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  • How can one describe a rock-paper-scissors relationship between 3 items?

    - by Madara Uchiha
    Let's say I have the following structure: abstract class Hand {} class Rock extends Hand {} class Paper extends Hand {} class Scissors extends Hand {} The goal is to make a function (or a method) Hand::compareHands(Hand $hand1, Hand $hand2), which would return the winning hand in a rock-paper-scissors match. That would be very easy with a bunch of ifs, but the point is to have a more robust structure, that's relying on polymorphism rather than on procedural code. P.S. this is done in actual production code, if someone is asking. This isn't some sort of challenge or homework. (It's not really rock-paper-scissors, but you get the point).

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  • In which controller do you put the CRUD for the child part of a relationship?

    - by uriDium
    I am using ASP.Net MVC but this probably applies to all MVC patterns in general. My problem, for example I have companies and in each company I have a list of contacts. When I have selected a company I can see its details and a list of the contacts for that company. When I want to add a new contact for that company, should the implementation of that action go into the company controller as an "AddContact" action or should it go into the contact controller into a "New" action and we pass the Company ID in the URL? What is the usual way of dealing with this sort of thing in ASP.Net MVC? Is there a better stategy?

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  • Lite-Javascript Gallery - Can I position the img absolutely in relationship to the lis?

    - by blackessej
    I have a lite-javascript run image gallery. The javascript grabs each element in the list and places it as a background in the parent element. Then the CSS styles the thumbnails as small blocks with a defined height/width. A click-event for each object toggles it’s child’s element’s visibility and adds an “active” class name to the . Using CSS, I'm trying to place the absolutely to make it appear at the same position for each thumb, but it's moving in relation to the thumbs. Here's the CSS: #jgal li { background-position:50% 50%; background-repeat:no-repeat; border:solid #999 4px; cursor:pointer; display:block; float:left; height:60px; width:60px; margin-bottom:14px; margin-right:14px; opacity:0.5; } #jgal li img { position:absolute; top:0px; left:210px; display:none; } And the site: http://www.erisdesigns.net Thanks in advance for any help!

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  • Django: How can I add weight/ordering to a many to many relationship?

    - by Klaas van Schelven
    I'm having Pages with TextBlocks on them. Text Blocks may appear on different pages, pages may have several text blocks on them. Every page may have these blocks in an ordering of it's own. This can be solved by using a separate through parameter. Like so: class TextBlock(models.Model): title = models.CharField(max_length=255) text = models.TextField() class Page(models.Model): textblocks = models.ManyToManyField(TextBlock, through=OrderedTextBlock) class OrderedTextBlock(models.Model): text_block = models.ForeignKey(TextBlock) product_page = models.ForeignKey(ProductPage) weight = models.IntegerField() class Meta: ordering = ('weight',) But I'm not very enthousiastic about the violations of DRY for my app. (There's a lot of ordered ManyToMany relations). Is there a recommended way to go about this?

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  • SQL Database Schema Design For Large 3 Billion Relationship Database.

    - by K-Bell
    Get your geek on. Can you solve this? I am designing a products database for SQL Server 2008 R2 Ed. (not Enterprise Ed.) that will be used to store custom product configurations for over 30,000 distinct products. The database will have up to 500 users at a time. Here is the design problem… Each Product has a collection of Parts (up to 50 parts per product). So if I have 30,000 Products and each of them can have up to 50 Parts, that’s 1.5 million distinct Product-to-Part relationships …or as an equation… 30,000 (Products) X 50 (Parts) = 1.5 million Product-to-Parts records. …and If… Each Part can have up to 2000 finish options (A finish is a paint color). NOTE: Only one finish will be selected by a user at run-time. The 2000 finish options I need to store are the allowed options for a specific part on a specific product. So if I have 1.5 million distinct product-to-part relationships/records and each of those parts can have up to 2,000 finishes that is 3 billion allowable product-to-part-to finish relationships/records …or as an equation… 1.5 million (Parts) x 2,000 (Finishes) = 3 Billion Product-to-Part-to-Finishes records. How can I design this database so that I can execute fast and efficient queries for a specific product and return its list of Parts and all the allowable Finishes for each part without 3 Billion Product-to-Part-to-Finish records? Read time is more important then write time. Please post your thoughts/suggestions if you have experience with large databases. Thanks!

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  • possible to show composition relationship in a diagram generated from PHP source code?

    - by ajsie
    i have tested several UML applications for whether they could show composition relationships in the UML diagrams generated form the PHP source code or not, and the result is they can't. i know that this is a typical problem for PHP cause we don't declare a data type when we code, so it's difficult for the UML applications to know if an instance variable is a reference to an object or not. i also tested nWire with the same result. will there never be applications that could show us a complete map over all object relationships more than just inheritance? i think it's a pity that you can't have a good view over all the relationships for an application. cause when i use an open source solution, i always want to know how the objects are related to each other. maybe we could make comments for the instance variable telling the software that this is an reference to an object? but that would mean that the mapping software is using this solution. i feel its a pity nWire/visual paradigm can't give us a complete map:(

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  • [app-engine-java-groovy] One-to-Many relationship. Select objects from datastore.

    - by Olexandr
    Hi. I've omitted some code(package declarations, imports, other fields) for shortness. I have here simple One-to-Many relation. It worked fine till this moment. @PersistenceCapable(identityType = IdentityType.APPLICATION, detachable="true") class Restaurant implements Serializable { @PrimaryKey @Persistent(valueStrategy = IdGeneratorStrategy.IDENTITY) Key id @Persistent(mappedBy = "restaurant") List<RestaurantAddress> addresses = new ArrayList<RestaurantAddress>() } //-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= @PersistenceCapable(identityType = IdentityType.APPLICATION, detachable="true") class RestaurantAddress implements Serializable { @PrimaryKey @Persistent(valueStrategy = IdGeneratorStrategy.IDENTITY) Key id @Persistent Restaurant restaurant } Now i need to get(select) all the Restaurants from DB: def getRestaurantsToExport(final String dst, final int count) { String field = restaurantExportFields[dst] return transactionExecute() { PersistenceManager pm -> Query q = pm.newQuery(Restaurant.class) q.filter = "$field == null" q.setRange(0, count) return q.execute() } } But there are on problem - query gives me 12 restaurants(as in DB) but every Restaurant has 0 Address but in Datastore every Restaurant has minimum 2 addresses. Have anyone the same problem or knows the solution ?

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  • Rails ActiveRecord - How to set association save order

    - by Altonymous
    I have a weird relationship that needs to be maintained for legacy processes. I'm trying to figure out how to create the relationship given the new model association. New Relationship Setup Machine has_many MachineReadings has_many Disks has_many DiskReadings Old Relationship Setup Machine has_many MachineReadings has_many DiskReadings has_many Disks The problem is data will come in on the Machine model as nested attributes using the new relationship setup. I need to update the machine_reading_id in the DiskReading model so the old association can continue to be used. I tried doing this via an after_save hook that would traverse back up to the machine and then down to the readings to get the machine_reading.id so I could populate the DiskReading model. However, the associations aren't being saved in the order I would expect. They are saving the Disks & DiskReadings before saving the MachineReadings. So when I go after the machine_reading.id it hasn't been written and thus I am unable to get access to it. For example: #machine_disk_reading.rb after_save :build_old_relationship def build_old_relationship self.machine_reading_id = self.disk.machine.readings.find_by_date_time(self.date_time).id end

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  • How to dynamically order many-to-many relationship with JPA or HQl?

    - by Indrek
    I have a mapping like this: @ManyToMany(cascade = CascadeType.PERSIST) @JoinTable( name="product_product_catalog", joinColumns={@JoinColumn(name="product_catalog", referencedColumnName="product_catalog")}, inverseJoinColumns={@JoinColumn(name="product", referencedColumnName="product")}) public List<Product> products = new ArrayList<Product>(); I can fetch the products for the catalog nicely, but I can't (dynamically) order the products. How could I order them? I probably have to write a many-to-many HQL query with the order-by clause? I though of passing the orderBy field name string to the query, or is there a better solution? Tables are: products, product_catalog, product_product_catalog (associative table) P.S. Using Play! Framework JPASupport for my entities.

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  • c# How to make linq master detail query for 0..n relationship?

    - by JK
    Given a classic DB structure of Orders has zero or more OrderLines and OrderLine has exactly one Product, how do I write a linq query to express this? The output would be OrderNumber - OrderLine - Product Name Order-1 null null // (this order has no lines) Order-2 1 Red widget I tried this query but is not getting the orders with no lines var model = (from po in Orders from line in po.OrderLines select new { OrderNumber = po.Id, OrderLine = line.LineNumber, ProductName = line.Product.ProductDescription, } ) I think that the 2nd from is limiting the query to only those that have OrderLines, but I dont know another way to express it. LINQ is very non-obvious if you ask me!

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  • How can I achieve this kind of relationship (inheritance, composition, something else)?

    - by Tim
    I would like to set up a foundation of classes for an application, two of which are person and student. A person may or may not be a student and a student is always a person. The fact that a student “is a” person led me to try inheritance, but I can't see how to make it work in the case where I have a DAO that returns an instance of person and I then want to determine if that person is a student and call student related methods for it. class Person { private $_firstName; public function isStudent() { // figure out if this person is a student return true; // (or false) } } class Student extends Person { private $_gpa; public function getGpa() { // do something to retrieve this student's gpa return 4.0; // (or whatever it is) } } class SomeDaoThatReturnsPersonInstances { public function find() { return new Person(); } } $myPerson = SomeDaoThatReturnsPersonInstances::find(); if($myPerson->isStudent()) { echo 'My person\'s GPA is: ', $myPerson->getGpa(); } This obviously doesn't work, but what is the best way to achieve this effect? Composition doesn't sond right in my mind because a person does not “have a” student. I'm not looking for a solution necessarily but maybe just a term or phrase to search for. Since I'm not really sure what to call what I'm trying to do, I haven't had much luck. Thank you!

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  • Use SQL to clone data in two tables that have a 1-1 relationship with each other

    - by AmoebaMan17
    Using MS SQL 2005, Table 1 ID | T1Value | T2ID | GroupID ---------------------------------- 1 | a | 10 | 1 2 | b | 11 | 1 3 | c | 12 | 1 4 | a | 22 | 2 Table 2 ID | T2Value ---------------- 10 | H 11 | J 12 | K 22 | H I want to clone the data for GroupID == 1 into a new GroupID so that I result with the following: Table 1 ID | T1Value | T2ID | GroupID ---------------------------------- 1 | a | 10 | 1 2 | b | 11 | 1 3 | c | 12 | 1 4 | a | 22 | 2 5 | a | 23 | 3 6 | b | 24 | 3 7 | c | 25 | 3 Table 2 ID | T2Value ---------------- 10 | H 11 | J 12 | K 22 | H 23 | H 24 | J 25 | K I've found some SQL clone patterns that allow me to clone data in the same table well... but as I start to deal with cloning data in two tables at the same time and then linking up the new rows correctly... that's just not something I feel like I have a good grasp of. I thought I could do some self-joins to deal with this, but I am worried in the cases where the non-key fields have the same data in multiple rows.

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  • Whats the relationship between Spring and javax.enterprise.inject?

    - by HDave
    I was reading a Wikipedia article about Java EE application servers here: http://en.wikipedia.org/wiki/Java_Platform,_Enterprise_Edition#Java_EE_5_certified It says that 2 APIs that Java App Services implement are: javax.enterprise.inject javax.enterprise.context These both relate to application context and dependency injection JSR-299. I had never heard of these APIs before. Does Spring implement these APIs? Would it matter to anyone if they did?

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  • Is there such a thing as a MemberExpression that handles a many-to-many relationship?

    - by Jaxidian
    We're trying to make it easy to write strongly-typed code in all areas of our system, so rather than setting var sortColumn = "FirstName" we'd like to say sortOption = (p => p.FirstName). This works great if the sortOption is of type Expression<Func<Person, object>> (we actually use generics in our code but that doesn't matter). However, we run into problems for many-to-many relationships because this notation breaks down. Consider this simple code: internal class Business { public IQueryable<Address> Addresses { get; set; } public string Name { get; set; } } internal class Address { public State MyState { get; set; } } internal class State { public string Abbreviation { get; set; } public int StateID { get; set; } } Is it possible to have this sort of MemberExpression to identify the StateID column off of a business? Again, the purpose of using this is not to return a StateID object, it's to just identify that property off of that entity (for sorting, filtering, and other purposes). It SEEMS to me that there should be some way to do this, even if it's not quite as pretty as foo = business.Addresses.SomeExtension(a => a.State.StateID);. Is this really possible? If more background is needed, take a look at this old question of mine. We've since updated the code significantly, but this should give you the general detailed idea of the context behind this question.

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  • Use SQL to clone data in two tables that have a 1-1 relationship in each table

    - by AmoebaMan17
    Using MS SQL 2005, Table 1 ID | T1Value | T2ID | GroupID ---------------------------------- 1 | a | 10 | 1 2 | b | 11 | 1 3 | c | 12 | 1 4 | a | 22 | 2 Table 2 ID | T2Value ---------------- 10 | H 11 | J 12 | K 22 | H I want to clone the data for GroupID == 1 into a new GroupID so that I result with the following: Table 1 ID | T1Value | T2ID | GroupID ---------------------------------- 1 | a | 10 | 1 2 | b | 11 | 1 3 | c | 12 | 1 4 | a | 22 | 2 5 | a | 23 | 3 6 | b | 24 | 3 7 | c | 25 | 3 Table 2 ID | T2Value ---------------- 10 | H 11 | J 12 | K 22 | H 23 | H 24 | J 25 | K I've found some SQL clone patterns that allow me to clone data in the same table well... but as I start to deal with cloning data in two tables at the same time and then linking up the new rows correctly... that's just not something I feel like I have a good grasp of. I thought I could do some self-joins to deal with this, but I am worried in the cases where the non-key fields have the same data in multiple rows.

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  • how do simple SQLAlchemy relationships work?

    - by Carson Myers
    I'm no database expert -- I just know the basics, really. I've picked up SQLAlchemy for a small project, and I'm using the declarative base configuration rather than the "normal" way. This way seems a lot simpler. However, while setting up my database schema, I realized I don't understand some database relationship concepts. If I had a many-to-one relationship before, for example, articles by authors (where each article could be written by only a single author), I would put an author_id field in my articles column. But SQLAlchemy has this ForeignKey object, and a relationship function with a backref kwarg, and I have no idea what any of it MEANS. I'm scared to find out what a many-to-many relationship with an intermediate table looks like (when I need additional data about each relationship). Can someone demystify this for me? Right now I'm setting up to allow openID auth for my application. So I've got this: from __init__ import Base from sqlalchemy.schema import Column from sqlalchemy.types import Integer, String class Users(Base): __tablename__ = 'users' id = Column(Integer, primary_key=True) username = Column(String, unique=True) email = Column(String) password = Column(String) salt = Column(String) class OpenID(Base): __tablename__ = 'openid' url = Column(String, primary_key=True) user_id = #? I think the ? should be replaced by Column(Integer, ForeignKey('users.id')), but I'm not sure -- and do I need to put openids = relationship("OpenID", backref="users") in the Users class? Why? What does it do? What is a backref?

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  • Is there limit of "join" or the "where" or length of SQL query ?

    - by Chetan sharma
    Actually i was trying to get data from elgg database based on multiple joins. It generated very big query with lots of JOIN statements and query never respond back. SELECT distinct e.* from test_entities e JOIN test_metadata m1 on e.guid = m1.entity_guid JOIN test_metastrings ms1 on ms1.id = m1.name_id JOIN test_metastrings mv1 on mv1.id = m1.value_id JOIN test_objects_entity obj on e.guid = obj.guid JOIN test_metadata m2 on e.guid = m2.entity_guid JOIN test_metastrings ms2 on ms2.id = m2.name_id JOIN test_metastrings mv2 on mv2.id = m2.value_id JOIN test_metadata m3 on e.guid = m3.entity_guid JOIN test_metastrings ms3 on ms3.id = m3.name_id JOIN test_metastrings mv3 on mv3.id = m3.value_id JOIN test_metadata m4 on e.guid = m4.entity_guid JOIN test_metastrings ms4 on ms4.id = m4.name_id JOIN test_metastrings mv4 on mv4.id = m4.value_id JOIN test_metadata m5 on e.guid = m5.entity_guid JOIN test_metastrings ms5 on ms5.id = m5.name_id JOIN test_metastrings mv5 on mv5.id = m5.value_id JOIN test_metadata m6 on e.guid = m6.entity_guid JOIN test_metastrings ms6 on ms6.id = m6.name_id JOIN test_metastrings mv6 on mv6.id = m6.value_id where ms1.string='expire_date' and mv1.string <= 1272565800 and ms2.string='homecity' and mv2.string LIKE "%dasf%" and ms3.string='schoolname' and mv3.string LIKE "%asdf%" and ms4.string='award_amount' and mv4.string <= 123 and ms5.string='no_of_awards' and mv5.string <= 7 and ms6.string='avg_rating' and mv6.string <= 2 and e.type = 'object' and e.subtype = 5 and e.site_guid = 1 and (obj.title like '%asdf%') OR (obj.description like '%asdf%') and ( (e.access_id = -2 AND e.owner_guid IN ( SELECT guid_one FROM test_entity_relationships WHERE relationship='friend' AND guid_two=5 )) OR (e.access_id IN (2,1) OR (e.owner_guid = 5) OR ( e.access_id = 0 AND e.owner_guid = 5 ) ) and e.enabled='yes') and ( (m1.access_id = -2 AND m1.owner_guid IN ( SELECT guid_one FROM test_entity_relationships WHERE relationship='friend' AND guid_two=5 )) OR (m1.access_id IN (2,1) OR (m1.owner_guid = 5) OR ( m1.access_id = 0 AND m1.owner_guid = 5 ) ) and m1.enabled='yes') and ( (m2.access_id = -2 AND m2.owner_guid IN ( SELECT guid_one FROM test_entity_relationships WHERE relationship='friend' AND guid_two=5 )) OR (m2.access_id IN (2,1) OR (m2.owner_guid = 5) OR ( m2.access_id = 0 AND m2.owner_guid = 5 ) ) and m2.enabled='yes') and ( (m3.access_id = -2 AND m3.owner_guid IN ( SELECT guid_one FROM test_entity_relationships WHERE relationship='friend' AND guid_two=5 )) OR (m3.access_id IN (2,1) OR (m3.owner_guid = 5) OR ( m3.access_id = 0 AND m3.owner_guid = 5 ) ) and m3.enabled='yes') and ( (m4.access_id = -2 AND m4.owner_guid IN ( SELECT guid_one FROM test_entity_relationships WHERE relationship='friend' AND guid_two=5 )) OR (m4.access_id IN (2,1) OR (m4.owner_guid = 5) OR ( m4.access_id = 0 AND m4.owner_guid = 5 ) ) and m4.enabled='yes') and ( (m5.access_id = -2 AND m5.owner_guid IN ( SELECT guid_one FROM test_entity_relationships WHERE relationship='friend' AND guid_two=5 )) OR (m5.access_id IN (2,1) OR (m5.owner_guid = 5) OR ( m5.access_id = 0 AND m5.owner_guid = 5 ) ) and m5.enabled='yes') and ( (m6.access_id = -2 AND m6.owner_guid IN ( SELECT guid_one FROM test_entity_relationships WHERE relationship='friend' AND guid_two=5 )) OR (m6.access_id IN (2,1) OR (m6.owner_guid = 5) OR ( m6.access_id = 0 AND m6.owner_guid = 5 ) ) and m6.enabled='yes') order by obj.title limit 0, 10 this is the query that i am running.

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  • Understanding the value of Customer Experience & Loyalty for the Telecommunications Industry

    - by raul.goycoolea
    Worried by economic woes and market forces, especially in mature markets, communications service providers (CSPs) increasingly focus on improving customer experience. In fact, it seems difficult to find a major message by a C-level executive in the developed world that does not include something on "meeting and exceeding customers' needs". Frequently in customer satisfaction studies by prominent firms, CSPs fall short of the leadership demonstrated by other industries that take customer-centric approaches to their bottom-line strategies. Consider the following:Despite the continued impact of global economic crisis, in July 2010, Apple Computer posted record revenue and net quarterly profit. Those who attribute the results primarily to the iPhone 4 launch should note that Apple also shipped around 30% more Macintosh computers than the same period the previous year. Even sales of the iPod line increased by 8% in a highly commoditized, shrinking media player market. Finally, Apple began selling iPads during the quarter, with total sales of more than 3 million units. What does Apple have that the others lack? Well, some great products (and services) to be sure, but it also excels at customer service and support, marketing, and distribution, and has one of the strongest brands globally. Its products are useful, simple to use, easy to acquire and augment, high quality, and considered very cool. They also evoke such an emotional response from many of Apple's customers, which they turn up their noses at competitive products.In other words, Apple appears to have mastered virtually every aspect of customer experience and the resultant loyalty of its customer base - even in difficult financial times. Through that unwavering customer focus, Apple continues to drive its revenues and profits to new heights. Other customer loyalty leaders like Wal-Mart, Google, Toyota and Honda are also doing well by focusing on customer experience as an essential driver of profitability. Service providers should note this performance and ask themselves how they might leverage the same principles to increase their own profitability. After all, that is what customer experience and loyalty are all about: profitability.To successfully manage all the critical touch points of customer experience, CSPs must shun the one-size-fits-all approach. They can no longer afford to view customer service fundamentally as an act of altruism - which mentality dates back to the industry's civil service days, when CSPs were typically government organizations that were critical to economic development and public safety.As regulators and public officials have pushed, and continue to push, service providers to new heights of reliability - using incentives and punishments - most CSPs already have some of the fundamental building blocks of customer service in place. Yet despite that history and experience, service providers still lag other industries in providing what is seen as good customer service.As we observed in the TMF's 2009 Insights Research report, Customer Experience Management: Driving Loyalty & Profitability there has been resurgence in interest by CSPs. More and more of them have stated ambitions to catch up other industries, and they are realizing that good customer service is a powerful strategy for increasing business performance and profitability, not an act of good will.CSPs are recognizing the connection between customer experience and profitability, as demonstrated in many studies. For example, according to research by Bain & Company, a 5 percent improvement in customer retention rates can yield as much as a 75 percent increase in profits for companies across a range of industries.After decades of customer experience strategy formulation, Bain partner and business author, Frederick Reichheld, considers "would you recommend us to a friend?" as the ultimate question for a customer. How many times have you or your friends recommended an iPod, iPhone or a Mac? What do your children recommend to their peers? Their peers to them?There are certain steps service providers have to take to create more personalized relationships with their customers, as well as reduce churn and increase profitability, all while becoming leaner and more agile. First, they have to define customer experience, we define it as the result of the sum of observations, perceptions, thoughts and feelings arising from interactions and relationships between customers and their service provider(s). Virtually every customer touch point - whether directly or indirectly linked to service providers and their partners - contributes to customer perception, satisfaction, loyalty, and ultimately profitability. Gaining leadership in customer experience and satisfaction will not be a simple task, as it is affected by virtually every customer-facing aspect of the service provider, and in turn impacts the service provider deeply - especially on the all-important bottom line. The scope of issues affecting customer experience is complex and dynamic.With new services, devices and applications extending the basis of customer experience to domains beyond the direct control of the service provider, it is likely to increase in complexity and dynamism.Customer loyalty = increased profitsAs stated earlier, customer experience programs are not fundamentally altruistic exercises, but a strategic means of improving competitiveness and profitability in the short and long term. Loyalty is essential to deriving long term profits from customers.Some of the earliest loyalty programs date back to the 1930s, when packaged goods companies offered embedded coupons for rewards to buyers, and eventually retail chains began offering reward programs to frequent shoppers. These programs continued for decades but were leapfrogged in the 1980s by more aggressive programs from the airlines.This movement was led by American Airlines, which launched the first full-scale loyalty marketing program of the modern era with the AAdvantage frequent flyer scheme. It was the first to reward frequent fliers with notional air miles that could be accumulated and later redeemed for free travel. Figure 1: Opportunities example of Customer loyalty driven profitOther airlines and travel providers were quick to grasp the incredible value of providing customers with an incentive to use their company exclusively. Within a few years, dozens of travel industry companies launched similar initiatives and now loyalty programs are achieving near-ubiquity in many service industries, especially those in which it is difficult to differentiate offerings by product attributes.The belief is that increased profitability will result from customer retention efforts because:•    The cost of acquisition occurs only at the beginning of a relationship: the longer the relationship, the lower the amortized cost;•    Account maintenance costs decline as a percentage of total costs, or as a percentage of revenue, over the lifetime of the relationship;•    Long term customers tend to be less inclined to switch and less price sensitive which can result in stable unit sales volume and increases in dollar-sales volume;•    Long term customers may initiate word-of-mouth promotions and referrals, which cost the company nothing and arguably are the most effective form of advertising;•    Long-term customers are more likely to buy ancillary products and higher margin supplemental products;•    Long term customers tend to be satisfied with their relationship with the company and are less likely to switch to competitors, making market entry or competitors gaining market share difficult;•    Regular customers tend to be less expensive to service, as they are familiar with the processes involved, require less 'education', and are consistent in their order placement;•    Increased customer retention and loyalty makes the employees' jobs easier and more satisfying. In turn, happy employees feed back into higher customer satisfaction in a virtuous circle. Figure 2: The virtuous circle of customer loyaltyFigure 2 represents a high-level example of a virtuous cycle driven by customer satisfaction and loyalty, depicting how superiority in product and service offerings, as well as strong customer support by competent employees, lead to higher sales and ultimately profitability. As stated above, this is not a new concept, but succeeding with it is difficult. It has eluded many a company driven to achieve profitability goals. Of course, for this circle to be virtuous, the customer relationship(s) must be profitable.Trying to maintain the loyalty of unprofitable customers is not a viable business strategy. It is, therefore, important that marketers can assess the profitability of each customer (or customer segment), and either improve or terminate relationships that are not profitable. This means each customer's 'relationship costs' must be understood and compared to their 'relationship revenue'. Customer lifetime value (CLV) is the most commonly used metric here, as it is generally accepted as a representation of exactly how much each customer is worth in monetary terms, and therefore a determinant of exactly how much a service provider should be willing to spend to acquire or retain that customer.CLV models make several simplifying assumptions and often involve the following inputs:•    Churn rate represents the percentage of customers who end their relationship with a company in a given period;•    Retention rate is calculated by subtracting the churn rate percentage from 100;•    Period/horizon equates to the units of time into which a customer relationship can be divided for analysis. A year is the most commonly used period for this purpose. Customer lifetime value is a multi-period calculation, often projecting three to seven years into the future. In practice, analysis beyond this point is viewed as too speculative to be reliable. The model horizon is the number of periods used in the calculation;•    Periodic revenue is the amount of revenue collected from a customer in a given period (though this is often extended across multiple periods into the future to understand lifetime value), such as usage revenue, revenues anticipated from cross and upselling, and often some weighting for referrals by a loyal customer to others; •    Retention cost describes the amount of money the service provider must spend, in a given period, to retain an existing customer. Again, this is often forecast across multiple periods. Retention costs include customer support, billing, promotional incentives and so on;•    Discount rate means the cost of capital used to discount future revenue from a customer. Discounting is an advanced method used in more sophisticated CLV calculations;•    Profit margin is the projected profit as a percentage of revenue for the period. This may be reflected as a percentage of gross or net profit. Again, this is generally projected across the model horizon to understand lifetime value.A strong focus on managing these inputs can help service providers realize stronger customer relationships and profits, but there are some obstacles to overcome in achieving accurate calculations of CLV, such as the complexity of allocating costs across the customer base. There are many costs that serve all customers which must be properly allocated across the base, and often a simple proportional allocation across the whole base or a segment may not accurately reflect the true cost of serving that customer;  This is made worse by the fragmentation of customer information, which is likely to be across a variety of product or operations groups, and may be difficult to aggregate due to different representations.In addition, there is the complexity of account relationships and structures to take into consideration. Complex account structures may not be understood or properly represented. For example, a profitable customer may have a separate account for a second home or another family member, which may appear to be unprofitable. If the service provider cannot relate the two accounts, CLV is not properly represented and any resultant cancellation of the apparently unprofitable account may result in the customer churning from the profitable one.In summary, if service providers are to realize strong customer relationships and their attendant profits, there must be a very strong focus on data management. This needs to be coupled with analytics that help business managers and those who work in customer-facing functions offer highly personalized solutions to customers, while maintaining profitability for the service provider. It's clear that acquiring new customers is expensive. Advertising costs, campaign management expenses, promotional service pricing and discounting, and equipment subsidies make a serious dent in a new customer's profitability. That is especially true given the rising subsidies for Smartphone users, which service providers hope will result in greater profits from profits from data services profitability in future.  The situation is made worse by falling prices and greater competition in mature markets.Customer acquisition through industry consolidation isn't cheap either. A North American service provider spent about $2,000 per subscriber in its acquisition of a smaller company earlier this year. While this has allowed it to leapfrog to become the largest mobile service provider in the country, it required a total investment of more than $28 billion (including assumption of the acquiree's debt).While many operating cost synergies clearly made this deal more attractive to the acquiring company, this is certainly an expensive way to acquire customers: the cost per subscriber in this case is not out of line with the prices others have paid for acquisitions.While growth by acquisition certainly increases overall revenues, it often creates tremendous challenges for profitability. Organic growth through increased customer loyalty and retention is a more effective driver of profit, as well as a stronger predictor of future profitability. Service providers, especially those in mature markets, are increasingly recognizing this and taking steps toward a creating a more personalized, flexible and satisfying experience for their customers.In summary, the clearest path to profitability for companies in virtually all industries is through customer retention and maximization of lifetime value. Service providers would do well to recognize this and focus attention on profitable customer relationships.

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