...Bidding on Hertz LBO case study The main issue of the case is that The Carlyle Group and its partners (Clayton, Dubilier & Rice, and Merrill Lynch Global Private Equity) must make a decision about the final terms of a bid to purchase the Hertz Corporation, a wholly owned subsidiary of The Ford Motor Company. Hertz had been put up for sale in June 2005. In order to initiate “consideration of strategic alternatives” Ford entered a dual-track process, which means pursuing an initial public offering and a private auction process simultaneously. A dual-track process gives several advantages to a Ford, a seller of Hertz, by affecting the bidding process: 1) a dual-track approach can maximize the price obtained by the seller because of competitive pressure of a viable IPO process that can drive bidders to put more money on the table; 2) a dual-track process can give a seller significant negotiating leverage over deal terms in a private sale; 3) a dual-track approach allows a seller to keep its options open until it becomes clear which route will yield the highest value, thus preserving flexibility and hedging against deal uncertainty; 4) while a dual-track approach magnifies the cost, complexity and management distraction inherent in any sale process, it may also enjoy useful synergies by providing to a seller a comparable information about value of the selling company. The seller can judge about fair ask price of its company based on due diligence of bidders...
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...9-208-030 REV: JANUARY 16, 2009 TIMOTHY A. LUEHRMAN DOUGLAS C. SCOTT The Hertz Corporation (A) We started banging on the door of Ford in 2002 and got very little reception at first. I wouldn’t say we got laughed out of the room, but close to it. — David Wasserman, Clayton, Dubilier & Rice In the first days of September 2005, partners at the private equity firm Clayton, Dubilier & Rice (CD&R) were preparing a final bid in pursuit of the largest and most complicated deal they had ever undertaken. For more than three years CD&R had patiently pursued an acquisition of The Hertz Corporation, the global car and equipment rental company owned by Ford Motor Company. A three-firm consortium led by CD&R made a preliminary bid in July 2005, as did two other interested buyers. In late July they began conducting due diligence and working on a complex financing scheme in preparation for a final bid due August 30. George Tamke, a CD&R operating partner, led a team that identified and quantified significant opportunities to improve Hertz’s operating efficiency. At the same time, CD&R financial partners David Wasserman, Michael Babiarz, and Nathan Sleeper worked on a capital structure that would lower Hertz’s capital costs substantially, partly by making use of asset-backed securities (ABS) secured by its large fleet of rental cars. The combination of potential operating and financing efficiencies made Hertz an attractive investment opportunity. However, the proposed transaction entailed...
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...The Hertz buyout is one of the largest private equity deals. It drew criticism in the media and from union members, after the company’s new owners paid themselves $1.3 billion in dividends not long after the transaction closed and ultimately financed the payments by selling stock to the public. The company has realized hundreds of millions of dollars in improved financial results annually, but also has cut thousands of jobs as it has sought to make operations more efficient. Background: Hertz says it is the world’s largest general use car rental company, with approximately 8,100 locations in about 145 countries. Hertz also operates an equipment rental company with about 380 locations worldwide, although car rentals accounted for 80 percent of 2007 revenues. Ford Motor Co. had purchased an ownership stake in Hertz in 1987 and purchased the company outright in 1994. CD&R executives said that the firm emphasizes making operational improvements in companies it acquires. The firm has long had an interest in multilocation service businesses, they said, as evidenced by investments including Kinko’s and ServiceMaster. The Carlyle Group is one of the biggest private equity firms and says it has demonstrated expertise in the automotive and transportation sectors. Its investments include Dunkin’ Brands, AMC Entertainment, Inc., and Grand Vehicle Works, which provides products and services to truck fleets and recreational vehicle users. Merrill Lynch Global Private Equity is the private...
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...ET1410 - Module 2 - Homework Wk3 Due: Beginning of Class Week 4 Student Name: _________________________ Instructions: This is a homework assignment that covers Week 3 of our ET1410 class. You should neatly show your steps in the middle column, and provide your final answer in the right hand column. NEATNESS COUNTS. You must work in pencil, not pen. Must include units of measure in your answer!!! If no units, or incorrect units, points will be deducted. ProblemNo. | Your Solution | Final Answer (Restate Your Final Answer Here) | 1 | The upper critical frequency of an op-amp’s open-loop response is 200 Hz. If the midrange gain is 175,000 what is: (a) ideal gain at 200 Hz? (b) the actual gain? (c) The op-amps open loop bandwidth? | (a)____________(b)____________(c)____________ | 2 | Determine the attenuation of an RC lag network with fc = 12 kHz for each of the following frequencies. (a) 1 kHz (b) 5 kHz (c) 12 kHz (d) 20 kHz (e) 100 kHz | (a)____________(b)____________(c)____________(d)____________(e)____________ | 3 | Determine the phase shift through the network at a frequency of 2 kHz: | | 4 | Determine the phase shift through the network at a frequency of 2 kHz: | | 5 | Determine the phase shift through the network at a frequency of 2 kHz: | | 6 | A certain op-amp has three internal amplifier stages with midrange gains of 30dB, 40dB, and 20dB. Each stage also has a critical frequency associated with it as follows: fc1= 600Hz...
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...One of the eye-popping techno wonders from the latest Batman installment involves the use of cellphones to “see” inside of buildings. When Lucius Fox plants a cellphone in Lau’s office building, the modified phone somehow creates a map of the buildings interior by using a “high frequency pulse” to create an image. Fox tells Bruce that it works just like sonar. Well to be more accurate he should have called it radar because that’s what it is. Sonar is when you reflect sound waves off of objects to determine their location. Radar is when you use electromagnetic radiation usually at microwave frequencies to do the same thing. Since cellphones emit and receive e-m radiation in the radio wave/microwave frequency range (for cellphones the frequencies are usually between about 800 and 2000 MHz) if you tried to use them as imaging devices then that’s radar. There are many different types of radar designed for different purposes including military defense, meteorological applications (the Doppler radar they use to image precipitation), police radar, geologic applications (ground penetrating radar); the list goes on. But, not surprisingly, none is capable of coming close to the level of 3-D resolution and the depth of penetration that Fox’s is able to achieve. The more layers the radiation penetrates the more it is absorbed and scattered. This makes resolving and interpreting the reflected signals extremely challenging. Nevertheless it’s an interesting idea extrapolated from existing technology...
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...ECET 340 Week 6 Homework 6 To Buy This material Click below link http://www.uoptutors.com/ecet-340-devry/ecet-340-week-6-homework-6 1. What are the four main functions of the HC12 Timer unit? 2. Two input capture events occur at counts 0×1037 and 0xFF20 of the free-running counter. How many counts (in decimal) have transpired between these two events? 3. What is the maximum time possible before the free-running counter overflows when the e MHz? 4. Two input capture events occur at 0×1037 and 0x002A. If the prescaler bits PR[2:1:0] are set to 101 and the e clock is 24 MHz, how much time as transpired between the two events? 5. Calculate the count that should appear in the timer capture register TC0 if a 125 kHz rectangular wave is inputted on timer pin PT0 while TCTL4 is preset for falling edge detection. Assume a 24 MHz e-clock, TMSK2 was programmed with the value $02, and the count of the 1st edge event has already been subtracted off from TC0. 6. Write down the name of the HC12 timer register that should be polled through software to determine whether or not an active input edge has been captured on one of the port T pins. 7. What is the duty cycle of a signal produced by the PWM when and ? a. 28.0% b. 29.8% c. 50.0% d. 72.0% 8. What values are required for period and duty cycle to generate a 6.0 kHz, 95% duty cycle waveform using the PWM function? Assume e-clock frequency is 24 MHz. 9. What is the slowest clock signal that can be generated from the PWM output...
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...Topics : Sound Generation Introduction: From 'O' level physics, you have learned that sound is the vibration of air within the frequencies of range of 20 Hz. to 20KHz. If we can generate these air waves, we can make sound. A loud speaker is made up of a moving coil wound round a permanent magnet as shown in Figure 1. When an alternating current passes through the moving coil (also known as voice coil), a force is transmitted to the paper cone. The moving cone produces air pressure waves and acoustic energy is radiated. The frequency of the alternating current through the coil will be the same frequency as the sound wave. Permanent Magnet Cone radiator X Flexible suspension Moving coil Figure 1. Dynamic loudspeaker construction Microcontroller Technology Page 1 If a program, when executed, can generate a square wave of say 440 Hz. (the A note) at the output where speaker/buzzer is connected, a tone radiated from the speaker will be heard. It is not difficult at all to generate such a frequency using a program. The principle is to have a delay routine of the right duration and have the logic state of speaker/buzzer output pin toggle every time when the delay routine expires. For example, to generate a 500 Hz. wave, a delay routine of 1 ms is needed. This 1 ms duration is derived from the following. Given, frequency, f= 500 Hz. 1 1 Period , T 2mS f 500 Since the signal needs to change state at every half cycle therefore the Delay...
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... For the second part of the experiment, choose your heaviest weight from the three that you use earlier. Then, move the laser 1 cm away from the string to obtain better results. Next turn on the wave driver and find the resonant frequency for the fundamental node of this string by playing with the frequency until the amplitude is maximized. Record this frequency. Now set up to plot Lissajous figures with the photodiode signal and the drive signal. This parametric plot will allow you to find the other frequencies for higher order nodes by observing the symmetry of the parametric plot. The Lissajous figures can be used to record the uncertainty as well by seeing how much the frequency can be changed before there is a visible change in the diagram. Estimate where the frequency might be for each value of n by multiplying the fundamental frequency by n. Find five values of frequency for any random five harmonics from n=1,2,3,…,9. Turning up the amplitude for higher frequency nodes may prove to help with collecting data. Finally, observe the 30th harmonic and record your observations. Analysis The period of the data can be found with Figure 1. To do this, find the time difference between the first and last peak upwards. Then divide this number by the number of upwards peaks subtracted by 1. This results in a period of 1.4s. Uncertainty for this can be found by finding the standard deviation of periods in-between consecutive peaks. This results in a value of (1.4 ± 0.4)s. This period...
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...Task 1B Frame Rates Frame Rates or FPS (Frames Per Second) is the amount of images that are being displayed each second, FPS would be used in videos, TV broadcasts, Video games, etc. The more frames that are put in per second will show a more smoother looking video and it will give better visuals. Higher FPS would be used for more professional use because it would take more to load comparing to a less FPS video, so a high FPS video can result in a bigger file size. YouTube has added a feature where if you can view a higher resolution video, it will player in a higher FPS. Some professional technologies that would have high FPS would be Smart TVs and Movie projectors. This is to give a more smoother look when watching something. Video Format Video formats are different types of files that provide different types of viewing when watching/listening to a file. There are many types of video formats being used in Multimedia, some of these would be MPEG, GIF and AVI. MPEGs would be used for Motion Graphics that have been compressed, it would be known globally as the best compatible lossy format. GIFs would be used in Motion Graphics as it is basically a mini video clip that is being played in a loop, this can be used graphically and creating a nice motion graphic. AVIs would be the best Motion Graphic option because it has a number of codec options, it would be mainly used as a media container. Screen Ratios Ratios is relating on the display that would be found on screens like...
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...that performs arithmetic and logic operations. Arithmetic instructions include addition, subtraction, and shifting operations, while logic instructions include AND, OR, XOR, and NOT operations. CPU (Central Processing Unit) CPU (Central Processing Unit) - otherwise known as a processor - is an electronic circuit that can execute computer programs. Both the miniaturization and standardization of CPUs have increased their presence far beyond the limited application of dedicated computing machines. Modern microprocessors appear in everything from automobiles to mobile phones. The clock rate is one of the main characteristics of the CPU when performance is concerned. Clock rate is the fundamental rate in cycles per second (measured in hertz, kilohertz, megahertz or gigahertz) for the frequency of the clock in any synchronous circuit. A single clock cycle (typically shorter than a nanosecond in modern non-embedded microprocessors) toggles between a logical zero and a logical one state. With any particular CPU, replacing the crystal with another crystal that oscillates with twice the frequency will generally make the CPU run with twice the performance. It will also make the CPU produce roughly twice the amount of...
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...Unit 1 Assignment 1 Blank Answer Sheet Name: Date: Electronics ET2530 1. Define Modulation. The process of varying one or more properties of a periodic waveform, called the carrier signal, with a modulating signal that contains information to be transmitted. 2. What is a carrier Frequency? A carrier frequency is a single radio frequency with steady amplitude. 3. Describe two reasons that modulation is used for communications transmissions. a) Signal integrity b) Power savings 4. List three parameters of a high frequency carrier that may be varied by a low frequency intelligence signal. Amplitude, frequency and phase v = Vp·sin(ωt + Φ) ω = 2πf (describe which parameters can be varied) 5. What are the frequency ranges included in the following frequency subdivisions: MF (medium frequency)300-3000kHz, HF (high frequency)3-30MHz, VHF (very high frequency)30-300MHz, UHF (ultra-high frequency)300-3000MHz, and SHF (super high frequency)3-30GHz? 7. A microwave transmitter typically requires a +8dBm audio level to drive the input fully. If a +10 dBm level is measured, what is the actual voltage level measured? Assume a 600Ω system. (2.45V) Hint: Use this equation and solve it for V2 10 dBm = 20 · log10 (V2/0.77459) 9. Convert the following power to their dBm equivalents: (a) p = 1 W = 30.0000dBm (b) p = 0.001 W = 0.00000dBm (c) p = 0.0001W = -10.0000dBm (d) p = 25 μW = dBm Use this formula: dBm = 10 · log10 [P2/(10-3)] 15. Define electrical noise...
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...206. What is Half Power Frequency? A) Voltage gain reduced to 50% B) Voltage gain reduced to 60.7% C) Voltage gain reduced to 70.7% D) Voltage gain reduced to 80.7% ANSWER: C 207. If the voltage gain drops to 70.7% of its midrange value, it is said to be _______ A) attenuated B) down 6 dB C) down 3 dB D) down 1 dB ANSWER: C 208. What is the gain that occurs between the lower and upper critical frequencies in amplifier? A) midrange gain B) critical gain C) bandwidth gain D) decibel gain ANSWER: A 209. What can be measured using the decibel? A) voltage gain B) power gain C) attenuation D) all of these ANSWER: D 210. Voltage gain in dB is ______ A) log Av B) 10 log Av C) 20 log Av D) Av ANSWER: C 211. Power gain in dB is ________. A) log Av B) 10 log Av C) 20 log Av D) Av ANSWER: B 212. For the low-frequency response of a BJT amplifier, the maximum gain is at ________. A) RB = 0 Ω B) RC = 0 Ω C) RE = 0 Ω D) RS= 0 Ω ANSWER: C 213. Where the Miller effect capacitance is not involved in BJT configuration? A) Common-emitter B) Common-base C) Common-collector D) All of the above ANSWER: B 214. What does positive and negative sign of dB represents? A) positive-amplification, negative-attenuation B) Positive-attenuation, Negative-amplification C) Zero value D) sign can be neglected ANSWER: A 215. At the midrange of the amplifier, the voltage gain of 100. If 6 dB gain decreases, what happens to the gain? A) 0 B) 70.7 C) 50 D) 20 ANSWER: C 216. If the frequency changed from 1 kHz to 10...
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...Based on the results, the maximum frequency that would not be aliased was the Nyquist frequency, which was 5000 Hz. In addition to that, the amplitude of the loud voice was higher than the amplitude of the low voice. The number of bits calculated based on the quantization was equal to 9.97 bits. The high cutoff and the low cutoff which are 110 Hz and 1100 Hz, were used to show the Fast Fourier Transform of the filtered voices, which results in producing better voice of the recording. Based on the FFT graphs, the amplitude of the 440 Hz was the highest at the same peak, and that is why the sound was directly clear, and that proves that the played sound was with the frequency of 440 Hz. For the low and loud frequency, the graph showed the background noise recorded with the voice signal. The use of the decimating factor is that it will resample the recorded voice signal at a lower sampling frequency. Therefore, by changing the value of the decimating factor, the between 1 to 8, the sounds became more clear and recognizable. That is because the decimating factor, decrease the sample rate and increase the frequency of the voice signal in order to avoid the signal to be aliased. Based on research, the frequency range that the human voice can produce is in the range of 300 to 3000 Hz. The Fast Fourier Transform of recorded voice in this experiment was within the range of the human voice frequency. If the electret microphone element was used outside on a windy day, the FFT graphs...
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...Checkpoint in your own words IT/241 Intro to W-Lan technologies FRED HOLMES IV April Sturgess Friday May 4, 2013 What is frequency? Frequency is the number of cycles of a repeating signal. The unit of frequency is measured in hertz. Each hertz is equal to one cycle of an event in seconds. For a frequency to be considered an audio frequency it has to be between 15 Hz and 20000 Hz. A radio frequency will range from 20 kHz to 1 terahertz. A medium for radio frequency will be air and metal. Air will not conduct electricity so the displacement won’t happen. When it comes to infinity value, air will become resistant and only some current flow will go through. Metal is an electricity conductor. This means that the flow will be based on the resistance of metals and the components with in the metal. What is analogue modulation? Is the information which is applied to the carrier signal. This is usually with radio signals. The amplitude modulation will change any frequency of the signal. Frequency Modulation is the frequency of the Carrier can wave at the input signal with the level use multiple methods When comparing the modulation we can show that both use the same method. The only difference is that digit. While this process is going on the digital modulation can use multiple times. Digital will also apply filters to interfering...
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...NTC 406 Week 3 Individual Assignment Bandwidth is a Rate at which electronic signals can travel through a medium, such as a wire, cable, or channel or capacity of network to transmit data on network. There are different transmissions medium such as audio video and voice. Bandwidth need is different in the all because these all are different data format. In analog transmission (such as of voice signals over copper telephone lines) bandwidth is measured in cycles per second (or Hertz); for example, a telephone conversation requires about 4,000 Hertz (4 KHz) of bandwidth. In digital transmission (such as of data from one computer to another) bandwidth is measured in bits per second (BPS); for example, modern modems can send and receive data at 56,000 bps (56 Kbps) over ordinary telephone lines. For the same amount of data, digital transmission requires more bandwidth than the analog transmission, and different types of data require very different bandwidths. For example, full motion video normally requires about 10 million bits per second (10 Mbps) bandwidth which is sufficient to carry 1,200 simultaneous telephone conversations. A typical voice signal has a bandwidth of approximately three kilohertz (3 kHz); an analog television (TV) broadcast video signal has a bandwidth of six megahertz (6 MHz) -- some 2,000 times as wide as the voice signal. Three bandwidth techniques involves are as like Traffic Management/QoS, Caching and Compression so these are different bandwidth...
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