How can faculty best telegraph work-life balance challenges under COVID to our administration?

[edited after 1st comment from: @chepner - thanks!]

/bin/bash allows hyphens in function names, /bin/sh (Bourne shell) does not. Here, the offending "some-function" had been exported by bash, and bash called yum which called /bin/sh which reported the error above.

fix: rename shell functions to not have hyphens

man bash says that bash identifiers may consist: "only of alphanumeric characters and underscores"

The /bin/sh error is much more explicit:

some-function () { :; }

sh: `some-function': not a valid identifier


The region you are interested is the blue shaded region shown in the figure below. enter image description here

The coordinates of the centroid denoted as $(x_c,y_c)$ is given as $$x_c = \dfrac{\displaystyle \int_R x dy dx}{\displaystyle \int_R dy dx}$$ $$y_c = \dfrac{\displaystyle \int_R y dy dx}{\displaystyle \int_R dy dx}$$ where $R$ is the blue colored region in the figure above.

Let us compute the denominator in both cases i.e. $\int_R dy dx$. Note that this is nothing but the area of the blue region. Hence, we get that \begin{align} \int_R dy dx & = \int_{x=0}^{x=1} \int_{y=0}^{y=x^3} dy dx + \int_{x=1}^{x=2} \int_{y=0}^{y=2-x} dy dx = \int_{x=0}^{x=1} x^3 dx + \int_{x=1}^{x=2} (2-x) dx\\ & = \left. \dfrac{x^4}{4} \right \vert_{0}^{1} + \left. \left(2x - \dfrac{x^2}2 \right)\right \vert_{1}^{2} = \dfrac14 + \left( 2 \times 2 - \dfrac{2^2}{2} \right) - \left(2 - \dfrac12 \right) = \dfrac14 + 2 - \dfrac32 = \dfrac34 \end{align}

Now lets compute the numerator for both cases.

To find $x_c$, we need to evaluate $\int_R x dy dx$. We get that \begin{align} \int_R x dy dx & = \int_{x=0}^{x=1} \int_{y=0}^{y=x^3} x dy dx + \int_{x=1}^{x=2} \int_{y=0}^{y=2-x} x dy dx = \int_{x=0}^{x=1} x^4 dx + \int_{x=1}^{x=2} x(2-x) dx\\ & = \left. \dfrac{x^5}{5} \right \vert_{0}^{1} + \left. \left( x^2 - \dfrac{x^3}{3}\right) \right \vert_1^2 = \dfrac15 + \left( 2^2 - \dfrac{2^3}3\right) - \left( 1^2 - \dfrac{1^3}3\right) = \dfrac15 + \dfrac43 - \dfrac23 = \dfrac{13}{15} \end{align}

To find $y_c$, we need to evaluate $\int_R x dy dx$. We get that \begin{align} \int_R y dy dx & = \int_{x=0}^{x=1} \int_{y=0}^{y=x^3} y dy dx + \int_{x=1}^{x=2} \int_{y=0}^{y=2-x} y dy dx\\ & = \int_{x=0}^{x=1} \left. \dfrac{y^2}{2} \right \vert_0^{x^3} dx + \int_{x=1}^{x=2} \left. \dfrac{y^2}{2} \right \vert_{0}^{2-x} dx\\ & = \int_{x=0}^{x=1} \dfrac{x^6}{2} dx + \int_{x=1}^{x=2} \dfrac{(2-x)^2}{2} dx = \left. \dfrac{x^7}{14} \right \vert_{0}^{1} + \left. \dfrac{(x-2)^3}{6} \right \vert_{1}^{2}\\ & = \dfrac1{14} + \left( \dfrac{(2-2)^3}{6} - \dfrac{(1-2)^3}{6} \right) = \dfrac1{14} + \dfrac16 = \dfrac5{21} \end{align}

Hence, $$x_c = \dfrac{\displaystyle \int_R x dy dx}{\displaystyle \int_R dy dx} = \dfrac{13/15}{3/4} = \dfrac{52}{45}$$ $$y_c = \dfrac{\displaystyle \int_R y dy dx}{\displaystyle \int_R dy dx} = \dfrac{5/21}{3/4} = \dfrac{20}{63}$$


I faced the same problem just recently.

And I've found a way of getting a better path. In my case, I was trying to visualise the effect of having the Panama and Suez canals.

My suggestion isn't going to help you find the exact distance, however - but it will trace a more realistic minimum cost path which should be closer in length to the real optimum.

You've got a uniform cost value for the sea, and a high cost value for land. So far, so good.

When I did the least-cost path from Bristol, England to Sydney in Australia, the suggested path hugged the west coast of Africa, before rounding the cape and going in a straight line (line shown in red)

enter image description here

As whuber hinted at, this is caused by lots of local optimums not adding up to a global optimum.

The way I did it was to add peturbation, so that some randomness comes into play.

  • add a random value raster in SAGA GIS ('Random field'), the same size as my cost surface
  • and add this to my cost raster (Grid Sum in SAGA)
  • trace the least cost route on the noisy raster.

In my case I added a value between 0 and 10, as I wasn't bothered about the actual distance, just a better path. I find uniform works better than guassian. I'm using 9999 as the land cost.

Now, with the randomised cost surface, I find the shortest path now crosses diagonally, rather than hugging coastlines. The orange route is the one based on the noisy raster...

enter image description here

In my case I'm using a python script (from the gdal python cookbook) to do the least-cost tracing as a raster, so you mileage may vary depending on how you do the point to point path tracing (I can't get r.drain to work at the moment with QGIS).