Ripng Ospfv3 And Eigrpv6 In Ipv6 Information Technology Essay

Ripng Ospfv3 And Eigrpv6 In Ipv6 Information Technology Essay A working model of the purpose is implemented in Packet Tracer and it illustrates how routing can be achieved between to dissimilar routing domains. It highlights both positive and negative aspects like full visibility and potential routing loops consequently. Findings Access lists can be used to avoid routing loops. More study is needed when OFPF is redistributed into RIP-EIGRP combination. Originality/value this paper addresses how different routing protocol domains can be interconnected and enjoy the full visibility of routers that belongs to different routing protocols Keywords EIGRPv6, RIPng, OSPFv3, redistribution, Administrative Distance, Access list, routing loops Introduction Though IPv4 is proved as one of the dominant network layer protocol which has been in use for last 3 decades since its development in 1981 (RFC 791), people still find a need for an enhanced and better protocol due to the ever growing networks. When IPv4 address scheme was made, developers did not give a thought that internet would explode and expand too rapidly as we see the current picture of networks today. To slow down the depletion of address space, IPv4 protocol was further updated in 1993 and started using as classless (CIDR) (RFC 1519) which was failed to be adequate enough as a long term solution for conserving IP addresses. As the architecture of IPv4 has been subjected to changes, so the underneath routing protocols like RIP, OSPF, IGRP, EIGRP, BGP, IS-IS etc. are also been updated or reformed in order to have better routing and converged networks. In 1994, NAT was came up with public address and private address concept and succeeded to a great extent in conserving the IPv 4 addresses (RFC 1631). A report generated by IANA predicted that IPv4 would completely run out of available addresses by 2011 (Potaroo, 2011). In 1998 December, a new internet protocol, IPv6, is proposed with Expanded Address Capabilities, support for hierarchy in addressing, simplified header format and support for real-time services like IPTV (RFC 2460). Just like in IPv4, IPv6 (routed protocol) is also dependent on routing protocols to route the packets. Since 1997, networking groups started developing routing protocols to support IPv6 and they successfully standardized RIPng, OSPFv3, EIGRPv6, BGPv6 and IS-ISv6 in the subsequent years. These protocols are developed for 128 bit address and better security, without having much difference to their counterparts in IPv4. We cannot expect entire network to be operated by a single type of dynamic routing protocol. Depending upon size of network and requirements, administrator may choose relevant routing protocol still maintaining full connectivity among the nodes in network. The possible solutions are using static routes or deploying mutual redistribution among different protoc ol domains. As IPv4 routing protocols, RIPv2, OSPFv2 and EIGRP are well established, there are minimal redistribution issues among them. Whereas in case of IPv6 there is good amount of scope to understand and research on practical issues that arise when mutual redistribution is applied among them. Following sections of document deals with review of RIPng, OSPFv3 and EIGRPv6 individually and a practical implementation in packet tracer is shown. Later sections demonstrate how Administrative Distance (AD) of a dynamic routing protocol create loops or route to infinity and how access-list can be applied in order to avoid them. RIPng RIPng is developed in 1997 and it uses Ford-Fulkerson algorithms. And nevertheless to mention, RIPng is a distance vector protocol which depends on route on rumor, just like its earlier versions. RIPng is best suited for smaller size networks having hop count limited to 15. Beyond 15, any router would be considered as unreachable (RFC 2080). For every 30 seconds, entire Database of routing information is being exchanged and this could be keep networks always processing data. It uses hold down timer prevent route loops when a router or a particular interface of a router, or a network goes down. Split horizon is another concept which is enabled by default in cisco routers help preventing routing loops. Trigger updates are generated when there is a change in the topology to have better convergence. One of the reasons for slower convergence when compared to link-state routing protocols is when an update is being multicasted to the neighbors, first they sniff the packet, analyze and then make changes to TTL field, metric and then forward the packet to next hop. Because of this update packet spends notable amount of time at every node and hence convergence is slower as shown below. Fig: forwarding process in RIPng The main drawbacks of this protocols is higher convergence time, limited number of router in routing domain and high amount of traffic for administration and maintenance. Below is another screenshot from the working model which depicts the RIPng routing domain. OSPFv3: RFC2740 is proposed and standardized for OSPF that supports IPv6 networks in 1999. As its previous versions, OSPFv3 came up with many of its fundamental concepts like support for areas, flooding, and algorithm (Dijkstra) for calculating shortest path first (SPF), Designated and Backup Designated router selection intact. Changes are done to accommodate new IP address format which is 128 bit one. OSPF in IPv6 networks run on per-link basis where as it is per-IP-subnet based in IPv4. And the main difference is authentication is removed from OSPF protocol itself as IPv6 got its own Authentication Header and Encapsulating Security Payload (RFC 2740). OSPF form adjacency with attached routers and work in areas. A router can have many instances of OSPF process and hence inter connecting more nodes which belongs to more than one area as shown in the below screenshot from the working model of packet tracer. Fig3: OSPFv3 and Inter-Area connections OSPF uses hello packets and to make sure of adjacency. For every 10 seconds, OSPF enabled router sends hello packets to neighbors and in case if does not get reply for hello in four times the hello interval that particular adjacent node is announced as dead and accordingly updates will be sent to the affected nodes only. OSPF does not process the update packet before forwarding. Hence convergence time is very low even in a big network. Even though SPF algorithm looks simple, it involves complex calculation and hence it requires large amount of resources i.e CPU memory and time. OSPF is best suited to deploy in larger networks. EIGRPv6: EIGRP is ciscos proprietary protocol and works only in cisco routers. EIGRPv6, like its previous version, uses Diffusion Algorithm (DUAL) to make network really loop-free. Though EIGRP is categorized as distance vector protocol, it carries features from link-state model and hence EIGRP deals with neighbor and topology databases. Like RIP, EIGRPv6 does not send its entire database to the adjacencies which would create lot of administrative traffic. EIGRPv6 maintains installs both successor and feasible successor routes in the topology table. Whenever the successor route goes down, within no time, it installs feasible successor into the routing table and hence convergence is very fast in EIGRPv6. It generates triggers whenever there is a change in the topology. Just like OSPF, EIGRP can run multiple instances of the process and hence work in multiple process domains. EIGRPv6 uses bandwidth and delay for calculating metric as default parameters and user can change this metric by includi ng reliability and load. Below is the screenshot of EIGRPv6 routing domain with backup routes (Netacad, 2011). Fig4: EIGRPv6 domain Redistribution: Using a routing protocol to advertise routes that are learned by some other means, such as by another routing protocol, static routes, or directly connected routes, is called redistribution (Cisco, 2011). Though it is recommended to use single routing protocol throughout the network, in some scenarios it may be required to advertise routes of a particular routing protocol domain to a different routing protocol domain, especially when organizations merge, or multiple departments merge. Every routing protocol has its own way of calculating and using metric for routing packets. RIPng uses hop count as metric, OSPF is based on Bandwidth and EIGRP use Bandwidth, delay, reliability, load and MTU to calculate metric. As metric plays key role in redistribution, it needs to be set along with CLI commands of redistribution. Care must be taken to advertise correct metric while redistributing. Following is the screenshot from working model. For illustration purpose let us focus on redistribution between EIGRP and RIP routing protocol domains. Fig: Mutual redistribution between EIGRPv6 and RIPng Routers in the EIGRP domain neither reach nor have the visibility of routers deployed in RIP domain. The common point between these routing domains is called edge router and this router run both RIPng and EIGRPv6. We need to enable mutual redistribution on these edge routers. When we configure the edge router with following configuration, routes of RIP protocol domain will be learned by routers in the EIGRP protocol domain with the specified metric. In the same way when we issue the following configuration on the edge router, routes of EIGRP will be distributed into RIP protocol domain and RIP speaking routers learns about EIGRP routes with specified metric. We can find if redistribution successfully imported routes into adjacent routing domain by observing the route entries of IPv6. All native routes are either marked as C or D representing directly connected and EIGRP routes respectively. Whereas distributed routes are marked as EX exterior EIGRP routes with administrative distance of 170 and calculated metric to the respective network. The same can be observed in the following screenshot. Each and every router in both the domains has full visibility to any other router and the same can be checked in the working model of packet tracer using ping command. In the same way multi-area OSPF protocol domain is configured with the below shown configuration having Area0 as backbone area and Area1 and Area2 are connected to backbone area on both the sides. Now configure mutual redistribution between OSPF into RIP. Routes should be updated accordingly and full connectivity must be established between any two nodes in the packet tracer. When OSPF routes are redistributed into a system having RIP and EIGRP, all the routers are not converged and formations of routing loops are suspected. This will be discussed in the following sections. Note: After launching the packet tracer, it is recommended to wait for 1-2 minutes before pinging any node as it takes some time for converging. Administrative Distance contributing to routing loops: As each dynamic routing protocol has its own way of calculating metric, they cannot be used to compare routes. This can be taken care by Administrative Distance as it represents the degree of reliability of a route. Most preferable routes are chosen based on the AD. Lower the AD, better the reliability and precedence will be given to that route. Static routes got preference over any other route as AD for static route is 0. In the same way Interior EIGRP got AD of 90, OSPF is of 110, RIP is of 120 and finally Exterior EIGRP is of 170. A close observation of EIGRP domain says that it got redundant paths introducing possibility for routing loops. R2 and R4 learn about network 2006::0/64 (which is highlighted in red) through RIPng and advertise this information into EIGRP domain when redistribution is applied. Using EIGRPv6, R2 learns about network 2006::0/64 from R1 or R4 and R4 learns the same network from R1, R5 or R2. As EIGRP has less Administrative Distance (90) than RIP (120), the EIGRP route is the one used in the routing table causing routing loop. Even techniques like split horizon is used, still these networks suffer from convergence problem. When EIGRP is redistributed into RIP this problem gets more worsen as R3 learns about 2006::0/64, which is a directly connected network, from other routers. Solution to this problem is using access lists to deny routing updates of 2006::0/64 into its own routing domain i.e RIP. Following shown configuration applied on both R2 and R4 describes how to set access-list. After access lists are applied on edge routers, both domains become fully converged with full reachability. Insufficient route updates will happen when redistribution is applied between RIP and OSPF domains. Potential reasons for this problem are yet to be known and further study is required. Packet tracer 5.3 does not support tracert for IPv6 networks and sometimes it causing the application to crash. CONCLUSION: In the stand still mode, each routing domain is fully converged and any node can ping any other node inside the routing domain. When RIP and EIGRP are mutually redistributed into each other, potential route loops and insufficient route tables are observed because of Administrative Distance. These problems are attended by using access list with permit and deny commands. After applying access lists on edge routers, EIGRP and RIP are fully converged and full connectivity is established. Insufficient route tables are observed with OSPF redistribution into RIP and the causes of this problem are yet to be studied. Simulation tool, Packet tracer 5.3, needs to be updated as when IPv6 networks are simulated two things are observed. One, application is getting crashed frequently without any reason and the second is tracert is not supported for IPv6 networks. This model could be further used to analyze QoS and path vector protocols like BGP and IS-IS and to examine how networks behave in Autono mous Systems (AS). [1]RFC 791, http://www.rfc-editor.org/rfc/rfc791.txt, Last accessed 14-01-2011 [2] RFC 1519, ftp://ftp.rfc-editor.org/in-notes/rfc1519.txt, Last accessed 14-01-2011 [3] RFC 1631, ftp://ftp.rfc-editor.org/in-notes/rfc1631.txt, Last accessed 14-01-2011 [4] Potaroo 2010, IPv4 Address report, http://www.potaroo.net/tools/ipv4/, Last accessed 14-01-2011 [5] RFC 2460, ftp://ftp.rfc-editor.org/in-notes/rfc2460.txt, Last accessed 14-01-2011 [6] RFC 2080, ftp://ftp.rfc-editor.org/in-notes/rfc2080.txt, Last accessed 14-01-2011 [7] RFC 2740, ftp://ftp.rfc-editor.org/in-notes/rfc2740.txt, Last accessed 14-01-2011 [8] Netacad 2011, https://auth.netacad.net/idp/Authn/NetacadLogin, Last accessed 14-01-2011 [9]Cisco 2011, http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a008009487e.shtml, Last accessed 14-01-2011

Paula and Christina’s Haunted Adventures Essay

The night was very chilly and all the kids were running, laughing, and having a lot of fun at the third grade’s fall festival. The Roosevelt Elementary school decided to do a haunted house maze. The haunted maze was pitched black and filled with gory decorations. One room had a bed with a dead woman laying on it and the walls were stained with blood. In another section of the maze was a cemetery, where zombies were walking by. The haunted house maze also had a garage where a car was parked and there was also room filled toys and with many clowns. The children of Roosevelt Elementary were thrill to go inside wondering what they would encounter and how scary it might be. There were two little girls, barely nine year’s old, seeking the frightening adventure, but they were also very scared. One of the little girls was shivering with fear and she slowly walked from room to room looking at the decorations and the scary people. She knew this was make-believe, but she was still very cautious. The other little girl had made her way into the room where the dead woman was laying on the bed. She slowly tried to make her way across the room to the other door. She never noticed the woman rising from the bed and was walking towards her very rapidly. The dead, bloody woman touch the little girl, which made her scream and take off running. The little girl ran as fast as she could, turning all corners until she ran straight into the other little girl. They both ended up falling to the ground. Looking up, they both smiled at one another. One little girl says, â€Å"Hi, my name is Paula Ramirez. † The other little girl replies, â€Å"Sorry for running into you, my name is Christina Lopez. Paula and Christina help each other get up from the ground and agree to finish the haunted house maze together. They felt less scared and even laugh at some of the scary people. As they are exiting the haunted house maze, Paula tells Christina â€Å"wow, this was so much fun†. Christina replies, Paula let’s make a pact, every year we will seek a scary haunted house and we will both go together†. Paula smiles and agrees. Since then Paula and Christina have become best friends and they search for scary haunted houses to visit every year. The family vanishes in 1925 leaving the sole survivor James Clark. Christina does not read the paragraph, but smiles and turns to Paula and says â€Å"perfect†. The two girls get together every day preparing for the trip they will make to the haunted house. They go to the mall and shop for costumes. They buy food so they can take for the road, and plan on when they should head to the haunted house. The night of the haunted house arrives and Christina goes to Paula’s house. Paula is sitting in her room when Christina enters and asks, Paula â€Å"Are you ready? Paula replies â€Å"of course† Then, the two girls go down the stairs, out the house and enter Paula’s car. Paula drives thirty miles and arrives and Saddle Creek Ranch where the haunted house is located. The yard is filled with many trees and with old rusted decorations on the lawn. The house has busted windows, and the painting of the house is falling apart. Paula notices that there is two boys standing in the entrance of the house. Paula and Christina exit the car and make their way to the boys. One boy says â€Å"Come on Erik, stop being scare and let’s go in already. Erik replied â€Å"Scott, if you are so tough you open the door†. Just as soon as Scott was going to touch the door knob the door swings open and a tall dark figure stands there looking at them. The tall dark figure steps forward and says â€Å"Do you dare to enter? Beware this may be you last hallows eve† â€Å"YES! † replied all four of them. Paula and Christina follow the two boys and the tall man into the house. The man leads them into the living room and directs them to stand in a circle. He looks at each one, studying them from head to toe. He says â€Å"My name is Mr. Clark; Saddle Creek Ranch is y family’s home. They lived here many years until they all vanish one night, but you can still hear them, and if they want you to, you can definitely see them†. He goes on to give directions and instructions how to enter the house, and to exit the house. When Paula and Christina turn to see each other, Mr. Clark vanishes. A light turns on and leads the group into the kitchen, you hear can hear the water running but the water is off; Paula grabs Christina’s hands and says â€Å"this house is different†. The sound of people talking suddenly starts and is very loud that makes Scott jump up. The group makes their way to the door but the light to the kitchen turns off and then turns on again and leads them to the dining room. As they enter the room the talking suddenly dies down, and a chair is thrown to the group. As the group separates to avoid being hit, the lights go out and it is so dark that Paula cannot see where Christina is at. BANG. Suddenly the lights turn on and Paula looks for Christina, Erik calls out to Scott but he is nowhere to be found. Paula and Christina start to get worried and scare sets in. Erik starts to whimper, and continues to call for Scott but his call is not answered. Paula and Christina continue to make their way out the dining room and Erik follows. The lights to a hallway turn on and the group passes many rooms, and walk down some stairs. They reach a door it creeks open, Christina turns towards Paula and says, â€Å"Paula let’s not go in, we should turn around and go back†, before Paula can reply, Erik replies â€Å"let’s just continue this could be the end†. Erik pushes the door open, the room is completely dark, and the lights do not turn on like before. Suddenly a rustling and whimpering sound begins Christina is so frightened that she begins to cry. Paula hold Christina’s hand so tight and tries to reassure her that everything will be okay. The sounds get louder and louder and suddenly â€Å"BANG†. Paula and Christina stop walking, the lights turn on, both girls try to search and look around to see where and what room they are in. Paula is the first to notice that they are in the basement but someone is missing, Erik is nowhere to be found. Christina falls to the floor and begins to sob very loud and yells â€Å"Paula, let’s get out, I do not want to be in here anymore! Paula holds back her tears and replies, â€Å"Christina we will be okay, I will get us out, I promise†. Paula begins to look for something that can be a used as weapon, and looks for an exit. Paula finds a hammer, and notices a window. She rushes to it, and then the lights go out again. Seconds pass by but to Paula it seems like an eternity, she calls Christina’s name, â€Å"I am here† replies Christina. The lights turn on again; Paula looks to find Christina, and sees Mr. Clark holding her with a knife to her throat. â€Å"Did you think I was being funny when I said this would be your last hollows eve? Mr. Clark asked Paula. Paula replies â€Å"this trip was meant to be fun I never imagine it would be like this. † Mr. Clark looks at her and gives her an evil smile. Then, he says to Paula, â€Å"Let’s play a game, if you win you and your friend may leave, but if you lose you and your friend must stay forever. † Paula thinks for a long minute and nods her head up and down. He looks at Paula for a moment and says â€Å"Answer me this if you are correct you and your friend may go. Am I alive or is this your imagination? † Paula stares at Mr. Clark and responses â€Å"you are alive†. Mr.  Clark laughs, â€Å"Child, how can I be alive if I died in 1963† replies Mr. Clark. Paula begins to cry along with Christina, and mouths the words â€Å"I am sorry†. The lights turn off again, bang, bang, bang. Paula holds her hands out trying to feel for Christina; the lights turn back on again. Mr. Clark is nowhere to be found. Paula grabs Christina’s hand and yells â€Å"hurry, the window, now! † Paula and Christina, crawl out the window, look to see where the car is parked and ran towards it. The girls quickly get in, crying and scare. Paula turns the car on, put it in gear and head out the ranch. As both Paula and Christina seat in the car crying for a long while Paula breaks the silence and says â€Å"Christina, promise me that this will be our last haunted house†. Christina replies â€Å"definitely, Paula this was our last one†. The girls continue to drive on the dark road, without realizing the road does not end. Then Christina turns to Paula and asks â€Å"Paula, are we getting closer to your house now? † Paula looks at the road and realizes it’s the road to Saddle Creek ranch. Paula realizes what has happened and turns to Christina and says â€Å"Christina, I am sorry but Saddle Creek ranch is our home now. †

Type 2 Diabetes Mellitus As A Disease State - 1816 Words

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An Exegesis Of Philippians 3 12-21 - 1822 Words

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Token Economies Have A Long History Of Being Identified As...

Token economies have a long history of being identified as evidence based practice (Simonsen, Fairbanks, Briesch, Myers, Sugai, 2008). Through the use of positive reinforcement and negative reinforcement, token economies function to manage student behaviors. Such economies are widely used in classrooms, especially in special education classrooms that tend to have students with more severe behavior problems. According to Cooper, Heron, and Heward (2007) significant research has demonstrated the effectiveness of the token economy as a means to change behavior. As previously stated, token economies rely on principles of positive reinforcement to increase the occurrence of target behaviors through the delivery of a token. Tokens can†¦show more content†¦The token economy was selected based on the fact that is uses many of the core principles of applied behavior analysis such as positive reinforcement. It was also selected based on its history of strong empirical evidence that s upports the effectiveness of such a behavior change system. A multiple stimuli without replacement (MSWO) preference assessment was conducted to identify tangible reinforcers to use as part of the token system as a means to reduce undesired student behavior. According to Daly, Wells, Swanger-Gagnà ©, Carr, Kunz, and Taylor (2009), multiple-stimulus without replacement (MSWO) preference assessments are helpful for identifying preferred common classroom activities as reinforcers with children with behavioral disorders. Using an MSWO, Daly et al. (2009) identified and used high, medium, and low preferred stimulus contingent on the completion of math problems. The researchers reported a high correlation between the preference ranking and number of problems the students completed. For this study, the MSWO assessment will be conducted over the course of three consecutive days. For a selected item to be ranked as preferred, it must be selected in at least 80% of opportunities (Tarbox, Ghezzi, Wilson, 2006). Literature Review According to Matson and Boisjoli (2009), token economies have been commonly used to support the improvement of target behaviors such as attention seeking and task avoidance behaviors. Token

Esr Experiment Free Essays

string(86) " may be introduced in the ampli\? cation stages of the spectrometer and oscilloscope\." Electron Spin Resonance Tabish September 2003 Aim: To determine the Land? g-factor using Electron Spin Resonance. e Apparatus: ESR setup which includes Helmholtz coils, R. F. We will write a custom essay sample on Esr Experiment or any similar topic only for you Order Now oscillator and the test sample, and in addition, a cathode ray oscilloscope (CRO). Theory Background Suppose a particle having a magnetic moment  µ is placed in a uniform magnetic ? eld of intensity B, then the Hamiltonian can be written as ? H=g e ? J  · B, 2mc where g is the Land? g-factor, which is 1 for orbital angular momentum, and 2 for spin angular e e? h momentum. The factor 2mc , sometimes written as  µB , is called Bohr magneton, if the particle in question is an electron. If the particle is a nucleon, then the factor is called the nuclear magneton. If the angular momentum J results from a combination of an orbital angular momentum and a spin, then g would be given by the Land? formula: e g =1+ j(j + 1) + s(s + 1) ? l(l + 1) , 2j(j + 1) where l, s and j represent the magnitude of the orbital, the spin and the total angular momenta, respectively. Remember that j can go from l ? s to l + s. Conventionally, the static magnetic ? eld is assumed to be pointing along the z? xis, which modi? es the above equation to e ? ? Jz B. H=g 2mc Let us now consider an atom which has an electronic ground state with total angular momentum j = 1/2 and an excited state with j = 3/2 (see ? gure 2). There is only a single transition which can be induced by the absorption of radiation of frequency ? 12 = (E2 ? E1 )/? . As the energy does not depend h on the angul ar momentum states, the ground state is doubly degenerate corresponding to eigenvalues  ±1/2 ? of Jz and the excited state is quadruply degenerate corresponding to eigenvalues +3/2, 1/2, ? 1/2, ? 3/2 of ? Jz . 1 Electronic excited state Electronic transition j=3/2 ESR Electronic ground state j=1/2 ESR Zeeman effect If one now applies a magnetic ? eld B along the z-axis, each of the angular momentum states acquires a di? erent energy. The ground state energy level thus splits into two sublevels and the excited state level into four sublevels. This is called Zeeman splitting. Now instead of a single transition of frequency ? 12 = (E2 ? E1 )/? , many transitions of frequencies close to ? 12 h are possible. Experimentally this is seen as a splitting a single absorption or emission line into several closely spaced lines. This is called Zeeman e? ect. As one would have noticed, transition should also be possible between the sublevels of the same energy level. It is indeed possible and this phenomenon is known as electron spin resonance (ESR). Electron Spin Resonance Let us try to understand the phenomenon of ESR in somewhat more detail. As ESR invloves transitions only between the sublevels of one energy level, we will not bother about the Hamiltonian of the atom/molecule which gives us the energy levels. We will only worry about the part of the Hamiltonian which is the result of the applied magnetic ? ld B, which gives us the sublevels. For simplicity, we will consider one electron with angular momentum j, in a magnetic ? eld B. In addition we have an electromagnetic ? eld of frequency ? in the direction perpendicular to B. The time-dependent Hamiltonian can thus be written as ? H=g eB ? ? ? Jz + V0 ei? t + V0†  e? i? t , 2mc ? where V0 represents the interaction of the electromagnetic ? eld wit h the electron. The electromagnetic ? eld is supposed to be very weak compared to the applied static ? eld B, and so one can use time-dependent perturbation theory to study this problem. The states ? hat we will use are the eigenstates of Jz : ? Jz |m = hm|m , ? where m will take 2j + 1 values, from ? j to +j. The energy of these levels is given by g where n eB ? Jz |n = 2mc n |n , = geB? n h 2mc = gB µB n. In time-dependent perturbation theory, we know that the time-dependent interaction can cause transition between various |m states. The transition rate per unit time, from i th level to j’th level is given by: 2? ? Wij = | j|V0 |i |2 ? ( j ? i ? h? ), ? h ? assuming that j i . This expression says that transition from state |i to |j is possible when the frequency of radiation ? ( j ? i )/? . This is the condition for resonance, or in our case, h electron spin resonance. ? ? There is one important point about the form of V0 . It happens to be such that j|V0 |i is nonzero only when j = i  ± 1. This means that transition is possible between, say, | ? 3/2 and | ? 1/2 , but not between, say, | ? 3/2 and |1/2 . Such restrtictions, imposed by the kind of interaction and the nature of states, are called selection rules. 2 The ESR setup Description of the ESR Spectrometer A block diagram of the ESR Spectrometer is given in the ? gure above. Basic circuit The ? st stage of the ESR circuit consists of a critically adjusted radio frequency oscillator. This type of oscillator is required here, so that the slightest increase in its load decreases the amplitude of oscillation to an appreciable extent. The sample is kept inside the tank coil of the oscillator, which in turn, is placed in the 50 Hz magnetic ? eld generated by the Helmholtz coils. At resonance, i. e. when the frequency of oscillation becomes equal to frequency corresponding to the energy splitting of the sublevels, the oscillator amplitude registers a dip due to the absorption of power by the samp le. This obviously, occurs periodically four times in each complete cycle of the supply voltage of the magnetic ? eld. The result is an amplitude modulated carrier which is then detected using a diode detector and ampli? ed by a chain of three low noise, high gain audio-frequency ampli? ers to suit the input requirement of any oscilloscope. Highly stabilized and almost ripple free power supply for the above circuit is obtained using an integrated circuit regulator. Phase shifter This can compensate the undermined phase di? erence which may be introduced in the ampli? cation stages of the spectrometer and oscilloscope. You read "Esr Experiment" in category "Papers" 0 Hz sweep unit A 50 Hz current ? ows through Helmholtz coils which provides a low frequency magnetic ? eld to the sample. As the resonance is observed at a few gauss only, no static magnetic ? eld is applied. R. F. Oscillator It is a transistorised radio frequency oscillator suitable for the determination of resonance frequenc y. Frequency range: 10 MHz to 18 MHz Accuracy: Better than 0. 5 % The Sample The sample used in our ESR setup is diphenyl-picryl-hydrazyl (DPPH). It is a widely used standard in ESR experiments. The structure of this organic molecule, shown in the ? gure, contains three benzene rings. Its important feature is that it contains a single unpaired electron, whose orbital angular momentum is 3 O2N N N NO2 O2N zero. So, the electron has only the spin angular momentum, and the material gives a g? factor which is close to 2. 0038. One thus has to deal with the simple situation where j = 1/2, and only two sublevels are involved. In conventional spectroscopy, absorption intensity is plotted against the frequency of radiation to get the absorption spectrum. In the present case, one should obtain a single abosorption geB peak at frequency ? = ( j ? i )/? , which is nothing but ? = 2mc . However, in this setup it is h di? ult to vary the frequency of radiation. So, what is done is that the frequency of radiation is ? xed at some ? 0 , and the normally static, magnetic ? eld is swept between the positive and negative extremes of a maximum ? eld value. This is done by supplying an alternating current to the Helmholts coils which are supposed to generate the magnetic ? eld. Durin g the AC cycle, 2mc whenever the strength of the magnetic ? eld (+ve or -ve) becomes equal to B0 = ? 0ge , there is a resonance condition, and radiation is absorbed. Origin of four peaks In this experiment, the CRO is used in the x-y mode. The signal from the AC source, which supplies current for the magnetic ? eld, is fed to the X plates of the CRO, and the absorption signal is B fed to the Y plates. The point on the extreme right on the CRO 2 4 3 1 screen represents the maximum positive value of the ? eld, and the point on the extreme left represents the maximum negative value ? B of the ? eld. The point at the center represents zero ? eld. Without Time the Y-plates, the point on the CRO screen goes from maximum negative value to zero, and the maximum positive value, and then back again to the mimimum value. As one can see from the ? gure, the ? eld strength becomes B0 four times in one single sweep cycle. 0 0 0 Now if the absorption signal is fed to the Y-plates, whenever the ? eld strength becomes B0 , the Y-axis will show a peak. So, one should see four peaks corresponding to points 1,2,3,4 in the ? gure. But one can see that on the X-axis of the CRO screen, points 2 and 3 are the same, because they correspond to the same value of the ? eld B0 , and points 1 and 4 are the same because they correspond to the ? eld ? B0 . So, the four peaks should overlap such that only two are visible. However, the absorption signal passes through some electronic circuitry before being fed to the Y-plates of the CRO, so it very di? cult to make sure that no phase change occurs in the process. If there is a small phase di? erence between the AC signal on the X plates and the signal on the Y plates, when points 3 and 4 are traced, the peaks do not overlap with those at 1 and 2. So, in practice one would see four peaks. If one has a way of changing the phase of, say, the Y signal, one can adjust the phase manually so that the four peaks merge into two. Getting the numbers We have the control over the current that is passing through the Helmholtz coils, and this can also be measured. But what we actually need for our calculation is, the magnetic ? eld B applied to the sample. Let us ? rst calculate the magnetic ? eld through the Helmholtz coils. This can be done easily 4 using the Biot-Savart law. B =  µ0 4 5 3/2 I N , r where:  µ0 = 4? ? 10? 1 (cgs units) N = number of turns in each coil. r = the radius of the Helmholtz coils in cm (which is equal to their separation when they are properly arranged). I = current passing through the coils. The value of B is obtained in gauss. As the current is measured by an AC ammeter, the value of the current, and thus the ? eld, is the r. m. s. value. The peak value of the ? eld will be given by v v 8 2 I N . Bmax = 2B =  µ0 v 125 r Suppose the peak value of the ? eld (= Bmax ) corresponds to P divisions from the center on the x-axis of the CRO screen. Then if Q be the distance of the observed resonances from the center (in the units of divisions), the ? eld corresponding to the resonance will be given by: B0 = Q But the resonance condition is given by: B0 = h ? 0 ? , g µB Bmax P hich can be used to determine the value of g, once B0 is known. Now, for a ? xed ? 0 , B0 is ? xed, although one can vary the current I and get various position of the absorption peaks. Let us write the expression for B0 and see what is most accurate way to calculate it: v N  µ0 8 2 v B0 = I  · Q. rP 125 The ESR spectrometer is such that P does not vary as one varies I. So, the best way to evaluate the above expression will be to plot a graph between 1/I and Q, and ? nd out the slope, which will give the average value of I  · Q. The ? eld at the absorption peaks can be calulated as: v N  µ0 8 2 v B0 = ? lope of graph between 1/I and Q. rP 125 Procedure Connections Connections are done as follows: †¢ ESR spectrometer and power supply are connected with connecting cables. †¢ Connect the coaxial cable of the induction coil to the oscillator through the socket marked â€Å"input†. 5 †¢ Connect the Helmoltz coils to the power supply terminal marked â€Å"H† coil. †¢ Connect the â€Å"Out-put† terminal marked X, Y, E on the ESR spectrometer to the X plate, Y plate input and ground of the oscilloscope respectively and switch on the oscilloscope. †¢ Connect the power supply with AC mains. Adjustments Adjust the current in the Helmholtz coils at 150 mA. The front panel controls of the ESR spectrometer are adjusted as follows: frequency, detector and phase, all centered. Experimental procedure The X plate of the CRO is callibrated in terms of magentic ? eld as follows: 1. X ampli? er of the CRO is adjusted to obtain the maximum X de? ection (e. g. P divisions. 2. Note the current ? owing in the Helmholtz coils. The magnetic ? eld can then be calculated from the formula for B given before. Number of turn in the coils N = 500 and the radius r = 7. 7cm. The positions of the two peaks of the ESR signal at resonance is measured. Let this be Q divisions from the center. The best possible resonance peaks are obtained by varying the frequency in the range of 12 to 14 MHz and the Y sensitivity of the oscilloscope. The pahse knob is adjusted to coincide one pair of peaks with the other. The current through the coils is then varied, keeping the frequency ? xed, and the corresponding position of the peaks from the center noted. A graph between 1/I and Q is then plotted and can be used in calculating the g-factor, as described earlier. Repeat the above procedure for di? erent values of frequency. Observations and calculation S. No. 1. 2. 3. 4. 5. 6. I(mA) 150 175 200 225 250 275 I(A) 1/I Distance of peaks from center (Q) 10 MHz 13 MHz 15 MHz 17 MHz 2. 4 1. 9 1. 9 1. 9 2. 0 1. 6 1. 6 1. 5 1. 4 1. 4 1. 4 1. 4 1. 2 1. 3 1. 2 1. 2 1. 1 1. 1 1. 1 1. 0 1. 0 1. 0 1. 0 1. 0 0. 150 6. 667 0. 175 5. 714 0. 200 5. 00 0. 225 4. 44 0. 250 4. 00 0. 275 2. 636 Slope of the graph (= I  · Q) = 0. 282, P = 5, N = 500 r = 7. 7cm,  µ0 = 0. 1 ? 4? ,  µB = 9. 2741 ? 10? 21 , h = 6. 626 ? 10? 27 . v N  µ0 8 2 v B0 = I  ·Q rP 125 v 500 ? . 1 ? 4? 8 2 v ? 0. 282 = 7. 7 ? 5 125 = 4. 657 6 ?0 = 13 MHz 2 1. 8 1. 6 Q 1. 4 1. 2 1 0. 8 3. 5 4 4. 5 5 1/I 5. 5 6 6. 5 7 g = h? 0  µB B 0 6. 626 ? 10? 27 ? 13 ? 106 = 9. 2741 ? 10? 21 4. 657 = 1. 9944 Precautions 1. The direction of the Helmholtz coils should be preferable adjusted so that the ? eld is perpendicular to earth’s magnetic ? eld, which is about 0. 3 Gauss. 2. Setup the experiment at a place free from electric and magnetic ? elds and mechanical disturbances. 3. Y-output from the ESR spectrometer should be through a good shielded cable. 7 How to cite Esr Experiment, Papers

Physical Distribution and Logistics Method

Question: Discuss about the Physical Distribution and Logistics Method. Answer: Introduction A distribution channel is a path on a way through which goods and services move until it reaches the consumer. Distribution channels are of different length. Some of the distribution channels can be short such as direct transactions from the producer to the consumer while others can be long involving many other intermediaries like retailers, distributors or wholesalers, agents among others. Distribution systems where the goods or services reaches the consumer without any middleman are called direct distribution, while in cases where the goods or services leave the producer but passes through intermediaries before reaching the consumer are called indirect systems. Distribution can also be said to be business-to-business distribution when one business interacts with another to carry out commercial business or can be business-to-consumer distribution when the producer transacts with the consumer (Krause 2007, pp 528-545). There are many challenges that face distribution channels. Some of these problems are major while others are minor. Examples of the significant operational challenges faced by these distribution channels include market regulations for different goods and services, the effects of labor conditions for various markets in the world, the cost of managing the various distribution channels among many others. Every business organization should work to understand and have a way to regulate or curb these problems for the smooth business operation. Failure to have an established strategy to handle these challenges will reduce the profitability of the organization and therefore reduce organizations competitive advantage (Partridge, 2010). Cost as a challenge faced by distribution channel Many business groups have come to a realization that most of the traditional distribution channels are not working for them, mostly because these channels have become so costly to the companies. In addition, these traditional approaches have added very little value to the business enterprise. Therefore, they neither meet most of the needs of the customers nor handle the manufacturers needs and expectations. This reduces the business profitability and reduces its competitive advantage of these business organizations (Baird et al. 2011). Reduction of the distribution channel cost is an important aspect of consideration for every business organization. They should, therefore, look for ways to achieve this. Organizations should be vigilant enough to identify and innovatively use the possible available channel options that are cost effective to the business group (Gartner 2013a). Use of the traditional distribution channel approaches has become obsolete, and an organization which does not explore new innovative options for the channel distribution is likely to lose its competitive advantage. It is, therefore, clear that every manufacture needs a fresh new template guide to lead them through exploring of the available distribution channel options (Dittmann 2012). Many business organizations have been presented with distribution channels coming as complete distribution channel packages. Businesses, therefore, have options to choose the package that seems appropriate to them in consideration of the firm situational factors. The options can be direct sales to the consumers, use of manufacturers representatives in the distribution channel, use of wholesalers and distributors, and even the retailers among others (Sweeney 2011, pp 30-48). Every package operates in a different way, and therefore business organizations always evaluate the most appropriate package depending on the context of the operation, the industry itself, the situation among other factors. There have been little possibilities for the organizations to make changes in the existing distribution channel packages. This has therefore made business organizations to consider traditional approaches perfect and have not made enough efforts to understand and evaluate the distribution channe l costs and value derived out of it (Stock Boyer 2009, pp 690711). The best way to evaluate the cost and value brought to the business organization by any distribution channel option is through a keen identification of the activities done by a given particular channel, find out the costs of performing activities and then comparing those actions with the values and needs of the consumer (Bradley 2013, pp 10051022). This will, therefore, help the manufacturers to be able to compare the kind of activities performed by an individual channel, the amount of cost incurred and see the value derived out of it. With this evaluation, the manufacturers will be able to identify the best channels that are efficient and cost effective o the organizations. How to reduce cost of distribution channels Every organization that wants to get more profit will always try to cut down the costs. Cost reduction in distribution channels lies around quality and process improvement. This is the only sure way of reducing the expenses incurred in the distribution process (Lawson 2008, pp 446-460). Every business organizations management should not view their company to be composed of organization units or solid entities like marketing unit or sales unit etc., but rather the management should narrow down up to the level of identifying the various activities performed by each unit. This is an inevitable and crucial step in quality and process improvement (Nike 2013). Given the fact that organizations management only gets packages of distribution channels available for them to choose from, identification and analysis of each activity performed by various specific units in an organization is a crucial aspect of consideration. This is because it brings an understanding of the best distribution channel to use regarding efficiency, cost-effectiveness, and value derived from the channel. This, in turn, becomes very useful for the organizations to choose the best channel to use for distribution in a cost-effective way (Coyle 2013). Narrowing down from the organizations subunits means that the organizations management establishes all the activities performed by each unit in the distribution. This will help them to come to an understanding of what is done on the ground and also be able to identify the specific party responsible for performing each particular activity. With this understanding, the company will be able to compare the value derived from each specific activity in comparison to the cost incurred for those activities. With so doing, the company will be able to reduce these costs by smoothening the existing process through narrowing down to the events and making appropriate changes and improvement on what is done or making changes to the party that does each of the particular activity (Institute for Supply Management 2010). A change for improvement in distribution channel process entails the use of innovative ways to replace or improve the various specific activities performed. Use of these creative options will consequently bring an improvement change concerning quality and process. At the moment, most organizations are applying technology in their operation process to improve the process and reduce the distribution channel costs (Gartner 2013b). With the changing world, technology has always been the best option to be integrated into process improvement. It brings efficiency, and also saves reasonable costs as a distribution channel. Organizations should, therefore, embrace technology in the distribution channel process to save on the costs of distribution (Cudahy 2012). Recommendations As a recommendation, the business organization should venture more into the use of technology to improve their distribution process. With the advent of technology, a business can now make more use of text messaging, emails, push notifications, websites among other technologies to smoothen the distribution channel process. E-commerce also plays a significant role in improving business efficiency and cost reduction in channel distribution. Customers can make orders and payments online from wherever they are. These technology integration possibilities, therefore, shows how technology can significantly reduce the costs. Every business organization can, therefore, look into ways of incorporating technology in operation to lower costs in channel distribution and therefore improve on organization's profit income. Conclusions In conclusion, every organization should manage well its logistic and supply chain to ensure that there are no losses incurred in this process. The organizations should understand its operational challenges associated with the channel distribution used. Cost as a problem in channel distribution should be handled effectively. Organizations should look into ways of cost reduction in channel distribution. The basic approach to cost reduction, in this case, is through choosing the right distribution channel depending on the situational factors. In additional to this, organizations should also work out to reduce costs through breaking down channel distribution units into activities and then handling each activity to ensure that the value derived out of each merges with the cost incurred for the particular activity else it is changed. The right application of these concepts can greatly improve the channel distribution process and therefore bring success to the organization. References Baird, N., Kilcourse, B. (2011). Omni-Channel Fulfillment and the Future of the Retail Supply Chain. Retrieved Match 27, 2017, from https://www.scdigest.com/assets/reps/Omni_Channel_Fulfillment.pdf. Bradley, P. (2013) Collaboration bears fruit. CSCMPs Supply Chain Quarterly, 7(2), 3436. Coelho and Easingwood, 2008 F. Coelho, C. Easingwood An exploratory study into the drivers of channel change European Journal of Marketing, 42 (10) (2008), pp. 10051022 Cooke, J. A. (2013) Kimberly-Clark connects its supply chain to the store shelf. DC Velocity, 11(5): 5355. Coyle, J. J., Langley, C. J., Novack, R. A., Gibson, B. J. (2013) Supply Chain Management: A Logistics Perspective. Mason, OH: South-Western Cengage Learning. Cudahy, G. C., George, M. O., Godfrey, G. R., Rollman, M. J. (2012) Preparing for the unpredictable. Outlook: The Online Journal of High-Performance Business. Retrieved March 27, 2017, from https://www.accenture.com/us-en/outlook/Pages/outlook-journal-2012-preparing-for-unpredictable.aspx. Dittmann, J. P. (2012) Start with the customer! CSCMPs Supply Chain Quarterly. Retrieved March 27, 2017, from https://www.supplychainquarterly.com/topics/Strategy/20121217-start-with-the-customer/. Gartner. (2013a) Gartner Announces Rankings of its 2013 Supply Chain Top 25. Retrieved March 27, 2017, from https://www.gartner.com/newsroom/id/2494115. Gartner. (2013b) IT Glossary. Retrieved March 27, 2017, from https://www.gartner.com/it-glossary/supply-chain-management-scm/. Institute for Supply Management. (2010) Supply Management Defined. Retrieved March 27, 2017, from https://www.ism.ws/tools/content.cfm?ItemNumber=5558. Krause, D. R., Handfield, R. B., and Tyler, B. B. (2007). The relationships between supplier development, commitment, social capital accumulation and performance improvement. Journal of Operations Management, pp 528-545. Lawson, B., Tyler, B.B Cousins, P.D. (2008). Antecedents and consequences of social capital on buyer performance improvement. Journal of Operations Management, Vol. 26(3), pp 446-460 Lawson Nike, Inc. (2013) Global Manufacturing. Retrieved March 27, 2017, from https://manufacturingmap.nikeinc.com/. Partridge, A. R. (2010) Managing a customer-driven supply chain. Inbound Logistics. Retrieved August, 8, 2013, from https://www.inboundlogistics.com/cms/article/managing-a-customer-driven-supply-chain/. Stock, J., Boyer, S. (2009) Developing a consensus definition of supply chain management: A qualitative study. International Journal of Physical Distribution Logistics Management, 39(8), 690711. Sweeney, E. (2011) Towards a unified definition of supply chain management. International Journal of Applied Logistics, 2(3), 3048.