Electricity and Magnetism Essay Example for Free

Electricity and Magnetism Essay In this investigation I will be burning a range of alcohol in a method known as calorimetry. This will allow me to see the amount of energy produced by each alcohol, and then look at the structure of the alcohol and investigate why an amount of energy is produced for each alcohol. Before I go on with the experiment there are several factors that must be first understood. What is an alcohol? The definition of an alcohol as taken from Richard Harwoods Chemistry textbook is a series of organic compounds containing the functional group OH. The OH group, called a hydroxyl group is what defines the compound as an alcohol. The alcohol compounds are very similar to the alkanes, however alcohols contain one oxygen atom, creating the hydroxyl group, and making the alkane an alkanol (alcohol). As the hydroxyl group makes the compound different to an alkane, the hydroxyl group is seen to be functional. The formula for alcohol is: In this investigation I will look at the first five alcohols. These are methanol, ethanol, propan-1-ol, butan-1-ol and pentan-1-ol. The classification of alcohols is similar to the classification of alkanes, where the name refers to the number of carbon atoms i. e.meth- one carbon atom, eth-=two carbon atoms, prop-=3 carbon atoms, but-=four carbon atoms, pent-=five carbon atoms and so on. The carbon atom that the hydroxyl group is bonded to may classify the alcohol further. If it is joined to the end carbon atom it is classified as -1-ol, if it is bonded to the second carbon atom it is classified as -2-ol and so on. The five alcohols I am testing are shown with formulas, and atomic structure diagrams below: Name RMM Formula Structure Boiling Point Methanol. What is combustion? The definition of combustion as taken from Richard Harwoods Chemistry textbook is a chemical reaction in which a substance reacts with oxygen the reaction is exothermic. Burning is a combustion reaction that produces a flame. The reactions that will take place in our investigation will be combustion reactions, as we burn the alcohol in calorimetry. The combustion of an alcohol produces carbon dioxide and water. This is represented by the following equation: ALCOHOL + OXYGEN i CARBON DIOXIDE + WATER e. g. The combustion of methanol produces carbon dioxide and water: METHANOL + OXYGEN i CARBON DIOXIDE + WATER 2CH3OH + 3O2 i 2CO2 + 4H2O What is calorimetry? Calorimetry is defined as the science of heat. It may be used in chemistry to measure the heat energy exchanged for a substance during a reaction, by allowing the substance reacting to heat a measured mass of liquid. The temperature change of this liquid is measured and recorded, and the following equation may be used to calculate the heat energy transferred to the liquid: Heat change = MC? T (where M = mass of liquid, C = critical temperature, and T = temperature change). The critical temperature is the proportionality between heat energy applied and the subsequent temperature rise. E. g. for 1g of water a temperature rise of 1 i C requires 4. 8 joules of heat energy. Using calorimetry may be used to measure both reactions in solution, and reactions not in solution. Wet reactions or those in solution are more accurate than those not in solution, as the actual energy release of the reaction is being measured directly as a thermometer is placed in the solution. For dry reactions, (a reaction not in solution such as the burning of an alcohol) the reaction must be used to heat a mass of water and the temperature increase measured from the water. As the combustion reaction must transfer its heat energy from the reaction to the water, calorimetry for dry reactions can be inaccurate due to heat loss. HYPOTHESIS I believe that the increased complexity of a molecule and the energy released by it are proportional. I found this idea upon the theory of breaking and making bonds. In a molecule, bonds hold atoms together. When these bonds are formed energy is given out to the surroundings (exothermic), and when these bonds are broken energy is absorbed from the surroundings (endothermic). This may be seen by the alcohols boiling points: Name Boiling Point Methanol 65 Ethanol 78 Propan-1-ol 97 Butan-1-ol 117 Pentan-1-ol 137 Using this theory of breaking/making bonds, we can say that there will be a greater amount of energy released from the combustion of a more complex alcohol, as more bonds will be formed. When more bonds are formed, more heat energy is released. The hypothesis may be seen on the flowing graph: TESTING To investigate the relationship between the structure and heat provided by combustion of a range of alcohols, we will use the method of calorimetry. We will do this by burning an alcohol, and allowing it to heat a mass of water. Measuring the temperature rise of this mass of water we can use the formula to find the heat energy released during the reaction. Apparatus == 1 can == 1 measuring cylinder == Water == 1 thermometer == 1 stand == 1 clamp == 1 boss == 1 heat mat == 1 alcohol burner/candle == 1 balance == Draught shields Fig1: Experiment setup. Method. Safety must be ensured at all times, as highly flammable substances are being burned here. All students must wear eye protection during the experiment, stay standing during the experiment, and be aware of the flames around them. 1. The apparatus is set up as shown in fig 1. 2. The alcohol candle is weighed on the balance, and the mass recorded. 3. The mass of water in the can measured and recorded 4. The height between the candlewick, and the base of the can must be set and recorded, measuring the distance with a ruler. 5. The temperature of the water is measured and recorded using the thermometer. 6. The candle is lit and the thermometer monitored until the water has risen by a given temperature. 7. The candle is blown out, the burnt wax on the base removed of the candle, and the candle is weighed. The data produced by this method will then be calculated by the formulas: ? H (j) = mass of water (g) X specific heat capacity of water (S. H. C) X temperature rise (? C) The result of this is found per gram of alcohol burnt by dividing it by the change of mass ? H per gram of alcohol burnt (J/g) = ? H (? C) / change in mass (g) The results of this are the found per mole of alcohol burnt. ? H per gram of alcohol burnt per mole (J/mole) = ? H per gram of alcohol burnt (g) X RMM of alcohol Variables There are two types of variables: dependant variables and independent variables. Dependant variables are those that are kept constant at a set value, so that they cannot cause variation in the results, by changing throughout the experiments. There should only be one independent variable in a test, this is what is being tested. Therefore the dependent variables for this investigation are as follows (see fig 2): 1. The distance of the wick from the base of the can The distance of the flame from the base of the can may vary the results greatly, as it will decide how much of the flames heat is transferred directly to the can and to the water. The experiment should allow the flame to contact the can at its hottest point. The most accurate and safe way to measure flame distance from the can is to measure the distance of the wick. For the purposes of our experiment this should be accurate enough. Measuring the distance from the wick and the base of the can each time will control this variable. 2. The mass of water heated. The volume of water can vary the experiment greatly, as the greater the mass of water used, the more alcohol will be combusted to reach the target temperature. Also when a greater mass of water is used, the test will go on for longer, and thus a greater amount of energy will be transferred to the surroundings as wasted energy. Measuring the amount of water used each time will control this variable. 3. The type of can used. The type metal the can is made from is extremely important to the experiment, as different metals conduct heat with different efficiency and at different rates. Therefore the type of can used will affect the heat transfer to the water. Using the same can each time will control this variable. 4. The heat increase The heat increase can vary the experiment, as the greater the energy increase, the greater the mass of alcohol burnt, and the greater amount of energy transferred to the surroundings as waste. The start temperature may affect the results, as the higher the temperature the equipment reaches, the more heat will be wasted in the surroundings, according to the theory that heat energy moves from areas of high heat energy to areas of low energy. By allowing the equipment to cool after each experiment to a set temperature this variable may be controlled. 5. Stirring of the water Stirring the water will affect the results, as it means that the whole mass accounted for in the calculation will be being heated, not just one volume of water. The type of stirring in terms of frequency and power may affect the results also, and must be kept constant. To control this variable, the same type of stirring will be used each time. The only variable in the experiment will be the changing type of alcohol being combusted, as this is what is being investigated.

The Color Purple - Character Analysis of Celie :: Alice Walker

In the color purple, we can see how Celie develops an identity for herself throughout the novel. At first we can appreciate how Celie does not longer believe in herself and looses all trust she had on herself. When Nettie gets older, about 12 years old, their father Fonso tries to get to Nettie, but Celie protects her and lets Fonso rape her instead of him raping Nettie. This at the beginning shows that Celie has enough strength to take decisions that will affect other people, however, this strength starts to disappear as the story continues. Celie is at most 14 when Mr.____ marries her. She does no talk about "making love" but instead refers to sex as "he gets on top of me and does his business". That shows how little Celie believes on herself as she thinks that Mr.____ could do sex without her. Then the story changes direction and the discrimination against Celie starts to be present. When Mr.____ beats her she does not try to fight back, but instead she remains passive to what is happening. On the other hand, Celie continues to work as the perfecto maid of the children and keeps the house perfectly clean and tidy. This shows that indirectly she does not like being beaten (obviously) but she does not say anything because of the fear she feels for men, she instead reacts to it in a way which shows the big heart Celie has. When Shug comes to Mr.____Â ´s house Celie starts to feel something that she had never felt. She start to feel emotions for Shug. As the relationship between Celie and Shug develops, Shug shows Celie that life, freedom and having an identity should be the best present for a black woman which is condemned and trapped for ever after on the cruel and racist society of those times. To begin with, Celie obtains freedom as she escapes from Mr.____ with Shug, then, she realises that life can be much better when you do not depend on any one else, and so Celie does no longer depend on any one but on herself.

Gender roles Essay

Gender roles refer to the set of social, attitudinal and behavioral roles, norms and expectations that, within a definite culture, are also formally or informally required or widely measured to be socially appropriate for persons of a precise gender identity. They are constructed for a variety of genders in order to channelize their energies towards some socially intended goals, which are either frequently shared or affixed from many of the experimental differences in behaviors, attitudes and personalities, amid various genders, come naturally, a lot of of these characteristics are, either in part or wholly, socially constructed, therefore, a product of socialization experiences. Qualities of a meticulous gender identity do not need to be imposed through rules and norms, while artificial roles have to be ‘enforced’ on people throughout some kind of psycho-social mechanism. Gender roles of a particular sex may not always be in accordance with the normal or biological traits of that gender, and they may turn out to be too strict or constricting so as to cause in the oppression of that gender. This is because of the potential of the gender roles to manage the behavior of people that these roles have been tremendously politicized and manipulated with the ruling forces, for several millenniums now, resulting in severe oppression of every gender of humans. Gender has numerous valid definitions, but its here in reference to an individual’s inside sex or psychological sense of being a male or female irrespective of one’s (outer) sex identity as determining one’s sexual organs. We find three major genders: masculine (inner male identity), feminine (inner female identity) and neutral (a balance of inner male and female identity). In conclusion gender roles of women have been enforced on them through force, and have thus been extra visible. Men gender roles are difficult to enforce but are indirectly enforced. References Bem, S. L. (1981). Gender schema theory: A cognitive account of sex typing. Psychological Review 352-365. Connell, Robert William: Gender and Power, Cambridge: University Press 1987.

Studying The Bankruptcy Of Orange County Finance Essay - Free Essay Example

Sample details Pages: 10 Words: 2920 Downloads: 2 Date added: 2017/06/26 Category Finance Essay Type Research paper Did you like this example? The case of bankruptcy of Orange County in 1994 emphasize the importance of using duration and Value at risk (VAR) to assess portfolio risk and avoid future bankruptcy. Duration and VAR analysis provide deeper understanding about the underlying risk of the Orange County Investment Pool which was heavily leveraged and interest-pledged through reverse repurchase agreements and other derivatives in the pool. Some VAR estimation, including historical simulation method, delta-normal method and Monte Carlo simulation will be used to calculate worst possible loss. Don’t waste time! Our writers will create an original "Studying The Bankruptcy Of Orange County Finance Essay" essay for you Create order The EWMA will be used to provide more accurate estimation of the volatility to improve the accuracy of VAR estimation. Background: On Dec 6, 1994, Orange County declared bankruptcy after suffering losses of around $1.6billion from a wrong way bet on interest rates 7.5 billion investment pool. This pool was intended to gain some returns from the investing the money which is raised from taxes and other government incomes. It was implemented a bet that the interest would decline or stay low by Citron (the portfolio manager). Because of the steadily declining interest rates from 1989 to 1992, the portfolio performed extremely well before 1994 and earned millions of above average profit. However, in 1994, the government suddenly declared policies which included raise the interest rates from 3.45% to 7.14% to prevent high inflation and overheating economy. This increase in interest rate caused the portfolio suffer 1.6 billion loss and further lead the bankruptcy of Orange County. Section 1: The heavy leveraged and interest-pledged portfolio In order to sustain above average returns, several investment tools are used by Citron to leverage the $7.5 billion funds into $20.5billion investment. In detail, reverse repurchase agreements allow Citron to use the securities which had already purchased as collateral on further borrowing and then reinvested the cash into new securities (Jameson, 2001). Besides the heavily leveraged risk, the portfolio also encounters significant risk from the unexpected interest movement. Firstly, these repurchase agreements values significantly depend on the change in interest rate. In detail, its value decrease as the interest rate increase and increase as the interest rate decrease (P=). Secondly, $2.8 billion of derivatives, including inverse floating-rate notes, dual index notes, floating-rate notes, index-amortizing notes and collateralized mortgage obligations, are used to increase the portfolio bet on the term structure of the interest rate (Jorion, 2009). Thirdly, median term maturities which had higher yields (5.2%) than the short term investments (3%) were used to increase the return of the portfolio (Jorion, 2009). However, by using longer term maturities, the portfolios sensitivity to interest change will significantly increase. Clearly, by doing these, the portfolios value will be significantly impacted by the movement of the interest. Section 2: Duration of the portfolio and its application Duration of the portfolio Hull (2009) defines the duration as a measure of how long, on average, the holder of the instrument has to wait before receiving cash payments. It measures sensitivity of price changes with changes in interest rates. Duration can be calculated by weighting average (the weight is the proportion of portfolios total present value of cash flow received at time t) of the times. In this case, the portfolio was heavily bet on the interest, therefore, duration might be a good measure for the portfolio. In the $7.5 billion portfolio, median term maturities (5 years), rather than short term maturities (1-3 years), were used to increase the return. By doing this, the duration of the portfolio significant increased. In other words, the portfolio exposed higher risk of interest rate movements. In December 1994, the average duration of the securities in the portfolio was 2.74 years. It means 1% change in interest would cause 2.74% change in portfolios prices. M oreover, Citron leveraged $7.5 billion equity into a $20.5 billion portfolio. This means that a 2.73 leverage ratio (20.5/7.5). In other words, for every dollar of the pool invested, the pool borrowed extra $1.73. For a leveraged portfolio, the effective portfolio duration = ordinary duration * leverage ratio. Thus, the effective portfolio duration of the portfolio is 7.4 (2.74*2.7). Estimation by using duration The response of portfolio prices to change in interest rate: In 1994, the interest rates went up by about 3.5 ( and the 5 years bond yield was 5%, therefore, the loss of the portfolio equals 1.85 (7.5*7.4*3.5%/1.05) which is slightly larger than the actual loss of 1.64 billion. This slightly difference between the loss estimated by duration and the actual loss might be caused by that the duration applies to only small changes in interest rate. As a first order approximation, duration cannot capture the information that two bonds with same duration can have different change in price for large change in interest rate (different convexity). So, convexity (second order approximation) which can capture this information should be added into the estimating model. Through adding this (convexity factor), the estimated loss will slightly less than before, and will more close to the actual loss (1.64 billion). Thus, duration seems to have the ability to accurate measure the portfolios sensitivity to interest rate change. Section 3: Value at risk (VAR) Value at risk (VAR) In order to estimate the underlying risk of the portfolio, VAR which measures the worst expected loss over a given horizon under normal market conditions at a given confidence level could be used (Jorion, 2001). Because the portfolio was heavily bet on the interest rate, its return and risk are significantly depending on the change of interest rate. In other words, the change of interest yield multiplies the modified duration and portfolio value could be used as an approximation of the change of portfolios value. Thus, the change of interest yield could be used in the 3 simulation methods as the only factor that contribute the change of portfolio value. Non-parametric approach (no need to identify variance-covariance matrix) Historical simulation approach The historical simulation accounts for non-linearity, income payments, and even time decay effects through using marking-to-market the whole portfolio over a large number of realizations of underlyi ng random variables. VAR is calculated from the percentiles of the full distribution of payoffs (Jorion, 2001). By using actual price, the method captures Greek risk (gamma, vega risk etc.) and corrections of securities (already exist in the real historical data) in the portfolio, and it does not rely on some specific assumption, such as the underlying stochastic structure of the market (the pre-requests of estimating volatility and mean). Moreover, it can account for fat tails distribution besides normal distributions (Jorion, 2001). (Figure 1) The root-T approach will be used to transfer the monthly VAR to yearly VAR in all the 3 approaches. Its success significantly relies on the some specific assumptions, including the monthly yield changes of the portfolio are identically and independently distributed (iid distribution) and the return has a constant variance (Cuthbertson and Nitzsche, 2001). However, in the real world, stock returns always has time varying variance and th ere are some autocorrelation factors exist (thus, not independent). Therefore, as the T increase, the error of the transformation will significantly increase. The VAR will be calculated through sorting the monthly yield change and picking the worst daily yield change at 5% percentile (see details in CD). However, in this case, the increase in yield will cause decrease in portfolio return, therefore, the worst daily yield change should be picked at the right hand side of the histogram (see figure 1). The VAR equals 1.24 billion annually (0.36 billion monthly) which is less than the actual loss (1.64 billion). This inaccuracy might be caused by the problems exist in historical simulation method. Firstly, the success of the method significantly relies on the assumption that the past price can represent the future price information. However, the assumption is not realistic to some extent because of the existence of market efficient. Secondly, simple historical simulation method may m iss the information of temporarily elevated volatility, such as structural breaks and extreme value (Butler and Schachter, 1996). In this case, the historical simulation method cannot capture the extreme value (1.64 billion loss) which is caused by 6 suddenly decreases of interest rate. Parametric approach (need to need to identify variance-covariance matrix) Delta normal approach The delta normal method is particularly simple approach to implement. It takes account simple variance-covariance matrix and then forecast the total variance of the portfolio (volatility). Then, The VAR can be calculated through the formula: VAR = MD*Portfolio Value*=7.4*7.5*0.4%*1.65/(1.005)=0.35 billion (monthly) = 1.21 billion (annually). Delta normal method is slightly less accurate than the historical in the case. This might caused by that the change in yield does is a fat tail distribution (Kurtosis =6.9, Skewness = -0.44) rather than a normal distribution (Kurtosis =6.9, Skewness = -0.44 ). Thus, the model based on the normal distribution will underestimate the proportion of outliers and hence the value at risk (Jorion, 2001). In addition, the portfolio contains a lot of derivatives instrument. This will cause the method inadequately measures the risk of nonlinearity. Monte Carlo simulation (MCS) (the theoretical most powerful method) Unlike historical simulation, through specifying and stimulating a stochastic process for financial variables, Monte Carlo simulation covers a wide range of financial variables (volatility and stochastic variables) and fully captures correlations of securities (unlike HS, need to define the matrix) in the portfolio (Jorion, 2001). It does not only account for a wide range of risks, such as nonlinear price, volatility and model risks (the same as historical simulation), but also incorporate time variation of volatility (structural breaks and extreme values), and fat tails. Moreover, it can capture the structure changes in the port folio as the time pass (Jorion, 2001). In theoretical way, MCS should be the best method in estimating VAR. The MCS VAR is about 0.295 monthly, through using the root-T rule, the annually VAR is about 1 billion (see detail calculation in CD). There are also some limitations of Monte Carlo simulation cause the estimated error between the estimated loss and actual loss. Its success significantly relies on the specific pricing model for underlying assets and stochastic processes for the underlying risk factors. In this case, the pricing formula is Brownian approach without drift may not accurately capture the actual value change of the portfolio. This might be one possible reason that the estimated loss is not equal to the actual loss. Moreover, the problems may exist in the sample used to derivate the underlying risk factors. For example, MCS will generate less accurate estimates then delta normal method when the risk factors are jointly normal and all payoffs are liner (Cuthbertso n and Nitzsche, 2001). Why MCS (theoretical best method) shows the worst estimation in this case MCS seems to have the least accurate estimation (more closer to the actual loss) in this case. This might be caused by the portfolio used in MCS are treated as one asset which is only impacted by the interest yield. Three factors, including the correlation between all the securities in the portfolio, the underlying risk factors of these securities and the different price formula should be used for each security, are ignored in the powerful approach (Tardivo, 2002). On the other hand, compared with the MCS, historical simulation does not need to define the correlation matrix, because the data has already captured the information. In addition, underlying risk factors also contains in the actual data. Thus, in the case with limited information, historical simulation provides more accurate estimation. Section 4: EWMA In realistic world, the variance of the time series is varying overtime. Thus, the simple unconditional variance (simple variance/standard deviation) may not provide unbiased estimation of the volatility. This will further result in inaccuracy estimation of the VAR. in the case, In the case, the simple variance (volatility) are calculating through assigning the same weight on all observations during Jan 1953 and Dec 1994. This may lead to biased forecasts of VAR because the Fed dramatically increased/decreased the interest rate during this time period. In order to improve the accuracy of estimating VAR, Exponentially weighted moving average (EWMA) will be used to provide more accurate estimation to the volatility at a specific time (conditional standard deviation) (Cuthbertson and Nitzsche, 2001). EWMA method allows more recent observations to have stronger impact on the forecast of volatility than the old observations. In practical way, the recent data are given more weights th an the old data. By applying this model, volatility in practice will be more impacted by recent events and the impacts on volatility will decline as time pass (smaller weights apple to the event) (Brooks, 2002). Through applying the EWMA model, the monthly standard deviation for the six months before December 1994 is 0.348%. The next 6 months volatility could be forecasted through using the formula:. In addition, the actual monthly volatility could use the change in yield as approximation. According to RiskMetrics, the optimalshould be 0.97 (Brock, 2002). ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ Forecast volatility (%) Actual volatility (%) Range of the possible volatility at 5% confidence level ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ Volatility at june 1994 0.35 ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ Left side (-1.65) Right side (1.65) Forecasted volatility ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ Jul-94 0.35 -0.26 -0.57 0.57 Aug-94 0.34 0.08 -0.56 0.56 Sep-94 0.35 0.47 -0.57 0.57 Oct-94 0.34 0.20 -0.56 0.56 Nov-94 0.34 0.31 -0.56 0.56 Dec-94 0.34 0.04 -0.55 0.55 Generally, the EWMA approach does not fully capture abnormal volatility change in 1994. In detail, the actual volatility change more volatile than the forecast one (table 1). The inaccuracy involve in estimating the volatility may result in that the calculated VAR is significantly different from the actual possible loss of the portfolio (table 2). If the forecast volatility is used to calculate VAR, manager should aware that the calculated VAR is only an approximation and it cannot capture all the volatility change information. For example, in this case, the actual volatility in Sep-94 is significantly larger than the forecast one. This may cause manager to underestimate the risk in the time period and then holding the portfolio unchanged as before. It is also support by Mahoney (1996) who empirically support that the EWMA volatility has inaccuracy problems. Table 1: ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ Forecast volatility (%) Actual volatility (%) Left side (-1.65) Right side (1.65) Volatility at June 1994 0.35 ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ Forecasted volatility ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ Jul-94 0.35 -0.26 -0.57 0.57 Aug-94 0.34 0.08 -0.56 0.56 Sep-94 0.35 0.47 -0.57 0.57 Oct-94 0.34 0.20 -0.56 0.56 Nov-94 0.34 0.31 -0.56 0.56 Dec-94 0.34 0.04 -0.55 0.55 On the other hand, VAR calculated based on EWMA volatility can still be used as a benchmark to assess the portfolios risk. All of the actual volatility is in the boundary of the forecast volatilitys 5% tail cut off (on both sides *1.65). That is to say, although there are significant differences between the forecast and the actual volatility in this case, portfolio manager may still not underestimate the underlying risk at 5% confidence level (normal distribution). In addition, if better models are used, including GARCH, EGARCH, and GJR , the VAR can provide more precise estimation of the worst possible loss. Table 2: ÃÆ' £Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬Ãƒ ¢Ã¢â‚¬Å¡Ã‚ ¬ Forecasted VAR(*-1.65) monthly Actual VAR monthly Forecasted VAR annually Actual VAR annually Jul-94 -0.302 0.227 -1.045 0.786 Aug-94 -0.297 -0.070 -1.030 -0.242 Sep-94 -0.301 -0.410 -1.044 -1.420 Oct-94 -0.298 -0.174 -1.034 -0.604 Nov-94 -0.298 -0.270 -1.031 -0.937 Dec-94 -0.293 -0.035 -1.015 -0.121 Section 5: Backtest EWMA model In order to test whether VAR can be used as s a benchmark to assess the portfolios risk, the backtest should be used to test whether EWMA can capture the actual change in interest yield at the 5% left tail cut off level (normal distribution). Practically, if all of the actual changes in interest yield are within the forecast volatilities boundary (the forecast volatility multiply 1.65 at right hand side and -1.65 at the left hand side), the EWMA model can be considered as providing accurate estimation at 5% confidence level. According to figure 2, there are 4 outliers (Aug-89, Jan-92, Feb-94 and Mar-94) are outside the forecast. This will cause manager to over/under estimate the underlying risk of the portfolio. Figure 2: forecast volatilities boundary and actual change in interest yield Section 6: Whether the portfolio should be liquidated in December 1994 Miller and Ross (1997) recommend that the portfolio should not be liquidated until the maturity of the structural notes. This is because after the Orange County bankruptcy, the interest rate fell from 7.8% to 5.25% during Dec 1994 to Dec 1995. If it did not announce the bankruptcy, this decrease in interest rate could help the County to recover 7.4*7.5*2.55%/1.05= 1.32 billion losses. However, the problem is that in Dec 1994, how the managers would know that there would be a decrease in interest in 1995. Jorion (1997) suggest that because it is impossible to predict suddenly interest rate decrease, holding the assets in order to recover value in the next years is speculative and risky. Given this change in yield is a normal distribution, the probability of 2.55% decrease in interest can be calculated through P(=P(-6.223). According to the normal statics table, the probability of such large decrease in interest is less than 1%. Thus, the rational managers would not expect suddenl y large decrease in interest rate. In order to minimize to further loss, it is reasonable to liquidate the portfolio on Dec 1994. In addition, as the portfolio is interest pledged, some interest futures, such as the T-bond futures, could be shorted to hedge the portfolio in Dec 1993. Long cap could also a good choice to generate profit when interest rate exceeds the strike rate. This could partially compensate the massive loss. Conclusion The orange countys heavy leveraged and interest-pledged portfolio suffer massive loss in 1994 because of the suddenly increase of interest rate. Through examining this case study, the Duration and VAR are important measurement of risk to avoid future bankruptcy. Compare the duration estimated loss with the actual loss, Duration (plus convexity) of the portfolio seems to have the ability to accurately measure the portfolios sensitivity to the change in interest rate. In addition, all of the VARs calculated through three approaches, including historical simulation, delta normal, and MCS, are less than the actual loss. The theoretical best approaches (MCS) does not provide the most accurate estimation because of ignorance of some important factors, such as the correlation between all the securities in the portfolio, the underlying risk factors of these securities and the different price formula should be used for each security. The backtest of EWMA (4 outliers) suggest that there are some risk in using VAR to measure the worst possible loss in the real world.