Cost and Pathway, Ex8

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Cost and Pathway, Ex8
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First part of the Problem

• In an order different from that in the problem
  statement makes more sense
• Avoid steep slopes
• Avoid intrafficable soils
• The closer to the roads the better
• Avoid Residential and Seasonal LUs
• Not cross Protected landuses
• Not cross Lake or Wetlands

Simple
Distance
CostGrow, -1
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So!
Let's take these up one by one!
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Friction

• Remember that Friction is an abstract idea.
• In the first part we consider friction the
  physical (energy) cost to cross a cell
• In the second part we consider friction to be the
  cost to cross a cell

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The pure friction layers

• Well, the notes say all friction components
  should have a maximum value of exactly 30
• Avoid steep slopes
• Slope in has a maximum of 31.5209789
• 30/31.5209789
• 0.95174709183920680839007826625587
• Use that value to normalize slope to 30

Where did that come from?
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Avoid InTrafficable soils

• Notes give key to this. What is the solution?
• Just reclass soils with the values given in the
  problem statement.
• Maximum is 30 so that fits

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Closer to the Road the Better

• Closer? What does that mean?
• That cost increases with distance from roads
• How do we make that happen?
• DISTANCE!!!
• How do we normalize it to a max of 30?
• Multiply by (30/max distance) Duh!

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Avoid Residential and Seasonal LUs

• Key word is Avoid. What does that mean? What
  IDRISI function do we use?
• Ahh this is different in fact just the
  opposite from the roads friction.
• Cost must be max near these LUs and decrease as
  we move away from them
• DISTANCE again, but we have to mess with it (next
  slide)
• OK, How do we make it have a maximum value of
  30? (same as for roads)

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Avoid Residential and Seasonal LUs

• This is how
• DISTANCE
• - max distance
• ABS

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Another way

• DISTANCE
• Normalize by x 30/max value
• X 1
• 30

Normalized to 30
X -1
30
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So to get total Friction
F1_Soil
F1_Slope
F1_Roads
F1_LUs

Max Possible 4x30120
F_Total
Will we actually get a value that high?
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Not cross Lake or Wetlands

• How do we prevent COST from crossing these
  landuses?
• Or, put another way, how do we prevent a PATHWAY
  from crossing them?
• Use COSTGROW
• Then 1 is a absolute barrier

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Not cross Lake or Wetlands

• Well, thats a piece of cake!
• What is the solution????
• Make Lake and Wetlands 1 with reclass
• Same for not crossing Protected landuses
• But how do we introduce this into the friction
  layer???
• Remember, there are four friction layers
• Which we totaled into one total friction layer

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Not cross Protected LUs

• Well, same thing
• Make them 1 also
• So we have two images with not crossable areas
  with values of _-1_
• How to combine them? What should the value of
  the crossable areas be?
• That is, where are we going with this?

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Not Cross Protected LUs

• Humm Have to combine the not crossable with
  the Friction layers so that not crossable is
  maintained in the result!
• So we use MINIMUM
• If that is the case then the value of crossable
  areas should be gt200
• But look ahead -----
• What is the total possible in Alternative 2?

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So

• -1 where not crossable
• 200 where crossable

Range is 1 to lt 120
F1_Total
Not_Cross
?
Min
F_Final1
So min will result in file with 1 in uncrossable
areas and total friction in rest of image
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And the Rest is a piece of Cake
F1_Total
Terminal
COST
Cost1
Well
PATHWAY
Path1
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The 2nd Alternative

• If you figure out what has to be done here before
  doing Alternative 2 then you can save yourself
  some time.
• Think it through
• How can you maximize the re-use of layers in 2
  from 1?
• If you make the 1 images have a background of,
  say, 300 or 400 then you can use them in the 2nd
  part.

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Dollar Costs

• How can the dollar costs of the two pipelines be
  computed?
• We need two parts
• Total length of the path in feet
• Total cost per cell for traversing the path
• Length in feet x cost/ft total cost
• One way to get length is to count the cells
• But remember that diagonal links are 1.414 x as
  long as orthogonal links and each cell is 165
  feet in dimension.
• Then make each cell reflect the total cost for
  crossing that cell

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How to get Total Cost

• Simple
• Add the cost images together
• Ok, Now you have total costs for every cell
• Now what?

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How to get Total Cost

• This is How (for three layers) ..

Total per foot of line. But each cell is 165
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1/foot fro each 100 feet?

• Distance again
• But /100 to get 100s of feet from road to that
  cell.

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And then

• Duh Multiply the total Cost by the pathway
  image!
• Now count cells of same total cost
• Bummer! ? Lots of work
• Wait there is a better way!!!!!
• Use COSTGROW again!!! WOW!
• But remember to make pathway background 1 AFTER
  ADDING THE COST TOGETHER!!!!

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The Better Way
TotalCost
Pathway1
Real, Binary w/ -1.0s not 1s
X
Total length of line in cells counting diagonals
in last cell
CostInPath
RECLASS to what?
RECLASS 0? -1
CostCell
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Aaaaaad Theeen
CostCell
Pathway1
X
PathCost
Terminal
And where isTOTALCOST??
COST
AccumCost
In the last cell, Where Else!!!!
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NOT quite that simple!!

• For cost, the files have to be real, binary
• So if you are making new initial files for
  pathway make sure that they are that
• And the only way to make them real, binary is to
• Select Binary file type
• PUT A REAL NUMBER IN INITIAL VALUE BOX!

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A wee perception problem

• Even though the values are really 1.0 and 1.0
  autoscale will provide all the nonsense numbers
• Autoscale is ON and you cant turn it off ?
• Histo will show 1.0 as 0 ?
• If there are no decimal values Structure will
  show the data as Integer ?

-1.0
1.0
And I had trouble with CONVERT producing noise
when converting files with 1 to real binary! ?
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DEMO
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What was done

• Created window (21 x 21) of slope and called that
  dollar cost
• Digitized vector from C10R2 SE at 45 degrees
  converted to raster (Real, binary with initial
  image all 1.0s)
• Ran COSTGROW from start to get total length of
  path in cells
• X 165 to get total length in feet

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Demo1
Pathway (value 1.0)from upper left to lower right
on a background of 1.0. Forces Autoscale so
legend is misleading
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Demo2
Running cost over the pathway in the previous
slide produces an accumulation of unit distances
in the pathway. The final cell says that the
pathway is 12.73 units long. Since each cell is
165 the total length of the path is 2,100.45
0.0
1.414
2.828
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Demo3
Multiplying the previous image by 165 produces
this image of the total length of the pathway is
2100.11 feet (why it is not 2100.45 I dont know
could be rounding error in the algorithm.)
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Demo4
Multiplying the pathway in Demo1 by slope which
we said was friction in this case (remember,
friction is an abstract idea, in this case we are
saying it is cost to cross the cell) we get this
image that says that the total cost for
transversing the path is 9.45 /foot. That times
the total length
of 2100.11gives a total cost 19,849.2522 for the
trip.
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What was done

• Length was 2,100.0714 feet
• Multiplied dollar cost times path
• Ran cost again
• Now total was 9.45875946492 /foot
• X 2,100.0714 feet 19,864.41 for building that
  2100 foot road.

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Questions???
Go For It!!