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'Nights in White Satin' Revisited
Moody Blues drummer Graeme Edge recalls 50 years of the iconic album Days of Future Passed, which the band will play in its entirety for its desert visit.
Mike Mettler May 30, 2017 Arts & Entertainment
(From left) The Moody Blues, featuring Justin Hayward, Graeme Edge, and John Lodge, open their summer tour June 3 at Agua Caliente Casino Resort Spa in Rancho Mirage.
PHOTOS COURTESY OF THE MODDY BLUES
Pairing rock bands with orchestras in both the recording studio and in live settings is a relatively common practice these days. But 50 years ago, when The Moody Blues were approached by Decca Records to record a "rock version" of Dvořák's 9th Symphony, it was considered to be a novel — and even risky — idea.
The Moody Blues had other plans, however, and along with conductor Peter Knight and The London Festival Orchestra, they put aside the Dvořák concept and instead went on to create the original seven-track song cycle known as Days of Future Passed.
The rest, as the saying goes, is history.
Days of Future Passed, which was released in November 1967, expertly fused sweeping orchestral movements with the envelope-pushing rock of those heady musical times. And two of the album's most enduring tracks, Nights in White Satin and Tuesday Afternoon, have been fan-favorite staples in the band's live repertoire ever since.
"We didn't know it at the time," marvels Moody Blues drummer Graeme Edge. "I think it's an example of 'ignorance is bliss.' If we had known how impossible it was, we wouldn't have done it!"
To celebrate the album's golden anniversary, The Moody Blues will perform Days of Future Past in its entirety and in its original running order for the first time ever on their upcoming 28-date summer tour. And Greater Palm Springs residents literally get to see it first, as the Days of Future Passed – 50th Anniversary Tour commences June 3 at Agua Caliente Casino Resort Spa in Rancho Mirage.
Besides being The Moody Blues' founding drummer, Edge also happens to be band's resident poet. He's the man who wrote the spoken-word ending of Nights in White Satin that's titled, Late Lament, the official name of the "Breathe deep the gathering gloom" segment that carries the Justin Hayward-penned song to its dramatic conclusion.
Palm Springs Life called Edge shortly before the 50th Anniversary Tour got underway to discuss how it feels to play the Days album in full, meeting Frank Sinatra but just missing Elvis, and his experiences driving up and down the Pacific coastline in an RV.
PSL: Your appearance in the desert will actually be the very first time you've ever played Days of Future Passed from beginning to end in front of an audience.
Graeme Edge: Absolutely, yeah, And there are three songs from it that have never, ever actually been played by us.
PSL: How did it feel revisiting the album's material to get ready for this tour?
GE: It's been quite amazing! About a month ago, I started putting the album through my cans [i.e., his headphones], and I played my electronic drum kit right along with it. I was amazed that, after 50 years, I knew all of the songs.
PSL: What song do you enjoy playing the most from that album?
GE: I like The Sun Set [the first section of the song called Evening]. I like the desert feel to that one. It all feels like [Maurice Jarre's score for] Lawrence of Arabia.
PSL: The album has two poetic bookends you wrote — Morning Glory at the end of The Day Begins, and Late Lament at the end of Nights in White Satin. I'd say those are two of the most iconic spoken pieces in rock music.
GE: I don't think about it too much, because that way lies to a bloated ego. I've had other people do it to me, and I know what I think of them. They're very self-centered.
PSL: What were your personal impressions of coming to Southern California for the first time in 1968, when the band was here supporting In Search of the Lost Chord?
GE: The first time I was in California, I thought the weather was so wonderful. I liked the open spaces and the length of time you could drive. In Britain, if you would drive 150 miles, you would pass through three towns at least. In California, you could drive 150 miles and still be in the same town, because I felt like L.A. was the biggest suburb on the planet!" (laughs)
PSL: Did you ever think much about all the Palm Springs lore related to Frank Sinatra and The Rat Pack?
GE: We thought about that, yes, but we thought of them more being in Las Vegas. I went there three times to see Frank Sinatra, and once to see Sammy Davis Jr."
PSL: Did you ever meet or speak with Frank or Sammy?
GE: I did meet both of them because we had the same promoter, Jerry Weintraub. I also went to see Elvis Presley, again in Vegas. I was talking with Col. Tom Parker [Elvis' manager], and he said, 'Come on up to my room, and you can meet him.' I've had people up to my room when I've been on the road who have turned out to be boors, and I didn't want to do that to him. So I said, 'No, but thank you.' I figured since we also had the same promoter that I'd be bound to run into him — but then, of course, he died three or four months later [in August 1977], so I never did get to meet him.
PSL: Ah, that's too bad. Frank Sinatra never covered any Moody Blues songs, did he? I would have loved to hear him sing Tuesday Afternoon, or even Late Lament.
GE: He never did, but his treatment of that would have been… (pauses) wow. And then he would have said [affects raspy American voice], Pretty deep, punk! (both laugh heartily)
The Moody Blues, 8 p.m., June 3 at Agua Caliente Casino Resort Spa, 32250 Bob Hope Drive, Rancho Mirage. 888-999-1995; www.hotwatercasino.com
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NOOR has a new vision of the future for all of us.
In order for a company or an organization to function in a successful and efficient way, its members have to adhere to a specific set of values. Based on these values, our employees can achieve the business successes in dealing with our clients. Our values are definitely business relevant and we firmly believe that no client would want to work with a company that is not perceived as trustworthy, outstanding, and creative.
Our vision at NOOR is to pursue infinite possibilities by allowing the smartest to interact in a non-restrictive working environment, so the unexpected can emerge. We don't follow a predictable and convenient pattern in our framework, but rather an unorthodox one. We also encourage new ideas, so that terrific concepts can be developed.
NOOR has been taking the bull by the horns for over a decade, and its quality service can easily testify to that.


The United Methodist Church has a beautiful history and theology of baptism. As we commit to rediscovering the power of baptism, first we can root ourselves in the understanding of what we believe. Here please find information that we have prepared for our churches, as well as links to important denominational resources, to deepen our understanding of baptism.
An act of divine grace to which we respond. God is the primary actor.
A moment when a person is given identity as a daughter or son of God.
An initiation into Christ's holy church and incorporation into the Body of Christ.
A commissioning into ministry to continue the work of Christ for the redemption of the world.
A ritual that is part of a lifelong process of salvation.
A sacrament, like the Lord's Supper, in which Jesus participated (Matthew 3:13-17; Mark 1:9-11; Luke 3:21-22).
Believer's baptism. You do not have to believe to receive God's gift of grace and be initiated into the community and ministry.
Rebaptism. We do not rebaptize Catholics or others who were baptized in the name of Father, Son and Holy Spirit. We do re-baptize Mormons and Jehovah's Witnesses, who were not baptized in the name of Father, Son and Holy Spirit.
Baptism must be done in community.
Baptism isn't the end of the journey, it is the beginning.
Use warm water, especially with babies to avoid shocking them.
Put the baptismal font in a location that is visible to the congregation with enough space for the family, sponsors and congregational representatives to stand around it.
There are options about who can hold the baby during the baptism– the pastor, the parents or the sponsors.
Baptizing an infant before the sermon can lessen the anxiety of parents trying to keep their child quiet during worship. The baptismal liturgy can also come after the sermon.
For infant baptisms, a parent or sponsor must be a member of a Christian church.
Sponsors vs. Godparents: functionally they are the same and there is no legal implications of either.
Want to learn more about the United Methodist understanding of baptism? The resources below are provided by Discipleship Ministries.
By Water and the Spirit is the official document outlining United Methodist beliefs on baptism.
This recent, free book from Mark Stamm provides a clear, easy to read description of baptism in the United Methodist Church.
This document outlines the baptismal liturgy, what it means, and where it came from.
These are the services approved by General Conference.


Guide to the William C. Dilger Papers,1950-1970.
William C. Dilger papers, 1950-1970.
Collection consists of research notebooks, photographs, slides, 16mm films and some additional papers and publications.
William C. Dilger '46, M.S. '51, Ph.D. '55, was a professor of neurobiology and behavior. He served as the director of research at the Cornell Lab of Ornithology. Dilger's research focused on the genus Agapornis, which included the small parrots known as lovebirds. He studied behaviors and whether they are learned versus inherited.
Collection consists of research notebooks, photographs, slides, 16mm films and some additional papers and publications. Much of the material relates to the study of birds in the genus Agapornis. The film collection consists of nearly 200 16mm films, mainly from the late 1950s and early 1960s. Films include several different species of lovebirds as well as a few others. Roseicollis, cana, personata fischeri, are all included. Other species include, Aequidens portalegrensis, Cichlasoma meeki, Trichogaster trichoptetus, Spinus tristis, Loriculus galgulus, Pterocnemia pennata, Betta slendens, and many, many others. Films depict behaviors such as preening, courtship, mating, fighting, etc.. A card index to all the films is included. Photographs, slides and negatives includes images of birds from various locations including South Africa, but also of individuals such as colleagues at the Cornell Laboratory of Ornithology and other scientists. Mainly Agapornis and Rhea birds, but also other birds and animals including amphibians. Slide trays include indices in lids, slides are numbered and organized in the trays. Slides of research facilities include the Max Planck Institute for Ornithology in Seewiesen, aviaries, zoos or other buildings in Leiden, Edinburgh, Cambridge, Oxford, Ludwigsburg, Ithaca, Groningen Extensive behavioral fieldnotes in binders, some taken at the Catskill Game Farm, include many notes on Rhea Americana as well as Agapornis and others. Sixteen numbered volumes of notes on Agapornis from 1954 to 1963. Audio reels are mainly of birds, but also some other items including reels from talks in Cambridge at the International Ethological Congress. Lectures by Konrad Lorenz, William Homan Thorpe and others. Additional material includes Data sheets on research, notes on gamopractic experience. Draft sketches and layout for a book on Canada geese will Dilger illustrations. Readings for classes, notebook from Zoology 126, Birds of the World.
William C. Dilger papers, #21-18-4214. Division of Rare and Manuscript Collections, Cornell University Library.
Several dozen specimens of Agapornis feces on filter papers collected and cataloged by Dilger were removed from the collection.
Loose notes and data sheets and two blue journal notebooks.
8 large slide boxes with indices inside, several loose packages of slides, several sets of photo negatives, and a number of loose photographs of various bird and non-bird subjects.
Contains descriptive information about the films, including species, date, location, behaviors, etc..
Small 16mm film containers. Some films missing.
14 7" audio reels, 10 5" audio reels, 2 16mm films, 1 8mm film.


text:
Archbishop Charles Scicluna and Lifeline captain Claus-Peter Reisch on Wednesday were among those who paid tribute to the 850 people who died in an attempt to reach Italy from Libya in 2015.
A long-awaited memorial plaque was laid down at the Addolorata Cemetery, where 24 dead migrants had been buried in April of that year.
"All humanity weighs the same at eternity's gates."
Our heartfelt solidarity with all those lost at sea who will never have a grave to mourn at. This morning, we stood shoulder to shoulder with Maltese civil society to lay a memorial for the victims of the 2015 shipwrecks. pic.twitter.com/LZYLldMLt4
— MISSION LIFELINE (@SEENOTRETTUNG) May 15, 2019
Watch: Captain fined over Lifeline registration; magistrate decries 'frightening levels' of racism
Despite an extensive rescue operation, only 28 people survived. Twenty-four corpses were brought to Malta and the remaining 800 were believed to have drowned when their crowded boat capsized some 30 miles off the Libyan coast.
On that fateful morning, Prime Minister Joseph Muscat had told party supporters in Marsa: "They are literally trying to find people alive among the dead floating in the water".
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redpajama_set_name: RedPajamaCommonCrawl


text:
The treaty that threatens to derail Europe's Green Deal
Economy and ecology 01.04.2021 | Nico Schmidt
When private courts of arbitration can award transnational energy firms billions in compensation for lost profits, climate protection becomes a farce
Steam rises from the cooling towers of the coal power plant of RWE in Niederaussem
Artikel auf Deutsch lesen
The British oil company Rockhopper Exploration is suing Italy in an international court of arbitration for around €225 million in damages – because of an Italian government decree that bans oil drilling. Meanwhile, German energy giant RWE is demanding €1.4bn from the Netherlands because it banned coal burning starting in 2030, which means RWE will have to close down its coal-fired power plant. Thus far, only a few corporations like Rockhopper and RWE have filed suits against EU climate measures. But these are probably just the beginning of a massive wave of lawsuits that could cause the EU and its member states to fall short of their climate targets.
Such lawsuits are permitted through a little-known investment protection agreement, the Energy Charter Treaty (ECT). All EU countries, including Germany, signed the treaty in December 1994. According to figures from Investigate Europe, the ECT protects the fossil-fuel infrastructure in the EU, Great Britain and Switzerland to the tune of €350bn. Fear of ECT-based lawsuits is leading EU countries to postpone and water down their climate laws – or abandon them altogether.
Right after the end of the Cold War, Western states sought to create legal certainty for energy companies wanting to invest in the new post-Soviet countries. Not trusting the local legal systems, they used investment protection treaties to transfer mediation to binding international arbitration. One such treaty is the Energy Charter to protect companies investing in the energy sector. Besides the EU, Afghanistan and post-Soviet countries like Azerbaijan and Kazakhstan also signed the treaty. So did other industrialised countries, including Turkey and Japan. But today, energy companies are using the Charter to sue for billions in compensation when nation states ban the use of coal, oil and gas to achieve their climate goals.
Such lawsuits are negotiated in private arbitration courts. In recent years, criticism of these common procedures for protecting investments has become very shrill: for example, during the Transatlantic Trade and Investment Partnership (TTIP) negotiations. While some details in current trade agreements like the TTIP are more modern than those in the Energy Charter –considered the very first investment protection treaty – they rely on the same system of private arbitration courts: Negotiations are largely shielded from public gaze and it's almost impossible for states to challenge compensation awards.
Can the ECT be reformed?
Many ECT lawsuits, including those of Rockhopper and RWE, are being handled by the International Centre for Settlement of Investment Disputes (ICSID). If one state refuses to accept a judgment and pay compensation, its property may be confiscated in other states that accept ICSID rules. That is, investors can use the ECT to sue states in courts of arbitration and enforce compensation claims even after they have left the treaty. So a breach of contract is no way out of the ECT. What options do the EU and its member states have?
EU countries have been trying to reform the Energy Charter Treaty for years. In 2018, the 55 contracting parties agreed to begin a modernisation process. In October 2020, the European Commission presented member states with an EU negotiating position, which was approved a few weeks ago. But EU consensus does not mean that the treaty can be reformed. That takes all contracting parties voting on a new version – and in 2019, the Japanese government opined that 'it is not necessary to amend the current ECT provisions'. Not doing that could thwart the goals of the European Green Deal.
The EU and its member states have one last hope for preventing a wave of ECT lawsuits.
Exasperated by the arduous negotiations, two EU members recently resorted to unusual means: Both the French and Spanish governments sent letters to the EU. The French letter states that the modernisation discussions 'are not likely to produce results for several years' and the EU's objectives are 'far from being achieved'. Therefore, '[t]he option of a coordinated withdrawal of the European Union and its member states should be raised publicly'. The Spanish government wrote that quitting the Treaty is the only effective long-term solution if the Charter cannot be reconciled with EU climate targets.
The EU Commission has long accepted the possibility that negotiations over modernisation could fail. Last December, EU Trade Commissioner Valdis Dombrovskis declared that the Commission 'may consider proposing other options, including the withdrawal from the ECT'. This would apply if 'core EU objectives, including the alignment with the Paris Agreement, are not attained within a reasonable timeframe'. The problem is: Parties are not allowed to quit right away.
'It will be difficult for EU member states to leave the Energy Charter Treaty immediately,' says legal scholar Markus Krajewski, who has been researching investment protection for many years. That's because the Energy Charter was designed to protect companies. Article 47 of the ECT provides that a state can be sued for 20 years after leaving the treaty. Scholars call this the 'sunset' clause, activists call it the 'zombie' clause.
The EU and its member states have one last hope for preventing a wave of ECT lawsuits. Two thirds of all recent ECT cases have involved EU investors suing EU states, as in RWE vs. the Netherlands. International treaty law, however, allows contracting parties to conclude supplementary agreements that affect only them. Thus, EU member states could simply agree to no longer apply the ECT to intra-European cases.
However, EU members are at loggerheads over this solution. A few years ago, the European Court of Justice (ECJ) ruled that investment protection treaties between EU states are incompatible with EU law. That led many states to conclude that the court's judgment also applies to the only investment protection agreement that exists between many EU countries, the ECT. But some governments vehemently disagree.
Even if EU governments agreed, however, it's possible that arbitrators would not recognise it. Investment protection expert Markus Krajewski says: 'There is certainly a perception that such an inter se agreement between the Parties would curtail investors' rights under the Energy Charter.' Inter-se (Latin for 'amongst themselves') agreements allow parties to derogate from an existing treaty.
Last December, the Belgian government requested the ECJ to clarify whether the Energy Charter Treaty is compatible with EU law. The ECJ could clarify if future intra-European ECT lawsuits will be permitted. That means that the Energy Charter Treaty and the arbitration regime that jeopardises the EU's climate targets could finally be checked – by a (public) court.
Nico Schmidt
Nico Schmidt is a member of the international journalist team Investigate Europe, which publishes reports on topics of Europe-wide relevance. His articles have appeared in Der Spiegel and Die Zeit, among others.
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text:
Disorders of Sexual Differentiation
Disorders of sexual differentiation program UCSF Pediatric Urology
UCSF Pediatric Urology has a comprehensive program to treat disorders of sex development. The multidisciplinary team is co-directed by Drs. Laurence Baskin from pediatric urology and Janet Lee from pediatric endocrinology. The multidisciplinary team meets consist of faculty from pediatric urology, endocrinology, genetics, psychiatry, social work, nursing and a nursing coordinator.
The team meets formally on a monthly basis to discuss and consult on specific patients with specific DSD diagnosis such as CAH, mixed gonadal dysgenesis, XY male severe hypospadias, androgen receptor defects, ovotesticular syndrome, vanishing testis syndrome and many other diagnoses. The program is complemented by active prospective NIH research grants studying the outcomes of children with disorders of sex development.
What are disorders of sex development?
Disorders of sex development (DSDs) are conditions in which the reproductive organs and genitals do not develop normally. People with DSDs have a mix of male and female sexual characteristics.
There are different types of DSDs:
A person may be genetically female (they have two X sex chromosomes), but have genitals that appear more masculine. They may, for example, have an enlarged clitoris that resembles a penis and their labia (lips surrounding the vagina) may be fused together. The most common cause for this type of DSD is a metabolic condition called congenital adrenal hyperplasia (CAH), which also affects other aspects of health.
Another type of DSD occurs when a person is genetically male (having one X and one Y sex chromosome), but the genitals appear more female. This type of DSD is most often caused by complete or partial androgen insensitivity syndrome (AIS), which causes the body to respond abnormally to male hormones.
People with male, female, or mixed chromosomes can be born with both ovarian and testicular tissue. In this type of DSD, the genitals can appear female, male or a mix of both.
Sometimes sexual organs are normally formed as either male or female, but growth and sexual development are disrupted at puberty. Individuals with this type of DSD have an abnormal chromosome arrangement, such as a missing or extra X chromosome. Females with a DSD called Turner syndrome, for example, are missing an X chromosome and are usually infertile and of shorter than average height.
What causes disorders of sex development?
DSDs occur because of variations in the genes that control sexual development or because a person has an underlying condition that affects how the body responds to sex hormones. For example, people with a DSD caused by congenital adrenal hyperplasia (CAH), do not produce an enzyme required to make two hormones, cortisol and aldosterone. As a result the body produces too much androgen (male sex hormone), resulting in male sex characteristics. If the DSD is caused by androgen insensitivity syndrome, the body does not process androgen in a normal way.
How are disorders of sex development diagnosed?
If sexual organs are ambiguous at birth, a child may receive a diagnosis right away. Other types of DSD do not become apparent until puberty. Partial androgen insensitivity syndrome, for example, is usually diagnosed at birth because the genitals appear different. In complete androgen insensitivity syndrome, the genitals look normal for a female, so the condition may not be picked up until a girl fails to menstruate.
When a DSD is suspected, tests are used to confirm the diagnosis. These include chromosome analyses, blood tests to measure hormone levels, and imaging studies to examine internal sex organs.
How are disorders of sex development treated?
It is important that a child with a DSD be carefully evaluated at a multidisciplinary clinic that specializes in these disorders. Pediatric endocrinologists, urologists, psychiatrists and other specialists work together to present treatment options. These may include medications, such as supplemental hormones, to encourage pubertal development and manage other growth and development problems. Surgical reconstruction of the genitalia may also be an option, although it is not necessary in all cases. With proper diagnosis and treatment, most individuals with DSDs live normal lives.
At UCSF we follow national guidelines for treatment of all patients with specific diagnosis of DSD's
Congenital adrenal hyperplasia due to steroid 21-hydroxylase deficiency: an Endocrine Society clinical practice guideline.
Speiser PW, Azziz R, Baskin LS, Ghizzoni L, Hensle TW, Merke DP, Meyer-Bahlburg HF, Miller WL, Montori VM, Oberfield SE, Ritzen M, White PC; Endocrine Society.J Clin Endocrinol Metab. 2010 Sep;95(9):4133-60. doi: 10.1210/jc.2009-2631.PMID: 20823466
Management of the infant with atypical genital appearance (difference of sex development)
Authors: Christopher P Houk, MD, Laurence S Baskin, MD, FAAP, Lynne L Levitsky, MD
Section Editor: Mitchell E Geffner, MD
Deputy Editor: Alison G Hoppin, MD
Contributor Disclosures
All topics are updated as new evidence becomes available and our peer review process is complete.
Literature review current through: Aug 2021.
This topic last updated: Aug 27, 2021.
Causes of differences of sex development
Authors: Yee-Ming Chan, MD, PhD, Lynne L Levitsky, MD
Section Editors: Laurence S Baskin, MD, FAAP, Mitchell E Geffner, MD
This topic last updated: Jan 31, 2021.
Disorders of Sexual Differentiation (DSD) Clinic
At UCSF we have a unique multidisciplinary clinic that provides care to families of children with ambiguous genitalia. Our experts include urologists, endocrinologists, reproductive endocrinologists, geneticists, child psychologists, a mental health nurse practitioner, nurses and social work professionals. The team meets with families to discuss diagnosis, management, goals and questions, and concerns surrounding genital surgery. The entire DSD team meets on the third Tuesday of every month to collaborate with individual families to provide a management plan for each child's medical and psychological needs. Please contact us 415 353-2083 to schedule an appointment.
Support groups, such as the Cares Foundation also provide help and support to young people, adults and families who are affected by CAH.
disorders_of_sex_development_080615.pdf


How to Make a Rice Heat BagUncooked rice retains heat well. It is easy to sew and fill a fabric bag. It can be useful for applying heat or cold to injuries. This is a guide about how to make a rice heat bag.
How to Make a Father's Day Bowtie CardMake a fun, personalized card to give to your dad this father's day. This is a guide about how to make a father's day bowtie card.
What Plant Is This?Can someone please tell me what plant this is?
Identifying a HouseplantI have this tiny little plant that isn't growing very well, but I don't know how to grow it because I don't know what it is. It has really tiny sad-looking crinkly dark green leaves. Do you know what it is?
Value of a Florence Doll by GoldenvaleI have a Florence porcelain doll made by Goldenvale; she is flawless. She stands 46 inches tall. I am curious on her value. Can anyone help me?


Comments Off on Cab Insurance That Is Affordable – Do You Find It?
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When you work for a company at a higher postcode you're going to pay more on your insurance. However, if you're able to lower this you're going to pay less.
Learn about which discounts your company applies for: There is various reductions available. These include low-mileage discounts, and safe-driver discounts. A number of businesses offer discounts for companies that pay their invoices online. This ends in less paperwork for the insurer.
Generally, any taxi that's several distinct drivers will have to pay more on their taxi cab insurance providers than someone who drives a single use taxi. Typically, it is better to be the only person that drives the taxi. Some of the other things that influence the rate that you pay are where you store the cab.
Understand what services you're looking for: Before you call around, understand what you're looking for in a limo service. For example, know the man count, time you'll need it for and the places and places that you will desire to see.
Finding a site that is dependable similar is your first step. It is vital that you work with a firm who will find the best deals based on the advice you provide. Many of these websites offer quotes that are counterfeit because their systems are outdated. You will need a company that'll give up-to-date advice. Plus, it's important to have plenty of alternatives to look over instead of only one. Take time to assess the credentials of the comparison sites you encounter. The good ones will be backed by accredited institutions. Best of all they will be in plain view, especially since they want to earn your trust.
Purchasing cheap cab insurance will not need to be a hassle. Then you can get past any type of problems that comes with improper research and running when you take the time to look.


Here comes another caregiver confession: sometimes I feel guilty.
OK, let me try that more honestly: most days I am guilt-ridden about something. Guilt is not an unusual feeling for me. I'd say I was born with a healthier dose of the stuff than most people. But it's also true that the experience of caregiving has amplified my sense of guilt, making me feel, at alternating times, undeserving, selfish, and inhumanly lucky.
Like many caregivers, one of my greatest sources of guilt stems from the fact that I'm not the one with the brain injury. No matter how hard my day has been, no matter how much my body hurts, or how overwhelmed my head feels, I know it will never compare to the challenges my husband faces in navigating this complicated world as a TBI survivor. My life, in comparison, will always be a little easier, a little more manageable – and for that, I am supremely grateful. And a lot bit guilty.
My husband has lost so much to brain injury that I know it's unfair to harp on the areas of my life that are less than perfect. But I'm also human. And as such, I can't help but surrender to the human condition of wanting more. Logically, I understand this is how we are built: to want. Just as TC's desire for more fueled his hard-earned TBI recovery, I too can't resist making demands from the universe. I want more time to pursue my interests. I want more help in managing the circus that is a four-person/one dog/one fish household, and I want to feel taken care of by the person I love.
They're the same wishes shared by many folks I know, but my situation is compounded by the silent yet significant fifth member of our household – brain injury. My head is home to a constant battle between greed and guilt, leading me to wrestle with one question: how much is it OK to ask for? In other words, how much is it fair to demand from someone who is already working harder than most just to make it through each day? And as a wife, how do I count on my husband to be a true and equal partner, when deep down I'm pinged with the feeling that I ought to just be grateful he's here at all? After all, what kind of shameless person asks for the cherry on top of a prayer so wonderfully answered?
Me, I guess. That's who.
While I don't have answers to these burning questions, I'm learning to accept the inevitability of my human desires and the duality of my emotional experience. Yes, I can be a grateful person and still hope for more. Yes, I can want more and not be selfish. Yes, it is natural to struggle with guilt, but it's also critical that guilt evolve into honesty, not repression or regret.
The idea of survivor guilt is not a new one. In fact, it's a term that was coined back in the 1960s as therapists began to observe a range of guilt-related symptoms among Holocaust and other trauma survivors. Later on, it was adopted under the umbrella of post-traumatic stress (PTSD). As I've learned over the past five years, PTSD is one wide-reaching umbrella. A lot of behavior and emotional experiences are encompassed within that classification – all of them completely natural. As one yoga teacher so wisely explained, post-traumatic stress disorder is not disorder at all. It's the body and mind's way of reestablishing order.
I think that's what TC and I are doing most of the time: reestablishing order. Rebalancing expectations. Five years in, I now understand the fallacy in believing we might achieve either state. The TBI journey does not point toward one outcome or destination. We are simply traveling. Evolving in the process.
The facts of brain injury can't be changed. I did not suffer the injury. My husband did. But we are both trauma survivors. No amount of guilt can rewrite those certitudes, but it is both human and empathic to wish things were different. To feel grateful we're all here. To be confused as hell about how to move forward.
There's enough to feel bad about in this world of brain injury. And while we can acknowledge guilt as a natural part of the process, we certainly don't need to let it eat us alive. We deserve more than that.
As always , it's like you're reading my mind :). Nicely written. Thank you for sharing!
No truer words spoken than what you wrote here. I battle the same feelings everyday. So nice to know I am not alone. Thank you for writing what a lot of us caregivers are feeling.


Authentic 55mm Ohio University watch with collector's box. Officially licensed by Ohio University.
The 55mm Ohio Bobcats watch by the College Watch Co. is a must have collector's item for super fans, alumni, and graduates alike. Prominently displaying the famous Bobcats logo, this watch exudes Ohio pride.


text:
Arkansas Teacher Resigns After Posting Racist Messages About Obamas On Social Media
The Little Rock teacher resigned after making the disparaging comments on Facebook.
NewsOne Staff
An Arkansas educator turned in his resignation after the deplorable comments he made about the Obamas on social media were brought to his employers attention, according to USA Today.
#BREAKING— Malvern High School teacher resigns after racist comments on THV11's facebook pagehttps://t.co/65p1huZGDq
— THV11 (@THV11) December 30, 2016
Trent Bennett, a teacher at Malvern High School in Little Rock, took to KTHV-TV's Facebook page on Christmas Eve to unleash a series of racist remarks about President Obama and First Lady Michelle Obama.
In the post, Bennett referred to Michelle Obama as "Michelle Obummer… America's First Chimp." He continued saying he will be overjoyed to see "that nasty chimp and her spider monkey hubby gone for good."
The former teacher also drummed up birther sentiment by claiming that the President was born in Kenya, although it has been proven on numerous occasions he was born in Hawaii.
Malvern High School denounced the comments in a Facebook post, calling Bennett's rhetoric, "inappropriate," "insensitive" and "racially charged."
Bennett offered an apology and acknowledged his statements were "disrespectful and offensive."
On Thursday during a special school board meeting, board members voted 6-0 to accept Bennett's resignation, which was made effective immediately, USA Today reports.
Bennett also told KTHV-TV that he deleted all of his social media accounts as the second step in his apology tour.
SOURCE: USA Today
Disgraced Trump Campaign Official Claims Racist Obama Email Sent By 'Mistake'
Nonprofit Director Loses Job After 'Ape In Heels' Comment About Michelle Obama
Arkansas Teacher Resigns After Posting Racist Messages About Obamas On Social Media was originally published on newsone.com
arkansas	, facebook	, racism	, social media	, Teacher


Throughout the years there has been many questions asked about when a car that drives its self will be made and now it's here!
Mercedes-Benz have made this happen. It was spotted in California and San Francisco being test driven. The electric car and is quoted by Mercedes to be a luxury to be inside and is like a private airplane. We are still yet to be told the price. This is could be our future!


You might recognize Blabey's work from Pig the Pug.
In this funny new book, a porcupine wants a hug. He asks a bear, a moose and a rabbit who all refuse to hug such a spikey creature. He then happens upon a slithery snake who wants a kiss but also keeps getting rejected. Together they are able to help each other and find a new friend.
Hmmm, this is actually a toughie. I'm currently trying to teach George that it's okay to say 'no' when he doesn't want to hug or kiss someone. At the same time, I'm teaching him that hugs are ways that friends can express their love for each other. We've recently had two interactions with affectionate boys who happen to have developmental disabilities and desperately wanted to hug George. He was afraid of them which made me sad – I know that their hearts only know love and they want to express it. This book actually helped me explain that when someone you don't know or someone who seems different asks for a hug, they want to be your friend. You don't have to hug anyone you don't want to – but you can still be kind to them. This is a BIG lesson for a book with 100 words.
I love Blabey's simple and bright illustrations along with his rhyming text. George thinks it's funny and is happy when the snake and porcupine hug each other.
In this book, Shorty the dinosaur happens upon a present that is for Clem. He can't handle the excitement of what could be in the box. Is it a race car or a drum or a trampoline? Poor Shorty just can't take his it and his curiosity and rips into the present. He then realized how upset Clem will be when he finds out that Shorty opened his present. Luckily for Shorty, Clem knows him all too well and the present was meant for him all along.
That you shouldn't open presents that aren't meant for you. Or that friendship is the best present of all.
A lot of this book happens in the pictures, which is always great with kids George's age to get them talking about what is happening. Plus it has lots of anticipation and guessing which is another fun way to get your child engaged in a book. Simple, clear illustrations that are cute and two unlikely friends. Great read!
George is obsessed, obsessed with horses (although he won't go near one) so I pick up every book I see that involves a horse. In this book, Steve the horse wants to be an exceptional horse. He happens upon a golden horn and ties it to his head in order to stand out. Steve shows off to all his forest friends until Racoon points out that there is no horn on his head (it has slipped to under his chin). He teaches high and low for his horn, eventually falling into a pond and losing it all together. When he surfaces, he notices that ALL of his friends have tied random things to their heads to be like him. With excitement, he realizes he is exceptionally different (and a trendsetter).
This is a very funny book about a goofy horse. He talks to himself, there are funny side notes and it has a great lesson told in a silly way. Plus it has a horse. Horses for the win.


text:
\section{Introduction}
\label{sec:typesetting-summary}

In the open online dial-a-ride problem, we are given a metric space $(M,d)$ and have control of a server that can move at unit speed.
Over time, \emph{requests} of the form $(a,b;t)$ arrive.
Here, $a\in M$ is the \emph{starting position} of the request, $b\in M$ is its \emph{destination}, and $t\in\mathbb{R}_{\geq 0}$ is the \emph{release time} of the request. We consider the online variant of the problem, meaning that the server does not get to know all requests at time~$0$, but rather at the respective release times.
Our task is to control the server such that it serves all requests, i.e., we have to move the server to position~$a$, load the request $(a,b;t)$ there after its release time~$t$, and then move to position~$b$ where we unload the request. The objective is to minimize the \emph{completion time}, i.e., the time when all requests are served.

We assume that the server always starts at time~$0$ in some fixed point, which we call the \emph{origin} $\o\in M$.
The server has a \emph{capacity} $c\in(\mathbb{N}\cup\{\infty\})$ and  is not allowed to load more than~$c$ requests at the same time.
Furthermore, we consider the \emph{non-preemptive} version of the problem, that is, the server may not unload a request preemptively at a point that is not the request's destination.
In the dial-a-ride problem, a distinction is made between the open and the closed variant.
In the \emph{closed} dial-a-ride problem, the server has to return to the origin after serving all requests.
By contrast, in the \emph{open} dial-a-ride problem, the server may finish anywhere in the metric space.
In this work, we only consider the open variant of the problem.
By setting $a=b$, we obtain the \emph{online travelling salesperson problem (TSP)} as a special case of the dial-a-ride problem.

In this work, we only consider deterministic algorithms for the online dial-a-ride problem. As usual in competitive analysis, we measure the quality of a deterministic algorithm by comparing it to an optimal offline algorithm.
The measure we apply is the completion time of a solution.
For a given sequence of requests~$\sigma$ and an algorithm $\textsc{Alg}$, we denote by $\textsc{Alg}(\sigma)$ the completion time of the algorithm for request sequence~$\sigma$.
Analogously, we denote by $\textsc{Opt}(\sigma)$ the completion time of an optimal offline algorithm.
For some $\rho\geq1$, we say that an algorithm $\textsc{Alg}$ is \emph{$\rho$-competitive} if, for all request sequences~$\sigma$, we have $\textsc{Alg}(\sigma)\leq\rho\cdot\textsc{Opt}(\sigma)$.
The \emph{competitive ratio} of $\textsc{Alg}$ is defined as $\inf\{\rho\geq1\mid\textsc{Alg}\textrm{ is }\rho\textrm{-competitive}\}$. The \emph{competitive ratio of a problem} is defined as $\inf\{\rho\geq1\mid\textrm{there is some $\rho$-competitive algorithm}\}$

\paragraph*{Our results.}

We consider the parametrized algorithm $\textsc{Lazy}(\alpha)$ that was presented in \cite{waoa} and prove the following result.
\begin{theorem}\label{mainthm}
For $\alpha=\frac{1}{2}+\sqrt{11/12}$, $\textsc{Lazy}(\alpha)$ has a competitive ratio of $\alpha+1\thickapprox2.457$ for the open online dial-a-ride problem on general metric spaces for any capacity $c\in\mathbb{N}\cup\{\infty\}$.
\end{theorem}
Prior to our work, the best known upper bound on the competitive ratio of the open online dial-a-ride problem was $\varphi+1\thickapprox 2.618$ \cite{waoa} where $\varphi$ denotes the golden ratio. The best known lower bound remains 2.05 \cite{BirxDisser/19b}.
In \cite{waoa} it was shown that the competitive ratio of $\textsc{Lazy}$ is at least $\frac{3}{2}+\sqrt{11/12}\thickapprox 2.457$ for any choice of $\alpha$, even if the metric space is the real line. Our upper bound tightly matches this lower bound.
Thus, $\alpha=\frac{1}{2}+\sqrt{11/12}$ is the best possible waiting parameter for $\textsc{Lazy}$, even on the line.

Furthermore, we analyze the algorithm for open online dial-a-ride on the half-line, i.e., where $M=\mathbb{R}_{\geq 0}$, and show that, in this metric space, even better bounds on the competitive ratio are possible.
More precisely, we show the following.

\begin{theorem}\label{half-line_result}
For $\alpha=\frac{1+\sqrt{3}}{2}$, $\textsc{Lazy}(\alpha)$ has a competitive ratio of $\alpha+1\thickapprox2.366$ for the open online dial-a-ride problem on the half-line for any capacity $c\in\mathbb{N}\cup\{\infty\}$.
\end{theorem}

This improves on the previous best known upper bound of 2.618 \cite{waoa} on the competitive ratio of the open online dial-a-ride problem on the half-line. The best known lower bound is 1.9 \cite{Lipmann/03}.

In the preemptive version of the dial-a-ride problem, the server is allowed to unload requests anywhere and pick them up later again. In this version, prior to our work, the best known upper bound on general metric spaces was 2.618 \cite{waoa} and the best known upper bound on the half-line was 2.41 \cite{BjeldeDisserHackfeldEtal/20}. As the non-preemptive algorithm $\textsc{Lazy}$ can also be applied in the preemptive version of the problem, our analysis of $\textsc{Lazy}$ also improves the state of the art in this case. The best known lower bound in the preemptive version on general metric spaces is 2.04 \cite{BjeldeDisserHackfeldEtal/20} and the best known lower bound on the half-line is 1.62 \cite{Lipmann/03}.

\begin{corollary}
The competitive ratio of the open online preemptive dial-a-ride problem with any capacity $c \in \mathbb{N} \cup \{\infty\}$ is upper bounded by
\begin{enumerate}[a)]
\item $\frac{3}{2}+\sqrt{\frac{11}{12}} \thickapprox 2.457$ on general metric spaces and this bound is achieved by $\textsc{Lazy}\left(\frac{1}{2}+\sqrt{\frac{11}{12}}\right)$,
\item $1+\frac{1+\sqrt{3}}{2}\thickapprox 2.366$ on the half-line and this bound is achieved by $\textsc{Lazy}(\frac{1+\sqrt{3}}{2})$.
\end{enumerate}

\end{corollary}

An overview of the state of the art and our results can be found in \cref{resultstable}.

\begin{figure}
\begin{center}
\begin{tabular}{cccc|c}
& & lower bound & upper bound & our results\\
\hline
\multirow{2}{*}{general metric space}&non-preemptive&2.05&2.618 \cite{waoa}&2.457\\&preemptive&2.04&2.618 & 2.457\\
\hline
\multirow{2}{*}{line}&non-preemptive&2.05 \cite{BirxDisser/19b}&2.618&2.457\\&preemptive&2.04 \cite{BjeldeDisserHackfeldEtal/20}&2.41 \cite{BjeldeDisserHackfeldEtal/20}\\
\hline
\multirow{2}{*}{half-line}&non-preemptive&1.9 \cite{Lipmann/03}&2.618&2.366\\&preemptive&1.62 \cite{Lipmann/03}&2.41&2.366\\
\end{tabular}
\end{center}
\caption{State of the art of the open online dial-a-ride problem and overview of our results: Bounds without reference are inherited. Our results are upper bounds.}
\label{resultstable}
\end{figure}

\paragraph*{Related work.}

For the open online dial-a-ride problem, several algorithm have been proposed.
The basic idea of the algorithm \textsc{Ignore} is as follows. Whenever there is an unserved request, start a schedule over all unserved requests and ignore all requests released during its execution. It is known that the competitive ratio of \textsc{Ignore} is exactly 4 due to Birx \cite{Birx/20} and Krumke~\cite{Krumke0}. The algorithm \textsc{Replan} starts a new optimal schedule over all unserved requests whenever a new request arrives. This algorithm is known to have a competitive ratio between 2.5 and 4 due to Ausiello et al.~\cite{Ausiello/01} and Birx \cite{Birx/20}. Krumke \cite{Krumke0} suggested the algorithm \textsc{SmartStart} which combines the idea of \textsc{Ignore} with a waiting routine before starting a schedule. This algorithm is known to have a competitive ratio of exactly 2.94 on the line and of at most 3 in general metric spaces due to Birx \cite{Birx/20}. \textsc{SmarterStart} is a variation of \textsc{SmartStart} with a more refined waiting routine. This algorithm has a tight competitive ratio of 2.67 on the line and of at most 2.7 in general metric spaces. The results are due to Birx \cite{Birx/20} and Birx et al. \cite{BirxDisser/19b}. Lipmann \cite{Lipmann/03} proposed the algorithm \textsc{WaitOrIgnore} and claimed a competitive ratio of 2.618, yet its analysis was incomplete.
Prior to our work, the best known upper bound on the competitive ratio of the open online dial-a-ride problem was $\varphi+1$ where $\varphi=\frac{\sqrt{5}+1}{2}\thickapprox1.618$ denotes the golden ratio.
This bound is due to Baligács et al.~\cite{waoa} and achieved by $\textsc{Lazy}(\varphi)$. This $(\varphi+1)$-bound was still the best known upper bound when the metric space is the line or the half-line.

In the preemptive version of the dial-a-ride problem, the best known upper bound on general metric spaces was again the $(\varphi+1)$-bound in \cite{waoa}. When the metric space is the line, Bjelde et al.~\cite{BjeldeDisserHackfeldEtal/20} have shown that a better upper bound of $1+\sqrt{2}\approx 2.41$ is possible. This result is inherited on the half-line.

In terms of lower bounds, Birx et al. \cite{BirxDisser/19b} were able to prove that any algorithm for the open online dial-a-ride problem has a competitive ratio of 2.05, even if the metric space is the line. This separates dial-a-ride from online TSP on the line, where, due to Bjelde et al. \cite{BjeldeDisserHackfeldEtal/20}, it is known that the competitive ratio is exactly 2.04. This result is inherited for the non-preemptive dial-a-ride problem on the line.
For open online dial-a-ride on the half-line, Lipmann \cite{Lipmann/03} was able to show a lower bound of 1.9 in the non-preemptive version and a lower bound of 1.62 in the preemptive version.

In the closed variant of the online dial-a-ride problem, the competitive ratio is known to be exactly 2. The upper bound of 2 is due to Ascheuer et al. \cite{AscheuerKrumkeRambau/00} and Feuerstein and Stougie~\cite{FeuersteinStougie/01}.
The lower bound of 2 is due to Ausiello et al. \cite{Ausiello/01} and already holds in the special case of closed online TSP.
Thus, the competitive ratio of closed online TSP is also known to be exactly 2.
When the metric space is the line, Birx \cite{Birx/20} gave a lower bound of 1.76 which separated this problem from closed online TSP
For that problem, Ausiello et al. \cite{Ausiello/01} showed a lower bound of 1.64, and Bjelde et al. \cite{BjeldeDisserHackfeldEtal/20} later developed an algorithm whose competitive ratio tightly matches the lower bound.
For closed online dial-a-ride on the half-line, a lower bound of 1.71 is known due to Ascheuer et al. \cite{AscheuerKrumkeRambau/00}.
The TSP variant in the same setting was tightly analyzed by Blom et al.~\cite{BlomKrumkePaepeStougie/01} with a competitive ratio of 1.5.

Also, other variants of the problem have been considered in the literature.
This includes the restriction of algorithms, the restriction of adversary models, and resource augmentation.
Regarding the restriction of possible algorithms, the class of \emph{zealous} or \emph{diligent} algorithms was investigated by Blom et al. \cite{BlomKrumkePaepeStougie/01}.
Algorithms from this class always have to move towards an unserved request or the origin while there are unserved requests.
Another class of algorithms that was investigated by Birx \cite{Birx/20} are \emph{schedule-based} that operate in schedules and are not allowed to interrupt a schedule they have started.
Regarding the restriction of the adversary, \emph{non-abusive} or \emph{fair} adversaries have been studied by Krumke et al. \cite{Krumke/02} and by Blom et al.~\cite{BlomKrumkePaepeStougie/01}.
In those models, the adversary is not allowed to leave the convex hull of all released requests.
Another model is one with \emph{reasonable load} considered by Hauptmeier~et~al.~\cite{HauptmeierKrumkeRabau/00}, where, at every point in time, it may not take longer than roughly the elapsed time in order to serve all released requests.
Regarding resource augmentation, Allulli et al.~\cite{AllulliAusielloLaura/05} and Ausiello et al. \cite{AusielloAllulliBonifaciLaura/06} investigated a variant of dial-a-ride where the request is known to the server some time before it will be released,
and Lipmann et al. \cite{Lipmann/04} investigated a version where the server gets to know the destination of a request only after picking it up.
Other forms of resource augmentation include the number of servers, the servers' speed, and the servers' capacities, which were considered by Bonifaci and Stougie \cite{BonifaciStougie/08} and Jaillet and Wagner \cite{JailletWagner/08}.

Krumke et al. \cite{KrumkePaepePoensgenStougie/03} and Bienkowski et al. \cite{BienkowskiKL/21,BienkowskiLiu/19} investigated a setting where the goal is to minimize the sum of completion times.
Krumke \cite{Krumke0} considered randomized algorithms with the objective to minimize the expected completion time,
and Krumke et al. \cite{Krumke/02,KrumkePaepePoensgenEtal/06} and Hauptmeier et al. \cite{HauptmeierKrumkeRabau/00} minimized the flow time.
Online TSP where the metric space is a circle was investigated by Jawgal et al. \cite{JawgalMuralidharaSrinivasan/19}.
In the \emph{online quota TSP}, the goal is to minimize the time it takes to serve only a given quota of requests.
This problem was investigated by Ausiello et al. \cite{AusielloDemangeLauraPaschos/04}, and Jaillet and Lu \cite{JailletLu/11,JailletLu/14} considered a variant where not all requests have to be served, but not serving one gives a penalty in the objective.
Jaillet and Wagner~\cite{JailletWagner/08} and Hauptmeier et al. \cite{HauptmeierKrumkeRambauWirth/01} investigated a model where the requests may be required to be served in a partial order.

\section{Algorithm description and notation}
In this section, we define the algorithm $\textsc{Lazy}$ introduced in \cite{waoa}. The rough idea of the algorithm is to wait until several requests are revealed and then start a schedule serving them. Whenever a new request arrives, we check whether we can deliver all currently loaded requests and return to the origin in a reasonable time. If this is possible, we do so and begin a new schedule including the new requests starting from the origin. If this is not possible, we keep following the current schedule and consider the new request later.

More formally, a \emph{schedule} is a sequence of actions specifying the server's  behaviour, including its movement and where requests are loaded or unloaded. By $\textsc{Opt}[t]$, we denote an optimal schedule beginning in $\o$ at time 0 and serving all requests that are released not later than time $t$. By $\textsc{Opt}(t)$, we denote its completion time.
Given a set of requests $R$ and some point $x \in M$, by $S(R,x)$, we denote a shortest schedule serving all requests in $R$ beginning in point $x$ at some time after all requests in $R$ are released. In other words, we can ignore the release times of the requests when computing $S(R,x)$. As waiting is not beneficial for the server if there are no release times, the \emph{length of the schedule}, i.e., the distance the server travels, is the same as the time needed to complete it and we denote this by $|S(R,x)|$.

Now that we have established the notation needed, we can describe the algorithm (cf.\linebreak \hyperref[lazy]{Algorithm 1}). By $t$, we denote the current time, by $p_t$ we denote the position of the server at time $t$, and by $R_t$, we denote the set of released but unserved requests at time $t$. The variable~$i$ is a counter over the schedules started by the algorithm. The waiting parameter $\alpha\geq 1$ specifies how long we wait before starting a schedule. The algorithm uses the following commands: $\textsc{deliver\_and\_return}$ orders the server to finish serving all currently loaded requests and return to $\o$ in the fastest possible way, $\textsc{wait\_until}(t)$ orders the server to remain at its current location until time $t$, and $\textsc{follow\_schedule}(S)$ orders the server to execute the actions defined by schedule $S$.
When any of these commands is invoked, the server aborts what it is doing and executes the new command. Whenever the server has completed a command, we say that it becomes \emph{idle}.

\begin{algorithm}\label{lazy}
	\caption{$\textsc{Lazy}(\alpha)$}
	initialize: $i \gets 0$

	\vspace{.1cm}
	\hrule
	\vspace{.1cm}

	\emph{upon receiving a request:}\\
	\If{server can serve loaded requests and return to $\o$ until time $\alpha \cdot \textsc{Opt} (t)$}{
		\textbf{execute} $\textsc{deliver\_and\_return}$
	}

	\hrule
	\vspace{.1cm}

	\emph{upon becoming idle:}\\
	\uIf{$t<\alpha\cdot\textsc{Opt}(t)$}{
		\textbf{execute} $\textsc{wait\_until}(\alpha \cdot \textsc{Opt} (t))$}
	{\ElseIf{$R_t\neq\emptyset$}{
			$i\gets i+1$, $R^{(i)}\gets R_t$, $t^{(i)}\gets t$, $p^{(i)}\gets p_t$ \\
			$S^{(i)}\gets S(R^{(i)},p^{(i)})$\\
			\textbf{execute} $\textsc{follow\_schedule} ({S^{(i)})}$
		}}

\end{algorithm}
The following observations were already noted in \cite{waoa} and follow directly from the definitions above and the fact that requests in $R^{(i)} \setminus R^{(i-1)}$ were released after time $t^{(i-1)}$.

\begin{observation}[\cite{waoa}]\label{defobs}
For any request sequence the following hold.
\begin{enumerate}[a)]
\item For any $i>1$, $\textsc{Opt}(t^{(i)})\geq t^{(i-1)}\geq\alpha\cdot\textsc{Opt}(t^{(i-1)})$.
\item For any $x,y \in M$ and set of requests $R$, we have
$|S(R,x)|\leq d(x,y)+|S(R,y)|$.
\item Let $i>1$ and assume that $S^{(i-1)}$ was not interrupted. Let $a$ be the starting position of the request in $R^{(i)}$ that is served first by $\textsc{Opt}(t^{(i)})$. Then,
\begin{equation*}
\textsc{Opt}(t^{(i)})\geq t^{(i-1)}+|S(R^{(i)},a)|\geq \alpha\cdot\textsc{Opt}(t^{(i-1)})+|S(R^{(i)},a)|.
\end{equation*}
\end{enumerate}
\end{observation}

For illustration of the definitions above, let us make a few more comments on it. If the server returns to the origin upon receiving a request, we say that the schedule it was currently following is \emph{interrupted}.
Observe that, due to interruption, the sets $R^{(i)}$ are not necessarily disjoint. Also, observe that $p^{(1)}=\o$, and if schedule $S^{(i)}$ was interrupted, we have $p^{(i+1)}=\o$ and\linebreak $t^{(i+1)}=\alpha\cdot\textsc{Opt}(t^{(i+1)})$. If $S^{(i)}$ was not interrupted, $p^{(i+1)}$ is the ending position of $S^{(i)}$.

\section{Analysis on general metric spaces}

This section is concerned with the proof of \cref{mainthm}. For the remainder of this section, let $(r_1,\dots,r_n)$ be some fixed request sequence. Let $k$ be the number of schedules started by $\textsc{Lazy}(\alpha)$, and let $S^{(i)}$, $t^{(i)}$, $p^{(i)}$, $R^{(i)}$ ($1 \leq i \leq k$) be defined as in the algorithm. Note that we slightly abuse notation here because $k$, $S^{(i)}$, $t^{(i)}$, $p^{(i)}$, and $R^{(i)}$ depend on $\alpha$. As it will always be clear from the context what $\alpha$ is, we waive this dependency in the notation.

\newcommand{r_{f,\wait}^{(i)}}{r_{f,\textsc{Lazy}}^{(i)}}
\newcommand{a_{f,\wait}^{(i)}}{a_{f,\textsc{Lazy}}^{(i)}}
\newcommand{b_{f,\wait}^{(i)}}{b_{f,\textsc{Lazy}}^{(i)}}
\newcommand{t_{f,\wait}^{(i)}}{t_{f,\textsc{Lazy}}^{(i)}}
\newcommand{r_{l,\wait}^{(i)}}{r_{l,\textsc{Lazy}}^{(i)}}
\newcommand{a_{l,\wait}^{(i)}}{a_{l,\textsc{Lazy}}^{(i)}}
\newcommand{b_{l,\wait}^{(i)}}{b_{l,\textsc{Lazy}}^{(i)}}
\newcommand{t_{l,\wait}^{(i)}}{t_{l,\textsc{Lazy}}^{(i)}}
\newcommand{r_{f,\opt}^{(i)}}{r_{f,\textsc{Opt}}^{(i)}}
\newcommand{a_{f,\opt}^{(i)}}{a_{f,\textsc{Opt}}^{(i)}}
\newcommand{b_{f,\opt}^{(i)}}{b_{f,\textsc{Opt}}^{(i)}}
\newcommand{t_{f,\opt}^{(i)}}{t_{f,\textsc{Opt}}^{(i)}}
\newcommand{r_{l,\opt}^{(i)}}{r_{l,\textsc{Opt}}^{(i)}}
\newcommand{a_{l,\opt}^{(i)}}{a_{l,\textsc{Opt}}^{(i)}}
\newcommand{b_{l,\opt}^{(i)}}{b_{l,\textsc{Opt}}^{(i)}}
\newcommand{t_{l,\opt}^{(i)}}{t_{l,\textsc{Opt}}^{(i)}}
\newcommand{a_{f,\opt}^{(i+1)}}{a_{f,\textsc{Opt}}^{(i+1)}}

As it will be crucial for the proof in which order $\textsc{Opt}$ and $\textsc{Lazy}$ serve requests, we introduce the following notation. Let
\begin{itemize}
\item $r_{f,\opt}^{(i)}=(a_{f,\opt}^{(i)},b_{f,\opt}^{(i)};t_{f,\opt}^{(i)})$ be the first request in $R^{(i)}$ picked up by $\textsc{Opt}(t^{(i)})$,
\item $r_{l,\opt}^{(i)}=(a_{l,\opt}^{(i)},b_{l,\opt}^{(i)};t_{l,\opt}^{(i)})$ be the last request in $R^{(i)}$ delivered by $\textsc{Opt}(t^{(i+1)})$,
\item $r_{f,\wait}^{(i)}=(a_{f,\wait}^{(i)}, b_{f,\wait}^{(i)}; t_{f,\wait}^{(i)})$ be the first request in $R^{(i)}$ picked up by $\textsc{Lazy} (\alpha)$,
\item $r_{l,\wait}^{(i)}=(a_{l,\wait}^{(i)}, p^{(i+1)};t_{l,\wait}^{(i)})$ be the last request in $R^{(i)}$ delivered by $\textsc{Lazy} (\alpha)$.
\end{itemize}

\pagebreak
\begin{definition}
We say that the $i$-th schedule is \emph{$\alpha$-good} if
\begin{enumerate}[a)]
\item $|S^{(i)}| \leq \textsc{Opt}(t_i)$ and
\item $t^{(i)} + |S^{(i)}| \leq (1+\alpha) \cdot\textsc{Opt}(t_i)$.
\end{enumerate}
\end{definition}

In this section, we prove by induction on $i$ that, for $\alpha \geq \frac{1}{2}+\sqrt{11/12}$, every schedule is $\alpha$-good. Note that this immediately implies \cref{mainthm}.

As our work builds on \cite{waoa}, the first few steps of our proof are the same as in \cite{waoa}. For better understandability and reading flow, we repeat the proofs of some important but simple steps here and mark the results with appropriate citations. The results starting from \cref{wrongorder} are new and improve the analysis in \cite{waoa}.

We begin with proving the base case.

\begin{observation}[Base case, \cite{waoa}]\label{basecase}
For every $\alpha \geq 1$, the first schedule is $\alpha$-good.
\end{observation}

\begin{proof}
Recall that $S^{(1)}$ begins in $\o$ and is the shortest tour serving all requests in $R^{(1)}$. $\textsc{Opt} [t^{(1)}]$ begins in $\o$ and serves all requests in $R^{(1)}$, too, so that $|S^{(1)}|\leq \textsc{Opt}(t^{(1)})$. Next, note that $t^{(1)}=\alpha \cdot \textsc{Opt} (t^{(1)})$ so that
$t^{(1)}+|S^{(1)}| \leq (1+\alpha) \cdot \textsc{Opt} (t^{(1)})$.
\end{proof}

Next, we observe briefly that the induction step is not too difficult when the last schedule was interrupted.

\begin{observation}[Interruption case, \cite{waoa}]\label{interrupted}
Let $\alpha \geq 1$. Assume that schedule $S^{(i)}$ was interrupted. Then, $S^{(i+1)}$ is $\alpha$-good.
\end{observation}

\begin{proof}
If schedule $S^{(i)}$ was interrupted, we have $p^{(i+1)}=\o$ and $t^{(i+1)}=\alpha \cdot \textsc{Opt} (t^{(i+1)})$. Therefore,
$|S^{(i+1)}|=|S(R^{(i+1)},\o)|\leq \textsc{Opt} (t^{(i+1)})$ and
$t^{(i+1)}+|S^{(i+1)}| \leq (1+\alpha) \cdot \textsc{Opt} (t^{(i+1)})$.
\end{proof}

For this reason, we will assume in many of the following statements that the schedule $S^{(i)}$ was not interrupted.

By careful observation of the proof in \cite{waoa}, one can see that the following fact already holds for smaller $\alpha$. For convenience, we have put the proof of the following Lemma with adapted $\alpha$ in Appendix A.

\begin{restatable}[\cite{waoa}]{lemma}{thmmainthm}\label{waoa}
Let  ${\alpha\geq \frac{1+\sqrt{17}}{4}\thickapprox 1.281}$ and $i\in \{1, \dots, k-1\}$. If $S^{(i)}$ is $\alpha$-good, then\linebreak
$|S^{(i+1)}| \leq \textsc{Opt} (t^{(i+1)})$.
\end{restatable}

Recall that the goal of this section is to prove that every schedule is $\alpha$-good and that we have already proven the base case (cf. \cref{basecase}). Moreover, in the induction step, we have already proven that $|S^{(i+1)}|\leq \textsc{Opt}(t^{(i+1)})$ (\cref{waoa}). Therefore, it remains to show that $t^{(i+1)} + |S^{(i+1)}| \leq (1+\alpha) \cdot \textsc{Opt} (t^{(i+1)})$ assuming that $S^{(1)}, \dots, S^{(i)}$ are $\alpha$-good. In \cref{interrupted}, we have already seen that this holds if $S^{(i)}$ was interrupted. To show that the induction step also holds if $S^{(i)}$ was not interrupted, we distinguish several cases for the order in which $\textsc{Opt}$ serves the requests. We begin with the case that $\textsc{Opt}[t^{(i+1)}]$ picks up some request in $R^{(i+1)}$ before serving $r_{l,\wait}^{(i)}$, i.e., that $\textsc{Opt}[t^{(i+1)}]$ does not follow the order of the $S^{(i)}$.

\begin{lemma}\label{wrongorder}
Let $\alpha \geq 1$. Assume that $S^{(i)}$ is $\alpha$-good and was not interrupted, and that $\textsc{Opt}[t^{(i+1)}]$ picks up $r_{f,\textsc{Opt}}^{(i+1)}$ before serving $r_{l,\textsc{Lazy}}^{(i)}$. Then, $t^{(i+1)}+|S^{(i+1)}|\leq (1+\alpha) \cdot \textsc{Opt} (t^{i+1})$.
\end{lemma}
\begin{proof}
Using the order in which $\textsc{Opt}$ handles the requests, we obtain the following.
After picking up $r_{f,\textsc{Opt}}^{(i+1)}$ at $a_{f,\opt}^{(i+1)}$ after time $t^{(i)}$, $\textsc{Opt}[t^{(i+1)}]$ has to serve $r_{l,\wait}^{(i)}$ at $p^{(i+1)}$ so that
\begin{equation}\label{eq:abefore}
\textsc{Opt}(t^{(i+1)})\geq t^{(i)}+d(p^{(i+1)},a_{f,\opt}^{(i+1)}).
\end{equation}
After finishing schedule~$S^{(i)}$, the server either waits until time $\alpha\cdot\textsc{Opt}(t^{(i+1)})$ or immediately starts the next schedule, i.e., we have
	\begin{equation*}\label{eq:ti1}
		t^{(i+1)} = \max\{\alpha\cdot\textsc{Opt}(t^{(i+1)}),t^{(i)}+|S^{(i)}|\}.
	\end{equation*}
If $t^{(i+1)}=\alpha\cdot\textsc{Opt}(t^{(i+1)})$, the assertion follows immediately from \cref{waoa}. Thus, assume\linebreak $t^{(i+1)}=t^{(i)}+|S^{(i)}|$. This yields
	\begin{align*}
		t^{(i+1)}+|S^{(i+1)}|
		\overset{\text{Obs \ref{defobs}b)}}&{\leq} t^{(i)}+|S^{(i)}|+d(p^{(i+1)},a_{f,\opt}^{(i+1)})+|S(R^{(i+1)},a_{f,\opt}^{(i+1)})| \\
		\overset{\text{$S^{(i)} \alpha$-good}}&{\leq}
		(1+\alpha) \cdot \textsc{Opt} (t^{(i)}) +d(p^{(i+1)},a_{f,\opt}^{(i+1)})+|S(R^{(i+1)},a_{f,\opt}^{(i+1)})|\\
\overset{\text{Obs \ref{defobs}a)}}&{\leq}
\frac{1+\alpha}{\alpha}t^{(i)} +d(p^{(i+1)},a_{f,\opt}^{(i+1)})+|S(R^{(i+1)},a_{f,\opt}^{(i+1)})|\\
\overset{\text{Obs \ref{defobs}c)}}&{\leq}
\frac{1}{\alpha}t^{(i)} +d(p^{(i+1)},a_{f,\opt}^{(i+1)})+\textsc{Opt}(t^{(i+1)})\\
\overset{\text{\cref{eq:abefore}}}&{\leq}
\frac{1}{\alpha}\left(\textsc{Opt}(t^{(i+1)})-d(p^{(i+1)},a_{f,\opt}^{(i+1)})\right)+d(p^{(i+1)},a_{f,\opt}^{(i+1)})\\
&\hspace{1cm}+\textsc{Opt}(t^{(i+1)})\\
&=\left(1+\frac{1}{\alpha} \right) \textsc{Opt}(t^{(i+1)}) + \left(1-\frac{1}{\alpha} \right)d(p^{(i+1)},a_{f,\opt}^{(i+1)})\\
&\leq 2 \cdot\textsc{Opt}(t^{(i+1)}),
\end{align*}
where we have used in the last inequality that $	d(p^{(i+1)},a_{f,\opt}^{(i+1)})\leq \textsc{Opt}(t^{(i+1)})$ as $\textsc{Opt}[t^{(i+1)}]$ has to visit both points.
\end{proof}

Next, we consider the case where $\textsc{Opt}$ handles $r_{l,\wait}^{(i)}$ and $r_{f,\textsc{Opt}}^{(i+1)}$ in the same order as $\textsc{Lazy}$.

\begin{lemma}\label{rightorder1}
Let $\alpha\geq 1$. Assume that schedules $S^{(1)}, \dots, S^{(i)}$ are $\alpha$-good, that $S^{(i)}$ was not interrupted, and that $\textsc{Opt}[t^{(i+1)}]$ serves $r_{l,\wait}^{(i)}$ before collecting $r_{f,\textsc{Opt}}^{(i+1)}$. If we have\linebreak $d(p^{(i+1)},a_{f,\opt}^{(i+1)})+\textsc{Opt}(t^{(i)})\leq\alpha\cdot\textsc{Opt}(t^{(i+1)})$, then $t^{(i+1)}+|S^{(i+1)}|\leq (1+\alpha) \cdot \textsc{Opt} (t^{i+1})$.
\end{lemma}

\begin{proof}
Similarly as in the proof of \cref{wrongorder}, we can assume
	\begin{equation}\label{eq:noWaiting}
		t^{(i+1)}=t^{(i)}+|S^{(i)}|.
	\end{equation}

We have
	\begin{align*}
		t^{(i+1)}+|S^{(i+1)}|
		\overset{\text{Obs \ref{defobs}b)}}&{\leq} t^{(i)}+|S^{(i)}|+d(p^{(i+1)},a_{f,\opt}^{(i+1)})+|S(R^{(i+1)},a_{f,\opt}^{(i+1)})| \\
		\overset{\text{$S^{(i)} \alpha$-good}}&{\leq}
		(1+\alpha) \cdot \textsc{Opt} (t^{(i)}) +d(p^{(i+1)},a_{f,\opt}^{(i+1)})+|S(R^{(i+1)},a_{f,\opt}^{(i+1)})|\\
		\overset{\text{Obs \ref{defobs}c)}}&{\leq} (1+\alpha)\cdot\textsc{Opt}(t^{(i)})+d(p^{(i+1)},a_{f,\opt}^{(i+1)})\\
		& \hspace{1cm}+\textsc{Opt}(t^{(i+1)}) - \alpha \cdot \textsc{Opt} (t^{(i)}) \\
		&= \textsc{Opt}(t^{(i+1)})  + d(p^{(i+1)},a_{f,\opt}^{(i+1)})+\textsc{Opt}(t^{(i)})\\
		&\leq (1+\alpha) \cdot \textsc{Opt}(t^{(i+1)} ),
	\end{align*}
	where the last inequality follows from $d(p^{(i+1)},a_{f,\opt}^{(i+1)})+\textsc{Opt}(t^{(i)})\leq\alpha\cdot\textsc{Opt}(t^{(i+1)})$.

\end{proof}
As we have proven now the case described in \cref{rightorder1}, we will assume in the following that
\begin{equation}\label{eq:dlarge}
d(p^{(i+1)},a_{f,\opt}^{(i+1)})>\alpha\cdot\textsc{Opt}(t^{(i+1)})-\textsc{Opt}(t^{(i)}).
\end{equation}
\begin{lemma}\label{rightorder2}
Let $\alpha\geq \frac{1+\sqrt{3}}{2}\thickapprox 1.366$. Assume that the $i$-th schedule is $\alpha$-good and was not interrupted, and that $\textsc{Opt}[t^{(i+1)}]$ serves $r_{l,\wait}^{(i)}$ before collecting $r_{f,\textsc{Opt}}^{(i+1)}$. If \eqref{eq:dlarge} holds, then $p^{(i)}=\o$ and $t^{(i)}=\alpha\cdot\textsc{Opt}(t^{(i)})$.
\end{lemma}
\begin{proof}
	If $i=1$, we obviously have $p^{(i)}=\o$ and $t^{(i)}=\alpha\cdot\textsc{Opt}(t^{(i)})$.
	Thus, assume that\linebreak $i\geq2$.
	If \mbox{$r_{l,\wait}^{(i)}\in (R^{(i-1)}\cap R^{(i)})$}, schedule $S^{(i-1)}$ was interrupted and, thus,  the statement holds.
	Otherwise, request $r_{l,\wait}^{(i)}$ is released while schedule $S^{(i-1)}$ is running, i.e.,\linebreak  $t_{l,\wait}^{(i)}\geq t^{(i-1)}\geq \alpha \cdot \textsc{Opt}(t^{(i-1)})$. Combining this with the assumption that $\textsc{Opt}[t^{(i+1)}]$ serves $r_{l,\wait}^{(i)}$ before collecting $r_{f,\textsc{Opt}}^{(i+1)}$, we obtain
	\begin{equation}\label{eq:t_x-upperBound}
		\textsc{Opt}(t^{(i+1)})\geq t_{l,\wait}^{(i)}+d(p^{(i+1)},a_{f,\opt}^{(i+1)}) \geq \alpha \cdot \textsc{Opt}(t^{(i-1)})+d(p^{(i+1)},a_{f,\opt}^{(i+1)}).
	\end{equation}
	Rearranging yields
	\begin{align*}
		\alpha\cdot\textsc{Opt}(t^{(i-1)}) \overset{\eqref{eq:t_x-upperBound}}&{\leq} \textsc{Opt}(t^{(i+1)})-d(p^{(i+1)},a_{f,\opt}^{(i+1)}) \\
		\overset{\eqref{eq:dlarge}}&{<} \textsc{Opt}(t^{(i)})-(\alpha-1)\textsc{Opt}(t^{(i+1)}) \\
		\overset{\text{Lem \ref{waoa}}}&{\leq} (1+\alpha-\alpha^2)\textsc{Opt}(t^{(i)}),
	\end{align*}
	which is equivalent to
	\begin{equation}\label{eq:estimate_t_i-1}
		\textsc{Opt}(t^{(i-1)}) < \bigl(1+\frac{1}{\alpha}-\alpha\bigr)\textsc{Opt}(t^{(i)}).
	\end{equation}
	By the assumption that $S^{(i-1)}$ is $\alpha$-good, the server finishes schedule $S^{(i-1)}$ not later than time $(\alpha+1)\cdot\textsc{Opt}(t^{(i-1)})$.
	Thus, at the time where request~$r_{l,\wait}^{(i)}$ is released, the server can serve all loaded requests and return to the origin by time
	\begin{equation*}
		\max\{(\alpha+1)\textsc{Opt}(t^{(i-1)}),t_{l,\wait}^{(i)}\}+\textsc{Opt}(t^{(i-1)}).
	\end{equation*}
	We have
	\begin{align*}
		(\alpha+2)\cdot\textsc{Opt}(t^{(i-1)})\overset{\eqref{eq:estimate_t_i-1}}&{<}(\alpha+2)\left(1+\frac{1}{\alpha}-\alpha\right)\textsc{Opt}(t^{(i)}) = \left(3+\frac{2}{\alpha}-\alpha^2-\alpha\right)\textsc{Opt}(t^{(i)})\\
		& \leq \alpha\cdot\textsc{Opt}(t^{(i)}),
	\end{align*}
	where the last inequality holds for $\alpha\geq 1.343$.
	Furthermore, as $r_{l,\wait}^{(i)}\in R^{(i)}$, it holds that
	\begin{equation*}
		t_{l,\wait}^{(i)}+\textsc{Opt}(t^{(i-1)}) \leq \textsc{Opt}(t^{(i)})+\textsc{Opt}(t^{(i-1)}) \overset{\eqref{eq:estimate_t_i-1}}{\leq} \left(2+\frac{1}{\alpha}-\alpha\right)\textsc{Opt}(t^{(i)})\leq\alpha\cdot\textsc{Opt}(t^{(i)}),
	\end{equation*}
	where the last inequality holds for $\alpha \geq \frac{1+\sqrt{3}}{2}\thickapprox 1.366$.
	This implies that the server can return to the origin by time $\alpha\cdot\textsc{Opt}(t^{(i)})$, i.e., we have $p^{(i)}=\o$.
\end{proof}

\begin{lemma}\label{rightorder3}
Let $\alpha\geq \frac{1}{2}+\sqrt{11/12} \thickapprox 1.457$. Assume that the $i$-th schedule is $\alpha$-good and was not interrupted, and that $\textsc{Opt}[t^{(i+1)}]$ serves $r_{l,\wait}^{(i)}$ before collecting $r_{f,\textsc{Opt}}^{(i+1)}$. If \eqref{eq:dlarge} holds, then $t^{(i+1)}+|S^{(i+1)}|\leq (1+\alpha) \cdot \textsc{Opt} (t^{i+1})$.
\end{lemma}
\begin{proof}
We begin by proving the following assertion.\\
\emph{Claim: $\textsc{Opt}[t^{(i+1)}]$ serves all requests in $R^{(i)}$ before picking up~$r_{f,\textsc{Opt}}^{(i+1)}$ in~$a_{f,\opt}^{(i+1)}$.}\\
To prove the claim, assume otherwise, i.e., that $\textsc{Opt}[t^{(i+1)}]$ serves $r_{l,\opt}^{(i)}$  after collecting $r_{f,\textsc{Opt}}^{(i+1)}$.
	The request~$r_{f,\textsc{Opt}}^{(i+1)}$ is released after schedule~$S^{(i)}$ is started, i.e., after time $\alpha\cdot\textsc{Opt}(t^{(i)})$. Thus,
	\begin{align}\label{eq:case221_opt_estimate1}
		\textsc{Opt}[t^{(i+1)}] &\geq \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},b_{l,\opt}^{(i)}) \nonumber \\
\overset{\triangle\text{-ineq}}&{\geq} \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},p^{(i+1)})-d(b_{l,\opt}^{(i)},\o)-d(\o,p^{(i+1)}).
	\end{align}
	Since $S^{(i)}$ starts in~$\o$, ends in $p^{(i+1)}$ and serves $r_{l,\opt}^{(i)}$, we obtain
	\begin{equation}\label{eq:d(0,x_ell)-estimate}
		d(\o,b_{l,\opt}^{(i)})+d(b_{l,\opt}^{(i)},\o) \leq |S^{(i)}|+d(p^{(i+1)},\o) \overset{\text{Lem \ref{waoa}}}{\leq} \textsc{Opt}(t^{(i)})+d(p^{(i+1)},\o).
	\end{equation}
	Furthermore, because $\textsc{Opt}[t^{(i+1)}]$ serves~$r_{l,\wait}^{(i)}$ at~$p^{(i+1)}$ before picking up~$r_{f,\textsc{Opt}}^{(i+1)}$ at~$a_{f,\opt}^{(i+1)}$, we have
	\begin{equation}\label{eq:d(0,x_i+1)-estimate}
		\textsc{Opt}(t^{(i+1)})\geq d(\o,p^{(i+1)})+d(p^{(i+1)},a_{f,\opt}^{(i+1)})\overset{\eqref{eq:dlarge}}{>} d(\o,p^{(i+1)}) + \alpha\cdot\textsc{Opt}(t^{(i+1)})-\textsc{Opt}(t^{(i)}).
	\end{equation}
	Combining all of the above yields
	\begin{align*}
		\textsc{Opt}[t^{(i+1)}]
		\overset{\eqref{eq:case221_opt_estimate1},\eqref{eq:d(0,x_ell)-estimate}}&{\geq} \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},p^{(i+1)})-\frac{\textsc{Opt}(t^{(i)})+d(p^{(i+1)},\o)}{2}\\
		&\hspace{1cm}-d(\o,p^{(i+1)}) \\
		\overset{\eqref{eq:d(0,x_i+1)-estimate}}&{>} \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},p^{(i+1)})-\frac{\textsc{Opt}(t^{(i)})}{2}\\
		&\hspace{1cm}-\frac{3}{2}\left(\textsc{Opt}(t^{(i)})-(\alpha-1)\textsc{Opt}(t^{(i+1)})\right) \\
		&= \left(\frac{3}{2}\alpha-\frac{3}{2}\right)\textsc{Opt}(t^{(i+1)})-(2-\alpha)\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},p^{(i+1)}) \\
		\overset{\eqref{eq:dlarge}}&{>}  \left(\frac{3}{2}\alpha-\frac{3}{2}\right)\textsc{Opt}(t^{(i+1)})-(2-\alpha)\textsc{Opt}(t^{(i)}) +\alpha\cdot\textsc{Opt}(t^{(i+1)})-\textsc{Opt}(t^{(i)}) \\
		&= \left(\frac{5}{2}\alpha-\frac{3}{2}\right)\textsc{Opt}(t^{(i+1)})-(3-\alpha)\textsc{Opt}(t^{(i)}) \\
		\overset{\text{Obs \ref{defobs}a)}}&{\geq} \left(\frac{5}{2}\alpha-\frac{3}{2}\right)\textsc{Opt}(t^{(i+1)})-\left(\frac{3}{\alpha}-1\right)\textsc{Opt}(t^{(i+1)})\\
&=\left(	\frac{5}{2}\alpha-\frac{1}{2}-\frac{3}{\alpha}\right)\textsc{Opt}(t^{(i+1)}) \\
		&\geq  \textsc{Opt}(t^{(i+1)})
	\end{align*}
	where the last inequality holds if and only if $\alpha\geq \frac{1}{10}\cdot(3+\sqrt{129})\approx 1.436$. As this is a contradiction, we have that $\textsc{Opt}[t^{(i+1)}]$ serves all requests in $R^{(i)}$ before picking up~$r_{f,\textsc{Opt}}^{(i+1)}$ in~$a_{f,\opt}^{(i+1)}$. This completes the proof of the claim. \\

Now that we have established the claim, we turn back to the proof of \cref{rightorder3}.
	Let~$T\geq 0$ denote the time it takes $\textsc{Opt}[t^{(i+1)}]$ until it has served $r_{l,\opt}^{(i)}$, i.e., all requests from $R^{(i)}$. First, observe that
	\begin{equation}\label{eq:Tlower}
	T \geq \textsc{Opt}(t^{(i)}).
	\end{equation}
	By the claim, we have
	\begin{equation}\label{eq:case222_opt_i+1_estimate}
		\textsc{Opt}(t^{(i+1)}) \geq T+d(b_{l,\opt}^{(i)},a_{f,\opt}^{(i+1)})+|S(R^{(i+1)},a_{f,\opt}^{(i+1)})|.
	\end{equation}
	The algorithm $\textsc{Lazy}(\alpha)$ finishes $R^{(i+1)}$ by time
	\begin{align}
	t^{(i+1)}+S^{(i+1)} \overset{\text{Lem \ref{rightorder2}}}&{=}
		 \alpha\cdot\textsc{Opt}(t^{(i)})+|S^{(i)}|+|S^{(i+1)}| \nonumber \\
		&\leq \alpha\cdot\textsc{Opt}(t^{(i)})+|S^{(i)}|+d(p^{(i+1)},a_{f,\opt}^{(i+1)})+|S(R^{(i+1)},a_{f,\opt}^{(i+1)})| \nonumber \\
		&\leq \alpha\cdot\textsc{Opt}(t^{(i)})+|S^{(i)}|+d(p^{(i+1)},b_{l,\opt}^{(i)})+d(b_{l,\opt}^{(i)},a_{f,\opt}^{(i+1)}) \nonumber \\
		& \hspace{1cm}+|S(R^{(i+1)},a_{f,\opt}^{(i+1)})| \nonumber \\
		\overset{\eqref{eq:case222_opt_i+1_estimate}}&{\leq} \alpha\cdot\textsc{Opt}(t^{(i)}) + |S^{(i)}| + d(p^{(i+1)},b_{l,\opt}^{(i)}) + \textsc{Opt}(t^{(i+1)}) - T.\label{eq:case222_alg_estimate1}
	\end{align}
	As $S^{(i)}$ visits $b_{l,\opt}^{(i)}$ before $p^{(i+1)}$ and $\textsc{Opt}[t^{(i+1)}]$ visits $p^{(i+1)}$ before $b_{l,\opt}^{(i)}$, we have
	\begin{align}
		|S^{(i)}|+T &\geq \left(d(\o,b_{l,\opt}^{(i)})+d(b_{l,\opt}^{(i)},p^{(i+1)})\right)+\left(d(\o,p^{(i+1)})+d(p^{(i+1)},b_{l,\opt}^{(i)})\right) \nonumber \\
		&= 2\cdot d(p^{(i+1)},b_{l,\opt}^{(i)})+d(\o,b_{l,\opt}^{(i)})+d(\o,p^{(i+1)})
		\geq 3\cdot d(p^{(i+1)},b_{l,\opt}^{(i)}). \label{eq:third}
	\end{align}
	Combined, we obtain that the algorithm finishes not later than
	\begin{align*}
		t^{(i+1)}+|S^{(i+1)}| \overset{\text{\cref{eq:case222_alg_estimate1}}}&{\leq} \alpha\cdot\textsc{Opt}(t^{(i)}) + |S^{(i)}| + d(p^{(i+1)},b_{l,\opt}^{(i)}) + \textsc{Opt}(t^{(i+1)}) - T \\
		\overset{\eqref{eq:third}}&{\leq} \alpha\cdot\textsc{Opt}(t^{(i)}) + |S^{(i)}| + \frac{|S^{(i)}|+T}{3} + \textsc{Opt}(t^{(i+1)}) - T \\
		\overset{\text{Obs \ref{defobs}a)}}&{\leq} 2 \cdot \textsc{Opt}(t^{(i+1)})+\frac{4}{3} |S^{(i)}|-\frac{2}{3}T\\
		\overset{\text{Lem \ref{waoa}, \cref{eq:Tlower}}}&{\leq} 2\cdot \textsc{Opt}(t^{(i+1)}) + \frac{2}{3} \textsc{Opt} (t^{(i)})\\
		\overset{\text{Obs \ref{defobs}a)}}&{\leq} \left( 2+ \frac{2}{3\alpha} \right) \cdot \textsc{Opt}(t^{(i+1)})\\
		&\leq (1+\alpha) \cdot \textsc{Opt}(t^{(i+1)})
	\end{align*}
	where the last inequality holds if and only if $\alpha \geq \frac{1}{2}+\sqrt{11/12}$.
\end{proof}

Let us summarize the results from this section. Our goal was to prove that every schedule is $\alpha$-good for $\alpha\geq \frac{1}{2}+\sqrt{11/12}$. In \cref{basecase}, we have proven the base case. In the induction step, we have distinguished several cases. First, we have seen in \cref{interrupted} that the induction step holds if the previous schedule was interrupted. Next, we have seen in \cref{waoa} that the induction hypothesis implies $S^{(i+1)}\leq\textsc{Opt}(t^{(i+1)})$. If the previous schedule was not interrupted, we have first seen in \cref{wrongorder} that the induction step holds if $\textsc{Opt}[t^{(i+1)}]$ loads $r_{f,\textsc{Opt}}^{(i+1)}$ before serving $r_{l,\opt}^{(i)}$. If $\textsc{Opt}[t^{(i+1)}]$ serves $r_{l,\opt}^{(i)}$ before loading $r_{f,\textsc{Opt}}^{(i+1)}$, we have seen in \cref{rightorder1} and \cref{rightorder3} that the induction step holds. This completes the proof of \cref{mainthm}.

\section{Analysis on the half-line}

In this section, we prove that $\textsc{Lazy}$ is even better if the metric space considered is the half-line. In particular, we prove \cref{half-line_result}.

Since Observations \ref{basecase} and \ref{interrupted}, as well as Lemmata \ref{waoa}-\ref{rightorder2} hold for all $\alpha\geq\frac{1+\sqrt{3}}{2}\thickapprox 1.366$, it remains to show that a similar version of \cref{rightorder3} holds for $\alpha\geq\frac{1+\sqrt{3}}{2}$ on the half-line.

\begin{lemma}\label{rightorder3_half_line}
	Let $2\geq\alpha\geq\frac{1+\sqrt{3}}{2}\thickapprox 1.366$, and let $M=\mathbb{R}_{\geq0}$ be the half-line. Assume that the $i$-th schedule is $\alpha$-good and was not interrupted, and that $\textsc{Opt}[t^{(i+1)}]$ serves $r_{l,\wait}^{(i)}$ before collecting $r_{f,\textsc{Opt}}^{(i+1)}$. If \eqref{eq:dlarge} holds, then $t^{(i+1)}+|S^{(i+1)}|\leq (1+\alpha) \cdot \textsc{Opt} (t^{i+1})$.
\end{lemma}
\begin{proof}
First, observe that, in \cref{rightorder3}, we have the same assumptions except that we worked on general metric spaces. Therefore, all the inequalities shown in \cref{rightorder3} hold in this setting, too, so that we can use them for our proof. Next, note that on the half-line, we have for any $x,y \in M$
\begin{equation}\label{eq:half-line}
d(x,y)\leq \max \{d(x,\o),d(y,\o)\}.
\end{equation}
We show that this implies that a similar claim as in \cref{rightorder3} holds.\\
	\emph{Claim: $\textsc{Opt}[t^{(i+1)}]$ serves all requests in $R^{(i)}$ before picking up~$r_{f,\textsc{Opt}}^{(i+1)}$ in~$a_{f,\opt}^{(i+1)}$.}\\
To prove the claim, assume otherwise, i.e., that $\textsc{Opt}[t^{(i+1)}]$ serves $r_{l,\opt}^{(i)}$  after collecting $r_{f,\textsc{Opt}}^{(i+1)}$.
	The request~$r_{f,\textsc{Opt}}^{(i+1)}$ is released after schedule~$S^{(i)}$ is started, i.e., after time $\alpha\cdot\textsc{Opt}(t^{(i)})$. Thus,
	\begin{align}\label{eq:case221_opt_estimate1_half-line}
		\textsc{Opt}[t^{(i+1)}] &\geq \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},b_{l,\opt}^{(i)}) \nonumber \\
		& \geq \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},p^{(i+1)}) - d(b_{l,\opt}^{(i)}, p^{(i+1)}) \nonumber\\
		\overset{\cref{eq:half-line}}&{\geq} \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},p^{(i+1)})-\max\{d(b_{l,\opt}^{(i)},\o),d(\o,p^{(i+1)})\}.
	\end{align}
	Combining the above with the results from the proof of \cref{rightorder3} yields
	\begin{align*}
		\textsc{Opt}(t^{(i+1)})
		\overset{\eqref{eq:case221_opt_estimate1_half-line},\eqref{eq:d(0,x_ell)-estimate}}&{\geq} \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},p^{(i+1)})\\
		&\hspace{1cm}-\max\Bigl\{\frac{\textsc{Opt}(t^{(i)})+d(p^{(i+1)},\o)}{2},d(\o,p^{(i+1)})\Bigr\} \\
		\overset{\eqref{eq:d(0,x_i+1)-estimate}}&{>} \alpha\cdot\textsc{Opt}(t^{(i)})+d(a_{f,\opt}^{(i+1)},p^{(i+1)}) \\
		&-\max\Bigl\{\frac{2\textsc{Opt}(t^{(i)})-(\alpha-1)\textsc{Opt}(t^{(i+1)})}{2},\textsc{Opt}(t^{(i)})-(\alpha-1)\textsc{Opt}(t^{(i+1)})\Bigr\} \\
		& = \frac{\alpha-1}{2}\textsc{Opt}(t^{(i+1)}) + (\alpha-1)\textsc{Opt}(t^{(i)}) + d(a_{f,\opt}^{(i+1)},p^{(i+1)}) \\
		\overset{\eqref{eq:dlarge}}&{>}  \Bigl(\frac{\alpha-1}{2}+\alpha\Bigr)\textsc{Opt}(t^{(i+1)}) + (\alpha-2)\textsc{Opt}(t^{(i)}) \\
		\overset{\text{Obs \ref{defobs}a)}}&{\geq} \Bigl(\frac{\alpha-1}{2}+\alpha+\frac{\alpha-2}{\alpha}\Bigr)\textsc{Opt}(t^{(i+1)}) \\
		& \geq \textsc{Opt}(t^{(i+1)})
	\end{align*}
	where the last inequality holds for all $\alpha\geq\frac{4}{3}$. As this is a contradiction, we have that $\textsc{Opt}[t^{(i+1)}]$ serves all requests in $R^{(i)}$ before picking up~$r_{f,\textsc{Opt}}^{(i+1)}$ in~$a_{f,\opt}^{(i+1)}$. This completes the proof of the claim.\\

	Now that we have established the claim, we turn back to the proof of \cref{rightorder3_half_line}. Let~$T\geq 0$ denote the time it takes $\textsc{Opt}[t^{(i+1)}]$ until it has served $r_{l,\opt}^{(i)}$, i.e., all requests from $R^{(i)}$. First, observe that
	\begin{equation}\label{eq:Tlower_hl}
		T \geq \textsc{Opt}(t^{(i)}).
	\end{equation}
	If $p^{(i+1)}\geqb_{l,\opt}^{(i)}$, as $\textsc{Opt}[t^{(i+1)}]$ visits $p^{(i+1)}$ before $b_{l,\opt}^{(i)}$, we have
	\begin{equation*}
		T \geq d(\o,p^{(i+1)}) + d(p^{(i+1)},b_{l,\opt}^{(i)}) \overset{\eqref{eq:half-line}}{\geq} 2\cdot d(p^{(i+1)},b_{l,\opt}^{(i)}).
	\end{equation*}
	Otherwise, if $p^{(i+1)}<b_{l,\opt}^{(i)}$, as $S^{(i)}$ visits $b_{l,\opt}^{(i)}$ before $p^{(i+1)}$, we have
	\begin{equation*}
		T \overset{\eqref{eq:Tlower_hl}}{\geq} \textsc{Opt}(t^{(i)}) \geq S^{(i)} \geq d(\o,b_{l,\opt}^{(i)}) + d(b_{l,\opt}^{(i)},p^{(i+1)}) \overset{\eqref{eq:half-line}}{\geq} 2\cdot d(b_{l,\opt}^{(i)},p^{(i+1)}).
	\end{equation*}
	Thus, in either case, we have
	\begin{equation}\label{half-line_distance2}
		d(b_{l,\opt}^{(i)},p^{(i+1)}) \leq \frac{T}{2}.
	\end{equation}
	Combined, we obtain that the algorithm finishes not later than
	\begin{align*}
		t^{(i+1)}+|S^{(i+1)}| \overset{\text{\cref{eq:case222_alg_estimate1}}}&{\leq} \alpha\cdot\textsc{Opt}(t^{(i)}) + |S^{(i)}| + d(p^{(i+1)},b_{l,\opt}^{(i)}) + \textsc{Opt}(t^{(i+1)}) - T \\
		\overset{\eqref{half-line_distance2}}&{\leq} \alpha\cdot\textsc{Opt}(t^{(i)}) + |S^{(i)}| + \frac{T}{2} + \textsc{Opt}(t^{(i+1)}) - T \\
		\overset{\text{Obs \ref{defobs}a)}}&{\leq} 2 \cdot \textsc{Opt}(t^{(i+1)}) + |S^{(i)}| - \frac{T}{2} \\
		\overset{\text{Lem \ref{waoa}, \cref{eq:Tlower_hl}}}&{\leq} 2\cdot \textsc{Opt}(t^{(i+1)}) + \frac{1}{2} \textsc{Opt}(t^{(i)})\\
		\overset{\text{Obs \ref{defobs}a)}}&{\leq} \left( 2+ \frac{1}{2\alpha} \right) \cdot \textsc{Opt}(t^{(i+1)})\\
		&\leq (1+\alpha) \cdot \textsc{Opt}(t^{(i+1)})
	\end{align*}
	where the last inequality holds if and only if $\alpha \geq \frac{1+\sqrt{3}}{2}\thickapprox 1.366$.
\end{proof}

Similar to the proof of \cref{mainthm}, combining Observations \ref{basecase} and \ref{interrupted} and Lemmata \ref{waoa}-\ref{rightorder2} with \cref{rightorder3_half_line} completes the proof of \cref{half-line_result}.
\newpage

redpajama_set_name: RedPajamaArXiv


Texas' high-speed train is closing out a year marked by significant progress, building a new industry with international expertise.
And just this week, a TV report by KERA in Dallas showcased the project's benefits, saying the train is helping to drive innovation as Texas' economic might is leading the country into the future.
It is hard to believe that just about a year ago the Federal Railroad Administration released its more than 5,000-page Draft Environmental Impact Statement, identifying a single preferred alignment for the entire route.
Since then, technical experts using the report's detailed information have put in thousands of hours, updating the train's low-impact building plan and ensuring it has minimal effect on communities and the environment.
The project also added several new powerhouse partners, international companies that competed vigorously to be part of the world-class team bringing the first high-speed train to North America and guiding Texas ever closer to a dynamic and much-needed transportation choice.
Among the highlights was the addition of multinational firm Salini Impregilo – operating in the U.S. market with The Lane Construction Corporation and WSP – to lead the civil construction consortium that will build the train's alignment. Salini and Lane have vast international experience, having built more than 4,000 miles of railway infrastructure in 50 countries.
Also, international railway company Renfe was chosen as the train's operating partner. Renfe already operates the second-largest railroad system in the world, bringing more than 25 years of experience in operating high-speed trains in Europe and other countries. Renfe operates 5,000 trains daily on 7,500 miles of track, from local commuter routes to high-speed national and international travel.
In May, we announced that Bechtel, a U.S.-based company with an operational headquarters in Houston, will be the project manager. It has global expertise in managing and constructing high-speed train infrastructure and complex, heavy civil projects. That includes more than 300 major train and subway projects, such as the Channel Tunnel High Speed 1, Crossrail in London and Dulles Corridor Metrorail in the Washington, D.C., area.
We also reached agreement with Amtrak, allowing passengers to use Amtrak's reservation system to buy tickets for through travel on both the high-speed train and Amtrak's national routes. Texas Central will offer a convenient transfer service connecting riders between Amtrak passenger stations and the high-speed train terminals.
Another highlight was the announcement of passenger stations. More details were unveiled about the Dallas site in the Cedars neighborhood, our preferred station location in Houston, and the Brazos Valley terminal. These developments will be a catalyst for local jobs and economic growth as well as providing a convenient, efficient and direct network for passengers to and from local transit systems in Houston and Dallas.
You can't look at 2018 without also recognizing the incredible growth we have seen in Texas, especially along this alignment. The state's population growth has blown past already aggressive predictions, boosting the demand for a safe, reliable and productive link between these exploding economies. Research shows Texans are ready to ride this train, and growth only drives higher the number of potential passengers needing the high-speed connection.
As a capstone to the year, the Texas train was cited as a top infrastructure project by the infrastructure advocacy group CG/LA in its annual strategic report, which identifies the "most imaginative and transformative" North American projects.
We're honored by that recognition and look forward to another year, making incredible progress as we work together to keep Texas moving.


text:
More Risk Aversion Lends Support to Dollar
The dollar has risen today by 0.8% against the Chinese yuan, 0.6% relative to sterling, 0.4% versus the euro, 0.3% vis-a-vis the loonie, and 0.1% against the Swiss franc and Mexican peso.
Japan (Sports Day) and Canada (Thanksgiving) are observing holiday closures, but China reopened for business following a one-week observance of the Communist Revolution of 1949.
Share prices in the Pacific Rim plunged 3.7% in China, 1.8% in Hong Kong and 1.4% in Australia. There were also drops of 0.6% in South Korea and Taiwan and 0.9% in New Zealand and Singapore. Markets in Europe have so far fallen 1.9% in Greece, 2.2% in Italy, 0.9% in France, 0.8% in Germany and Switzerland, and 0.5% in the U.K. and Spain.
The 10-year Italian sovereign debt yield jumped 16 basis points and is near a 4-1/2 year peak. Its Greek counterpart rose nearly as much (14 basis points). 10-year sovereign debt yield declines of 4 basis points in Germany and 3 bps in France and Great Britain further contributed to widening long-term interest rate differentials.
Commodity prices reflect the latest tide of risk aversion. WTI oil is down 1.3% at $73.38, while Comex gold fell 0.7% to $1,197.30 per troy ounce.
Several factors are encouraging investors to be risk averse.
An intensifying dispute between the Italian government and the EU over Italy's fiscal policy.
Worries that Chinese growth will slow because of a trade war. Chinese foreign exchange reserves fell $22.7 billion last month, most in 7 months, and officials at the Peoples Bank of China injected fresh liquidity via a lower reserve requirement in hopes of encouraging more bank lending.
Britain is running out of time to negotiate Brexit, and this issue is dividing the country's political cohesion.
The Kavanaugh confirmation was a huge win for President Trump. The Republicans are playing a weak hand masterfully, proving that democracy governments do not necessarily deliver policies that majorities prefer. With each political win, Trump will be emboldened to pursue his agenda of divide and conquer both domestically and in foreign policy. Trade tensions will swell.
The Sentix measure of investor sentiment toward the euro area slid to a 4-month low in October of 11.4 from a reading of 12.0 in September and 32.9 at the start of 2018.
German industrial production recorded a second straight month-on-month decline in August, falling 0.3% after July's slide of 1.3%. Production was 0.1% lower than in August 2017, compared to a 6.1% on-year increase recorded last January.
Ireland's construction purchasing managers index dropped 2.1 points to an 11-month low in September of 56.2, which still highlights a decent pace of improving operating conditions.
Czech industrial production and retail sales posted monthly advances and also were 1.9% and 4.2% higher than a year earlier in August. Norwegian industrial production climbed 2.3% between August 2017 and August 2018.
French business sentiment improved last month according to the Bank of France in manufacturing, services and construction. Confidence among manufacturers hit a 7-month high.
Russian CPI inflation accelerated 0.3 percentage points to a 14-month high in September of 3.4%. That compares with 2.3% just 3 months earlier.
Swiss unemployment edged down 0.1 percentage point as expected to 2.5% in September on a seasonally adjusted basis.
China's services PMI rebounded 1.6 points to a 3-month high of 53.1 in September but an 11-month low in that economy's manufacturing PMI limited the composite PMI's improvement to just 0.1 point. At 52.1, such was still down from 53.0 at mid-year.
Officials at the National Bank of Serbia's Executive Board agreed to hold their benchmark interest rate at 3.0%. Such was last changed back in April when the second of two back-to-back 25-basis point easings was engineered. 2017 and 2016 each also experienced 50 basis points of rate reduction. A released statement highlighted several external risks that warrant policy caution: volatile oil prices, further likely Fed tightening, the end of ECB bond buying, and world protectionism.
The Bank of Israel is also conducting a policy review today and expected to keep its interest rate steady at 0.1%.
Copyright 2018, Larry Greenberg. All rights reserved. No secondary distribution without express permission.
Tags: German industrial production, National Bank of Serbia
This entry was posted on Monday, October 8th, 2018 at 6:46 am	and is filed under New Overnight Developments Abroad - Daily Update. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed.


text:
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