0Angle and Magnitude conditions:| F ( z ) |= 1 F ( z ) = 180 0 (2k + 1)General procedure for
Swinburne - ENG - HET489
Digital Compensator Design based on the frequency response method Frequency response of a linear time invariant discrete time system The frequency response of G (z ) can be obtained by substituting z = e jT :G(e jT )M = M ( ) = G (e jT ) = ( ) = G (e j
Swinburne - ENG - HET489
Analysis digital control systems in state spaceBlock diagram of state space model representationx(t ) = Ax(t ) + Bu (t ) y (t ) = Cx(t ) + Du (t )1x(k + 1) = Gx(k ) + Hu (k ) y (k ) = Cx(k ) + Du (k )Example 1.Solving discrete-time state-space equat
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Design in State SpaceControllability Complete state controllability: A control system is controllable if every state variable can be controlled in a finite time period by some unconstrained control signal. It is possible to transfer the system from any a
Swinburne - ENG - HET489
Overview of analogue compensator designDesign a compensator using root locus method Determine the ideal closed loop poles Lead compensator Lag compensatorDesign tacho feedback system Root Locus method Characteristic equation1Design a compensator us
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Cornell - CHEM - 8
UPenn - CHEM - 241
DEPARTMENT OF CHEMISTRYUNIVERSITY OF PENNSYLVANIA CHEMISTRY 241 , SUMMER SESSION I 2010SYLLABUS(updated 05.24.10) Instructor: Bryan Roberts Professor of Chemistry emeritus 610-565-1033 broberts@sas.upenn.edu Office hours: M-F 10:00-10:30 extended by ap
UPenn - CHEM - 241
CHEMISTRY 241 PROBLEMS CHAPTER 1 ANSWERS1a. 1b.H H H C H C N O H C H H H HH H C C O H H HH H1c.O C H H H C H C O N C H H1d.H HH HC HH C N C H H H2a.ClHOHClHOHClHOHClHOHthese three are less importantmost important2b.H H C C H O
UPenn - CHEM - 241
CHEMISTRY 241 PROBLEMS CHAPTER 11. Provide a Lewis structure for each of the following. Include all nonbonding pairs of valence electrons and assign any formal charges. a. c. [CH3CH2OH2]+ CH3CONHCH(CH3)CHO b. d. (CH3)3NO [(CH3)2N]-2. For each of the fol
UPenn - CHEM - 241
CHEMISTRY 241 PROBLEMS CHAPTER 21. a. b. Provide line angle formulas for all stereoisomers of the following compounds: CH3CH=CHCH2CH2CH=CHCH3 H2C=CHCH=CHCH2CH2CCCH32. For each of the following, (a) provide a full Lewis structure, (b) indicate the hybrid
UC Davis - AHI - ahi1dv
AHI 1DV WINTER 2010 FinalTA _NAME _ 1PLEASE READ THE INSTRUCTIONS CAREFULLY AS SOME ITEMS HAVE CHANGED. Instructions The examination is due in person on Thursday, March 18, at 3:30 PM in Art 204, sharp. Please bring a picture ID with you. This is an op
UC Davis - PHYSICS - phy7a
Physics 7ADLM 3 OverviewCG 1.2 Getting Quantitative with ModelsAct 1.2.2 Mostly quantitative application of the energy-interaction Model to thermal and bond systems (DLM 2 FNTs) (~45 mins)Learning Goals: More practice applying the Energy-Interaction M
UC Davis - ART - art24
GUY DEBORD'S"The Society of the Spectacle"_ 90 minutes. French videocopy with English subtitles by Keith Sanborn. _If we ever get out of this mess, future generations will look back on Guy Debord as the person who contributed to that liberation more th
NJIT - MGMT - 121
4/29/2010 Personality Profile of Serial Killers Serial killer- person who attacks and kills victims one by one in a series of incidents usually killing more than 3 people on different occasions within 14 days Although that are many different types of seri
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4/15/2010 Chapter 10 Karen Horney Karen Horney Believed that normal personality development occurred when the social environment allows for children to develop basic confidence in self and others- How does this occur? Parents who are: (1) Genuine (2) Pred
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Ashay Mehta Humanities 101-H05 Essay 3 Draft IVBasketball = Theory + PracticeBasketball is a game that strains a person mentally and physically. But some people believe that one can just learn the game in theory and master the skills using just that kno
NJIT - MGMT - 121
The time period that followed World War II presented Americans with a feeling of social unrest as a result of the threat of communism both in and outside of the United States, the ongoing arms race with the Soviet Union, and the domestic problems such as
NJIT - MGMT - 121
Ashay Mehta Chemistry 124 Honors March 24, 2009Colorimetric Method for Quantification of ProteinsIntroduction In this experiment, we used the Bradford method. This method works by protein which binds to the dye resulting of an alter in the absorption ma
NJIT - MGMT - 121
Ashay Mehta America is home to at least one hundred different kinds of cultures. Ever since the early 1900s, America has invited into her arms all different kinds of ethnicities. This melting pot that was brewed in the past has greatly effected the presen
NJIT - MGMT - 121
The right message is never Effects of Drunken Textsexpressed.+= The message is TOO blunt. The wrong person receives the text. The text tends to be garbled and jumbled. Be smart. Don't text after drinking alcohol. (Simon Cowell
NJIT - MGMT - 121
Essay I Draft IVAshay Mehta Humanities 101-H05 Mr. Burt KimmelmanObservations of Today's SocietyOften in society, how one person views a person, place or event is quite unlike how another person views that same thing. Each individual's mind thinks diff