Case 6.1 †Wild Horses Adm-624 Essay - 830 Words

Case 6.1 – Wild Horses Grand Canyon University: ADM-624 01/16/2013 In my opinion the Bureau of Land Management has had a great impact with all the services they have provided over the time. As the years have past it has become triple the coast to keep up with maintaining the care, services, shelter, and feeding of these horses. I understand the need and want to keep the horses population from dying out being instinct and letting them run freely in the wild as they are known to do. Not all animals should be restricted to where they can roam to just as humans. Also like the human race the more you roam and run wild the more it is going to cost you along with you reproducing. That is another mouth to take care of and provide for†¦show more content†¦Just as with us humans you cannot make everyone use some form of birth control but you can let them know it is out there and that it would be better to use it and be safe rather than not use it and add on additional mouths to feed, cloth or better yet not want abort. Now with all these know facts, acts, opinions all viewed as good and bad a decision can be made by looking at what is helping and what is not helping. What was the main goal and purpose of the program? Has it been met and if so are all the requirements and rules being kept and met to. Also have to look and what is going to help keep the funding up and enough of it to maintain the BLM up and running for years to come. This means abiding by the required rules regardless of what the public or anyone else thinks. The key factor here is that the BLM is making sure they are keeping these horses race alive, well, free to run, and very populated. By sticking to destroying and selling off with no limits then the BLM’s would not even be in this situation. Going through all those steps one at a time including feed back and input from all involved everyone will be able to see while agree on what needs to be done. Even with going step by step with the decision making process you may not come up with a resolution right at that moment or even a resolution that will make everyone happy. But know that in order to make any

I Had Been Out Of The Hospital For Two Weeks Essay

I had been out of the hospital for two weeks. I was staying on track with my medicine, with the help of my husband Bryon. I have been dealing with depression for a while now and I eventually ended up being emitted to the hospital. I decided to take time off work for a while until I got back in the swing of things. Me and my husband Bryon decided that was the best thing to do. Every day Bryon would make sure I took my medicine in the morning before he went to work and every evening when he got back. That’s what my psychiatrist recommended we do. I had been suicidal, and they wanted to make sure didn’t try anything to take my life. Bryon removed all of the sharp objects from the house, like knives and scissors. I didn’t feel like all of that was necessary, but I could see why he did that with the state of mind I was in before I was emitted to the hospital. Battling with severe depression took a toll on me. There was no definite cause for my depression either. It’s like no matter how good things were going in my life, there would always be a little bit of sadness that lingered. I was feeling confident that I was going to really get better and more stable. It is just so hard sometimes to find the medicine that worked. It is like trial and error. My medicine would have to be adjusted to get me feeling the best I could, but I was hoping that this was a real fix. I mean I knew that I would not always be happy because in life we all go through ups and downs, but I just wanted toShow MoreRelatedThe Crisis Center For South Suburbia Essay1275 Words   |  6 Pages†¢ This past week at the Crisis Center for South Suburbia was not as eventful as the past two weeks. The clients that were causing frequent verbal altercations have either exited on their own, exited involuntarily, or been given a written warning in which they are following the conditions of. 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Database Slides on Normalization Free Essays

Chapter 11 Relational Database Design Algorithms and Further Dependencies Chapter Outline ? ? ? ? ? ? ? 0. Designing a Set of Relations 1. Properties of Relational Decompositions 2. We will write a custom essay sample on Database Slides on Normalization or any similar topic only for you Order Now Algorithms for Relational Database Schema 3. Multivalued Dependencies and Fourth Normal Form 4. Join Dependencies and Fifth Normal Form 5. Inclusion Dependencies 6. Other Dependencies and Normal Forms DESIGNING A SET OF RELATIONS ? Goals: ? Lossless join property (a must) ? Algorithm 11. 1 tests for general losslessness. Algorithm 11. decomposes a relation into BCNF components by sacrificing the dependency preservation. 4NF (based on multi-valued dependencies) 5NF (based on join dependencies) ? Dependency preservation property ? ? Additional normal forms ? ? 1. Properties of Relational Decompositions ? Relation Decomposition and Insufficiency of Normal Forms: ? Universal Relation Schema: ? A relation schema R = {A1, A2, †¦, An} that includes all the attributes of the database. Every attribute name is unique. ? Universal relation assumption: ? (Cont) ? Decomposition: ? ? Attribute preservation condition: ? The process of decomposing the universal relation schema R into a set of relation schemas D = {R1,R2, †¦, Rm} that will become the relational database schema by using the functional dependencies. Each attribute in R will appear in at least one relation schema Ri in the decomposition so that no attributes are â€Å"lost†. (Cont) ? ? Another goal of decomposition is to have each individual relation Ri in the decomposition D be in BCNF or 3NF. Additional properties of decomposition are needed to prevent from generating spurious tuples (Cont) ? Dependency Preservation Property of a Decomposition: ? Definition: Given a set of dependencies F on R, the projection of F on Ri, denoted by pRi(F) where Ri is a subset of R, is the set of dependencies X Y in F+ such that the attributes in X U Y are all contained in Ri. Hence, the projection of F on each relation schema Ri in the decomposition D is the set of functional dependencies in F+, the closure of F, such that all their left- a nd right-hand-side attributes are in Ri. (Cont. ) ? Dependency Preservation Property of a Decomposition (cont. ): ? Dependency Preservation Property: ? ? A decomposition D = {R1, R2, †¦ Rm} of R is dependency-preserving with respect to F if the union of the projections of F on each Ri in D is equivalent to F; that is ((? R1(F)) U . . . U (? Rm(F)))+ = F+ (See examples in Fig 10. 12a and Fig 10. 11) ? Claim 1: ? It is always possible to find a dependency-preserving decomposition D with respect to F such that each relation Ri in D is in 3NF. Projection of F on Ri Given a set of dependencies F on R, the projection of F on Ri, denoted by ? Ri(F) where Ri is a subset of R, is the set of dependencies X Y in F+ such that the attributes in X ? Y are all contained in Ri. Dependency Preservation Condition Given R(A, B, C, D) and F = { A B, B C, C D}    Let D1={R1(A,B), R2(B,C), R3(C,D)} ? R1(F)={A B} ? R2(F)={B C} ? R3(F)={C D} FDs are preserved. (Cont. ) ? Lossless (Non-additive) Join Property of a Decomposition: ? Definition: Lossless join property: a decomposition D = {R1, R2, †¦ , Rm} of R has the lossless (nonadditive) join property with respect to the set of dependencies F on R if, for every relation state r of R that satisfies F, the following holds, where * is the natural join of all the relations in D: (? R1(r), †¦ , ? Rm(r)) = r ? Note: The word loss in lossless refers to loss of information, not to loss of tuples. In fact, for â€Å"loss of information† a better term is â€Å"addition of spurious information† Example S s1 s2 s3 P p1 p2 p1 D d1 d2 d3 = S s1 s2 s3 P p1 p2 p1 * P p1 p2 p1 D d1 d2 d3 Lossless Join Decomposition NO (Cont. ) Lossless (Non-additive) Join Property of a Dec omposition (cont. ): Algorithm 11. 1: Testing for Lossless Join Property Input: A universal relation R, a decomposition D = {R1, R2, †¦ , Rm} of R,and a set F of functional dependencies. 1. Create an initial matrix S with one row i for each relation Ri in D, and one column j for each attribute Aj in R. 2. Set S(i,j):=bij for all matrix entries. (/* each bij is a distinct symbol associated with indices (i,j) */). 3. For each row i representing relation schema Ri {for each column j representing attribute Aj {if (relation Ri includes attribute Aj) then set S(i,j):= aj;};}; ? (/* each aj is a distinct symbol associated with index (j) */) ? CONTINUED on NEXT SLIDE (Cont. ) 4. Repeat the following loop until a complete loop execution results in no changes to S {for each functional dependency X ? Y in F {for all rows in S which have the same symbols in the columns corresponding to attributes in X {make the symbols in each column that correspond to an attribute in Y be the same in all these rows as follows: If any of the rows has an â€Å"a† symbol for the column, set the other rows to that same â€Å"a† symbol in the column. If no â€Å"a† symbol exists for the attribute in any of the rows, choose one of the â€Å"b† symbols that appear in one of the rows for the attribute and set the other rows to that same â€Å"b† symbol in the column ;}; }; }; 5. If a row is made up entirely of â€Å"a† symbols, then the decomposition has the lossless join property; otherwise it does not. (Cont. ) Lossless (nonadditive) join test for n-ary decompositions. (a) Case 1: Decomposition of EMP_PROJ into EMP_PROJ1 and EMP_LOCS fails test. (b) A decomposition of EMP_PROJ that has the lossless join property. (Cont. ) Lossless (nonadditive) join test for n-ary decompositions. (c) Case 2: Decomposition of EMP_PROJ into EMP, PROJECT, and WORKS_ON satisfies test. (Cont. ) ? Testing Binary Decompositions for Lossless Join Property ? ? Binary Decomposition: Decomposition of a relation R into two relations. PROPERTY LJ1 (lossless join test for binary decompositions): A decomposition D = {R1, R2} of R has the lossless join property with respect to a set of functional dependencies F on R if and only if either ? ? The FD ((R1 ? R2) ? (R1- R2)) is in F+, or The FD ((R1 ? R2) ? (R2 – R1)) is in F+. 2. Algorithms for Relational Database Schema Design Algorithm 11. 3: Relational Decomposition into BCNF with Lossless (non-additive) join property Input: A universal relation R and a set of functional dependencies F on the attributes of R. 1. Set D := {R}; 2. While there is a relation schema Q in D that is not in BCNF do { choose a relation schema Q in D that is not in BCNF; find a functional dependency X Y in Q that violates BCNF; replace Q in D by two relation schemas (Q – Y) and (X U Y); }; Assumption: No null values are allowed for the join attributes. Algorithms for Relational Database Schema Design Algorithm 11. 4 Relational Synthesis into 3NF with Dependency Preservation and Lossless (Non-Additive) Join Property Input: A universal relation R and a set of functional dependencies F on the attributes of R. 1. Find a minimal cover G for F (Use Algorithm 10. ). 2. For each left-hand-side X of a functional dependency that appears in G, create a relation schema in D with attributes {X U {A1} U {A2} †¦ U {Ak}}, where X ? A1, X ? A2, †¦ , X Ak are the only dependencies in G with X as left-hand-side (X is the key of this relation). 3. If none of the relation schemas in D contains a key of R, then create one more relation s chema in D that contains attributes that form a key of R. (Use Algorithm 11. 4a to find the key of R) 4. Eliminate redundant relations from the result. A relation R is considered redundant if R is a projection of another relation S Algorithms for Relational Database Schema Design Algorithm 11. 4a Finding a Key K for R Given a set F of Functional Dependencies Input: A universal relation R and a set of functional dependencies F on the attributes of R. 1. Set K := R; 2. For each attribute A in K { Compute (K – A)+ with respect to F; If (K – A)+ contains all the attributes in R, then set K := K – {A}; } (Cont. ) 3. Multivalued Dependencies and Fourth Normal Form (a) The EMP relation with two MVDs: ENAME — PNAME and ENAME — DNAME. (b) Decomposing the EMP relation into two 4NF relations EMP_PROJECTS and EMP_DEPENDENTS. (Cont. ) c) The relation SUPPLY with no MVDs is in 4NF but not in 5NF if it has the JD(R1, R2, R3). (d) Decomposing the relation SUPPLY into the 5NF relations R1, R2, and R3. (Cont. ) Definition: ? A multivalued dependency (MVD) X — Y specified on relation schema R, where X and Y are both subsets of R, specifies the following constraint on any relation state r of R: If two tuples t1 and t2 exist in r such that t1[X] = t2[X], then two tuples t3 and t4 should also exist in r with the following properties, where we use Z to denote (R -(X U Y)): ? t3[X] = t4[X] = t1[X] = t2[X]. t3[Y] = t1[Y] and t4[Y] = t2[Y]. t3[Z] = t2[Z] and t4[Z] = t1[Z]. An MVD X — Y in R is called a trivial MVD if (a) Y is a subset of X, or (b) X U Y = R. ? ? ? Multivalued Dependencies and Fourth Normal Form Definition: ? A relation schema R is in 4NF with respect to a set of dependencies F (that includes functional dependencies and multivalued dependencies) if, for every nontrivial multivalued dependency X — Y in F+, X is a superkey for R. ? Informally, whenever 2 tuples that have different Y values but same X values, exists, then if these Y values get repeated in separate tuples with every distinct values of Z {Z = R – (X U Y)} that occurs with the same X value. Cont. ) (Cont. ) Lossless (Non-additive) Join Decomposition into 4NF Relations: ? PROPERTY LJ1’ ? The relation schemas R1 and R2 form a lossless (non-additive) join decomposition of R with respect to a set F of functional and multivalued dependencies if and only if ? (R1 ? R2) — (R1 – R2) (R1 ? R2) — (R2 – R1)). ? or ? (Cont. ) Algori thm 11. 5: Relational decomposition into 4NF relations with non-additive join property ? Input: A universal relation R and a set of functional and multivalued dependencies F. Set D := { R }; While there is a relation schema Q in D that is not in 4NF do { choose a relation schema Q in D that is not in 4NF; find a nontrivial MVD X — Y in Q that violates 4NF; replace Q in D by two relation schemas (Q – Y) and (X U Y); }; 1. 2. 4. Join Dependencies and Fifth Normal Form Definition: ? A join dependency (JD), denoted by JD(R1, R2, †¦ , Rn), specified on relation schema R, specifies a constraint on the states r of R. ? ? The constraint states that every legal state r of R should have a non-additive join decomposition into R1, R2, †¦ Rn; that is, for every such r we have * (? R1(r), ? R2(r), †¦ , ? Rn(r)) = r (Cont. ) Definition: ? A relation schema R is in fifth normal form (5NF) (or Project-Join Normal Form (PJNF)) with respect to a set F of functional, multivalued, and join dependencies if, ? for every nontrivial join dependency JD(R1, R2, †¦ , Rn) in F+ (that is, implied by F), ? every Ri is a superkey of R. Recap ? ? ? ? ? D esigning a Set of Relations Properties of Relational Decompositions Algorithms for Relational Database Schema Multivalued Dependencies and Fourth Normal Form Join Dependencies and Fifth Normal Form Tutorial/Quiz 4 Q1) Consider a relation R with 5 attributes ABCDE, You are given the following dependencies: A B, BC E, ED A a) List all the keys, b) Is R in 3 NF c) Is R in BCNF Q2) Consider the following decomposition for the relation schema R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {A, B} {C}, {A} {D, E}, {B} {F}, {F} {G, H}, {D} - {I, J} }. Preserves Lossless Join and Dependencies? a) D1 = {R1, R2, R3, R4, R5}, R1={A,B,C} R2={A,D,E}, R3={B,F}, R4 = {F,G,H}, R5 = {D,I,J} b) D2 = {R1, R2, R3} R1 = {A,B,C,D,E} R2 = {B,F,G,H}, R3 = {D,I,J} How to cite Database Slides on Normalization, Essay examples

Capital Maintenance Doctrine and Securities

Question: Discuss about the Capital Maintenance Doctrine and Securities. Answer: Introduction The doctrine of capital maintenance enjoins a company to safely maintain its capital for the benefit of creditors and allow the courts to ensure check if it has been lawfully spent (Saidul, 2013). It is an important principle in company law that requires the company to obtain a lawful consideration fro all the shares that it may seek to issues. The main principles that underpin this doctrine include the prohibition of a company purchasing its own shares and the payment of dividends to shareholders (Saidul, 2013). According to the doctrine, the profit that is made by the company is not to be recognizes unless the amount of the net assets owned by the company is maintained. Origin and Rationale of the Doctrine It is imperative to note that the doctrine has been developed as result judicial interpretation. In Flitcrofts Case (1882) the court highlighted two components of the doctrine; that the creditors have the right to check and ensure that the capital of the company is not used or shared unlawfully and that the capital of the company should not be incongruously shared to the members of the company by way of shares. In the case of Trevor v Whitworth (1887) a company bought a significant amount of its own shares from the company. It was held that such an action would lead to the reduction of the capital owned by the company and therefore the company was first obligated to pay the shareholder the amount of his contribution upon liquidation. As a matter of course, it was held in Aveling Barford Ltd. V. Perion Ltd (1989) that the shareholders of a company are entitled to their contribution to the capital upon liquidation, but the creditors will be given priority during the payment. It bears n oting that the doctrine of capital maintenance has mainly originated from the development of English case laws. The rationale for the existence of the doctrine is largely two fold. First it seeks to protect the interest of the creditors and secondly it ensures that the capital of the company is lawfully used. The courts have been vigilant in protecting the capital of the company so that it remains intact for the creditors (Zahir, 2000 p 50). Application of the Doctrine in Australia The capital maintenance doctrine is a weakening phenomenon in Australia. This has been demonstrated by the fact that most of the financial institutions such as banks do not maintain an intact capital that will prevent them from the adverse effects of a financial crisis (Gluyas, 2014 p 23). According to Roman (2016) there has been limited regulation on the use of capital in Australia. Australia has recorded an upsurge in flexibility in the freedom of use of capital but the same time the protection of the creditors has also been a top priority (Roman, 2016). It can thus be conceded that the doctrine of capital maintenance is debilitating in Australia and its application is losing relevance in most companies and financial institutions. References Aveling Barford Ltd. V. Perion Ltd (1989) BCLC 626 at p. 630-3. Flitcrofts Case (1882) 21 Ch. D. 519. Gluyas, R (2014) Capital hijacks Murray inquiry as regulators put stability before credit creation, The Australian, Roman, T (2016) The Rise and Fall of the Capital Maintenance Doctrine in Australian Corporate Law, Commercial Law in the Twenty-first Century Forum, Tsinghua University, Beijing Saidul, I. ( 2013) The Doctrine of Capital Maintenance and its Statutory Developments: An Analysis , The Northern University Journal of Law Trevor v Whitworth (1887) 12 App. Cas. 409. Zahir, M. (2000) Company and Securities Laws, , The University Press Limited, Dhaka