EXPENDITURE ON MOBILE PHONES Essay Example | Topics and Well Written Essays - 1500 words

EXPENDITURE ON MOBILE PHONES - Essay Example rates and tariff. The behavioural pattern is more obvious in groups of consumers than in individual consumers. The mobile phone market is expected to grow by 2% in retail volume compared to its review period which witnessed a 40% growth in the earlier period. Owing to the already high penetration rate of mobile phones in the household sector which accounts for 261% in 2013, the future CAGR of the mobile phone market is expected to remain low. Australia experienced a high penetration rate in mobile phone usage. The household penetration of mobile phones grew owing to the rise of smart phones. Consumers in Australia had the tendency to retain the old phones for spare use. This also led to the high penetration rate of mobile phone usage. This trend was observed more in case of simple mobile phones, but in case of smart phones the scenario was different. Smart phones were evolving at a high rate and consumers were less inclined to retain it for spare use as it failed to match the superior performance and feature of the new smart phones. This led to the drop in the penetration rate of smart phone in the Australian market. The penetration rate dropped from 86% in 2010 to 33% in 2013. Samsung retained its leader position in the smart phone market in Australia with more than 32% retail market share. 1 The smart phone market in US has shown significant growth owing to the existence of low value contracts of older models of smart phones. The total shipping value in units for smart phones accounted for 134 million units in 2014. It witnessed an annual growth rate of 9% in the smart phone market in 2013 and an absolute growth rate of 37% from 2008 to 2013. The smart phone industry in US has observed the shift more in case of low priced phones than high value phones. The key focus was on volume and not on value. This was mainly observed when retailers in US offered smart phones for a contract of less than $200 for two years. The older version of the

Sports marketing Research Paper Example | Topics and Well Written Essays - 500 words

Sports marketing - Research Paper Example They may also sponsor individual players or in some cases, teams. (Huguelet, 2010). Planning of the whole process of sports marketing commences with the objectives of the corporation and its mission statement. (Kriemadis and Terzoudis, 2007, p. 32). The process of marketing sports fundamentally involves use of 4 Ps, namely, product, pricing, promotion and place (Summers and Morgan, 2005, p. 6). Indeed, these are the 4 most critical factors that decide the success of a sports event. â€Å"The level of support for a football club is a key variable for matchday revenue and also determines most other club revenue streams† (Kase, 2007, p. 278). Sports and business have many traits in common. For example, â€Å"sport and business share values such as the magnitude of teamwork, line of attack and striving also toward a goal† (Hameed, n.d.). The professionals involved in sports marketing also work to address the concerns of an individual country by promoting its team. They may also work for organizers of some professional tournament. Either way, they promote the individual team or event. Sports marketing professionals cardinally make use of advertisements to achieve their objects. Radio, internet, newspapers and television are commonly chosen as the means to announce games coming in near future. Many times, the sports marketing professionals make use of unusual strategies to gather large number of visitors to see the match. They do so by offering to give away their products to the public via lottery or such other schemes. Also, many sports marketing professionals carry out market research to investigate the popularity of individual athletes among public. They require this information so that they may be able to launch a successful advertisement by making that athlete, part of the advertisement of their product. Entities like Fantasy Frontline remain on the forefronts of sports news reporting, and hence, play an

Involvement of the U.S Military in Border Protection and the Drug Essay

Involvement of the U.S Military in Border Protection and the Drug Policy Since 1960s - Essay Example Involvement of the U S Military in Border Protection and the Drug Policy Since 1960s Introduction The United States’ borders are extensively guarded following effective legislations that target safeguarding Americans. The U.S Customs and Border Protection (CBP) present the federal enforcement agency in the department of homeland security that regulates foreign trade, trade customs, immigration, import duty collections, and external regulations enforcement (Guerette & Clarke, 2005). The agency has a good number of officers, over 4600 well-trained officers who guard the borders of the U.S to ensure border security. However, the United States military strongly claim involvement in the border security and international security maintenance in the country. With the changes in the drug policy in the United States, the body stands at the pole position to ensure that illegal drugs do not get into the country easily. Enforcement of the immigration rules solely falls in the docket of th e United States military. The immigration rules in the United States contribute much on the security of the people. Since 1900s, the U.S government has adopted strict regulations on immigration that has led to incorporation of the state’s military to stiffen security at the border. Ceasar (2008) argues that tougher rules led to growth of the illegal market especially drugs and narcotics trade. The border patrol thus was adopted to boost security at the border and prevent further turmoil. Later in the 1920s, the United States feared that the countries ravaged with war threatened their position thus adopting tighter immigration rules. For example, the adopted legislations ensured that foreigners could not get into the country without being cleared at the border. Particularly, current policies on immigration state that people should not enter the United States unless they get clearance at the border. Consequently, the United States military assumes the role of ensuring that no p erson gets into the country illegally, and curtailing black market at the borders by conducting constant patrols. The United States military takes part in border protection to prevent entry of illegal drugs in the country. The war against illegal drugs dates back to the 1920s when the congress passed the drugs act that blocked importation of drugs from outside the country. Many people perceived this as a move to protect the people, but there existed certainty as to who could oversee the same. The American government tightened drug policy in 1960s thus the military came into force to ensure that the prohibited drugs do not enter the country. However, the black market expanded because the people found no other way to get to the illegal drugs. In this sense, it is apparent that activities of the military officers at the borders related directly with the prevalence of the black market. Rising concerns on the health of the United States plant and animal resources heightened level of invo lvement of the United States military in border protection in the country (Ceasar, 2008). Many people in the country raised concerns that their plants and animals got infections from foreign countries, which threatened human health. Any concerns about the health of plants and animals meant that the level of involvement for the military increased to enhance a safer environment. In this respect, the U S military embanked on undertaking passenger operations, targeting and analysis, and

Self managed teams Essay Example for Free

Self managed teams Essay Using self-managed teams within the Sandwich Blitz company would help the company grow. This will also allow Dalman to have more time to concentrate on the expansion as well. Self-managed teams empower employees most often in manufacturing, workers are trained to do all most of the jobs in the unit, they have no immediate supervisor, and they make decisions previously made by the first-line supervisors (Bateman Snell, 2012). Self-managed teams can offer several possible advantages that could be used at Sandwich Blitz; including stronger commitment, employee and customer fulfilment, improved quality, improved productivity, and faster product and service development. On the other hand there will be some drawbacks. Self-managed teams are tough to execute, and failure is a risk when used in an unsuitable situations, or lacking appropriate leadership. A number of organizations have been dissatisfied with the results from self-managing teams. In the case of Sandwich Blitz, self-management teams may work to a sure extent, for example, having the team members cross-trained to do various jobs. This will enhance the flexibility of the team in dealing with staff shortages and in addition to help out with Dalman having to help fix every problem managers have. Thus applying Dalman and Lei’s knowledge of work procedures with the teams would help the team members to find a solution to problems and build improvements. It will also help decrease any absences that maybe have occurred in the past which would help free up some of Dalman and Lei’s time. Nonetheless, Dalman would need to choose team members he feels secure enough with the size of issues that would need to be handled, and continued presence of the team. As said above, self-managed teams can be hard to put into effect, and Dalman needs to carefully think it through, as well as, being ready to devote some of his time into creating the team and then removing himself once he feels confident with the team. By creating the self-managed teams within sandwich blitz managers will be able to participate and contribute to productivity. This will alter the distribution of the workload for Dalman. It also allows the managers to feel empowered as power, authority, and responsibility is re-distributed amongst the team. This eventually is beneficial to the company as the people closest to the customer; for example the location managers, have the decision making power and which ultimately effects the outcome. References: Bateman Snell 3rd Edition 2012 Introduction to Management https://www.boundless.com/management/groups-teams-and-teamwork/types-of-teams/self-managing-teams/

Implement Synthesizable Square Root Algorithm On Fpga Engineering Essay

Implement Synthesizable Square Root Algorithm On Fpga Engineering Essay The main objective of this paper is to implement synthesizable square root algorithm on FPGA. As square root function is not synthesizable on Silicon, this paper proposes optimized non restoring square root algorithm for unsigned 8 bit number on ED2C20F484C7 device in Cyclone II family. This algorithm is implemented in gate level abstraction of Verilog HDL. The basic building block of the design is CSM (Controlled Subtract Multiplex) block. It makes use of only subtract operation and append 01 which is an improvement over restoring algorithm. Keyword: FPGA,CSM,Verilog HDL,fixed point Introduction The square root function is a basic operation in computer graphic and scientific calculation application. Due to its algorithm complexity, the square root operation is hard to be designed in digital system. As known, digital system has been used in daily life or industrial purpose that may have been in need of square root operation to fully execute its functions. Scientists have developed various algorithms for square root calculation. But the implementation of algorithms is difficult because of their complexities and thus results into long delays for its completion. There are two main families of algorithms that can be used to extract square roots. The first family is that of digit recurrence, which provides one digit (often one bit) of the result at each iteration[6]. Each iteration consists of additions and digit-by-number multiplications (which have comparable cost) Such algorithms have been widely used in microprocessors that didnt include hardware multipliers. Most of the FPGA implementations in vendor tools or in the literature use this approach. Second family of algorithms uses multiplications. It includes quadratic convergence recurrences derived from the Newton-Raphson iteration [5]. The digit recurrence approaches allow one to build minimal hardware, while multiplicative approaches allow one to make the best use of available resources when these include multipliers. Also there are estimation method and digit-by-digit method. Digit-by-digit method is classified into two distinct classes: restoring and non- restoring algorithm [1]. In restoring algorithm, remainder is restored in the regular flow. So its implementation needs more hardware. Compared to the restoring algorithm, the non restoring algorithm does not restore the remainder, which can be implemented with fewest hardware resource and the result is hardware simple implementation. It is most suitable for FPGA implementation. Restoring and non restoring square root calculation Restoring Algorithm Step 1: If it is a 2n bit number then divide it in a group of 2 bits Step2: Subtract 1 from the first 2 digits (starting from MSB) Step3: Whenever the result of the subtraction is positive then the developed root is 1 otherwise 0 Step4: Whenever the result is negative, write it as it is. We have to restore the wrong guess by appending 01 and guessed square root. Step5: Now take the next two digits Step6: Append 01 (to be subtracted from next two digits of dividend) and guessed square root to subtract from the remainder. Step7: If the result of subtraction is negative then restore previous remainder by adding wrong guess by appending 01 and guessed square root. Step8: Every time guessed square root has to be updated while appending 01. Step9: Continue the steps until the group of two digits end 1 0 0 1.1 0 1 0 01 01 11 01.00 00 00 00 01 00 01 take next two digits from dividend 1 01 Append 01 Negative value 11 00 + 1 01 0 0 01 11 -10 01 11 10 Negative value + 10 01 01 11 01 10 00 01 11 00 00 10 01 01 01 01 1 00 10011 01 1011111 + 1001101 010110 00 010011001 000010111 00 100110101 1100100111 Figure 1: The example of restoring algorithm to solve square root B. Proposed Modified Non Restoring Algorithm A little modification in non restoring algorithm makes calculation faster. It uses only subtract operation and appends 01. It uses n stage pipelining to find square root of 2n bit number. The following algorithm describes the modified non restoring square root algorithm. Step1: Start Step2: Initialize the radicand (p) which is 2n bit number. Divide the radicand in two bits beginning at binary point in both directions. Step3: Beginning on the left (most significant), select the first group of one or two digit (If n is odd then first group is one digit ,else two bits) Step4: Select the first group of bits and subtract 01 from it. If borrow is zero, result is positive and quotient is 1 else it is 0. Step5: Append 01(to be subtracted next two digits of dividend) and guessed square root to subtract from remainder of previous stage Step6: If result of subtraction is negative, write previous remainder as it is and quotient is considered as 0, else write the difference as remainder and quotient as 1. Step7: Repeat step 5 and step 6 until end group of two digits. Step8: End 1 0 0 1.1 0 1 0 01 01 11 01.00 00 00 00 00 01 00 01 take next two digits from dividend 1 01 Append 01 11 10 01 01 11 01 100 01 11 00 00 1001 01 001011 00 1001101 00101100 00 10011001 0000010111 00 100110101 001011100 Figure 2: The example of modified non restoring algorithm to solve square root Basic Building Block for Non restoring algorithm Inputs of the building block are x,y,b and u while d and b0(borrow) are outputs. If b0=0, then d b0=( ~ x .y)+(b.~x)+(by); d= (~x.y.~b.~u)+(~x.~y.b.~u)+(x~y.~b)+(x.u)+(x.y.b); csmblock.jpg Figure 3: RTL schematic of CSM block The generalization of simple implementation of non restoring digit by digit algorithm for unsigned 6 bit square root by array structure is shown in Fig.4. Each row of the circuit executes one iteration of non restoring digit by digit square algorithm, where it only uses subtract operation and appends 01. Figure 4: Pipelined structure of 6 bit unsigned square root number The design can be optimized by minimizing the logic expressions and can be implemented by modifying CSM block. The specialized entities A,B,C,D,E,F,G and H are derived from CSM block and are defined as follows: For csmA, ybu = 100 b0 = ~x d = ~x For csmB, yu = 00 b0 = ~x.b d = ~x.b + ~b.x For csmC, u = 0 b0 = ~x.y + ~x.b + y.b d = ~x.y.~b + ~x.~y.b + x.~y.~b + x.y.b For csmD, yb = 10 b0 = ~x d = ~x.~u + x.u For csmE, y = 0 b0 = ~x.b d = ~x.b.~u + ~b.x + x.u For csmF, xy = 00 b0 = b d = b.~u For csmG, xyb = 010 b0 = ~x d = ~u For csmH, xyu = 000 b0 = b Figure 5: Optimized Pipelined structure of 8 bit unsigned square root number Results and analysis The Non Restoring algorithm can be implemented with least hardware resources and the result will be the faster than restoring square rooting techniques. The source code is implemented in such a way that it can be extended according to users requirement to calculate complicated square root in FPGA. Figure 6: Simulation result of 8 bit square root using non restoring algorithm The DE1 kit has 4 seven segment displays only so the maximum number which can be displayed is 9999d and also it doesnt have a decimal point. Hence output obtained is less precise if one of the displays is considered as a decimal point. Table 1 shows the list of Logic Elements usage for 8 bit implementation. This indicates the size of the implemented circuit hardware resource. Table 1: Comparison of LEs usage in 8 bit implementation No Implementation of non restoring algorithm for 8 bit LEs 1 8 bit (with seven segment) 85 2 8 bit (without seven segment) 71 3 optimized 8 bit (with seven segment) 64 4 optimized 8 bit (without seven segment) 50 Table 2: PowerPlay Power Analyzer Status No PowerPlay Power Analyzer Status 8 bit with optimization (mW) 8 bit without optimization (mW) Low Medium Low 1 Total Thermal Power Dissipation 71.65 447.96 72.84 2 Core Dynamic Thermal Power Dissipation 0 190.47 0 3 Core Static Thermal Power Dissipation 47.36 48.06 47.36 4 I/O Thermal Power Dissipation 24.29 209.44 25.48 Conclusion: This implementation and analysis shows that proposed method is most efficient of hardware resource. This is reasonable, because it only uses subtract operation and append 01. The result shows that the proposed algorithm is easy to implement and also uses less resources. The result is extended for square root implementation of 8 bit floating point number and also it can be expanded to larger numbers to solve complicated square root problem in FPGA implementation.

Prostitution Should be Legalized Essay -- Argumentative Persuasive Ess

Prostitution Should be Legalized I think that prostitution should be legalized because it is no different than any other service that we pay to receive. Besides, there are far more serious crimes that require the full attention of our police force than prostitution; therefore, policing it is a costly waste of time and police resources. Furthermore, prostitution is already legal in Singapore, Denmark, and a part of the United States as well. In this Essay, I will discuss these ideas; thereby, proving why prostitution should be legalized. Prostitution should be legalized because it should be treated like any other consensual, in-demand, and legitimate service. A prostitute performs sexual acts in exchange for money or gifts; therefore she sells her body and talents just like any other service. This system of selling sexual favors can be a profitable business and occupation for some. For example, a masseuse provides her service, massages, for people who want and are willing to pay for them. And, she uses her talent to create a source of income. This ...

Process Safety And Loss Prevention Plant Engineering Essay

The system in figure 1 schematic of a nomadic incineration unit. The equipment is arranged as a skid mounted bundle, recess and out pipes have been disconnected from unit.for the care purpose unit can be skiding out to open infinite and accessing needed constituents straight, or subsequently taking constituents from the unit in order to derive the entree. All supply and waste connexion are from the unit. Because of cramped conditions. Figure 2 it shows the forepart and side positions of the unit is 2.5m tallness, 5m deep, 2m broad. [ 1 ] Components: Heat money changer ( EX ) Rotary kiln ( RK ) Scrubing unit ( SC ) Temperature accountant ( TC ) Fan motor ( FM ) Screw feeder ( SW ) Screw motor ( SM ) Feed hopper ( FH ) The kiln, heat money changer, and scrubber are each secured to border by 6 bolts and there are 4 connexions to each of the motors. The whole unit can be slid out to let care utilizing raising cogwheel and this requires 20 proceedingss to hale out and 40 proceedingss to return. The clip takes to take nuts and bolts 2 proceedingss and the clip takes to replace 5 proceedingss [ 1 ] MTTR ( Average Time To Repair ) is besides known as Mean Corrective Tim – Mct, or TC. is colored norm of the fix times for the system. ( a ) ( I ) Calculation of MTTR when the unit is slid out for fix: Here failure constituents are removed from unit and it will be repaired and replaced to unit. Components: Heat money changer ( EX ) Rotary kiln ( RK ) Scrubing unit ( SC ) Temperature accountant ( TC ) Fan motor ( FM ) Screw feeder ( SW ) Screw motor ( SM ) Feed hopper ( FH ) Formula for MTTR: TE† c = [ a?‘ni=1 ( I »i.Tc ( I ) ) ] / a?‘ni=1 ( I »i ) Where: TE† c ( I ) is the disciplinary clip for the i'th unit. I »i is the failure rate of the i'th unit. N is the figure of unit. [ 2 ] Failure informations ( I » ) : Heat exchanger failure rate ( I » ) = 40 ( failure per 10^6hours ) or 40A-10^-6hours [ 3 ] Rotary kiln ( I » ) basic constituents of a rotary kiln are the shell, the furnace lining liner, support tyres, rollers, driven cogwheel and internal heat money changer. So rotary kiln failure rate we may gauge amount of all constituents which are utilizing to do rotary kiln. Under technology premise rotary kiln failure rate ( I » ) = 30 ( failures per 106hours ) or 30A-10-6 hours Under technology premise Scrubbing unit failure rate ( I » ) = 45 ( failures per 106hours ) or 45A-10-6hours Under technology premise fan failure rate ( I » ) = 57 ( failures per 106 hours ) or 57A-10-6 Corrective clip for constituents ( Tc ) : ( Tc ) = Tdet + Tloc + Tpla + Tsel + ( Tpre / Tlog ) + ( [ Trem + Trep ] /Trip ) + Tver + Tstu Tdet = observing mistake Tlo = placement failure Tpla = be aftering the work Ts = select the failed point Tpre = shutdown & A ; readying Tlog = logistics clip Trem = remotion of failed point Trep = replacing of failed point Trip = repair-in-place Tver = verify the repaired point Tstu = re-start [ 4 ] Corrective clip for heat money changer ( Tc ) Heat money changer has four connexions in the unit and heat money changer framed by 6 bolts and nuts so clip to take take that constituent ( heat money changer ) Entire nuts and bolts for the heat money changer in the unit = 6 Time taking to take bolts and nuts at each connexion = 2 proceedingss So clip taking to take heat exchanger = 6A-2 = 12 proceedingss Time taking to replace bolts and nuts at each connexion = 5 proceedingss Time taking to replace heat money changer = 6A-5 = 30 proceedingss And we have to unplug the connexions here we have entire 4 connexion Time taking to unplug pipe line the unit line from whole unit Unpluging pipe line from temperature accountant it will take clip = 20 proceedingss Unpluging pipe line from fan it will take clip = 25 proceedingss Unpluging pipe line from rotary kiln it will take clip = 40 proceedingss Unpluging pipe line from another connexion it will take clip = 20 proceedingss Connecting pipe line to temperature accountant it will take clip = 25 proceedingss Connecting pipe line to fan it will take clip = 35 proceedingss Connecting pipes line to rotary kiln it will take clip = 45 proceedingss Connecting pipe line to another connexion it will take clip = 30 proceedingss Corrective clip for heat money changer ( Tc ) = 12+30+20+25+40+20+25+35+45+30 =282 proceedingss or 4.7 hours Corrective clip for rotary kiln ( Tc ) Rotary kiln has four connexions connexions in the unit and rotary kiln framed by 6 bolts and nuts so clip to take take that constituent ( rotary kiln ) Entire nuts and bolts for the rotary kiln in the unit = 6 Time taking to take bolts and nuts at each connexion = 2 proceedingss So clip taking to take rotary kiln = 6A-2 = 12 proceedingss Time taking to replace bolts and nuts at each connexion = 5 proceedingss Time taking to replace rotary kiln = 6A-5 = 30 proceedingss And we have to unplug the connexions here we have entire 4 connexion Time taking to unplug the unit line from whole unit Unpluging pipe line from prison guard motor it will take clip = 23 proceedingss Unpluging pipe line from heat money changer it will take clip = 30 proceedingss Unpluging pipe line from another connexion it will take clip = 25 proceedingss Unpluging pipe line from another connexion it will take clip = 20 proceedingss Connecting pipe line to sleep together motor it will take clip = 28 proceedingss Connecting pipe line to heat exchanger it will take clip = 35 proceedingss Connecting pipe line to another connexion it will take clip = 25 proceedingss Connecting pipe line to another connexion it will take clip = 40 proceedingss Corrective clip for rotary kiln ( Tc ) = 12+30+23+30+25+20+28+35+25+40 = 268 proceedingss or 4.46 hours Scrubing unit has four connexions in the unit and framed by 6 bolts and nuts so clip to take take that constituent ( scouring unit ) Entire nuts and bolts for the scouring unit in the unit = 6 Time taking to take bolts and nuts at each connexion = 2 proceedingss So clip taking to take scouring unit = 6A-2 = 12 proceedingss Time taking to replace bolts and nuts at each connexion = 5 proceedingss Time taking to replace scouring unit = 6A-5 = 30 proceedingss And we have to unplug the connexions here we have entire 4 connexion Time taking to unplug the unit line from whole unit Unpluging pipe line from fan it will take clip = 25 proceedingss Unpluging pipe line from another connexion it will take clip = 30 Unpluging pipe line from another connexion it will take clip = 35 Unpluging pipe line from another connexion it will take clip = 25 Connecting pipe line to fan it will take clip = 30 proceedingss Connecting pipe line to another connexion it will take clip = 33 Connecting pipe line to another connexion it will take clip = 38 Connecting pipe line to another connexion it will take clip = 30 Corrective clip for scouring unit ( Tc ) = 12+30+25+30+35+25+30+33+38+30= 288 proceedingss or 4.80 hours Fan has besides four connexions with whole unit Unpluging pipe line from heat money changer it will take clip = 25 proceedingss Unpluging pipe line from temperature accountant it will take clip = 30 Unpluging pipe line from scouring unit it will take clip = 33 Unpluging pipe line from fan motor it will take clip = 27 Connecting pipe line to heat exchanger it will take clip = 30 proceedingss Connecting pipe line to temperature accountant it will take clip = 33 Connecting pipe line to scouring unit it will take clip = 38 Connecting pipe line to fan motor it will take clip = 30 Corrective clip for fan unit ( Tc ) = 25+30+33+27+30+33+38+30= 246 proceedingss or 4.10 hours Table 1: Technetium for the when the unit is slid out for fix Component I » ( failures per 106or A-10-6hours ) Tc ( hours ) I » . Tc Heat money changer 40 4.70 188 Rotary kiln 30 4.46 133.8 Scrubing unit 45 4.80 216 Fan 57 4.10 233.7 a?‘I »= 172 a?‘I »Tc= 771.5 Tc = a?‘I »Tc / a?‘I » = 771.5 /172 = 4.48 hours The MTTR ( Average Time To Repair ) when the unit is slid out for fix = 4.48 hours ( a ) ( two ) Calculation of MTTR when the unit is repaired in topographic point: Here we have to cipher MTTR ( Average Time To Repair ) whole unit in topographic point Components: Heat money changer ( EX ) Rotary kiln ( RK ) Scrubing unit ( SC ) Temperature accountant ( TC ) Fan motor ( FM ) Screw feeder ( SW ) Screw motor ( SM ) Feed hopper ( FH ) Formula for MTTR: TE† c = [ a?‘ni=1 ( I »i.Tc ( I ) ) ] / a?‘ni=1 ( I »i ) Where: TE† c ( I ) is the disciplinary clip for the i'th unit. I »i is the failure rate of the i'th unit. N is the figure of unit. [ 5 ] Failure informations ( I » ) : Heat exchanger failure rate ( I » ) = 40 ( failure per 10^6hours ) or 40A-10^-6hours [ 6 ] Rotary kiln ( I » ) basic constituents of a rotary kiln are the shell, the furnace lining liner, support tyres, rollers, driven cogwheel and internal heat money changer. So rotary kiln failure rate we may gauge amount of all constituents which are utilizing to do rotary kiln. Under technology premise rotary kiln failure rate ( I » ) = 30 ( failures per 106hours ) or 30A-10-6 hours Under technology premise Scrubbing unit failure rate ( I » ) = 45 ( failures per 106hours ) or 45A-10-6hours Under technology premise fan failure rate ( I » ) = 57 ( failures per 106 hours ) or 57A-10-6 Corrective clip for constituents ( Tc ) : ( Tc ) = Tdet + Tloc + Tpla + Tsel + ( Tpre / Tlog ) + ( [ Trem + Trep ] /Trip ) + Tver + Tstu Tdet = observing mistake Tlo = placement failure Tpla = be aftering the work Ts = select the failed point Tpre = shutdown & A ; readying Tlog = logistics clip Trem = remotion of failed point Trep = replacing of failed point Trip = repair-in-place Tver = verify the repaired point Tstu = re-start [ 7 ] here we do n't necessitate to take constituents from unit for fix Corrective clip for heat money changer ( Tc ) : Heat money changer has four connexion in the whole unit Time taking to unpluging the unit line from whole unit Unpluging pipe line from temperature accountant it will take clip = 20 proceedingss Unpluging pipe line from fan it will take clip = 25 proceedingss Unpluging pipe line from rotary kiln it will take clip = 40 proceedingss Unpluging pipe line from another connexion it will take clip = 20 proceedingss Connecting pipe line to temperature accountant it will take clip = 25 proceedingss Connecting pipe line to fan it will take clip = 35 proceedingss Connecting pipes line to rotary kiln it will take clip = 45 proceedingss Connecting pipe line to another connexion it will take clip = 30 proceedingss Corrective clip for heat money changer unit ( Tc ) = 20+25+40+20+25+35+45+30 = 240 minute or 4 hours Corrective clip for rotary kiln ( Tc ) : Unpluging pipe line from prison guard motor it will take clip = 23 proceedingss Unpluging pipe line from heat money changer it will take clip = 30 proceedingss Unpluging pipe line from another connexion it will take clip = 25 proceedingss Unpluging pipe line from another connexion it will take clip = 20 proceedingss Connecting pipe line to sleep together motor it will take clip = 28 proceedingss Connecting pipe line to heat exchanger it will take clip = 35 proceedingss Connecting pipe line to another connexion it will take clip = 25 proceedingss Connecting pipe line to another connexion it will take clip = 40 proceedingss Corrective clip for rotary kiln ( Tc ) = 23+30+25+20+28+35+25+40 = 226 minute or 3.76 hours Corrective clip for scouring unit ( Tc ) : Unpluging pipe line from fan it will take clip = 25 proceedingss Unpluging pipe line from another connexion it will take clip = 30 Unpluging pipe line from another connexion it will take clip = 35 Unpluging pipe line from another connexion it will take clip = 25 Connecting pipe line to fan it will take clip = 30 proceedingss Connecting pipe line to another connexion it will take clip = 33 Connecting pipe line to another connexion it will take clip = 38 Connecting pipe line to another connexion it will take clip = 30 Corrective clip for scouring unit ( Tc ) = 25+30+35+25+30+33+38+30 = 246 proceedingss or 4.10 hours Corrective clip for fan ( Tc ) : Unpluging pipe line from heat money changer it will take clip = 25 proceedingss Unpluging pipe line from temperature accountant it will take clip = 30 Unpluging pipe line from scouring unit it will take clip = 33 Unpluging pipe line from fan motor it will take clip = 27 Connecting pipe line to heat exchanger it will take clip = 30 proceedingss Connecting pipe line to temperature accountant it will take clip = 33 Connecting pipe line to scouring unit it will take clip = 38 Connecting pipe line to fan motor it will take clip = 30 Corrective clip for fan unit ( Tc ) = 25+30+33+27+30+33+38+30= 246 proceedingss or 4.10 hours So based on computations and observation MTTR ( Mean To Time Repair ) for unit is slid out for fix is significantly more than unit is repaired in topographic point. Table 2: Technetium for the when the unit is repaired in topographic point Component I » ( failures per 106or A-10-6hours ) Tc ( hours ) I » . Tc Heat money changer 40 4.0 160 Rotary kiln 30 3.76 112.8 Scrubing unit 45 4.10 184.5 Fan 57 4.10 233.7 a?‘I »= 172 a?‘I »Tc= 691.0 Tc = a?‘I »Tc / a?‘I » = 691 /172 = 4.01 hours The MTTR ( Average Time To Repair ) when the unit is slid out for fix = 4.01 hours Mentions: ( 1 ) ( a ( I ) ) ( a ( two ) ) [ 1 ] Plant dependability and maintainability, assignment inquiry paper, faculty ( CPE6250 ) held on November 30 to December 3 2009. [ 2 ] [ 4 ] [ 5 ] [ 7 ] Cris Whetton, ility technology. Maintainability. [ Lecture press release ] .from works dependability and maintainability, faculty ( CPE6250 ) held on November 30 to December 3 2009. [ 3 ] [ 6 ] Frank P. Lees, 1996, Loss bar in the procedure industries, 2nd edition, volume 3. 1b ) Design alterations to cut down Mean Time To Repair ( MTTR ) : To accomplish optimal MTTR the undermentioned design consideration are recommended: The heat exchanger stuff must be considered based on the operating temperature of the liquid More dependable and maintainable stuff used in the rotary kiln Better we have one more scouring unit to cut down the Mean Time To mend MTTR Motor capacity must designed based on chilling demands All the pipe parametric quantities must be based on the operating temperature of the liquid throwing it Material which is utilizing to do all constituents should be defy all status The temperature accountant must be calibrated for the liquid temperature 1c ) Instrumentality which has system is utile to find the mistakes.so instrumentality in this system temperature accountant ( TC ) : Here TC maps to modulate the temperature of the liquid come ining the heat money changer that is, it pre-controls the liquid come ining the heat money changer. As shown in the figure, the temperature accountant modulate the temperature of the liquid released from the heat money changer and before being cooled by the fan which is control by fan motor. So temperature accountant is utile to observing the mistake which may happen in the heat money changer. Based on the given figure it can be likely assume that degree index may be used for the rotary kiln. a flat index is placed at the top of the rotary kiln. This is used is indicate the maximal degree of the mixture that can be accommodated in a rotary kiln. So this may be indicated the mistakes if anything occur. A flow rate valve is placed in the scrubber unit, so as to command the flow rate alkalic solution into the scouring unit. This flow rate valve allows merely the coveted sum of solution in to the scouring unit. Once the coveted degree is reached the valve will automatically close off the flow of liquid into the unit. And we have some detector dismay at the fan and fan motor and screw motor why because if these have any jobs will gives the signals so we can easy find the mistakes. Due to the incorporation of these instrumentality into the chief system the opportunities of failure is significantly reduced 2 ) Question description: Procedure works to respond liquid A and liquid B to bring forth merchandise C. liquid A passing into storage A utilizing liquid accountant. From storage it will pump to reactor. Liquid B go throughing into storage B utilizing liquid accountant from storage B to pumping to reactor. From reactor merchandise C coming out. Acid gas from reactor pumping to scouring unit. In scouring unit acid gas is cleaned utilizing alkalic solution which is go throughing into scouring unit. Scrubing unit leaves impersonal waste watercourse. Liquids ever available at the recesss to the procedure. There is at least two scouring units working right for the procedure. Stand-by pumps switch over automatically. Pipe work failures can be ignored. [ 1 ] Available informations: The computing machine system has a dependability of 0.9997 over one twelvemonth The operator dependability over one twelvemonth is 0.85 for indicated mistakes and 0.95 for mistakes which raise an dismay Scrubber unit has a weilbull failure characteristic with I · = 600 yearss, I? = 60days, and I? = 1.8 Reactor failures can affect the fomenter which has two failure manners. Shaft break failure rate = 0.1/year Motor failure rate = 0.3/year [ 1 ] 2 ( a ( I ) ) Fault tree analysis here merchandise fails to run into specification is the top event Alarm failure Liquid control LAL fails Liquid control Low degree High degree Agitator failure Coking job Motor failure Shaft break High degree Low degree Excess flow of liquid Angstrom Excess flow of liquid B Reactor Pump failure 2 ( a ( two ) ) Fault tree analysis here liquid waste watercourse composing outside bounds is the top event Low degree High degree Internal mal maps failure Connection fails between scrubbers Improper cleansing temperature Improper alkaline solution pumping to scrubber unit Scrubber unit failure Improper flow reactor to scrubber High degree Low degree Low degree High degree 2a ) computation of dependability of parts of the system Here parts of the system: Storages Reactor Agitator Pumps Scrubing unit Dependability of reactor: Here reactor failure can affect the fomenter failure. First one is shaft break and 2nd one is motor failure Failure rate of shaft break = 0.1/year Failure rate of the motor = 0.3/year Scrubber unit has a weilbull failure characteristic with I · = 600 yearss, I? = 60days, and I? = 1.8 [ 1 ] Failure rate of pump ( I » ) = 13A-10-6hours [ 2 ] Dependability of shaft break: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 [ 3 ] Failure rate of shaft break = 0.1/year So utilizing this we are happening T Z ( T ) = I?/I ·I? ( t-I? ) I?-1 0.1/year = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 Here one twelvemonth = 365 yearss 0.1/365 = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 T = 90.11 yearss Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 = 0.995 So dependability for shaft break = 0.995 Dependability of motor: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 Failure rate of the motor = 0.3/year So utilizing this we are happening T Z ( T ) = I?/I ·I? ( t-I? ) I?-1 0.3/year = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 Here one twelvemonth = 365 yearss 0.3/365 = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 T = 177.29 yearss Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 = 0.948 So dependability for motor = 0.948 Dependability for scouring unit: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^I? Here we have the T = 133.6 yearss Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Z ( T ) = ( 1.8/ ( 600 ) 1.8 ) A- ( 133.6-60 ) 1.8-1 Z ( T ) = 0.2/year Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^I? = 0.996 So dependability for scouring unit R ( T ) = 0.996 Dependability of pump: Failure rate of pump ( I » ) = 13A-10-6hours Dependability of pump R ( T ) = e-I »t Surviving clip t = 70 yearss One twenty-four hours = 24 hours Surviving clip T = 1680 hours Dependability of pump R ( T ) = e-I »t = vitamin E ( -13A-10^-6A-1680 ) Dependability of pump R ( T ) = 0.978 Mentions: [ 1 ] Plant dependability and maintainability, assignment inquiry paper, faculty ( CPE6250 ) held on November 30 to December 3 2009. [ 2 ] Frank P. Lees, 1996, Loss bar in the procedure industries, 2nd edition, volume 3. [ 3 ] Cris Whetton, ility technology. Failure information analysis. [ Lecture press release ] .from works dependability and maintainability, faculty ( CPE6250 ) held on November 30 to December 3 2009. 2b ) Reliability block diagram for the complete system Pump 1 Storage A Pump2 Scrubing unit Reactor Pump Storage B computation of dependability of the complete system over one twelvemonth: Here parts of the system: Storages Reactor Agitator Pumps Scrubing unit Dependability of reactor: Here reactor failure can affect the fomenter failure. First one is shaft break and 2nd one is motor failure Failure rate of shaft break = 0.1/year Failure rate of the motor = 0.3/year Scrubber unit has a weilbull failure characteristic with I · = 600 yearss, I? = 60days, and I? = 1.8 [ 1 ] Failure rate of pump ( I » ) = 13A-10-6hours Failure rate of fan ( I » ) = 57A-10-6hours [ 2 ] Dependability of shaft break: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 [ 3 ] Failure rate of shaft break = 0.1/year So utilizing this we are happening T Z ( T ) = I?/I ·I? ( t-I? ) I?-1 0.1/year = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 Here one twelvemonth = 365 yearss 0.1/365 = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 T = 90.11 yearss Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 = 0.995 So dependability for shaft break = 0.995 Dependability of motor: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 Failure rate of the motor = 0.3/year So utilizing this we are happening T Z ( T ) = I?/I ·I? ( t-I? ) I?-1 0.3/year = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 Here one twelvemonth = 365 yearss 0.3/365 = ( 1.8/ ( 600 ) 1.8 ) A- ( t-60 ) 1.8-1 T = 177.29 yearss Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^6 = 0.948 So dependability for motor = 0.948 Dependability for scouring unit: Equation for failure rate: Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Here I? = form factor I · = characteristic life I? = location parametric quantity T = lasting a clip Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^I? Here we have the T = 133.6 yearss Z ( T ) = I?/I ·I? ( t-I? ) I?-1 Z ( T ) = ( 1.8/ ( 600 ) 1.8 ) A- ( 133.6-60 ) 1.8-1 Z ( T ) = 0.2/year Equation for the dependability: R ( T ) = e- ( ( t-I? ) /I · ) ^I? = 0.996 So dependability for scouring unit R ( T ) = 0.996 Dependability of pump: Failure rate of pump ( I » ) = 13A-10-6hours Dependability of pump R ( T ) = e-I »t Surviving clip t = 70 yearss One twenty-four hours = 24 hours Surviving clip T = 1680 hours Dependability of pump R ( T ) = e-I »t = vitamin E ( -13A-10^-6A-1680 ) Dependability of pump R ( T ) = 0.978 Dependability of the complete system over twelvemonth R ( T ) = norm of system parts dependability = ( 0.995+0.948+0.996+0.978 ) /4 = 0.979 Therefore dependability of the complete system over twelvemonth = 0.979 Mentions: [ 1 ] Plant dependability and maintainability, assignment inquiry paper, faculty ( CPE6250 ) held on November 30 to December 3 2009. [ 2 ] Frank P. Lees, 1996, Loss bar in the procedure industries, 2nd edition, volume 3. [ 3 ] Cris Whetton, ility technology. Failure information analysis. [ Lecture press release ] .from works dependability and maintainability, faculty ( CPE6250 ) held on November 30 to December 3 2009. 2c ) To accomplish a mark dependability of 0.90 over one twelvemonth: Reliability mark is a nothing failure mark. This is an of import mark implied for those low acting workss, such workss does non accomplish certain ends designed by applied scientists. So we have to put appropriate mark to accomplish works design. the dependability of the system must be improved to accomplish the mark. to accomplish the dependability mark or to better dependability three basic ways must be employed. By system design By component specification By preventative care By system design: – The basic regulation of our system design is to maintain the design has simple as possible. the system is more dependable if the system is simple. Some of the stairss include, System simplification: To cut down the complexnesss in procedure works at the design phase its ego Decrease in the usage of Complex parts by replacing them with more cardinal parts The design should be made simple and easy to under base Decrease in constituent count: The figure of constituents used in the works must be reduced. complex constituents must be avoided for the simpleness of the design. Mistake tolerance: The basic features of mistake tolerance require: No individual point of failure No individual point of repair- the system must run without any break during the procedure of fix when the system experiences any jobs. Mistake isolation to the neglecting component- in instance of failures the failed portion of the system must be isolated from the pained system. This requires necessary failure sensing mechanism. Fault containment to forestall extension of the failure Handiness of reversion modes- some failures may do cripples to the full system, to avoid the full procedure system must force to the safe manner By component specification: For the dependability of a constituent it must be adequately specified for their full length of service. Extra dependability can be provided by runing the constituents at lower emphasis so their operating emphasiss. By making so early failures of the constituents can be reduced. in a procedure industry it is really hard to better dependability merely by specification. This is attributed to the deficit of necessary informations sing the affect of emphasiss on the constituents. Components of high quality can non be used ever for economic grounds. Normally the parametric quantities required to better dependability frequently contradict with procedure demands. Some of the dependability betterments include: Use of disciplinary maintenance- it is defined as the care which is required to mend and convey merchandise after the fix is carried out. it is carried out in constituents who is failures does n't impact of the overall working of the procedure system significantly. This activity chiefly involves fix, Restoration or replacing of constituents. Design improvement-the design of any high quality procedure works is based on the design parametric quantities and proficient specifications. the reactor design must be improved for high rates of efficiency. Temperature, force per unit area and other external considerations must be included in the design of reactor and storage armored combat vehicles. Quality control-Quality control assures conformity to specifications. quality control checks whether measurings of the constituents like reactors, storage armored combat vehicle, scrub units as in this instance conform to the demands. Preventive care: Is defined as a care carried out to forestall failure or warring out of constituents in the procedure works. This is carried out by supplying systematic review, sensing and bar of inchoate failure. The preventive care attempts are aimed at continuing the utile life of equipment and avoiding premature equipment failures, minimising any impact on operational demands. In add-on to the everyday facets of cleansing, adjusting, lubricating and proving. it is carried out merely on those points where a failure would hold expensive or unacceptable effects e.g. reactors, storage armored combat vehicles, scouring units. Many of these points are besides capable to a statutory demand for review and preventative care. [ 1 ]