Benefits of Risk Management, Risk Management Frameworks Assignment

Benefits of Risk Management, Risk Management Frameworks - Assignment Example Importance of Risk Management Risk Management is extremely crucial because it helps the decision-making process regarding viability and competitiveness between organizations. It also helps in the creation of value which is the main factor to manage the business better in global markets when all organizations have equal access to available resources because then business processes to congregate on international standards. Operational efficiency can be increased by having plans for risk management already in place because it enables a business to do more for less. This means putting aside resources from regular expenditures in making strategic investments which will support company goals. A large percentage of the budget is spent on keeping the business running. Progressive organizations must have contingency plans to develop and expand resource deployment and implementation which will spare resources to concentrate on risk management strategies. (Mes 2010). Benefits of Risk Management Risk is always associated with insecurity and improbabilities with the possibility of things not turning out as expected. The benefits of risk management are that the organization is fully prepared for such eventualities and has a mechanism in place to handle risks and minimize losses. It is not possible to totally eliminate all risks, so good risk management develops awareness of risks when times are good, and perpetuates regulations and self-control during crises (Raz and Michael 2001). The benefits of risk management can be both long and short term. Accordingly, each stage of risk management efforts beginning from risk identification and evaluation and formulating alleviation or improvement strategies has its own benefits (Miller 1992). Question 2: Compare and contrast Management of Risk with another risk management framework (such as that offered in chapter 7 of Project Management by Larson and Gray), highlighting the similarities and differences between them.  

The Backscatter X-Ray Essay Example for Free

The Backscatter X-Ray Essay The backscatter X-ray is the right choice when compared to traditional X-rays or metal detectors. On December 27, 2006, the author, Thomas Frank,whom is the aviation security correspondant for USA Today, wrote, X-ray Tests Both Security, Privacy, and made valid points concerning the fear of radiation, how easily it finds metal guns and knives, and the provacy issue that surrounds X-ray technology (5). Throughout this essay Thomas Frank was able to write about the facts surrounding X-rays and keep hs opinion out of the article therefore giving his readers the chance to form their own belief. This essay was written to analyze which belief to choose. Should the public be against the backscatter X-ray because of radiation exposure or is safety against terrorists the better option? The fear of over-exposure to radiation is laughable at best. Its like being out in the sun for 15 to 20 minutes (5). Being out in the sun can in reality be a good thing. Through sunlight the body absorbs vitamin D which is needed for strong bones and teeth. Everyone gets sun exposure on an everyday basis. Its not a bad thing to have a little sun. Take people whom use artificial light to tan. A waiver must be signed before exposing the body to the light. This waiver states that skin cancer may occur with use; eye damage is possible with exposure. Tanners sign this form without another thought because artificial light is the same as being out in the sun. A waiver is not availabe for people to sign at the airport. Why is this? Is it because radiation damage is higher with artificial light thereby being higher with sunlight versus the backscatter X-ray? If so, should the public stay inside their house and avoid sunlight at all cost? They are foolproof when it comes to finding metal guns and metal knives (6). Clark Kent Ervin whom is the former Homeland Security Department inspector said during the interview. He stated, screeners repeatedly missed hidden weapons (6). If screeners are missing hidden weapons wouldnt the obvious choice be to use the backscatter X-ray? Isnt the publics safety more important? Our airport security has already been compromised once with September 11 . If there is anything that can prevent that from happening again then it should be done. Our safety is top priority. Backscatter X-rays dont show anything on the body deemed indecent. It only shows whats carried on the body. People really need to understand that the Transportation Security Administration (TSA) is putting in place technologythat does in fact do a digital strip search and stores those images at least temporarily, said Marc Rotenberg of the Electronic Privacy Information Center (5). Temporarily is the key. Whom cares about a picture being taken as long as its temporary? Its not as if the image is going to be stored forever and displayed worldwide. The TSA, however, says the images are cartoonlike sketches that show only outlines of each passenger and are never stored (5). The images arent stored so the public doesnt need to worry about the images being released. Cartoonlike sketches means that its not even a real human body. It doesnt look like a person at all. Its just an outline that looks cartoonlike at that. Theres nothing wrong with cartoons. I grew up watching them as a child as did many others. $100,000 backscatter machine finds many weapons missed by screeners and metal detectors (6). Finding weapons as soon as possible can be the difference between life and death. A machine that finds weapons more often than not should be used in place of a machine that is likely to miss. Although the backscatter X-ray is not going to catch everything that it scans it will make it harder for terrorists to sneak weapons on a plane. The object is not to prevent everything that can conceivably happen, he said, the object is to make it difficult (6). The fact claim the author is making is supported by the examples given in the article. Some examples include quotes from Marc Rotenberg and Frank Cerra. The warrant that connects the fact claim and support are digital strip search (5) and being out in thesun for 15 to 20 minutes (5). It seems these details were said to persuade the interviewer and readers of this article. Someone would need 2,500 backscatter scans a year before reaching the limits of safe radiation exposure (5). These guidelines are provided by the National Council on Radiation Protection and Measurements so therefore even frequent flyers will be protected from over-exposure to radiation. In closing, the authors fact claim of backscatter X-ry versus regular X-ray and metal detectors have been clearly stated. He went into detail giving both sides concerning radiation, worth, and privacy. Overall the essay was an effective argument that gave information from both sides equally. With public safety a concern the conflicts that arise concerning this matter should be of no importance. The public has a right to be protected no matter what it takes.

Complement Serum Activity by Lysing Sheep Erythrocytes

Complement Serum Activity by Lysing Sheep Erythrocytes Introduction The Immune system is a series of complex processes which has evolved to protect the body from attack by foreign pathogens. These pathogens are able to enter our body through the skin or lining of the internal organs. The immune system is able to protect us from intracellular and extracellular organisms as well as from ourselves, stopping malignancies and autoimmune diseases from spreading in our bodies (Bastian, 1993). There are two lines of defence, the adaptive (specific) and innate (non-specific immunity), though both are united in their goal to destroy pathogens they have different ways to tackle this. Innate immunity is the 1st line of defence while adaptive immunity is the 2nds line and thus takes longer to act (Clancy, 1998). The complement system is part of the immune system and can be bought into action by the adaptive system if required. Complement is a group of proteins working together within the immune system; once stimulated by one of many triggers, proteases begin to c leave protein in the system, bringing a cascade of enzyme reactions in order to fight off foreign pathogens and activate the inflammatory response. Within the complement cascade there are many proteins that play a role but C3 is a protein critical to the effector functions of the system (Abbas, 1994). There are many paths for immune mediated lysis and the one we will be looking at is intravascular haemolyse and occurs when the complement has been triggered through the classical pathway. When the antibody binds to the antigen on the surface of the erythrocyte, a complement component triggers the membrane attack complex to form pores in the cell membrane resulting in cell lysis (Chapel, 1990). The intensity and speed at which cells lyse is dependent upon the rate at which the complement cascades to enable complete cell lysis. Experiments like these are able to provide us with an understanding of how the complement immune system functions. It can also increase our understanding of autoimmunity and perhaps lead to ways in which the effects of immunity can be prolonged or inhibited according to the disease. Systemic Lupus Erythematosus (SLE) is an autoimmune disease, in which complement is analysed, as getting SLE is dependent upon the gene which is responsible for producing MHC, a component used in haemolysis (American, 1993), patients with other immunological disorders can require their complement activity to be monitored and thus this assay would be able to show how efficiently the complement component of the immune system is working to defend their bodies. Aims To determine complement serum activity by lysing sheep erythrocytes To determine the volume of complement required for 50% lysis. Materials 20 Cuvettes 1.0ml 20 test tube plastic disposable Automatic pipette 200-1000 ÂÂ µl 6 tips Automatic pipette 0-200 ÂÂ µl 6 tips Water bath at 37Â °C Spectrophotometer Test tube rack Centrifuge Ice bucket Ice Method Wash 4ml of erythrocyte suspension three times with barbitone saline solution. Prepare a 6% stock solution of erythrocytes In one test tube mix 3.0ml of sheep anti-erythrocyte antiserum, diluted 1/50 3.0ml of the 6% SRBC Mix and gently by capping and inverting several times Incubate at 37Â °C for 15min in the water bath, mix every 5min. Set up the test tubes on ice in duplicates and label Add the reagents in order as shown in table 1 below Incubate the tubes for 60 minutes at 37Â °C mixing gently every 15minutes Place the tubes on ice and then centrifuge at 200g for 10 minutes at 4Â °C Remove the samples and put into cuvettes and read the absorbance at 541nm, with ammonia solution as blank record the results in a table. Results Discussion When carrying out the experiment raw data was recorded, and presented in table 1. However the results obtained during the practical were not used as the erythrocytes lysed before complement was added and therefore complement activity could not be observed as adding complement to lysed cells is not able to produce results, therefore the ideal data provided was used and analysed. From table 1 it is clear that absorbance levels increased as serum volume increased, this is due to the fact that as volumes of complement increase more red blood cells are lysed which in turn allows haemoglobin to be let out, this is of a dark colour and as more cells are lyses the darker the resulting sample will be, and so the absorbance as read the spectrophotometer will increase. After the guinea pig serum has been mixed with the sensitised erythrocytes, it produces anti-body coated cells with complement attaching to the antibody, and activating this attracts the MAC molecules to take action and lyse the cell (Kuby, 1994). Following the pattern seen in table one table 3 shows a progressive % lysis of cells as the volume of serum is increased, however for the 100% lysis an ammonia buffer was used to ensure that all cells are lysed during the experiment. Further to this graph 1 produced a sigmoid curve, from which it was possible to estimate CH50. However calculating the 50% lysis from this graph is not very accurate. Thus a log graph 2 was constructed, with the use of van Krogh equation to determine the actual value of 50% lysis. The equation was provided by the lecturer. Van Krogh equation: x= k[ y ]1/n 100-y Where: x= amount of complement (ml of undiluted serum) y= proportion of cells lysed k=50% unit of compliment n=inclination of graph (ideally 0.2) This resulted in table 4 giving a volume of 133.5 CH50/ml. However when calculating CH50 the x values were all in the negative. Moreover, it was not possible to compare data sets obtained against ideal data as the experiment did not yield results due to lysis of erythrocytes before complement was added. This could have occurred due to improper pipetting, handling or transporting of the cells as shaking them too much could have lysed them due to shock, as the cells were sensitized and thus prone to quick lysis. Further to this it was reported by Inglis, et al, 2007, that the use of erythrocytes from different sheep can yield inaccurate results and thus produce different CH50. Although there are many inaccuracies present within the experiment, it also gives scope to further improve the method as well as explore other area of the subject at hand such as factors which affect the performance of complement like temperature or PH. This assay is a good way to measure the activity of the immu ne system within patients, such as patients with LSE as mentioned earlier, other patients with low immunity can also be tested to see how the complement system is or isnt aiding their recovery, thus steps can be taken by medical professionals to either boost or monitor the progress of the patients immunity as fundamentally the immune system is required to work at its optimum to keep humans and animals from dying of disease(Inglis,et al, 2007). Conclusion Overall this experiment has shown how complement is important in aiding white blood cells to lyse foreign bodies. Though in the experiment carried out the blood cells lysed before complement was added the method was presented and the ideal set of data, showed what results should have been obtained. Also the hypothesis that as the complement concentration increases so will the absorbance proved positive.

The Use Of The Word Love :: essays research papers

Six months after I met a young man, he expressed to me how much he loved me. Being sixteen years old, I thought it to be very flattering but I could not accept him saying this to me. The word, love in the romantic sense, is something that would take so much out of me to say to a person. Love is something that you express to someone that you can not, in any way, see living your life without. The last time I saw the young man who supposedly loved me, was on my seventeenth birthday when he told me I was a waste of his time. Love is the strongest emotion and most powerful word anyone can say to someone else. Some people use the word love everyday as though it is not as big of a deal as it really is. Jewel's song, "Always," illustrates a definition of love. The first line of the song totally defines love in the same way I do with, "Please don't say 'I love you,' those words touch me much too deeply." With this one line, my entire definition of love is presented. There are other things people can love: a pet, family members, certain kinds of food but these are all loved by a different kind of love. There are millions of ways you can love, but this kind of love does not happen all the time to a person. It is the kind of love that is not there at the beginning, but grows inside you. This kind of love is for another person that you can not see your life without, even though at one time it was without that person. To express this to someone is to propose to commit your life to that other person. The other person may not feel the same way and that is why it is hard to accept someone saying that. One expressing their love to another is the greatest compliment one could ever receive for they are wanted in someone else's life forever.There are some people that do not understand the meaning of love and use it all the time as though it was just some word that makes people happy. The young man who told me he loved me also told me a month later that I was a waste of his time. He is an example of someone using the word just to make others or even himself happy.

Mental Health and the Prison System

Who are the victims of mental health and how they are treated within the legal and prison system? Mental health or mental illness, base on the question, ‘is concern with illnesses of the mind, or with treating illnesses of the mind.’ (Longman, p.890) These illnesses, which affect the mind, create hallucinations that can lead the people who suffer from them to cause harm to the innocent people who fall victim to them. According to the Longman dictionary, the word ‘victim’ refers to ‘someone who has been attacked, robbed or murdered’. (Longman, p.1593) It can also refer to someone who suffers because they are affected by an illness. In January 1999, a young woman was pushed from behind, in front of a New York City subway train, to her death. Her murderer was a mentally ill patient who had refused to consume his medication. This young woman’s name was Kendra Webdale and the Kendra’s law was named after her. She was considered a victim of mental health. There are many other cases who were fall victim to these illnesses. Due to their mental illnesses, many people questioned whether these ‘criminals’ should be charged and face the consequences or should they be treated in the hospital for the mentally ill under strict control and supervision. They are being view as the victims of the mental health for the reason that they are not in their ‘right’ minds to judge. This occurs because they are suffering from the illness affecting their mind. Looking from the patients’ point of view, they are undergoing a lot of stress due to their conditions as they have fears that they are not accepted by the family, friends and society. With the stress they are undergoing, they might not be able to face the fact and would not want to take their medication. This had led to intervention of the legal system to control the situation. According to Kendra’s Law, a procedure has to be followed for obtaining court orders for certain individuals with mental illness to receive and accept assisted outpatient treatment (AOT). (Office of Mental Health, 2006) The mentally ill individual who can qualify for AOT must be at least 18 years old and shows a sign of being incapable of surviving on his own. In view of the concern that the patient may do serious harm to others in the society, an AOT is more likely to benefit him. The request for AOT can be done by the parent, spouse, sibling, director of a hospital, licensed psychologist or a probation officer. However, there are some lawyers who contest that the law will only serve to violate the patient’s process protection. In addition, many wonder if the implementation of this law is strong enough to force a person to take medication. There is always the possibility of the patient having tried to seek treatment but failed. As such, the government should not simply laws to force the taking of medication but for the state to provide medication as well. In the case of the prison system, there has been a significant increase in the number of inmates who are found to be severely mentally ill. Despite the increase in the number of inmates, the medical treatment necessary for this group of inmate did not increase. Moreover, the staff handling them are not properly trained and thus, many of these inmates are being victimized in the prison cells due to their disorganized speech and behaviour. Their inability to communicate well with others may in turn antagonize their officers or fellow inmates. In conclusion, the victims of mental health do not necessarily refer to only the ones suffering from the illness but those who are indirectly affected as well. The patients should be given proper and better treatment and the society should also be educated on how these patients need to be treated in order to survive in the society. Bibliography 1.  Ã‚  Ã‚  Ã‚  Ã‚   Harold E. Shabo. 2001. Social Costs: Criminal Justice and Mental Health System Gaps which Contribute to the Criminalization of Mentally Disordered Persons. California. 2.  Ã‚  Ã‚  Ã‚  Ã‚   Longman. 1999. Longman Dictionary of Contemporary English: International Students Edition. Pearson Education Limited. Spain. 3.  Ã‚  Ã‚  Ã‚  Ã‚   Office of Mental Health. 2006. An Explanation of Kendra’s Law. 4.  Ã‚  Ã‚  Ã‚  Ã‚   About: Mental Health. 2006. Forced Mental Treatment has a Place.   http://www.mentalhealth.about.com/cs/schizophrenia/a/commit204.htm 5.  Ã‚  Ã‚  Ã‚  Ã‚   Mental Health Services. 2006. Bureau of Mental Health Services. http://www.drc.state.oh.us/web/mentalhealth.htm

Slack Bus And Slack Generator Engineering Essay

The Table below shows input informations of each busbar in the system used to work out the power flow and the simulation consequence harmonizing to direction described in inquiry 1.BusInput Data[ Simulation Result ] BUS 1 plutonium P ( burden ) 100 MW Q ( burden ) 0 Mvar BUS 2 P ( burden ) 200 MW Q ( burden ) 100 Mvar CB of Generation Open BUS 3 1 plutonium P ( Gen ) 200 MW P ( burden ) 100 MW Q ( burden ) 50 Mvar AVR On AGC OffSlack coach and slack generatorIn power flow computation, alone numerical solution can non be calculated without mention electromotive force magnitude and angle due to unequal figure of unknown variables and independent equations. The slack coach is the mention coach where its electromotive force is considered to be fixed voltage magnitude and angle ( 1a? 0A ° ) , so that the assorted electromotive force angle difference among the coachs can be calculated regard. In add-on, the slack generator supplies as much existent power and reactive power as needed for equilibrating the power flow sing power coevals, load demand and losingss in the system while maintain the electromotive force changeless as 1a? 0A ° . In existent power system, when comparatively weak system is linked to the larger system via a individual coach, this coach can stand for the big system with an tantamount generator maintaining the electromotive force changeless and bring forthing any necessary power like sla ck coach. [ 1 ]Bus type ( PQ coach or PV coach )BusBus typeRemarksBUS 2 PQ Bus Generator is disconnected to Bus 2 BUS 3 PV Bus Generator is connected to Bus 3 and the magnitude of electromotive force of generator support invariable by utilizing AVR In general, each coach in the power system can be categorized into three coach types such as Slack Bus, Load ( PQ ) Bus, and Voltage Controlled ( PV ) Bus. The definition and difference between PQ Bus and PV Bus are described as follows ; [ 2 ] PV Bus ( Generator Bus or Voltage Controlled Bus ) : It is a coach at which the magnitude of the coach electromotive force is kept changeless by the generator. Even though the coach has several generators and burden, if any generators connected to the coach modulate the coach electromotive force with AVR, so this coach is referred to PV Bus. For PV coach, the magnitude of the coach electromotive force and existent power supplied to the system are specified, and reactive power and angle of the coach electromotive force are consequently determined. If a preset upper limit and minimal reactive power bound is reached, the reactive end product of the generator remains at the limited values, so the coach can be considered as PQ Bus alternatively of PV Bus. [ 2 ] PQ Bus ( Load Bus ) : It is a coach at which the electromotive force is changed depending on entire net existent power and reactive power of tonss and generators without electromotive force regulator. Therefore, in the power simulation and computation, the existent power and reactive power of the tonss are specified as input informations and consequently the electromotive force ( magnitude and angle ) is calculated based on the above input. The following table specifies input and end product of each coach type in the power system simulation and computation. Bus Type Phosphorus Q ( Magnitude ) I? ( Angle ) PQ Bus Input signal Input signal End product End product PV Bus Input signal End product Input signal End product Slack Bus End product End product Input signal Input signalSystem BalanceEntire Generation & A ; Load DemandBusReal Power ( MW )Fanciful Power ( Mvar )CoevalsLoadCoevalsLoadBUS 1 204.093 100 56.240 0 BUS 2 0 200 0 100 BUS 3 200 100 107.404 50 Entire 404.093 400 163.644 150DifferencePgen – Pdemand = 4.093Qgen – Qstored in burden = 13.644Reason: Real power loss due to opposition of transmittal line and fanciful power storage due to reactance of transmittal line are the grounds for the difference between power coevals and load demand in the system.P ( Losses ) & A ; Q ( Storage ) over the transmittal lineBusReal Power ( MW )Fanciful Power ( Mvar )SendingReceivingLosingssSendingReceivingStoredBUS 1 – Bus 2 102.714 100.650 2.064 56.653 49.773 6.88 BUS 1 – Bus 3 1.379 1.378 0.001 0.4141 ) 0.4131 ) 0.001 BUS 3 – Bus 2 101.378 99.350 2.028 56.990 50.227 6.763 Entire Palestine liberation organizations =4.093Qstored in burden =13.6441 ) Imaginary power flows from Bus 3 to Bus 1. The summing up of existent power losingss and fanciful power storage over the transmittal line are precisely same with entire difference between coevals and burden. Therefore, it is verified that the difference is shown over the transmittal line. ‘Kirchoff ‘ balance as each coach [ 4 ] Bus1 I? P1 = + Pgen1 – Pload1 – P12 – P13 = 204.093 – 100 – 102.714 – 1.379 = 0 I? Q1 = + Qgen1 – Qload1 – Q12 – Q13 = 56.24 – 0 – 56.653 + 0.413 = 0 Bus2 I? P2 = + Pgen2 – Pload2 – P21 – P23 = 0 – 200 + 100.65 + 99.35 = 0 I? Q2 = + Qgen2 – Qload2 – Q21 – Q23 = 0 – 100 + 49.773 + 50.227 = 0 BUS3 I? P3 = + Pgen3 – Pload3 – P31 – P32 = 200 – 100 + 1.378 – 101.378 = 0 I? Q3 = + Qgen3 – Qload3 – Q31 – Q32 = 107.404 – 50 – 0.414 – 56.99 = 0 Harmonizing to the computation supra, as summing up of incoming & A ; surpassing existent power and fanciful power at each coach become zero, it is verified that each busbar obeys a ‘Kirchoff ‘ balance. In add-on, the entire power system is wholly balanced, because entire coevals power ( existent & A ; fanciful ) are equal to summing up of entire load demand and existent power loss & A ; stored fanciful power over the transmittal ( i.e. Pgen – Pdemand = Plosses, Qgen – Qstored in burden = Q stored in system ) as shown above.Voltage Angle and Angle DifferenceAs a consequence of the Powerworld, the electromotive force angle and angle difference are shown in the tabular array below.BusVoltage AngleVoltage Angle DifferenceBUS1 I?1 = 0.00A ° BUS1- BUS2 I?1 – I?2 = 0.00A ° – ( -2.5662A ° ) = 2.5662A ° BUS2 I?2 = -2.5662A ° BUS2- BUS3 I?2 – I?3 = -2.5662A ° – ( -0.043A ° ) = -2.5232A ° BUS3 I?3 = -0.043A ° BUS3- BUS1 I?3 – I?1 = -0.043A ° – 0.00A ° = -0.043A °Power System Analysis -1The tabular array below summarizes coevals and electromotive force angle fluctuation at each coach as coevals at Bus 3 varies from 0 MW to 450 MW by 50MW.Simulation Consequences and ObservationP3 = 0 MW P3 = 50 MW P3 = 100 MW P3 = 150 MW P3 = 250 MW P3 = 300 MW P3 = 350 MW P3 = 400 MW P3 = 450 MW Reactive Power Generation at Bus 3: It is found that reactive power coevals Q3 ( gen ) lessening while existent power coevals P3 ( gen ) addition because Bus 3 as a PV Bus regulates the changeless coach electromotive force magnitude by commanding excitement of the coevals through the AVR. Power Generation at Bus 1: It is found that P1 ( gen ) decreases and Q1 ( gen ) increases at the same time, while P3 ( gen ) additions and Q3 ( gen ) lessening. As the entire load demand in the system keeps changeless ( i.e. Ptotal ( burden ) = 400 MW, Qtotal ( burden ) = 150Mvar ) , any necessary existent power and reactive power for the system balance demand to be supplied by generator ( loose generator ) at Bus 1. Therefore, power coevals P1 ( gen ) and Q1 ( gen ) at Bus 1 alteration reversely compared to power coevals alteration at Bus 3. Voltage Angle Difference: In general, existent power flow is influenced by electromotive force angle difference between directing coach and having coach harmonizing to PR = . Therefore, it is observed that every bit existent power coevals P3 ( gen ) increases existent power flow from Bus 3 to Bus2 addition, consequently voltage angle difference ( I?3 – I?2 ) between Bus 3 and Bus 2 additions. However, lessening in existent power from Bus 1 to Bus 2 due to increase of P3 ( gen ) consequence in lessening of electromotive force angle difference ( I?1 – I?2 ) . In add-on, Real power between Bus 1 and Bus 3 flows from Bus 1 to Bus 3 until P3 ( gen ) range to 200 MW and as P3 ( gen ) addition more than 200 MW the existent power flows from Bus 3 to Bus 1. So, it is besides observed that electromotive force angle difference ( I?3 – I?1 ) is negative angle when P3 ( gen ) is less than 200MW and the difference addition while P3 ( gen ) addition.Power System Analysis -2The tabular array below summarizes the fluctuation of power coevals and electromotive force angle difference at each coach when the burden demand at Bus 3 varies by 50MW and 25Mvar.Simulation Consequences and ObservationP2 = 0 MW Q2 = 0 MW P2 = 50 MW Q2 = 25 MW P2 = 100 MW Q2 = 50 MW P2 = 150 MW Q2 = 75 MW P2 = 250 MW Q2 = 125 MW P2 = 300 MW Q2 = 150 MW P2 = 350 MW Q2 = 175 MW P2 = 400 MW Q2 = 200 MW P2 = 450 MW Q2 = 225 MW Power Generation at Bus 1 and Bus 3: It is observed that as the entire load demand in the system increases due to increase of load demand P2 ( burden ) & A ; Q2 ( burden ) at Bus 2, any necessary existent power for the system balance is supplied by generator ( loose generator ) at Bus 1 sing changeless P3 ( gen ) , so P1 ( gen ) increases. In add-on, any necessary reactive power for the system balance is supplied from Bus 1 every bit good as Bus 3, so both Q1 ( gen ) and Q3 ( gen ) addition. Voltage Angle Difference: It is found that existent power flow addition both from Bus 1 to Bus 2 and from Bus 3 to Bus 2 due to increase of load demand at Bus2. Consequently, both electromotive force angle difference I?1 – I?2 and I?3 – I?2 addition when the power flow P12 and P32 addition. In add-on, when P2 ( burden ) is less than 200 MW, P1gen is comparatively low. Therefore existent power between Bus 3 and Bus 1 flows from Bus 3 to Bus 1 at lower P2 ( burden ) ( less than 200MW ) . On the other manus, while P2 ( burden ) addition more than 200 MW, the existent power flow way alterations ( Bus 1 to Bus 3 ) and the existent power flow additions. Consequently, the electromotive force angle difference I?1 – I?3 alteration from negative to positive and addition. Voltage Magnitude at Bus 2: It is observed that magnitude of coach electromotive force at Bus2 beads due to increase of the load demand at Bus 2.Question 2System Model & A ; Admittance MatrixIn order to build the entree matrix of Powerworld B3 instance, individual stage tantamount circuit can be drawn as below ;omega = R + jx ( r = 0, x = 0.05 )z12 = z21= j0.05 plutonium, y12 = 1/ z12 = 1/j0.05 = -j20 plutonium = y12 z13 = z31= j0.05 plutonium, y13 = 1/ z13 = 1/j0.05 = -j20 plutonium = y31 z23 = z32= j0.05 plutonium, y23 = 1/ z23 = 1/j0.05 = -j20 plutonium = y32 Admittance matrix can be defined as follows ; BUS = Diagonal elements Y ( I, I ) of the entree matrix, called as the self-admittance [ talk slide ] [ 6 ] , are the summing up of all entree connected with BUS I. = y12 + y13 = -j20 – j20 = -j40 plutonium = y21 + y23 = -j20 – j20 = -j40 plutonium = y31 + y32 = -j20 – j20 = -j40 plutonium Off diagonal elements Y ( I, J ) of the entree matrix, called as the common entree [ talk slide ] [ 6 ] , are negative entree between BUS I and BUS J. = – y12 = – ( -j20 ) = j20 plutonium = – y13 = – ( -j20 ) = j20 plutonium = – y21 = – ( -j20 ) = j20 plutonium = – y23 = – ( -j20 ) = j20 plutonium = – y31 = – ( -j20 ) = j20 plutonium = – y32 = – ( -j20 ) = j20 plutonium Therefore, the concluding entree matrix BUS is ; BUS = = The undermentioned figure shows the BUS of the Powerworld B3 instance and it is verified that the deliberate entree matrix is consistent with the consequence of the Powerworld.Power Flow CalculationNodal equation with the entree matrix can be used to cipher electromotive force at each coach if we know all the current ( i.e. entire coevals power and load demand at each BUS ) and eventually the power flow can be calculated consequently. , hence, In this inquiry, nevertheless, simulation consequences of the electromotive force at each coach from the Powerworld are used for the power flow computation as follows ; [ Simulation consequence ]Voltage at each Bus and Voltage DifferenceV1 = 1 a? 0.00A ° plutonium ( BUS1 ) V2 = 1 a? -0.48A ° plutonium ( BUS2 ) V3 = 1 a? 0.48A ° plutonium ( BUS 3 )Voltage difference between BUS 1 and BUS 2V12 = V1 – V2 = 1 a? 0.00A ° – 1 a? -0.48A ° = 3.5 x 10-5 + J 8.38 ten 10-3 = 8.38 ten 10-3 a? 89.76A ° plutonium V21 = V2 – V1 = – V12 = – 3.5 ten 10-5 – J 8.38 ten 10-3 = 8.38 ten 10-3 a? -90.24A ° plutoniumVoltage difference between BUS 3 and BUS 2V32 = V3 – V2 = 1 a? 0.48A ° – 1 a? -0.48A ° = J 16.76 ten 10-3 = 16.76 ten 10-3 a? 90A ° plutonium V23 = V2 – V3 = – V32 = – J 16.76 ten 10-3 = -16,76 x 10-3 a? -90A ° plutoniumVoltage difference between BUS 3 and BUS 1V31 = V3 – V1 = 1 a? 0.48A ° – 1 a? 0.00A ° = – 3.5 ten 10-5 + J 8.38 ten 10-3 = 8.38 ten 10-3 a? 90.24A ° plutonium V13 = V1 – V3 = – V31 = 3.5 ten 10-5 – J 8.38 ten 10-3 = 8.38 ten 10-3 a? -89.76A ° plutoniumLine CurrentCurrent flow from BUS I and BUS J can be calculated by utilizing electromotive force difference and interrelated entree of the line between coachs. [ Iij = yij * ( Vi – Vj ) ]Line current between BUS 1 and BUS 2I12 = y12 x ( V1 – V2 ) = -j20 x 8.38 ten 10-3 a? 89.76A ° = 167.6 ten 10-3 a? -0.24A ° plutonium ( BUS 1 a† Ã¢â‚¬â„¢ BUS 2 ) I21 = y21 x ( V2 – V1 ) = -j20 x 8.38 ten 10-3 a? -90.24A ° = 167.6 ten 10-3 a? -180.24A ° plutonium ( BUS 2 a† Ã¢â‚¬â„¢ BUS 1 )Line current between BUS 3 and BUS 2I32 = y32 x ( V3 – V2 ) = -j20 x 16.76 ten 10-3 a? 90A ° = 335.2 ten 10-3 a? 0.00A ° plutonium ( BUS 3 a† Ã¢â‚¬â„¢ BUS 2 ) I23 = y23 x ( V2 – V3 ) = -j20 x 16.76 ten 10-3 a? -90A ° = 335.2 ten 10-3 a? 180A ° plutonium ( BUS 2 a† Ã¢â‚¬â„¢ BUS 3 )Line current between BUS 3 and BUS 1I31 = y31 x ( V3 – V1 ) = -j20 x 8.38 ten 10-3 a? 90.24A ° = 167.6 ten 10-3 a? 0.24A ° plutonium ( BUS 3 a† Ã¢â‚¬â„¢ BUS 1 ) I13 = y13 x ( V1 – V3 ) = -j20 x 8.38 ten 10-3 a? -89.76A ° = 167.6 ten 10-3 a? -179.76A ° plutonium ( BUS 1 a† Ã¢â‚¬â„¢ BUS 3 )Apparent Power FlowApparent flow from BUS I and BUS J can be calculated by electromotive force at the directing coach and line current. [ Sij = Vi * I*ij ]Apparent Power from BUS 1 to BUS 2S12 = V1* I*12 = 1 a? 0.00A ° ten 167.6 ten 10-3 a? 0.24A ° = 167.6 ten 10-3 a? 0.24A ° = 0.1676 + J 7.02 ten 10-4 plutoniumApparent Power from BUS 2 to BUS 1S21=V2* I*21=1a? -0.48A ° x 167.6 ten 10-3a? 180.24A °=167.6 ten 10-3a? 179.76A ° = -0.1676 + j7.02 x 10-4 plutoniumApparent Power from BUS 3 to BUS 2S32 = V3* I*32 = 1 a? 0.48A ° ten 335.2 ten 10-3 a? 0.00A ° = 335.2 ten 10-3 a? 0.48A ° = 0.3352 + J 2.81 ten 10-3 plutoniumApparent Power from BUS 2 to BUS 3S23=V2* I*23=1 a? -0.48A ° x 335.2 ten 10-3 a? 180A °= 335.2 ten 10-3 a? 179.76A ° = -0.3352 + J 2.81 ten 10-3 plutoniumApparent Power from BUS 3 to BUS 1S31 = V3* I*31 = 1a? 0.48A ° ten 167.6 ten 10-3a? -0.24A ° = 167.6 x 10-3 a? 0.24A ° = 0.1676 + J 7.02 ten 10-4 plutoniumApparent Power from BUS 1 to BUS 3S13=V1* I*13=1a? 0.00A ° x 167.6 ten 10-3a? 179.76A °= 167.6 ten 10-3a? 179.76A ° = -0.1676 + J 7.02 ten 10-4 plutoniumComparison with simulation consequencesThe unit of the above computation consequences is pu value, so in order to compare the consequences with simulation consequences pu value of current and power flow demand to be converted to existent values by utilizing the undermentioned equation sing Sbase = 100MVA and Vline_base = 345kV. [ 3 ] Sactual = Sbase A- Spu = 100 MVA A- Spu Iactual = Ibase A- Ipu = A- Ipu = A- Ipu = 167.3479 A A- IpuCalculation Result and Simulation ResultFlow way & A ; ValueCalculation ConsequenceSimulation ConsequenceBUS 1 a† Ã¢â‚¬â„¢ BUS 2|S12| 0.1676 A- 100 = 16.76 MVA 16.67 MVA P12 16.76 MW 16.67 MW Q12 0.0702 Mvar 0.07 Mvar |I12| 0.1676 A- 167.3479 = 28.0475 A 27.89 ABUS 3 a† Ã¢â‚¬â„¢ BUS 2|S32| 0.3352 A- 100 = 33.52 MVA 33.33 MVA P32 33.52 MW 33.33 MW Q32 0.281 Mvar 0.28 Mvar |I32| 0.3352 A- 167.3479 = 56.0950 A 55.78 ABUS 3 a† Ã¢â‚¬â„¢ BUS 1|S31| 0.1676 A- 100 = 16.76 MVA 16.67 MVA P31 16.76 MW 16.67 MW Q31 0.0702 Mvar 0.07 Mvar |I31| 0.1676 A- 167.3479 = 28.0475 A 27.89 ABUS 2 a† Ã¢â‚¬â„¢ BUS 1|S21| 0.1676 A- 100 = 16.76 MVA 16.67 MVA P21 -16.76 MW -16.67 MW Q21 0.0702 Mvar 0.07 Mvar |I21| 0.1676 A- 167.3479 = 28.0475 A 27.89 ABUS 2 a† Ã¢â‚¬â„¢ BUS 3|S23| 0.3352 A- 100 = 33.52 MVA 33.33 MVA P23 -33.52 MW -33.33 MW Q23 0.281 Mvar 0.28 Mvar |I23| 0.3352 A- 167.3479 = 56.0950 A 55.78 ABUS 1 a† Ã¢â‚¬â„¢ BUS 3|S13| 0.1676 A- 100 = 16.76 MVA 16.67 MVA P13 -16.76 MW -16.67 MW Q13 0.0702 Mvar 0.07 Mvar |I13| 0.1676 A- 167.3479 = 28.0475 A 27.89 A It is found that computation consequences of current flow and evident power flows ( i.e. 28.0475 A and 56.0950 A/ 33.52 MVA and 16.76MVA ) are about 0.5 % higher than simulation consequence ( i.e. 27.89 A and 55.78 A / 33.33 MVA and 16.67 MVA ) which can be considered somewhat different. Difference of the electromotive force angle at each coach between computation ( 0.48A ° ) and simulation ( 0.4775A ° ) could be the ground for this minor difference.Question 3Admittance Matrix and Nodal EquationEntree between two coachsy12 = y21 = -j8 plutonium y13 = y31 = -j4 plutonium y14 = y41 = -j2.5 plutonium y23 = y32 = -j4 plutonium y24 = y42 = -j5 plutonium y30 = -j0.8 plutonium ( BUS3-Neutral BUS ) y40 = -j0.8 plutonium ( BUS4-Neutral BUS )Admittance MatrixYbus ( Admittance Matrix ) = Diagonal elements Y ( I, I ) of the entree matrix, called as the self-admittance [ 2 ] [ 4 ] , are the summing up of all entree connected with BUS I. = y12 + y13 + y14 = -j8 -j4 – j2.5 = -j14.5 = y21 + y23 + y24 = -j8 -j4 – j5 = -j17 = y30 + y31 + y32 = -j08 -j4 – j4 = -j8.8 = y40 + y41 + y42 = -j0.8 -j2.5 – j5 = -j8.3 Off diagonal elements Y ( I, J ) of the entree matrix, called as the common entree [ 2 ] [ 4 ] , are negative entree between BUS I and BUS J. = – y12 = – ( -j8 ) = j8 plutonium = – y13 = – ( -j4 ) = j4 plutonium = – y14 = – ( -j2.5 ) = j2.5 plutonium = – y21 = – ( -j8 ) = j8 plutonium = – y23 = – ( -j4 ) = j4 plutonium = – y24 = – ( -j5 ) = j5 plutonium = – y31 = – ( -j4 ) = j4 plutonium = – y32 = – ( -j4 ) = j4 plutonium = – y34 = 0 plutonium = – y41 = – ( -j2.5 ) = j2.5 plutonium = – y42 = – ( -j5 ) = j5 plutonium = – y43 = 0 plutonium Therefore, entree matrix Ybus is as follows ;Ybus = =Power Flow AnalysisPower flow disregarding transmittal line electrical capacityNodal EquationCurrent from the impersonal coach to each coach are given and entree matrix ( Ybus ) is calculated above. Therefore, concluding nodal equation is as follows ; Ibus = Ybus * Vbus a†¡Ã¢â‚¬â„¢ Vbus = Y-1bus * Ibus = Ybus a†¡Ã¢â‚¬â„¢ ==Voltage AnalysisVoltage at each coach can be derived from the equation ( Vbus = Y-1bus * Ibus ) and Matlab was used for calculate matrix division. ( Source codification is attached in Appendix-1 ) Vbus == V12 = 0.0034 + J 0.0031 plutonium V13 = -0.0277 – J 0.0257 plutonium V14 = 0.0336 + J 0.0311 plutonium V21 = -0.0034 – J 0.0031 plutonium V23 = -0.0311 – J 0.0288 plutonium V24 = 0.0302 + J 0.0280 plutonium V31 = 0.0277 + J 0.0257 plutonium V32 = 0.0311 + J 0.0288 plutonium V41 = -0.0336 – J 0.0311 plutonium V42 = -0.0302 – J 0.0280 plutoniumCurrent flow in the systemCurrent flow from BUS I and BUS J can be calculated by utilizing electromotive force difference and interrelated entree of the line between coachs. [ Iij = yij * ( Vi – Vj ) ] The computation consequence from Matlab is as follows ; I12 = 0.0249 – J 0.0269 plutonium I13 = -0.1026 + J 0.1108 plutonium I14 = 0.0777 – J 0.0840 plutonium I21 = -0.0249 + J 0.0269 plutonium I23 = -0.1151 + J 0.1243 plutonium I24 = 0.1399 – J 0.1511 I31 = 0.1026 – J 0.1108 plutonium I32 = 0.1151 – J 0.1243 plutonium I34 = 0 plutonium I41 = -0.0777 + J 0.0840 plutonium I42 = -0.1399 + J 0.1511 plutonium I43 = 0 plutoniumPower flow in the systemApparent flow from BUS I and BUS J can be calculated by electromotive force at the directing coach and line current. [ Sij ( plutonium ) = Vi * I*ij = Pij + jQij ] The computation consequence from Matlab is as follows ; S12 = 0.0311 + J 0.0175 plutonium S13 = -0.1283 – J 0.0723 plutonium S14 = 0.0972 + J 0.0548 plutonium S21 = -0.0311 – J 0.0174 plutonium S23 = -0.1438 – J 0.0803 plutonium S24 = 0.1749 + J 0.0977 plutonium S31 = 0.1283 + J 0.0780 plutonium S32 = 0.1438 + J 0.0875 plutonium S34 = 0 plutonium S41 = -0.0972 – J 0.0496 plutonium S42 = -0.1749 – J 0.0892 plutonium S44 = 0 plutoniumAdmittance Matrix sing transmittal line electrical capacityHarmonizing to the direction of the Question 3, power system theoretical account can be drawn by utilizing Iˆ tantamount circuit of the lines with capacitive shunt entree ( yc ) of 0.1 plutonium at each side as shown below.Admittance MatrixContrary to tantamount theoretical account in Question 3-1, the current flow through the capacitance in the transmittal line needs to be considered to happen the entree matrix. Therefore, sing the capacitances the current equation with Kirchhoff ‘s current jurisprudence at each coach is as follows ; [ 2 ] [ 5 ] Bus 1: I1 = I12 + I13 + I14 + Ic12 + Ic13 + Ic14 I1 = y12 ( V1-V2 ) + y13 ( V1-V3 ) + y14 ( V1-V4 ) + yc12V1 + yc13V1 + yc14V1 Bus 2: I2 = I21 + I23 + I24 + Ic21 + Ic23 + Ic24 I2 = y21 ( V2-V1 ) + y23 ( V2-V3 ) + y24 ( V2-V4 ) + yc21V2 + yc23V2 + yc24V2 Bus 3: I3 = I30 + I31 + I32 + Ic31 + Ic32 I3 = y30V3 + y31 ( V3-V1 ) + y32 ( V3-V2 ) + yc31V3 + yc32V3 Bus 4: I4 = I40 + I41 + I42 + Ic41 + Ic42 I4 = y40V4 + y41 ( V4-V1 ) + y42 ( V4-V2 ) + yc41V4 + yc42V4 Equation above can be rearranged to divide and group single merchandises by electromotive force. Bus 1: I1 = ( y12 + y13 + y14 + yc12 + yc13+ yc14 ) V1 – y12V2 – y13V3 – y14V4 = Y11V1 + Y12V2 + Y13V3 + Y14V4 Bus 2: I2 = ( y21 + y23 + y24 + yc21 + yc23+ yc24 ) V2- y21V1 – y23V3 – y24V4 = Y21V1 + Y22V2 + Y23V3 + Y24V4 Bus 3: I3 = ( y30 + y31 + y32 + yc31+ yc32 ) V3 – y31V1 – y32V2 = Y31V1 + Y32V2 + Y33V3 + Y34V4 Bus 4: I4 = ( y40 + y41 + y42 + yc41+ yc42 ) V4 – y41V1 – y42V2 = Y41V1 + Y42V2 + Y43V3 + Y44V4 Finally, Diagonal elements Y ( I, I ) and off diagonal elements Y ( I, J ) of the entree matrix are calculated as follows ; = y12 + y13 + y14 + yc12 + yc13+ yc14 = -j8 -j4 – j2.5 + j0.1 + j0.1 +0.1j = -j14.2 plutonium = y21 + y23 + y24 + yc21 + yc23+ yc24 = -j8 -j4 – j5 + j0.1 + j0.1 +0.1j = -j16.7 plutonium = y30 + y31 + y32 + yc31+ yc32 = -j08 -j4 – j4 + j0.1 +0.1j = -j8.6 plutonium = y40 + y41 + y42 + yc41+ yc42 = -j0.8 -j2.5 – j5 + j0.1 +0.1j = -j8.1 plutonium = – y12 = – ( -j8 ) = j8 plutonium = – y13 = – ( -j4 ) = j4 plutonium = – y14 = – ( -j2.5 ) = j2.5 plutonium = – y21 = – ( -j8 ) = j8 plutonium = – y23 = – ( -j4 ) = j4 plutonium = – y24 = – ( -j5 ) = j5 plutonium = – y31 = – ( -j4 ) = j4 plutonium = – y32 = – ( -j4 ) = j4 plutonium = – y34 = 0 plutonium = – y41 = – ( -j2.5 ) = j2.5 plutonium = – y42 = – ( -j5 ) = j5 plutonium = – y43 = 0 plutonium Therefore, entree matrix Ybus is as follows ;Ybus = =Annex-1: Matlab beginning codification and Calculation consequences with MatlabMatlab Source Code% define ego entree and common entree by utilizing admittace between % the coachs ( y12=y21=-j8, y13=y31=-j4, y14=y41=-j2.5, y23=y32=-j4, % y24=y42=-j5, y34=0, y43=0, y30=-j0.8, y40=-j0.8 y12=-8i ; y21=-8i ; y13=-4i ; y31=-4i ; y14=-2.5i ; y41=-2.5i ; y23=-4i ; y32=-4i ; y24=-5i ; y42=-5i ; y34=0 ; y43=0 ; y30=-0.8i ; y40=-0.8i ; Y11=-8i-4i-2.5i ; Y12=8i ; Y13=4i ; Y14=2.5i ; Y21=8i ; Y22=-8i-4i-5i ; Y23=4i ; Y24=5i ; Y31=4i ; Y32=4i ; Y33=-0.8i-4i-4i ; Y34=0 ; Y41=2.5i ; Y42=5i ; Y43=0 ; Y44=-5i-2.5i-0.8i ; % Bus 3 and Bus 4 is non connected, so admittance Y34 and Y43 are equal to zero % define the 4Ãâ€"4 entree matrix ( Ybus ) Ybus= [ Y11 Y12 Y13 Y14 ; Y21 Y22 Y23 Y24 ; Y31 Y32 Y33 Y34 ; Y41 Y42 Y43 Y44 ] ; % In order to specify the nodal equation ( I = Ybus*V ) , the given I needs to specify. i1=0 ; i2=0 ; i3=-i ; i4=-0.4808-0.4808i ; Ibus= [ i1 ; i2 ; i3 ; i4 ] ; % Each coach electromotive force can be calculated by utilizing matrix division ( V= YbusI ) Vbus=YbusIbus ; v1=Vbus ( 1,1 ) ; v2=Vbus ( 2,1 ) ; v3=Vbus ( 3,1 ) ; v4=Vbus ( 4,1 ) ; % Calculate electromotive force difference between coachs v12=v1-v2 ; v13=v1-v3 ; v14=v1-v4 ; v21=v2-v1 ; v23=v2-v3 ; v24=v2-v4 ; v31=v3-v1 ; v32=v3-v2 ; v34=v3-v4 ; v41=v4-v1 ; v42=v4-v2 ; v43=v4-v3 ; % current flow between coachs can be calculated by i12 = y12* ( v1-v2 ) i12=y12*v12 ; i13=y13*v13 ; i14=y14*v14 ; i21=y21*v21 ; i23=y23*v23 ; i24=y24*v24 ; i31=y31*v31 ; i32=y32*v32 ; i34=y34*v34 ; i41=y41*v41 ; i42=y42*v42 ; i43=y43*v43 ; % evident power can be calculated by s12 = v1 * conj ( i12 ) s12=v1*conj ( i12 ) ; s13=v1*conj ( i13 ) ; s14=v1*conj ( i14 ) ; s21=v2*conj ( i21 ) ; s23=v2*conj ( i23 ) ; s24=v2*conj ( i24 ) ; s31=v3*conj ( i31 ) ; s32=v3*conj ( i32 ) ; s34=v3*conj ( i34 ) ; s41=v4*conj ( i41 ) ; s42=v4*conj ( i42 ) ; s43=v4*conj ( i43 ) ; % Real power and Reactive power can be derived by following p12=real ( s12 ) ; p13=real ( s13 ) ; p14=real ( s14 ) ; q12=imag ( s12 ) ; q13=imag ( s13 ) ; q14=imag ( s14 ) ; p21=real ( s21 ) ; p23=real ( s23 ) ; p24=real ( s24 ) ; q21=imag ( s21 ) ; q23=imag ( s23 ) ; q24=imag ( s24 ) ; p31=real ( s31 ) ; p32=real ( s32 ) ; p34=real ( s34 ) ; q31=imag ( s31 ) ; q32=real ( s32 ) ; q34=imag ( s34 ) ; p41=real ( s41 ) ; p42=real ( s42 ) ; p43=real ( s43 ) ; q41=imag ( s41 ) ; q42=real ( s42 ) ; q43=imag ( s43 ) ; % terminalMatlab Calculation Results

A History of Soviet and American Interests essays

A History of Soviet and American Interests essays In the system of international politics almost everything is strategic. Even the most basic analysis of international historical events can seem like decoding the rules to a complex board game. This is most evident when analyzing grand wars, such as WWI and WWII. Although states and state leaders may attempt to rewrite world order in the wake of such disruptive events, history has proven these efforts extraneous. It is highly argued, for example, that restructuring efforts following WWI only contributed to WWII. So why is it that the unplanned system following WWII lasted twice as long as the carefully planned efforts after the First World War? This period of history, formally known as the Cold War, is sometimes referred to as the Long Peace because despite four decades of hostility, the Cold War never manifested into direct military confrontation. According to John Lewis Gaddis, war was prevented as a result of the bipolar configuration of the international system at the time, geographical advantages, advanced technology, shifting ideologies, and game theory. In order to understand why the international system remained stable in the post-WWII world, it is necessary to understand Systems Theory. This theory provides the criteria for stable or unstable international configurations. In this way, stability is characterized by the structure of states and not their behavior, even though both play a role in stabilizing state relations. In systems theory, international politics can take on a bipolar or multipolar configuration. Bipolarity happens when two states dominate the international political system. Post- WWII settlements arbitrarily divided the sovereign nations that could have rivaled the U.S. or Soviet Union in population or economic might. This created an avenue for the domination of the American and Soviet spheres of inf luence the first true polarization in modern history was created (Gaddis 45). Unl...