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Variable valve timing in ICE engineVariable valve mechanism for ICE engineAutomotive engine variable valve timingShort Quiz in Math 7 solves linear equation or inequality in one variable involving absolute valueMultiple choice questions on the following : Context: Park A (Sharjah): 120, 122, 121, 119, 118 Park B (Dubai): 90, 130, 150, 110, 120 Both parks have the same mean number of visitors. What is the correct value of the mean? A. 118 B. 119 C. 120 D. 121 Without calculating variance fully, which park is most likely to have the larger population variance? A. Park A, because its values are closer to the mean B. Park B, because its values are more spread out C. Park A, because it has a smaller range D. Park B, because it has more visitors overall The mean number of visitors for both parks is 120. Which value contributes the MOST to the variance of Park B? A. 110 B. 120 C. 130 D. 150 What is the squared deviation of 90 visitors from the mean? A. 30 B. 60 C. 900 D. 30² Why does the value 150 increase the variance more than 130? A. It increases the mean B. Its deviation from the mean is larger before squaring C. It occurs later in the data set D. Variance depends only on the largest value Which statement best describes the visitor pattern in Park A (Sharjah)? A. Highly variable with extreme values B. Random and unpredictable C. Consistent with small deviations from the mean D. Skewed due to an outlier Both parks have the same mean. Why is variance still needed to compare them? A. The mean does not show how spread out the data is B. Variance changes the data values C. Variance affects the mean D. Variance replaces the mean A municipality wants to plan staffing using predictable visitor numbers. Based on variance, which park should be chosen? A. Park B, because it has higher peaks B. Park A, because it has lower variance C. Park B, because it has more visitors D. Either park, since the means are equal A student says: “Park B has greater variation because it has the largest value.” Which response best evaluates this claim? A. Correct, because variance depends only on the maximum value B. Partially correct, because variance depends on all deviations from the mean C. Incorrect, because variance ignores extreme values D. Incorrect, because variance only uses the mean If the value 150 were removed from Park B’s data, what would most likely happen to its variance? A. It would increase B. It would stay the same C. It would decrease D. It would become zeroIntroduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles. Introduction to Hedging Instruments: Forwards, Futures, Options, and Swaps Hedging instruments are financial tools used by businesses and investors to mitigate risk. These instruments help protect against adverse price movements in assets such as commodities, currencies, interest rates, or securities. The four main hedging instruments are forwards, futures, options, and swaps. 1. Forwards A forward contract is a customised agreement between two parties to buy or sell an asset at a predetermined price on a specified future date. Key Characteristics: Over-the-counter (OTC): Traded directly between parties, not on an exchange. Customisation: Can be tailored to suit the needs of the parties involved. Settlement: Occurs at the end of the contract, which may involve physical delivery or cash settlement. Risk: Forwards carry counter-party risk, as there is a possibility one party may default. Example: A company that needs to import raw materials in six months may enter into a forward contract to lock in the current price, avoiding the risk of price increases. 2. Futures A futures contract is similar to a forward, but it is standardised and traded on an exchange. This standardisation eliminates counter-party risk. Key Characteristics: Standardised: Contract size, expiration, and other terms are fixed by the exchange. Mark-to-market: Gains and losses are settled daily. Liquidity: Futures are highly liquid because they are traded on exchanges. Regulation: As they are traded on formal exchanges, they are more regulated than forwards. Example: A wheat farmer may sell futures contracts to hedge against a possible decline in wheat prices before harvest. 3. Options Options provide the right, but not the obligation, to buy or sell an asset at a specified price on or before a certain date. There are two types of options: call options and put options. Call Option: Gives the holder the right to buy an asset at a predetermined price. Put Option: Gives the holder the right to sell an asset at a predetermined price. Key Characteristics: Premium: The buyer pays a premium upfront to obtain the option. Limited Risk: The maximum loss is limited to the premium paid. Flexibility: Options can be used for speculative or hedging purposes. Example: An investor holding stocks may buy a put option to protect against potential declines in the stock's price. 4. Swaps A swap is a contract in which two parties agree to exchange cash flows or liabilities over a specific period. The most common types are interest rate swaps and currency swaps. Key Characteristics: Customizable: Like forwards, swaps are often tailored to meet the needs of the parties involved. Counterparty Risk: Swaps are typically OTC instruments, exposing parties to default risk. Common Uses: Used to manage interest rate risk or currency risk. Example: A company with a variablerate loan may enter into an interest rate swap to exchange its variable payments for fixedrate payments, thus locking in stable costs. Hedging instruments are essential for managing financial risk in volatile markets. Each instrument serves different purposes, with varying levels of complexity, risk, and customization. Whether through forwards, futures, options, or swaps, businesses can better plan for the future by reducing exposure to uncertain price fluctuations. Hedging Strategies for Market Risk, Credit Risk, and Currency Risk 1. Hedging Strategies for Market Risk Market risk (also known as systematic risk) arises from fluctuations in asset prices, such as stocks, bonds, commodities, and interest rates, due to economic factors or market volatility. Key Hedging Instruments for Market Risk: Derivatives (Options, Futures, and Forwards): These instruments allow investors to hedge against unfavorable price movements in stocks, commodities, or interest rates. Example: An investor holding a large stock portfolio might buy a put option to protect against a potential market downturn. If the market declines, the put option increases in value, offsetting losses in the portfolio. Short Selling: Investors can sell borrowed assets with the expectation of buying them back at a lower price, profiting from the decline. Example: A fund manager expecting a market decline may short sell stocks to hedge a portfolio against losses. Common Hedging Strategies: Portfolio Diversification: Reducing market risk by spreading investments across various asset classes (stocks, bonds, commodities) and sectors. Using Index Futures: Large portfolios can be hedged using index futures that track the performance of the overall market. If the market declines, profits from the short position in the futures contract will offset losses in the portfolio. Risk Parity: Allocating assets based on the level of risk rather than the dollar amount invested, balancing risk exposure across asset classes. 2. Hedging Strategies for Credit Risk Credit risk refers to the possibility that a borrower will default on a debt obligation. This is especially important for banks, lenders, and institutions dealing with bonds and loans. Key Hedging Instruments for Credit Risk: Credit Default Swaps (CDS): A financial derivative where the buyer of a CDS pays a premium to the seller in exchange for protection against a default on a loan or bond. Example: A bank holding corporate bonds can buy a CDS to ensure they are compensated if the issuing company defaults. Collateralised Debt Obligations (CDOs): These instruments pool together various debt instruments and allow risk to be distributed among multiple investors. Credit Insurance: Companies may use insurance to protect against the risk of a customer defaulting on payments. Common Hedging Strategies: Diversification of Loan Portfolio: Spreading out credit exposures across various industries, geographies, and borrower profiles reduces the overall risk of default. Tightening Lending Standards: Limiting exposure to highrisk borrowers by implementing stringent credit assessments. AssetBacked Securities: Banks can sell loans or bonds packaged as assetbacked securities to reduce their exposure to credit risk. 3. Hedging Strategies for Currency Risk Currency risk (or exchange rate risk) arises from fluctuations in foreign exchange rates, which can affect companies involved in international trade or with investments in foreign countries. Key Hedging Instruments for Currency Risk: Forward Contracts: A firm agrees to exchange a specified amount of currency at a predetermined exchange rate on a future date. Example: A U.S. exporter expecting payment in euros might enter into a forward contract to sell euros and lock in a favorable exchange rate. Currency Options: These give the right, but not the obligation, to buy or sell currency at a specific price. Example: A U.S.based company buying goods from Japan might buy a call option on the yen to hedge against the risk of yen appreciation. Currency Swaps: Two parties exchange interest payments and principal in different currencies to hedge against exchange rate fluctuations. Common Hedging Strategies: Natural Hedging: Companies can offset currency risk by balancing foreign revenue with costs in the same currency. For example, if a company generates revenue in euros, it can also incur expenses in euros, reducing exposure to exchange rate fluctuations. Multi-Currency Invoicing: Firms can invoice in their home currency, shifting the currency risk to the buyer. Currency Diversification: Holding a diversified basket of currencies can reduce exposure to large fluctuations in any one currency. Effective hedging strategies are crucial for managing various types of risks in financial markets. Market risk can be managed using instruments like futures and options, while credit risk can be mitigated through diversification and credit derivatives. Currency risk, often faced by multinational firms, can be hedged using forward contracts, options, or swaps. Each strategy helps firms and investors protect their portfolios, ensure financial stability, and reduce the impact of adverse movements in the financial markets. Portfolio Risk Management Techniques: Diversification, Asset Allocation, and Risk Budgeting Managing risk is a fundamental aspect of portfolio management. Investors use various techniques to control and reduce the risks inherent in investing. Three key techniques used in portfolio risk management are diversification, asset allocation, and risk budgeting. Each of these techniques helps in mitigating potential losses while aiming to achieve the desired return. 1. Diversification Diversification is a risk management strategy that involves spreading investments across different assets, sectors, or geographic regions to reduce exposure to any single risk. The idea is that different assets perform differently under various market conditions, so losses in one investment can be offset by gains in others. Key Benefits of Diversification: Reduction of Unsystematic Risk: Unsystematic risk, which is unique to a specific company or industry, can be reduced by holding a variety of investments that respond differently to market conditions. Improved Stability: A diversified portfolio is less volatile, as the negative performance of one asset can be balanced by the positive performance of others. Methods of Diversification: Across Asset Classes: Investing in a mix of asset classes such as stocks, bonds, commodities, and real estate. Example: A portfolio with 60% equities, 30% bonds, and 10% commodities is more diversified than one solely consisting of stocks. Within Asset Classes: Diversifying within a single asset class (e.g., holding stocks from different sectors like technology, healthcare, and energy). Geographic Diversification: Investing in assets across various countries or regions to mitigate country-specific risks. Example: Holding U.S. stocks along with emerging market equities can reduce risks related to a downturn in one country's economy. 2. Asset Allocation Asset allocation refers to the process of dividing investments among different asset classes (such as stocks, bonds, and cash) to align with an investor's risk tolerance, time horizon, and financial goals. Asset allocation plays a crucial role in portfolio risk management by determining the overall risk-return profile of the portfolio. Key Elements of Asset Allocation: Strategic Asset Allocation: A longterm approach that involves setting target allocations for different asset classes based on financial goals and risk tolerance. Example: A young investor with a longterm horizon might allocate 70% to stocks, 20% to bonds, and 10% to cash. Tactical Asset Allocation: A more active approach that involves adjusting the asset mix in response to short-term market conditions. Example: If the investor expects an economic downturn, they might temporarily reduce exposure to equities and increase exposure to bonds. Types of Asset Allocation Models: Conservative: Focuses on preserving capital with a larger allocation to bonds and cash (e.g., 20% stocks, 80% bonds). Balanced: A moderate risk approach with an equal focus on growth and income (e.g., 50% stocks, 50% bonds). Aggressive: Targets higher returns by investing predominantly in equities, accepting higher risk (e.g., 80% stocks, 20% bonds). Example of Asset Allocation: A 40 year old investor with moderate risk tolerance may allocate their portfolio as follows: 50% equities, 40% bonds, and 10% in alternative investments such as real estate or commodities. The equities provide growth potential, while the bonds and alternative assets offer stability and income. 3. Risk Budgeting Risk budgeting is a method of allocating risk across different components of a portfolio, rather than focusing solely on returns. The goal is to optimise the portfolio’s risk-return profile by distributing risk in a way that aligns with the investor’s objectives and risk tolerance. Key Concepts of Risk Budgeting: Risk Contribution: Each asset class or investment in the portfolio contributes a certain amount of risk (measured by metrics such as volatility or Value at Risk). Risk budgeting ensures that no single asset class dominates the overall risk of the portfolio. Example: A portfolio may contain 60% stocks and 40% bonds, but if the stocks are highly volatile, they may contribute 90% of the portfolio's risk. Target Risk: Investors set a maximum acceptable level of risk (e.g., a portfolio volatility of 10%) and allocate investments so that the total risk remains within this target. Techniques in Risk Budgeting: Risk Parity: Allocates risk evenly across asset classes, rather than allocating capital based solely on return expectations. Example: In a risk-parity portfolio, both bonds and stocks might be balanced in such a way that they contribute equally to the overall portfolio risk, even though the dollar investment in bonds may be larger due to their lower volatility. Value at Risk (VaR): This technique measures the potential loss in a portfolio over a specific time period, under normal market conditions, at a given confidence level. The risk budget ensures that the potential loss stays within acceptable limits. Example of Risk Budgeting: An investor targets an overall portfolio risk of 8% volatility. After analyzing the risk contribution of each asset class, they determine that equities, which currently make up 60% of the portfolio, contribute 70% of the risk. To adhere to the risk budget, the investor may reduce their equity exposure and increase their allocation to bonds or other less volatile assets. Diversification, asset allocation, and risk budgeting are complementary techniques used in portfolio risk management. Diversification reduces unsystematic risk by spreading investments across various assets. Asset allocation ensures that investments align with an investor's goals and risk tolerance. Risk budgeting focuses on managing the contribution of risk from each asset class to create a balanced and efficient portfolio. Together, these strategies help investors achieve a balance between risk and return, ensuring longterm portfolio stability. Risk Mitigation Through Insurance, Securitisation, and Other Financial Engineering Techniques Risk mitigation is a core objective in financial management, and various strategies can be employed to reduce or manage risks. Three major approaches are insurance, securitisation, and financial engineering techniques. Each of these methods helps firms and individuals transfer, reduce, or eliminate certain financial risks. 1. Insurance as a Risk Mitigation Tool Insurance is a traditional risk transfer method that protects against financial losses by shifting the risk to an insurance company in exchange for premium payments. It is widely used to mitigate various forms of risk, such as operational, liability, and property risks. Key Aspects of Insurance for Risk Mitigation: Risk Transfer: The insurer takes on the risk in exchange for a premium, thus protecting the insured party from unexpected financial losses. Indemnity: In the event of a loss, the insurance policy compensates the insured based on the terms of the contract. Customisable Coverage: Insurance policies can be tailored to address specific risks, such as property damage, business interruption, liability, or cyber risks. Types of Insurance for Businesses: Property and Casualty Insurance: Covers physical assets like buildings, machinery, and inventory from risks like fire, theft, or natural disasters. Liability Insurance: Protects businesses against legal liabilities arising from accidents, negligence, or professional errors. Business Interruption Insurance: Compensates for lost income if a business has to halt operations due to unforeseen events. Credit Insurance: Shields companies from losses due to the nonpayment of trade receivables. 2. Securitisation as a Risk Mitigation Technique Securitisation is a financial engineering process that involves pooling various financial assets (such as loans, mortgages, or receivables) and converting them into marketable securities. This process allows firms to transfer risk to investors, thereby reducing their exposure. Key Elements of Securitisation: Risk Transfer: By securitising assets, companies can transfer the risk of default or nonpayment to investors who purchase the securities. Liquidity Creation: Securitisation converts illiquid assets (like mortgages or loans) into liquid, tradeable securities, improving cash flow for the originating firm. Diversification of Risk: Pooling assets with different risk profiles reduces the impact of individual defaults, spreading the risk across multiple investors. Common Forms of Securitisation: MortgageBacked Securities (MBS): Pools of mortgages are bundled and sold as securities to investors, transferring the risk of mortgage defaults. Example: A bank that issues home loans can bundle those loans into MBS and sell them to investors, transferring the credit risk of potential defaults. Asset-Backed Securities (ABS): Similar to MBS, but backed by other types of assets like credit card receivables, auto loans, or student loans. Collateralised Debt Obligations (CDOs): Structured financial products that pool different types of debt, such as loans and bonds, and sell them as securities with varying risk levels. Example: A bank may issue a portfolio of auto loans and then pool these loans into an assetbacked security (ABS). The ABS is sold to investors, who take on the risk of loan defaults. By securitising the loans, the bank reduces its exposure to credit risk and generates immediate cash flow. 3. Financial Engineering Techniques for Risk Mitigation Financial engineering involves the use of complex financial instruments, derivatives, and structured products to manage or mitigate financial risks. These techniques allow firms to hedge against specific risks, optimize capital structure, and improve financial stability. Common Financial Engineering Techniques: Derivatives: Financial instruments like futures, forwards, options, and swaps are used to hedge against price fluctuations, interest rate changes, or currency movements. Example: A company with significant foreign exchange exposure may use currency forwards or options to hedge against exchange rate fluctuations, ensuring predictable cash flows. Options and Futures: Options: Provides the right (but not the obligation) to buy or sell an asset at a predetermined price, allowing firms to hedge against unfavorable price movements. Example: An airline company can buy options on jet fuel to hedge against rising fuel prices. Futures: Standardized contracts to buy or sell an asset at a set price on a future date, commonly used to hedge commodities or financial assets. Example: A wheat producer may use futures contracts to lock in a favorable price for its crop, hedging against a potential price drop. Swaps: These involve the exchange of cash flows between two parties, often used to manage interest rate risk or currency risk. Interest Rate Swaps: Firms can exchange floatingrate interest payments for fixedrate payments to hedge against rising interest rates. Currency Swaps: Used to hedge exchange rate risk in crossborder transactions by exchanging principal and interest payments in different currencies. Example: A company with a variablerate loan may enter into an interest rate swap to exchange its variable payments for fixedrate payments, locking in stable costs. Structured Products: These are customised financial instruments designed to achieve specific riskreturn objectives. They often combine derivatives with other securities to create tailored risk exposures. Example: A structured note that combines a bond with an embedded option, offering downside protection while allowing for potential upside linked to the performance of an equity index. Credit Derivatives: Tools like credit default swaps (CDS) allow investors to transfer credit risk to other parties. Example: A bondholder worried about a company’s potential default may purchase a CDS, which pays out in case of a default event. Example: A company may issue a bond with an embedded call option, allowing it to repurchase the bond if interest rates decline. This financial engineering tool enables the company to mitigate the risk of rising interest rates, reducing future borrowing costs. Risk mitigation through insurance, securitisation, and financial engineering offers businesses a variety of tools to manage and transfer risks. Insurance allows for the direct transfer of risk to an insurer, while securitisation helps companies offload risk by packaging and selling assets as securities. Financial engineering techniques, including derivatives, swaps, and structured products, provide sophisticated ways to hedge market, interest rate, and currency risks. Each approach helps organizations improve financial stability, enhance liquidity, and manage potential losses in a volatile market environment.In this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level