Consider a discrete random variable X with moment generating functio.pdf

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Consider a discrete random variable X with moment generating function as: M(t)=0.3/(1-0.7e^t) ; where all t<.3567 .....What is the value of E[x] Solution E(X) = M\'(0) M\'(x) = (0.21e^t)/(1-0.7e^t)^2 E(X) = M\'(0) = 0.21/(1-0.7)^2 = 0.21/(0.3^2) = 2.33.

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Consider a family tree that consists of just you, your parents,their parents, etc. (no siblings). Assume you are Generation 0, your parents are Generation 1, yourgrandparents are Generation 2, etc. How many ancestors do you haveat Generation n, for any n 0? Give a proof using induction. Solution At generation n, there are 2n ancestors. Proof by induction. Base case: at generation 0, there is just you, so there isjust 2^0=1 people. Inductive step: assume generation n- 1 has 2n-1people. Each has two parents, so there are2*2n-1=2n people in the generation n..

Consider a family tree that consists of just you, your parents,their.pdf

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Consider a dinner table where n dining philosophers are seated. Each philosopher has a plate of food; however, there is only a single eating-utensil placed in the center of the table. Eating is done at discrete rounds. At the beginning of each round, if a philosopher wishes to eat, she may attempt to obtain the utensil from the center of the table. If the philosopher obtains the utensil, she eats for the duration of the round (i.e., only one philosopher may eat during any given round) and then places the utensil back at the table center at the end of the round. If two or more philosophers attempt to obtain the utensil at the beginning of the same round, then no philosopher will eat during that round (i.e., no one obtains the utensil); thus, if all the philosophers try to access the utensil on every round, no philosopher would ever eat. One way to avoid the starvation of these philosophers is to use randomization. A possible simple randomized algorithm is to have each philosopher attempt to obtain the utensil at any given round with probability p > 0 (independently of the decisions of the other philosophers). a) What is the probability (in terms of p) that a philosopher is able to successfully eat during any given single round? b) What value of p maximizes this probability? c) Using the probability from Part b, what is the probability that a philosopher does not successfully obtain the utensil after k consecutive rounds? Solution a.) p*(1-p)^n b.) p = 1/n c.) [(1- 1/n)^n ]/n.

Consider a dinner table where n dining philosophers are seated. Each.pdf

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Consider a $6500 piece of machinery , with a 5 year depreciable like and an estimated $1200 salvage value. The projected utilization of the machinery when it was purchased , and its actual production to date, are as follows: Year Projected Production (Tons) Actual Production (Tons) 1 3500 3000 2 4000 5000 3 4500 [Not 4 5000 Yet 5 5500 Known] Compute the depreciation using : (a) straight line; (b) sum of years digits; (c) double declining balance; (d) Unit of production (for the first 2 years only); (e) Modified accelerated cost recovery system Year Projected Production (Tons) Actual Production (Tons) 1 3500 3000 2 4000 5000 3 4500 [Not 4 5000 Yet 5 5500 Known] Solution CCA rate=30% Year Class No. Undep. capital cost at beginning of year Cost of acq. during the year Proceeds of disp. during the year Undep. capital cost 50% rule Reduced undep. capital cost CCA rate % Capital cost allowance Undep. Capital Cost at end of year 1 8 0 6,500.00 0 6,500.00 3,250.00 3,250.00 30 975.00 5,525.00 2 8 5,525.00 0 0 5,525.00 0.00 5,525.00 30 1,657.50 3,867.50 3 8 3,867.50 0 0 3,867.50 0.00 3,867.50 30 1,160.25 2,707.25 4 8 2,707.25 0 0 2,707.25 0.00 2,707.25 30 812.18 1,895.08 5 8 1,895.08 0 0 1,895.08 0.00 1,895.08 30 568.52 1,326.55 Salvage value=$1,200, therefore loss on disposal=$126.55 Year SL SOYD DDB UOP* CCA Year 1 $1,060 $1,767 $2,600 $707 $975.00 1 2 $1,060 $1,413 $1,560 $1,178 $1,657.50 2 3 $1,060 $1,060 $936 $1,160.25 3 4 $1,060 $707 $204 $812.18 4 5 $1,060 $353 $0 $568.52 5 Sum $5,300 $5,300 $5,300 $5,174.45 *UOP Depreciation is based on actual production. Year Class No. Undep. capital cost at beginning of year Cost of acq. during the year Proceeds of disp. during the year Undep. capital cost 50% rule Reduced undep. capital cost CCA rate % Capital cost allowance Undep. Capital Cost at end of year 1 8 0 6,500.00 0 6,500.00 3,250.00 3,250.00 30 975.00 5,525.00 2 8 5,525.00 0 0 5,525.00 0.00 5,525.00 30 1,657.50 3,867.50 3 8 3,867.50 0 0 3,867.50 0.00 3,867.50 30 1,160.25 2,707.25 4 8 2,707.25 0 0 2,707.25 0.00 2,707.25 30 812.18 1,895.08 5 8 1,895.08 0 0 1,895.08 0.00 1,895.08 30 568.52 1,326.55.

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Conduct a company review of Starbucks. What are their unique growth and implementation strategies? Explain why they give the company a competitive advantage. Compared to other firms, Starbucks has been very successful in China, why is that? How does their strategy differ from the competition? Solution The two main growth strategies that Starbucks has used include growth by concentration anddiversification.Concentration StrategyIn concentration Strategy Company focus on their core business and want to expand their corebusiness as Starbucks did when it went global and opened thousands of stores in US andworldwide. They focused on their primary business by expanding more outlets each year. Diversification StrategyStarbucks used both related and unrelated diversification. The example of relateddiversification is when it entered from coffee beans to food items, flavored coffee ice cream,ready to drink beverage. And the example for unrelated diversification is when it launchedprepaid cards, launch of music caf.

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Components of computer systems often have dependencies -- other components that must be installed before they will function properly. These dependencies are frequently shared by multiple components. For example, both the TELNET client program and the FTP cli ent program require that the TCP/IP networking software be installed before they can operate. If you install TCP/IP and the TELNET client program, and later decide to add the FTP client program, you do not need to reinstall TCP/IP. For some components it would not be a problem if the components on which they depended were reinstalled; it would just waste some resources. But for others, like TCP/IP, some component configuration may be destroyed if the component was reinstalled. It is useful to be able to r emove components that are no longer needed. When this is done, components that only support the removed component may also be removed, freeing up disk space, memory, and other resources. But a supporting component, not explicitly installed, may be removed only if all components which depend on it are also removed. For example, removing the FTP client program and TCP/IP would mean the TELNET client program, which was not removed, would no longer operate. Likewise, removing TCP/IP by itself would cause the fa ilure of both the TELNET and the FTP client programs. Also if we installed TCP/IP to support our own development, then installed the TELNET client (which depends on TCP/IP) and then still later removed the TELNET client, we would not want TCP/IP to be remo ved. We want a program to automate the process of adding and removing components. To do this we will maintain a record of installed components and component dependencies. A component can be installed explicitly in response to a command (unless it is alrea dy installed), or implicitly if it is needed for some other component being installed. Likewise, a component, not explicitly installed, can be explicitly removed in response to a command (if it is not needed to support other components) or implicitly remov ed if it is no longer needed to support another component. Input The input will contain a sequence of commands (as described below), each on a separate line containing no more than eighty characters. Item names are case sensitive, and each is no longer th an ten characters. The command names ( DEPEND , INSTALL , REMOVE and LIST ) always appear in uppercase starting in column one, and item names are separated from the command name and each other by one or more spaces. All appropriate DEPEND commands will appear before the occurrence of any INSTALL dependencies. The end of the input is marked by a line containing only the word END . Command Syntax Interpretation/Response DEPEND item1 item2 [item3 ...] item1 depends on item2 (and item3 ...) INSTALL item1 install item1 and those on which it depends REMOVE item1 remove item1 , and those on whch it depends, if possible LIST list the names of all currently - installed components Output .

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