On Verbal and Non-verbal Communication Essay Example for Free

On Verbal and Non-verbal Communication Essay It is but human nature to try to understand another person. It normal for people to try to interpret another persons actions or words. Interpreting these types of communication means is however, difficult. Each has its own way to be interpreted. Non verbal communication is when person utilizes not his lips and voice when relaying information to another person. It is refers to actions gestures or movements a person does in order to send a message to another person. Usually, this kind of communication is utilized when meeting a new individual. Because too much speech seem to be inappropriate in first meetings, people tend to gesture via facial expression, arm movements, or even bodily gestures. Thus, people interpret this differently, via cues, unlike when the words are spoken (Brunswick Piscataway, 2009). On the other hand, verbal communication relies on the lips or mouth gestures as well as the voice of the person who spoke the words or uttered the information. To some verbal communication is more reliable as people are given the exact information needed to be received. However, there is a tendency for people to lie. Thus, there are those who listen to verbal utterances while at the same time keeping watch of non-verbal gestures. This points that interpreting verbal communication may be literal or based as well to the actions and facial expression that come along with the words (Brunswick Piscataway, 2009). It may thus be concluded that each kind of communication is interpreted differently. Verbal communication sends literal meanings at times, while non-verbal communication is sometimes vague. There is not exact way to point which tells more accurately, however if combined the message will be conveyed easier and faster. Both are also more trustworthy if done side by side. The gestures support the words and the words support all the actions.. While the quote insists that action speaks louder than words, knowing the nature man, it is still better to rely in both words and actions for a clearer interpretation. References Brunswick Piscataway. (2009). Social Perception: How we come to Understand People. Social Psych Lecture. Rutgers University.

The Moving Image Essay Example for Free

The Moving Image Essay I will be looking closely at two particular programmes involving families: The Simpsons (Homer Alone) and The Royle Family (Sunday Dinner) both families have been labelled dysfunctional yet both programmes have attracted huge TV audiences. I will be comparing the portrayal of family life in both these programmes and will try to account for their popularity. The Simpsons is an American cartoon set in Springfield. The family consists of: Homer and Marge whom are married and parents to Bart, Maggie and Lisa. In the episode Homer Alone Marge is feeling depressed because the rest of the family depend on her to do everything for them, she is feeling under a lot of pressure. After a while she decided shes had enough of doing everything for the rest of the family, therefore, she drives her car across the motorway and remains inside the car when it is parked up. This results in other motorists having no access to the road. By doing this act Marge soon gets noticed. Police cars and reporters arrive questioning Marge about the event. Soon after Homer arrives announcing that he loves her and wants her to come home to her family. It works; she comes out of the car and goes home. After a long think she decided she needs a break to get away from everything for a short while, she needed to clear her mind. So the next day she goes away for a while, and leaves the rest of the family behind. Bart and Lisa went to stay with their aunties, whereas Maggie is left with Homer. During the short period of time when Homer was responsible for Maggie, he manages to lose her! He finally finds her just prior to the return of Marge. The Royle Family is a British sit-com set in Manchester. The family consists of Jim and Barbara, the parents to Denise and Anthony. Norma, the mother to Barbara, Dave, Denises boyfriend and Twiggy a family friend. In the episode Sunday Dinner the family is round at the Royles house for their Sunday dinner. Jim and Dave are late as usual due to them being at the pub. At the pub Jim invites Twiggy back round to their house to join them for Sunday dinner without consulting Barbara. They eat lunch and chat about usual family issues. When an audience sit down to watch a cartoon, for example The Simpsons, they expect it to be funny and continuously revolving around a family or a group of friends. It is usually set in the same village or area. Cartoons are usually played on set days during the week and last the same period of time, it usually tends to be a short amount of time, ranging from 20-30 minutes. Cartoons have a very busy atmosphere, there are barely any moments when nothing is happening or no-one is speaking. The appearance of a cartoon character is often very different to human beings. In The Simpsons the characters have yellow skin and hair that remains exactly the same all the time. The Simpsons is a very typical cartoon. The characters are abnormal as they have yellow skin and are not real people. The colours are very bright and unrealistic, for example the sky is all blue apart from a few clouds which are just white. Music is used very frequently throughout cartoons. In The Simpsons music is used to create an effect on the event or character to make the audience have opinions of the situation. For example, we see Bart in detention, therefore we automatically suspect him to be quite mischievous. The beat of the music would fit in with the atmosphere of the event. When an audience sit down to watch a sit-com they would expect a television series usually lasting about an hour. It would feature the reactions of a regular cast of everyday, realistic characters to unusual situations, such as misunderstandings or embarrassing coincidences. The Royle Family is a typical sit-com because it has people playing characters which relate closely to everyday life. The characters are involved in common situations and the language they use to speak to each other is very informal, they use slang words and occasionally swear. The cameras are hand-held which gives a real-life feeling and no added lighting or special effects have been used. The clips shown are not always the best, clearest quality picture which you would get in a top quality DVD. They are just what you would expect to see when using a video camera. An opening sequence to a programme acts like an overture it has a dramatic effect on the audience. After watching the opening sequence you may think that the programme is going to be boring and not worth watching. Your viewing of the opening sequence depends on whether it appeals to you or not. Usually if the opening sequence doesnt appeal to you, you decide against watching it. It gives a taster of the programme and shows the type of programme that it is and a little about the characters involved. The opening sequence of The Simpsons begins with clouds across the screen and a bright blue sky. The title The Simpsons appears across the screen in bright yellow bubble writing. This already suggests that it is a cartoon because the writing it bubbly bright and colourful. The camera then zooms into the letter P to show a birds-eye view of Springfield, it is very colourful. You can be definite that it is a cartoon now because the village is in two dimensional form not three dimensional and all the buildings are blue or purple. The first character to be introduced is Bart; he has yellow skin and hair. Hes in detention writing on the black board which immediately gives us the impression that hes mischievous. As soon as the bell goes, he rushes out of school as quickly as possible. We then meet Homer; he is working with radioactive materials at his work and as soon as his bell goes he rushes out to go home as soon as possible being extremely careless as he leaves. Marge and Maggie are then in the car on the way home. Lisa is at band practice, she is playing her instrument proudly. We then go back to see Marge and Maggie in the car, Marge obviously has a big influence on Maggie because she copies everything she does. When Marge beeps her horn so does Maggie. She has her own toy steering wheel in the passenger seat, she steers exactly the same ways as Marge pretending to be driving just like her mum. The credits appear and the programme begins. In the opening sequence there are 23 shots of different situations in a short time of one minute and fourteen seconds. During a cartoon there are never silences or gaps, something is always happening. During the opening sequence of the The Royle Family the background is dark blue and it is set in the Royles living room. The family members are introduced one by one with them sitting on the sofa. The way they are introduced is exactly how they are throughout the sitcom. Jim is sat on his own chair in full view of the TV. Everyone else has a seat on the sofa except for Anthony, who is left to sit on the floor by himself. The music throughout the opening sequence is Oasis half the world away. This song could be used to suggest that there is one world inside the Royles house and the outside world is something completely different from their own. The lighting is just plain and ordinary like everyday life, no brighter or duller colours have been added for any special effects. Characters are very similar in both programmes in different ways. There are few characters living and working together, there will be extra minor characters put into the programme on odd occasions. The characters in both programmes can express their personality without informing us about themselves; it is easy to tell their opinions through watching their actions and reactions. Marge, in The Simpsons is similar in character to Barbara in The Royle Family. They are both the person in each family whom keeps them together; they do everything for the family, and receive little in return. They dont get appreciated enough from the rest of the family. Just Also, we have Homer and Jim who are alike in many ways. They are both idle and expect everything to be done for them by their wives. They can be very rude and inconsiderate to the rest of their family. Jim makes several complaints to Barbara about her cooking, he makes jokes about other people in the room but will not accept rude remarks made about him. Homer expects Marge to do everything for him; she makes his sandwiches in the morning, goes to the shop with his bowling bowl straight after, food shops for the family. In Homer Alone Bart and Lisa make it clear that they dont get on in the beginning, but as time goes on we see them holding hands in fear whilst at their aunties house. This shows that they do love it each other but dont show it until they need to. Denise and Anthony act in exactly the same way at Bart and Lisa. Both The Simpsons and The Royle Family have been labelled as dysfunctional, however, I do not agree with this. I believe that a real family should consist of whom they are closest to. It could include friends and animals! Each family member should respect each other and support them whenever they require it most. They will stick by each other no matter what and rely each other to help them through tough situations. Some families dont show their love and support for one another until a real crisis occurs but you can be sure that they will be there for you. I think that both The Simpsons and The Royle Family are not dysfunctional. They act as if they dont care about each other but that changes when it comes down to it. The Simpsons dont realise how close they are until Marge goes away, when she returns the family all sleep in the same bed and discuss how much they have all missed her and how bad life could be without her. In The Royle Family they all ask about each other and are interested in each other as soon as Twiggy leaves they all talk about him as if he wasnt part of their close family and they all knew that if they expressed their own personal feelings it wouldnt leave that room. The target audience is whom the programme is directed to. I think that The Simpsons is directed to anyone as it has bright, funny cartoons, which is what children would like to see. They dont have to understand the speaking to understand the programme; you can tell whats happening by the actions and reactions of the characters. It is also targeted at adults because there are jokes about political issues and famous people that adults would understand and find funny. I think that The Royle Family is targeted at adults. It doesnt appeal to children as there arent bright colours, it hasnt got a busy atmosphere and there isnt enough slapstick, simple humour throughout for them to laugh at. Adults would find it funny because its just like real life; everything they say or do is just like a typical family, like our own. Both programmes are extremely popular in different ways. The Simpsons is funny because of the familys continuous sayings and jokes. The actions they do are so unrealistic that its just so funny and the family always have terrible things happening to them, but always manage to get through it. Homer Alone Homer loses Maggie but she is luckily found safe and sound just prior to the return of Marge. The Royle Family is popular because of the way it is filmed and presented. No extra lighting or special effects have been used to create a more exciting set. It is dull and just how it would be in a normal house. Its so realistic that it looks like there are hidden cameras in house and theyre just filming them without them knowing. It is popular because people enjoy watching people like themselves, they can see what real family life is like.

Weldon Hand Tools Case Study

Weldon Hand Tools Case Study This report discusses designing a production process to produce Weldon hand tool, the product detail, the time for assembling each element as well as the sales focus of the product are all given in the case study. The report first start with calculating the number of staff required for each volume of work per quarterly for the two years period of the sales forecast ,then looks at the required technology and facility to assemble the product, it then further discuss proposed design layout for assembling operation, lastly the report discusses the likely adjustment of the propose layout should the demands increases and finally the report ends with conclusion recommending the need to balance between demands and capacity in a well organised production that supply the market with a reliable, and fairly price tool. Staffing In terms of determining how many staff is needed to work in assembling of smoothing plane manufacturing plant, given the sales forecast for the next two years we assume that each employerto works 35 hours per week for 48 weeks a year excluding the holidays. For simplicity this report will only show the calculation of the 1st quarter, the 2nd and the last quarter of the 2nd year of sales forecast of the staff required and the rest is provide as shown in the table in fig 1 below. 2.1- Calculation 1st Quarter The required number of staff for the sales volume in the next two years.We first calculate the total time available per quarterly. The sales forecast for the first quarter is 98,000 units and given that it takes 1.60 minutes to assemble one unit, the time needed to assemble the 98,000 units will be 98,000 X 1.60= 156800 minutes. In assuming the full time workers work 35 hours a week (7 hour a day for 5 working days) one worker working 12 weeks will be 35 hrs X 12weeks X 60 minutes=25200 mins Therefore number of worker that is needed for manufacturing 98,000 units will be 156800/25200= 6.222 This will be equivalent to 7 people as there is no fraction of people 2.2- Calculation 2nd Quarter The sales forecast for this quarter is 140,000 units, therefore the time requires to assemble it is 140,000 X 1.6 = 224000 From equation 1 the time available for one worker will be 35 hrs X 12weeks X 60 minutes= 25200 mins Hence the number of staff required will be 224000/25200 = 8.888 equivalent to 9 people Subsequently every quarter is calculated the same till the last quarter. Therefore the sales forecast for the last quarter of the 2nd year is 230,000 units and given that it takes 1.60 minutes to assemble one unit the time needed to assemble will be 230,000 units will be 230,000 X 1.60= 368,000 minutes. Therefore the number of people required to work for the quarter will be. Since it was assume that full time workers working 35 hours a week (7 hour a day for 5 working days) one worker working for 12 weeks will be 35 hrs X 12weeks X 60 minutes=25200 mins Therefore number of worker that is needed for manufacturing 98,000 unit will be 368,000/25200= 14.603 This will be equivalent to 15 people as there is no fraction of people. In conclusion we see the demand is not constant and there is inflation and deflation in demand to solve this we either employ more worker or ask the existing staff to do some extra over time when ever demand increase Facilities and technology Since the sales forecast shows a high demand selecting the required facilities and technology in designing the manufacturing process for Weldon hand tools is very crucial therefore the first and foremost things to consider is the size of the facility and if there is room for expansion .the operation process and the required work force We need to have in place capital and labour, proper planning of the process, research on customer requirement, seminars and training to compete in the market as the sales forecast shows increase and decrease in demand and finally managing the inventory by having warehouse for the stocks Assembling smoothing plane can be divided into three major area, fly pressing, bench operation andwrapping, other than fly pressing which requires pressingmachine,the general operation is reasonably simple so some of the technology and facilities required are Press machines for press operation Customised kits for efficient bench operation Stapler, truck, and wrapping bench for easy handling of tools and wrapping Design layout Design layout is significant process in operation management ,the flow of the layout very much relies on the volume and variety characteristic which shapes the overall design (Slack et al, 2010).The decision as to which layout type to adopt will be influenced by an understanding of their relative advantages and disadvantages (Slack et al, 2010). Therefore in order to design the layout we need first to identify the type of manufacturing process that smoothing planes is. As per the case study and according to its sales focus there is indication that this is high volume output and low variety operation and so this can be class as a mass process manufacturing process and so product base layout is the appropriate design layout.Since that the new product has a similar design with little variation, the design layout can be the same as the small variation does not affect the fundamental production process. Finally the main factors that determine the design layout in this case study is the ease in which expansion can be made to meet the increasing demand, focusing on the process to combine flexibility with a minimum lead timeare the main The layout therefore can be organised in simple and flexible manner so as to enhance the speed of production and reduce inventory while meeting the expected increase of demand The first task is solely a fly press operation in order to minimise idle time between the work stations. The operation will be organised such as each unit recovers the goods from the previous unit 4.1- Layout stages S0 by numbering the element available to assemble the task 1-12 so as to calculate the number of e stages required for the product layout. This is worked out by requiring the cycle time, even though the cycle time for each quarter was shown in the above table in Fiq1, we will show the calculation of how it was arrive , but in term of calculating the stages we only need the sales of the first quarter Since the sales forecast for the first quarter is 98,000 unit and given that it takes 1.60 minutes to assemble one unit, its assume that full time workers work 35 hours a week (7 hour a day for 5 working days)the net time available for work in 12 weeks(1st quarter) will be 35 hrs. X 12weeks X 60 minutes=25200 minutes Therefore the required cycle time =total time available/work volume 25200/98000= 0.257 mins No of stages required will be = total volume /cycle time 1.6/0.257 = 6.222= 7 stages Having known the number of stages we then list the element 1-2 and draw the precedence diagram to calculate the balancing loss Element 1: Assemble poke subassembly 0.12 Element 2: Fit poke subassembly to frog 0.10 Element 3: Rivet adjusting level to frog 0.15 Element 4: Press adjusting nut screw to frog 0.08 Element 5: Fit adjusting nut to frog 0.15 Element 6: Fit frog screw to frog 0.05 Element 7: Fit knob to base 0.15 Element 8: Fit handle to base 0.17 Element 9: Fit frog subassembly to base 0.15 Element 10: Assemble blade subassembly 0.08 Element 11: Assemble blade subassembly, clamp and label to base and adjust 0.20 Element 12: Make up box and wrap plane, pack and stock 0.20 Fiq3below displays the final distribution of work after breaking down the process into different stages of the long thin arrangement, the advantage is that its an ideal way of arranging flexible and expandable operation., makes moving and handling materials between unit effective and as well as providing adequate amount of floor space . 0.15 à ¢Ã¢â‚¬ËœÃ‚ ¢ 0.08 à ¢Ã¢â‚¬ËœÃ‚   à ¢Ã¢â‚¬ËœÃ‚ ¡ à ¢Ã¢â‚¬ËœÃ‚ £ à ¢Ã¢â‚¬ËœÃ‚ ¤ à ¢Ã¢â‚¬ËœÃ‚ ¥ à ¢Ã¢â‚¬ËœÃ‚ ¦ à ¢Ã¢â‚¬Ëœ à ¢Ã¢â‚¬ËœÃ‚ ¨ à ¢Ã¢â‚¬ËœÃ‚ ª à ¢Ã¢â‚¬ËœÃ‚ «Ã‚   0. 12 0.10 0.08 0.15 0.05 0.15 0.17 0.15à ¢Ã¢â‚¬ËœÃ‚ © 0.20 0.20 Fig2 Element listing and precedence diagram 4.2- Calculating balancing loss 0.15 à ¢Ã¢â‚¬ËœÃ‚ ¢ 0.08 à ¢Ã¢â‚¬ËœÃ‚   à ¢Ã¢â‚¬ËœÃ‚ ¡ à ¢Ã¢â‚¬ËœÃ‚ £ à ¢Ã¢â‚¬ËœÃ‚ ¤ à ¢Ã¢â‚¬ËœÃ‚ ¥ à ¢Ã¢â‚¬ËœÃ‚ ¦ à ¢Ã¢â‚¬Ëœ à ¢Ã¢â‚¬ËœÃ‚ ¨ à ¢Ã¢â‚¬ËœÃ‚ ª à ¢Ã¢â‚¬ËœÃ‚ «Ã‚   0. 12 0.10 0.08 0.15 0.05 0.15 0.17 0.15à ¢Ã¢â‚¬ËœÃ‚ © 0.20 0.20 Stage1 stage2 satge3 stage4 satge5 stage6 stage7 stage8 Cycle time = 0.26 0.23 0.23 0.22 0.20 0.20 0.20 0.17 0.15 1 2 3 4 5 6 7 8 Idle time of every cycle =(0.26-0.22)+(0.26-0.25)+(0.26-0.20)+(0.26-0.15)+(0.26-0.17)+(0.26-0.23)+(0.26-0.20)+0.26-0.20) = 0.45 Balance loss = 0.45/8 0.26 = 21.6 % Fig 3 Allocation of element to stages and balancing loss for Weldon hand tool fig2 Meeting the demand In meeting increasing demand the above design layout can be adjusted by rearranging the stages. The conventional arrangement of the eight stages was to arrange them in one line and each stage takes 0.257 minutes (cycle time) worth of work; however this can be rearranged in to four shorter lines each stage with 0.514 minutes worth of work will give the same output. So following this conclusion the stages could be arranged in 8 parallel stages each responsible for the whole worked content, the advantage is that it will save time for product transportation both within the site and the nearby building where facilities are located, create plenty of space within the site hence enhancing easy expansion to meet increase in demand and finally the inventory will also assist to place the product well in stock Lastly the above layout has concentrated on the labour time effectiveness of assembling one unit per hour per person,however concentrating on speed and quickness of machines may encounter any competition of demand and capacity Conclusion In operational management capacity planning such knowing what is the size of the facilities, the geographical location and the row material for the production process is fundamental, crucial and necessary To successful manage the Weldon hand tool manufacturing process its desirable that the layout be design in a flexible way to rearrange stages that will create more rooms for expansion so as to meet future demands and capacity. The flexibility also makes the transportation (moving and handling) of materials between unit effective Finally at the moment the sales forecast does not predicts smooth distribution of demand to encounter this problem of inflation and deflation of the sales, marketing the product through sales promotion could improve the demand distribution to a smooth increase throughout the quarters.

Scarlet Letter :: essays research papers

For a person to be able to make a change in their life is a monumental task. To be able to make a change that can be life changing is a true test of a person’s will and desire. In the novel the Scarlet Letter, Hester Prynne changes from an adulteress that is despised by the community, to an able woman that the community depends on through her repentance of her sin, faith in transcendentalism, and her daughter Pearl.   Ã‚  Ã‚  Ã‚  Ã‚   Hester Prynne is cast out as an adulteress from the Puritan society, because she had a baby with another man than her husband. â€Å"She may cover it with a brooch, or such like heathenish adornment, and so walk the streets as brave as ever!†(Hawthorne 49). The city is shocked by Hester’s embodiment of the scarlet letter on her chest; because the scarlet letter was suppose to represent her punishment for her sin. This was an insult to the community, and it shows how the community despised Hester’s presence of an adulteress. â€Å"Madam Hester absolutely refuseth to speak, and the magistrates have laid their heads together in vain†(58). When Hester mounted the scaffold she refused to speak to the ministry, and she refused to tell them who the father of the baby was. By not telling the ministry who the father was Hester was being defiant and took the burden of the punishment on herself, this is another reason why the community despised H ester when she was considered an adulteress. Her strong will and silence lead the community to hate Hester, because she would not bend to the community and show weakness.   Ã‚  Ã‚  Ã‚  Ã‚  The change that Hester experiences occurs through the aid of her daughter Pearl, her strong belief in transcendentalism, and her repentance of her sin through aiding the community. â€Å"†¦that this brook is the boundary between two worlds†(119). Pearl crosses the brook to get away from Hester and Dimmesdale, after they have decided to go to Europe since it is better for his health. The brook and Pearl represent transcendentalism, while Dimmesdale represents the church, which shows how Hester has changed due to her belief in Transcendentalism. â€Å"I have no heavenly father!†(90). Pearl continually questions her mother about her father, and Hester tells her that she came from the heavenly father. By having Pearl, Hester is trying to raise her without having her make the same mistakes that she did, and in the transcendental belief, this is how Pearl is able to have a significant change on Hester.

The Industrial Revolution and Great Britain Essay -- The Industrial Re

Since the advent of man, the human race has gone through many changes throughout history. One of the greatest and most crucial changes was the Industrial Revolution of Great Britain. Although the Industrial Revolution did have a few drawbacks, the positive outcomes of the Revolution far outweighed the negative effects. It pushed Great Britain fifty years ahead of other European countries and morphed the country into one of the strongest nations of its time. The Revolution improved the overall state of Great Britain mainly through the innovation and invention of new technologies, improvement in communication and transportation, and enhancing the lifestyles of the British commoner. During the time period of the Industrial Revolution, Great Britain saw rapid change. Many new inventions and innovations drastically improved both the economic state of Britain and the lives of the people. The introduction of better farming practices and inventions of new farming tools such as the wheat drill, which was a â€Å"sowing device that positioned and covered seeds in the soil† , made farming a lot more efficient and required less human labor. As a result people moved to cities in search of jobs, thus providing a large workforce for factories. During the industrial revolution â€Å"many people left their rural communities to work in towns and cities...working in a factory† . These factories became the powerhouses of the industrial revolution. â€Å"Never before had people been put to work in such a well-organized way† as they had been in factories. This new method of manufacturing goods exponentially increased the economy of Britain as new machines we re introduced. These new machineries enabled cheaper labor and mass production of goods at lower costs, a... ... Heath, 1983. Print. 3) "British Industrial Revolution." British Industrial Revolution. Web. 28 May 2012. . 4) "Canal History." Canal History. Web. 28 May 2012. . 5) "The Fabulous Story of the Postage Stamp." Arpin Philately. Web. 28 May 2012. . 6) "How the Steam Engine Changed the World." LiveScience.com. Web. 28 May 2012. 7) Mooney, Carla. The Industrial Revolution, White River Junction, Nomad Press, 2011. print 8)"The Open Door Web Site : History : The Industrial Revolution : The Development of Roads." The Open Door Web Site : History : The Industrial Revolution : The Development of Roads. Web. 28 May 2012. .

Miwok Social Life :: essays research papers

History Final Miwoks set many standards for themselves and the tribe. They were usually very hard workers, working harsh numbers of hours a day to keep the tribe alive. There were hunters, fishers, and many other jobs for people. A non-nomadic people, the Miwoks settled in the Yosemite Valley. My report is on the Miwok Social Life. Games, customs, jobs, and many other things about the Miwok Indians will all be covered in this. The way they lived, what they ate, and what they farmed. They all had a job, some of the women wove baskets, and some of them cooked. The men hunted, fished, made canoes, and fought. The first step of stepping into the life of a Miwok is to understand their dress. They were generally very lenient on clothing, some children going completely naked. Many wore flaps and when the men hunted, they camouflaged themselves in deerskins and grasses. Now you must know how they lived. What they lived in, how they built their homes, and such. U-ma-cha is the named of the home they lived in. Much like a "tee-pee", These homes were made of the thick bark of the Sequoia Redwoods. Mud and dirt was piled on the bottom of these homes, to keep water and rodents out, and heat in. They were at times coated with a layer of pine needles. About 8-15 feet in diameter, these homes were small. They did not bathe, but instead sat in sweat houses until they sweat the smell off and then ran out and jumped in a cold stream (Chilly Willy). Most slept on Deerskins, but a few slept on willow frames lifting them only inches from the ground, while the chief slept on a bearskin. The fire, at the center of each U-ma-cha, was used for cooking and heat. It was vital to have this fire burning constantly. There was also a ceremonial sweat house used for special ceremonies. This had a roof of 5 inches thick and was in the center of the village. The foods the Miwoks ate are also another step. With plentiful amounts of food in the Yosemite Valley, the Miwok stayed where they were. With out the need to move about in a nomadic fashion, their villages grew. This meant for a large need of food. Their main food was acorns. The women prepared this by cracking and shelling the acorns, then drying them.

Database Slides on Normalization

Chapter 11 Relational Database Design Algorithms and Further Dependencies Chapter Outline ? ? ? ? ? ? ? 0. Designing a Set of Relations 1. Properties of Relational Decompositions 2. 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 t he decomposition D is the set of functional dependencies in F+, the closure of F, such that all their left- and 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 Decomposition (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"aâ €  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 decomposi tion 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 a nd 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 schema 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 SAlgorithms 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 { Compu te (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] a nd 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. ) Algorithm 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 ? ? ? ? ? Designing a Set of Relations Properties of Relational Decompositions Algorithms for Relational Database Schema Multivalued Dependencies and Fourth Normal Form Join Dependencies and Fifth Normal FormTutorial/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 }