Ubuntu 16.04.1 was released on 21st July 2016. According to      Today is also the day that users of Ubuntu 14.04 LTS (Trusty Tahr) are notified that a new Ubuntu release is ready.   I am still getting "No new release found" when trying to update my 14.04.4 desktop and a 14.04.4 server via the Software Updater and do-release-upgrade. Both have been set to notify me when new LTS versions are available. The 16.04.1 release is available from the mirror I am using. I switched to the default mirrors on the desktop but the message was the same.  Why is this the case and what can I do to fix it?
So, today is July 21st and . gives me an option to download 16.04.1 LTS. However, attempting to upgrade gets me this message -  $ sudo do-release-upgrade Checking for a new Ubuntu release No new release found   What is wrong? Does it matter that I am running apt-get update behind apt-cacher-ng?  Note: Commenting out the files in apt.conf.d that used apt-cacher-ng did not make a difference.  Edit: Explanation for possible duplicates:  deals with upgrade from 12.04LTS to 14.04LTS. My question is specific to 14.04 LTS to 16.04.1 LTS upgrade. Old answers, while good for archival purposes are less relevant in a fast changing package like ubuntu. Moreover none of the answers answer what is wrong. The link  pointed out is mentioned in one of the comments. This Q &amp; A is more helpful for people upgrading today than the work around mentioned in "".
Change Data Type Varchar To Varbinary(max) In SQL Server Is there way to replace value of symbolic link? Is there way to replace value of symbolic link?  For example I want to replace symbolic link like this:  first -&gt; /home/username/foo/very/long/directories/that/I/do/not/want/to/type/again    to   second -&gt; /home/username/bar/very/long/directories/that/I/do/not/want/to/type/again   I want to change only foo to bar.  Of course I can create a link again, but if it is possible to replace value of the link, it become easier. Why does a cork float to the side of a glass? Why does a cork ball float to the side of a glass as illustrated in the following GIF? What is the physical phenomenon behind this observation and why does it happen? 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I am getting the following errors while updating.  sudo apt-get update  Err http://ppa.launchpad.net natty/main Sources                                      404  Not Found Err http://ppa.launchpad.net natty/main i386 Packages                                404  Not Found ......  E: Some index files failed to download. They have been ignored, or old ones used instead. How to add a class to a given element? Is there a way of moving unwanted villagers out faster? Is there any way to speed up this process or indirectly 'force a villager to move out'?  I know they move out if it's time for a new one to move in, and I know you can't actually 'force' them persay.  I have a villager who has caused nearly all my other villagers to wear his.. rather hideous 'kiddie shirt'. ((the character's default shirt)) Aside from that his home blocks the perfect spot, very close to my river, for building a convenient bridge...  I've taken to pushing him, hitting him with a net, and I've been trying to find a Toy Hammer... He wanted to leave once and I told him to go, but then he said he wouldn't leave until I, "made an effort to get to know him" or some BS..  Are there any other ways I can get an unwanted, 'pesty', villager to leave town? Can I configure it to allow:   select right-click to copy right-click or ctrl-v to paste? My code follows:  \documentclass{book} \usepackage{color,colortbl} \begin{document}  \begin{tabular}{ll&gt;{\color{red}}p{20pc}} a &amp;1 &amp;Test \end{tabular}  \end{document}   The output was:    Some unnecessary vertical space came before the last column. If I remove the tag \color{red} then everything works fine, how can I fix this?  Above said issue was fixed based on the suggestion given by David Carlisle and CarLaTeX, but the same problem was there if I use the supertabular environment, I used supertabular package for auto breaking table and the code follows:  \documentclass{book} \usepackage{color,colortbl,supertabular} \begin{document}  \begin{supertabular}{ll&gt;{\leavevmode\color{red}}p{4pc}&gt;{\leavevmode\color{black}}l&gt;{\leavevmode\color{black}}l&gt;{\leavevmode\color{red}}p{4pc}} a &amp;1 &amp;Test &amp;a &amp;1 &amp;Test\\ a &amp;1 &amp;Test &amp;a &amp;1 &amp;Test\\ \end{supertabular}  \end{document} Here's a corollary of AC stating that: if there exists an onto mapping $f:A\to B$, then there exists a 1-1 mapping $g:B\to A$. What I'm confused about is that whether or not  AC can be derived from this proposition, in return? I have tried but haven't found a proof. Any help would be appreciated. biblatex – et al. beginning from second citation? Basically all I need to know is how I can use "et. al" beginning from the second citation of a source. I am using the following  \usepackage[ natbib=true, style=authoryear ]{biblatex}   I need citations to look like this:  First citation of a source     Drobetz, Schillhofer, and Zimmermann (2004)   Second and further citations     Drobetz et al. (2004)   Right now, every citation looks like the first one. With the biblatex option maxcitenames I can only adjust all citations and not all but the first one. Any help? Protect Google Spreadsheet formatting while allowing changes to data I would like to protect the formatting of a cell (or a range of cells) in a Google Spreadsheet, but still allow changing the data in that (or those) cell(s).  For example, I want to add a validation rule on a cell that says the cell has to contain a number between 1 and 10, and I would like others to be able to change the value in that cell, but not the validation rule itself.  How do I do that?  So far I was only able to completely protect the cell (content + formatting) or not at all (the user can change the content as well as the formatting).  To me this seems to be an important functionality when adding access rights to spreadsheets. Is it possible that it is completely missing from Google Spreadsheets? Worldwide Dissertation Database? or French, German, Italian ones? Is there a global/worldwide database of dissertations? If not, how can I access French, German, Italian, etc. dissertations online? I am trying to understand how to find the orthogonal complement of a subspace $M$ of a vector space $V$.   From my understanding, $M^\perp$ is also a subspace of $V$ where all its vectors are perpendicular (orthogonal) to the columns of $M$, which would mean that the dot product of those vectors with each column of $M$ is $0$.  However, how do I find the $M^\perp$ subspace, when $M = \{ \vec{x} \in \mathbb{R}:  2x_2 + 3x_1 = 0\}$? In general, if there's a general approach, how does one attempt to find $M^\perp$ of an arbitrary $M$?  From my understanding, in this case, $M$ is the set of vectors in a plane that satisfy that equation, where $x_1$ and $x_2$ should respectively be the first and second components of the vector $\vec{x}$. How long does mana from Burning-Tree Emissary stay around? Does mana added to your pool from  get destroyed when the card that added it is destroyed? I'm looking for a movie that was set in the US, and all the characters (including the robot) spoke English.  I can't recall the exact plot, but mostly it was about an innocent game robot that shoots small objects for fun and becomes dangerous, because its AI made a wrongful choice about 'winning the game'. The robot starts to arm itself with more dangerous objects (not sure but knives and stuff) and starts to attack people.  It was overall supposed to be fun and adventure movie for kids. Color movie, and probably made in 80~90's.   There is a recurring scene when the robot detects an aggressor, it scans and loudly announces "enemy one, enemy one". The robot was somehow looking similar to Wall-E, and had a saucer-like head part. Why is "union" an exception to the "a/an" rule? We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$  Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?