An Overview of Digital Terrain Modelling and Multi-Criteria Analysis

Transcription

1 Introduction An Overview of Digital Terrain Modelling and Multi-Criteria Analysis BC EnviroConsultants was recently awarded a contract from Environment Canada to produce a report on a rare plant species (Spiranthes diluvialis) that is known only from the southern Okanagan. They have asked you, their GIS analyst, to prepare a map that will help their botanist in her field work. The botanist has identified that the species prefers low-slope, low-elevation areas with a southern aspect. Using that knowledge you can produce a multi-criteria model that highlights potential habitat and helps direct her search efforts. Your goal is to identify 725 hectares that are most likely to support habitat for this species In order to prepare the potential habitat map you need to, first, produce an accurate digital elevation model (DEM) of the area. Once you have created the DEM you can derive a slope and an aspect map. Then, after appropriately transforming the three surfaces (elevation, slope and aspect) and determining their respective weights, you can use a weighted overlay method to produce a potential habitat map. + + = Elevation Slope Aspect Potential habitat In this lab you will explore GIS techniques for modelling characteristics of the earth's surface and be introduced to multi-criteria methods (e.g., how to normalize layers to a consistent scale, and how to produce a weighted overlay). You will also work with Geodatabases and ArcToolbox models. In Part I of the lab you will create a TIN, and from that a DEM, while in Part II of the lab you will conduct a simple MCE in order to identify potential habitat for the rare species. Data for this lab covers a small portion of the southern Okanagan Valley of BC (e.g., The three datasets you will be using for this lab water.shp, transportation.shp and elevation.shp have been derived from BC TRIM data ( -- search for TRIM). The results of this lab will be due in two weeks (Oct 9 th ), and should be handed in at the beginning of the lab, or to me at the beginning of the class, on Tuesday. Part I: Creation of Digital Terrain Models: A TIN and a DEM Two of the differing solutions used to represent 3-dimensional surfaces and elevations in a GIS will be considered in this lab: vector Triangulated Irregular Networks (TINs) and raster Digital Elevation Models (DEMs). Each solution possesses inherent strengths, weaknesses and application possibilities. The Triangulated Irregular Network model is a model that recreates a 1

2 topographic surface by connecting measured points of elevation with lines of slope and planes of aspect. A Digital Elevation Model uses a raster (cell-based) approach to the representation of topography. Contours the traditional vector approach to representing 3-dimensional surfaces, will not be considered in this lab. 1. Can you suggest some reasons why contours are not typically used in GIS analyses? [1] Tasks such as wildlife habitat mapping, site visibility analysis, hazards prediction (e.g., avalanches), satellite imagery classification, and modelling human and environment interactions depend upon quality digital terrain models (DTMs). A digital terrain model can be used both as a means of modelling the three-dimensional character of an area, and as base for deriving other characteristics (such as slopes and aspects). The DTM you chose (i.e., TIN or DEM), and the level of detail of that model, will affect the data uncertainty and any additional information derived from that model. Errors and accuracy in interpolation and surface-derivative calculations (e.g., calculating the slope from a DEM) depend upon the selection of an appropriate cell size [DEM] or point spacing [TIN] for the topographic data set, and upon the availability of relevant ancillary information. The Association for Geographic Information defined a TIN as: A form of the tesseral model based on triangles. The vertices of the triangles form irregularly spaced nodes. Unlike a grid, a TIN allows dense information in complex areas, and sparse information in more homogeneous areas. The TIN dataset includes topological relationships between points and their neighbouring triangles. Each sample point has an X,Y co-ordinate and a surface, or Z-Value. These points are connected by edges to form a set of non-overlapping triangles used to represent the surface. TINs are also called irregular triangular mesh or irregular triangular surface model. ( [not accessible anymore]). As noted in the ArcMap definition, a TIN is a surface representation derived from irregularly spaced sample points and breakline features. To find out more about TINs and the very important role that breakline features can play in the creation of a TIN, consult the ArcMap help file on TINs; Creating a TIN surface; Fundamentals of creating TIN surfaces. (Available online at: Unlike the standard interpolation processes used in the creation of raster DEMs (e.g., splines, inverse distance weighting topics we will cover in a future lecture), which only use the elevation spot heights or mass points in the surface creation process, TIN interpolation can make use of ancillary information such as the location of roads, rivers and lakes (and, in particular, the elevation of the lakes) and other important topographic features, such as cliffs and ridge lines, that can directly influence the topography. Therefore, a TIN should provide a better representation of a topographic surface if such information is available. Because of the benefits provided by TINs, the BC government collected TRIM data in a manner that would allow for the creation of TINs. Furthermore, once a TIN has been created it is a very simple process to create a DEM. (It should be recognized that in some cases it is better to use a point-based interpolation routine on the point data we will discuss such issues in the class on spatial interpolation.) 2

3 Use GetData to download the data for this lab into C:\data\lab2. You should end up with three shapefiles containing some BC TRIM data (used with permission) from the southern Okanagan region: elevation.shp (spot heights or mass points) water.shp (rivers and lakes; a line file) transportation.shp (roads; a line file) As well, you will find an ArcMap document file, lab2.mxd, that you must use (as it establishes the BC Albers Standard Projection as the projection associated with the layers, and sets various other parameters, such as the location of My Toolboxes ). (File Open Lab2.mxd) The first step in this lab will be to convert the three shapefiles into three ArcMap Geodatabase Feature Classes, all held within a File geodatabase. You will first need to create a new File Geodatabase in ArcMap using Catalog and then to convert the shapefiles into feature classes. Within Catalog, select C:\data\lab2, right-mouse click, and select New File Geodatabase. Rename the new Geodatabase.gdb to Okanagan. (As always, ensure that you also set Okanagan to be the default geodatabase, and that the Current and Scratch Workspaces [under Geoprocessing Environments] are also explicitly set to Okanagan [and confirm these selections every time you run ArcMap].) In ArcMap, within ArcToolbox, select Conversion Tools / To Geodatabase. Select Feature Class to Geodatabase (multiple). Select the three shapefiles as the input features, and select Okanagan.gdb as the Output Geodatabase (use the ADD button, don t double-click and open the geodatabase itself) within which the three new feature classes will be placed. Click on OK. Note that the conversion process can take some time. Once the conversion process is complete, delete the three shapefiles, and add the three feature classes from the Okanagan geodatabase (this time, double-click on Okanagan.gdb). As noted above, water contains arcs that define both rivers and lakes. When creating a TIN we should have the lakes represented as polygons in order to force the TIN to model lakes as flat surfaces. Because two types of features are currently included in the same feature class (rivers, which are linear features, and lakes, which should be represented as polygonal features but are represented as unclosed polyline features in this feature class), we need to extract those arcs that define lakes, and then convert that data from lines to polygons (i.e., a [topological] feature class can only contain one type of geometry, be it point, line or polygon). To extract the arcs that define lakes from the water feature class, we need to be able to identify those arcs. Open the attribute table for the water layer, and note the field names in the table. Two of the right-most columns of the table contain two important pieces of information the FCODE and the ELEVATION. The FCODEs represent the TRIM Feature Codes (i.e., each arc has a unique feature code associated with it) that can be used to select only those arcs that relate to lakes. Once they have been identified, we can create a new feature class that only contains arcs associated with lakes, and from that create a polygon feature class. You will need to use the elevation values later on when building the TIN (i.e., the entire lake surface will be assigned a single elevation). 3

4 In order to determine what FCODEs relate to lakes you will need to go to the BC government web pages that explain the TRIM feature codes ( and from there determine which codes describe water bodies. You should use the Integrated Land Management Bureau s searchable Geographic Feature Catalog database in order to identify those codes that describe water bodies (click on ). In the Geographic Feature Catalogue query form enter lake as the Feature, and click on Next. A listing with 17 feature codes should appear. For the purposes of this lab we are not interested in those lakes that have feature codes beginning with GB953, so that leaves 13 feature codes. Note that all of those codes begin with the same 7 letters. Write down the codes you will need that information later. 2. Based on the feature codes listed in your Geography Feature Catalogue query results, what type of lake is unlikely to be found in this area? What types of lake are especially known from this area? Of the 13 possible feature classes, which ones are actually present in the water layer? [2 marks] To create a new layer containing only the lakes we can use the ArcToolbox Conversion Tools / To Geodatabase / Feature Class to Feature Class tool. The Input Feature will be water, the Output Location would be the Okanagan geodatabase and the Output Feature Class name should be lakes. In order to create a feature class that contains only the lakes (i.e., we should exclude all of the arcs related to rivers), we must use the optional Expression. Click on the SQL button in order to invoke the Query Builder. Rather than typing in all of the possible TRIM feature codes, and noting that all feature codes for lakes start with the same 7 letters (and only lake features start with those 7 letters), we can use wildcards to make the selection process easier. When working with file geodatabases, the multiple wildcard character is the percentage symbol (%). Since only the last three digits are different, we can substitute a single % for those digits; that should enable you to use only one query statement to select all of the arcs that delimit lakes. (To find out more about wildcards and SQL statements, look at the Help file on building a query.) Within the query builder, you ll create an SQL statement by, first, double-clicking on the field name FCODE. ( FCODE should appear in the query box.). Then, select the operation LIKE. (LIKE is the appropriate SQL operator to use when using wildcards to select multiple entries, since it indicates that all values like the one presented should be selected.) You must then enter in the appropriate FCODE text, including the % wildcard. The FCODE value must be surrounded by single quotes ( ), should have 7 characters, and will end in %. There should be a space between LIKE and the quoted sample value. Click on Verify to ensure that there are no typos or other problems with the SQL statement. If the statement passes the verification test, click on OK. Leaving the other entries in the Feature Class to Feature Class tool as is, click on OK. You should examine the attribute table for lakes to ensure that out of 2726 records were selected. (If not, you ll have to redo the last steps.) 1 A careful look at the resulting layer will show that while some arcs are labelled as lake arcs, they do not, in fact, form a lake (assuming that a lake should be an enclosed body of water). 4

5 3. What does your SQL statement look like? What would the SQL statements look like if you were working with a personal geodatabase, or a shapefile (DBF) rather than with a file geodatabase? What would the three statements look like if you were to use a wildcard that represented single characters rather than multiple characters? [3] You now have a feature class layer that only contains lines that define lakes. However, two major issues remain the data is still just polylines (i.e., the lines are not connected; it lacks polygon topology), and Lake Osoyoos is not closed (i.e., the southern end of the lake is not present the lake extends into the United States and therefore the end of the lake was not included in the set of data we have). This lack of closure may be a common occurrence in many datasets, in that often a polygonal feature may extend outside of the area of interest and, as a result, you may have to manually close the feature in order to ensure that it can be polygonalized. If you do not close the feature it will not be present in the polygonal file you create later. Therefore, your first task will be to add a line that closes the lake. You will then convert the polylines into polygons. (Other unclosed lakes occur in the lakes layer, but for the purposes of this lab we will ignore them.) With lakes the only visible layer, zoom into the south end of Lake Osoyoos. You will manually add a line that closes Lake Osoyoos. Right-mouse click on Lakes within the Table of Contents and select Edit Features, and Start Editing. The Editor toolbar should appear, along with the Create Features tab 2. In the Create Features window, select Lakes. In the Construction Tools subwindow (bottom of the Create Features tab), click on Line. If you place the cursor over the end of the polyline that defines one side of the lake, a box with the words lakes: Endpoint should appear (when you are over the end point of the polyline that defines the lake the box should have a + in the center, as illustrated). Click once to start the new line. Move the cursor to the other end of the lake and, once lake: Endpoint appears, click to complete the line. To finish your digitizing, right-mouse click and select Finish Sketch. (If you make a mistake in digitizing the line you can select Delete Sketch and start over again.) Now, from the Editor toolbar pulldown menu, select Save edits, and Stop editing. (If the line that you just digitized disappears after you save your edits you will have to redo the digitizing.) Now that you have closed Lake Osoyoos, your next step will be to convert the lines to polygons. To create polygons from the lines (arcs) contained in the feature class lakes is a fairly simple process. Go to ArcToolbox / Data Management Tools / Features, and select Feature to Polygon. The Input Feature will be Lakes, the Output Feature Class will be lakesp (p for polygon), set the XY Tolerance to 2 meters. Accept the defaults for the other elements. Click on OK. Note that if Lake Osoyoos isn t a closed polygon in the lakesp layer (i.e., filled in) you will have to redo the digitizing process. 4. What is XY Tolerance? What could happen if you specified too large or too small a tolerance? (Use a figure in your answer to help clarify the issues.) [3] 2 If the create features tab doesn t automatically appear, you can open it by clicking on Editing (in the Editor toolbar), and then selecting Editing Windows, and then Create Features. 5

6 Open the attribute table for the lakesp feature class. A problem should become apparent somewhere in the process of converting the line feature class to a polygonal feature class the most important attribute (for our purposes) was lost the elevations (so, obviously, Preserve Attributes doesn t preserve what we hope it would!). We need to add those values back to the attribute table since they will be required in building the TIN. We know that the polylines in the lakes feature class overlay the lines that form the polygons in the lakesp polygon feature class (since we used one layer to create the other), so we can use a spatial join to bring over the attributes from one feature class to the other. Select the feature class Lakesp. Right-mouse click, and go to Joins and Relates, and select Joins. You will want to Join data from another layer based on spatial location. Select Lakes as the layer to join, and then select Average as the way to summarize the attributes extracted (joined) from Lakes. Name the output Lakespz [z for elevation] and save it within your geodatabase. Open the attribute table of lakespz to confirm that the field Avg_ELEVATION is present. (You could remove lakes and lakesp from your table of contents.) Now that we have all of the required input layers prepared, we are ready to produce the TIN! You will produce a TIN using all of the ancillary data (lakes, rivers and roads) and then produce a DEM from the TIN. The steps to create a TIN: ArcToolbox 3D Analyst Tools 3 Data Management TIN Create TIN. To create the TIN that includes all of the ancillary features: 1. Create a TIN name it tin_full 2. Input Feature Class: Add the layers lakespz, elevation, transportation and water (using the values presented in the figure below). Important: Note that while there is an elevation associated with the transportation and water features, you should not use those values in building the TIN; doing so will produce a malformed TIN. 3. Click on OK. Look carefully at the TIN, and note what happens across Lake Osoyoos and the other lakes (they should appear as flat polygons if not, then you will need to recreate tin_full). 3 You may need to enable the 3D Analyst extension (Tools Extensions 3D Analyst). 6

7 Note: IF you wish to change the symbology of the TIN that is created, double-click on the layer name to bring up the layer properties window click on the Symbology tab. Click on ADD, and select one of the symbology choices. Click on Add, then Dismiss. Uncheck the prior symbology type, and then click on OK. 5. Create a layout showing the TIN you just created. You should include the lakes, rivers and roads (appropriately symbolized) on your map. [3] For the next part of the lab a raster DEM will be used, so we need to convert the TIN to a raster DEM: ArcToolbox 3D Analyst Conversion From TIN TIN to Raster Input TIN: Tin_full Output Raster: dem50 Output Data Type: Integer (Int) Method: Linear (default) Sample distance: Select CellSize and change the numeric value to 50.0 Z Factor: 1 (default) Part II: Identifying potential habitat for a rare species As noted in the Introduction, BC EnviroConsultants has been hired to produce a report on a rare species that is known to prefer areas with low slopes and southern aspects, and is typically associated with lower elevation sites. We will create a multi-criteria model that will help direct the botanist s search efforts to areas of habitat that are more likely to support the species. The factors in our model will be elevation, slope and aspect; we have no constraints (although, if we had a layer showing property ownership, we would exclude private properties from the results). The goal of the model will be to limit the area we need to survey for the rare species given the total value of the contract it has been determined that she can only survey approximately 725 hectares. As in any MCE model, we need to create the factors, normalize them, determine the weights, and then conduct the weighted linear overlay. Thus, the first step in developing the model will be to create the slope and aspect layers from the DEM we just created. In ArcMap, slope can be calculated directly from a DEM using the Spatial Analyst Tools Surface Slope operation. The Input raster should be DEM50, the Output raster should be named Slope50. The Output measurement can remain as Degree, and the Z factor as 1. Aspect can be calculated using the Aspect operation. The Input raster should be DEM50 and the Output raster Aspect50. We now have the three layers (factors) required for our model (elevation [DEM50], slope and aspect). The next step will be to normalize the three layers to a common scale (e.g., 0 1). There are several ways of doing this, as demonstrated below. (Other possible approaches include the Reclass tools Reclassify and Slice.) 7

8 The rare species is more likely to be found at lower elevations, and has not been observed at elevations above 1500 m. We can use the Spatial Analyst Tools / Overlay / Fuzzy Membership tool to transform the elevations into an appropriate normalized factor. The Input raster will be DEM50. The Output raster should be named FLidem. The Membership type should be set to Linear. (You should carefully read over the Tool Help and, in particular, learn more about how the different types of Fuzzy Membership transformations work.) In order to create a normalized factor in which low elevations are good (i.e., have values closer to 1) and high elevations (especially those above 1500 m) are rated poor (i.e., have values of or closer to 0), we need to reverse the minimum and maximum values that linear uses to transform the actual values to the normalized values. So, set the Minimum value to 1500 and the Maximum value to the minimum elevation found in dem50 (the minimum should be 276 m, but you should confirm this). The Hedge value should be left as None. Click on OK. 6. What effect does setting the minimum value to 1500 rather than to the actual maximum elevation have on the transformed values? [1] Soon a new raster should appear (FLidem) with minimum values of 0 and maximum values near 1. You should confirm that low elevation areas have been assigned high scores (near 1) while high elevation areas are associated with low scores (near 0). We will use a different methodology to transform the slope values. The result will be similar to that produced using the Linear Fuzzy Membership function, but done by hand using the Raster Calculator. (Under Spatial Analyst Tools / Map Algebra) You will be building a Map Algebra expression that transforms the Slope values (actual min to max) to a range of 0 to 1 (recalling that low slopes are preferred by this species, so we need to reverse the high and low values when we make the transformation). In words, the expression will be: 1 ( (Slope50 Minimum value of (Slope50) ) / (Maximum value of (Slope50) Minimum value of (Slope50)). This equation creates a simple linear transformation with values that range from 1 (areas with the lowest slopes) to 0 (areas with the steepest slopes). Before starting to build the expression you should determine the actual minimum and maximum slope values. Right-mouse click on Slope50 and select Layer Properties and then Source. Scroll down to Statistics and note down the actual minimum and maximum slope values (rounding slightly down for the minimum value (if it is not 0), and slightly up for the maximum value). With the Raster Calculator open, click in the large empty horizontal box you will enter your Map Algebra expression there. Type in 1 ( (. Then, double-click on Slope50, and Slope50 should appear in the expression window. Then, type in [the minimum value] ) / [the maximum value the minimum value] ). The Output Raster should be name LiSlope. Your expression should look something like: To convert aspect into a normalized factor requires a more sophisticated transformation, since aspect ranges from 0 (north) to 180 (south) back to 360 (north). Therefore, one cannot simply linearly transform the data as we did with elevation and slope. Luckily we can use one of the Fuzzy Membership transformations to convert the aspect values into normalized scores, where 1 would be associated with southern aspects (e.g., 180 ) and 0 would be associated with northern 8

9 aspects (e.g., 0 and 360 ) 4. Return to the Fuzzy Membership tool (Spatial Analyst / Overlay). Set Aspect50 as the Input raster, name the Output raster FGaAspect. Select Gaussian as the Membership type. The Midpoint should be set to 180, and the Spread to Leave the Hedge as None, and click on OK. 7. Why was 180 selected as the midpoint (rather than, say, 45 )? What effect does changing the spread from 1 to 0.1 to have on the transformation of input values to output values? Including a figure in your answer would be helpful. [2] We now have the three normalized factors ready for the MCE. However, we still need to determine the weights for each factor (using an analytical hierarchical approach). I have found a company that does work on Collaborative Decision Making 5 ( and they have a demo tool that we can use to calculate the AHP weights. Go to their demo ( and click on the orange Run Online Demo button. On the Projects page, click on Open in the Blank project demo tab. Name the project Rare Plant AHP (don t worry about a description), and select 1. Alternatives (along the top tab). Name the first alternative Habitat, and leave the other two as is. Move onto 2. Criteria. The goal will be Best habitat. Select Criterion 1 (and 2 and 3) and fill in the name fields as listed below. Ensure that the Weighting method associated with each criterion is the Pairwise comparisons, and the Status is Active. Criteria 1 will be Elevation, Criteria 2 will be Slope, and Criteria 3 should be Aspect. Move onto 3. Evaluation. Click on the Expand all checkbox so that all three pairwise comparisons are displayed. Based on research that our botanist has conducted on the ecology of the rare species, we know that slope is slightly less important than aspect, aspect is slightly more important than elevation, slope is equally important as elevation. By moving your mouse over the blue arrows you can see how the weights progress, from equal [1 to 1] in the middle (yellow) square to absolute domination [9 to 1] in the outermost (9 th ) arrows (noting that the direction the arrow is pointing is indicating which factor is the more important of the two being compared). You need to select the appropriate values based on the botanist s findings. If you select inconsistent weights the program will indicate it by displaying symbols such as and the percentage value shown by Best habitat will be greater than 0%. Correctly selected values should result in a percentage value of 0 beside Best habitat. 4 If the best aspect were not due south (180 ) we would have had to perform a much more complex transformation in order to account for the fact that the values are aspect angles (i.e., an aspect of 350 is only 20 away from an aspect of 10, not 330 apart). 5 They provide useful introductions to MCE and to the use of AHP, so it is worth reviewing their pages. 9

10 Move onto 4. Results. You should save the chart as an image (to be included in your lab report) and note down the percentages associated with Best habitat (L = x%] [ G= y%]) as well. The values shown on the pie chart reflect the AHP weights assigned to each criterion When using the three values in the MCE steps below you should (if necessary) round the numbers off to integers (such that they sum to 100). To conduct the actual weighted linear combination we will use the Weighted Sum tool (under Spatial Analyst Tools / Overlay). Select the three appropriate rasters (FLidem, LiSlope, FGaAspect) and then ensure that the appropriate Field (Value) and Weight (as determined above) is associated with each raster. Name the Output raster MCEWtd. Click on OK. The resulting raster should have values that range from 0 to 100 (theoretically). The consulting contract only provides enough funds to conduct searches over (approximately) 725 hectares. Therefore, within the MCEWtd layer we need to identify ~725 hectares with the highest scores. The first step is to identify how many cells are contained in 725 hectares. (If each cell is 50 x 50 m, and a hectare is 100 x 100 m, how many cells would be contained in 725 hectares?) We can use the Classify tool (Layer Properties / Symbology / Classification) to identify those cells. Set the number of Classes to 2, and make the Break Values show % (by clicking on the immediately to the right of Break Values. 10

11 By clicking on and moving the lower (left) break line you should be able to identify a break point that leaves the upper class with the appropriate number of cells click and move the left break line, and then click on the right break line in order to see how many cells remain in the upper class (once you have identified an approximate break, round off the % break value to an even percentage [e.g., a number such as 86, 88 or 90]). Once you have determined the correct break value, click on OK. You should set the symbol colour for the upper class (selected cells) to black, and set the symbol colour for the lower class (excluded cells) to No Color or Null (so that, when mapped, the lower class is transparent). As will be discussed in a later class, in any MCE you should always conduct a sensitivity analysis. That is, if a different normalization approach was used, how would that affect the results? If a different set of weights was used, how would that affect the results? In order to explore one component of a sensitivity analysis, we can compare how using a different set of weights might affect the results. To make this comparison easy for you to make, and to demonstrate the use of ArcMap s Model Builder, I have prepared two simple Models for you to use (located within the Lab2 toolset with ArcToolBox). The two models will automate the process of performing an equally-weighted analysis and of overlaying the two sets of MCE results and determining how they differ. Open the Lab2 toolset now. You should see two models present. The first model you will use is called MCE Using Equal Wts. Right-mouse click on MCE Using Equal Wts and click on Edit. You should see the model diagram illustrated below: Model for 'MCE Using Equal Wts' The blue ovals on the left identify the input files (note that if you have named your three factors something other than LiSlope, FGaAspect and Flidem you will need to change the name of the input files accordingly click on each blue oval, select Open, and select your file using the drop-down menus). The next box represents an operation in this case, the Weighted Sum operation. If you Open that box you will see that in this case I have given each of the input factors a weight of 33. The output of the Weighted Sum operation is directed to the Okanagan 11

12 geodatabase. Note that the output oval has a P associated with it. That indicates that there is a Parameter associated with that layer. In this case it means that you will be asked for the name of the output file (which will be the MCE result based on equally-weighted factors) when you run the model. This ensures that the output will be saved with the unique name that you specify. However, since the results of this model will be used in the next model, you need to name the output file MCE33 so that the next step works properly. Once you have explored the model, close it (only Save the model if you made meaningful changes to it). Then, double-click on it to Run it. You should name of the output file as MCE33 (i.e., C:\Data\lab2\Okanagan.gdb\MCE33). Model for 'Overlay of the two MCE results' If you right-mouse click on Overlay of the two MCE results and click on Edit. you should see the model diagram illustrated above (note that the final model is slightly different than the one illustrated above since I have added a final reclassification in order to remove some outlier values). As before, the blue (left-most) ovals in the model represent the input layers (the results of the two differently-weighted MCE s). In over to combine those two sets of results, we first have to reclassify them so that the same sets of values are being compared (i.e., that the highestranking cells that make up the ~725 hectares our botanist can survey are identified in both sets of layers). That is the purpose of the two Reclassify operations (yellow boxes). The first (upper) reclassification identifies the cells--the highest ranking cells in MCEWtd that make up ~725 ha. as belonging to class 1, and labels all other cells as class 0. The second (lower) reclassification identifies the same set of cells--the highest ranking cells in MCE33 that make up ~725 ha. and places them in class 2; all other cells are placed in class 0. (You can confirm this by Opening each of the Reclassify boxes 6.) The results of the reclassify operations are sent to two temporary files (green ovals). Those two files are then combined using a Weighted Sum (with both input factors assigned weights of 1) and the final result send to on output raster that you will be asked to name (MCE_Comp). 6 You will note that the reclassification rule used for the equally-weighted MCE results is different from that used for the weighted MCD results. This is to be expected, since changing the weights changes the ranking of each cell, and it is probable that a different cut-off value will have to be used when identifying an area-based goal (~725 ha). I have set the reclassify level such that the number of cells selected should be approximately 725 hectares.. 12

13 Once you have explored the model, close it (only Save the model if you made meaningful changes to it). Then, double-click on it to Run it. You should name of the output file MCE_Comp (i.e., C:\Data\lab2\Okanagan.gdb\MCE_Comp). This final layer MCE_Comp should have only four values present: 0, which are background cells not included in either MCE results (should be set to transparent), and cells numbered 1, 2 and 3. In your final map you need to provide appropriate labels for each class (e.g., class 1 represents MCE Weighted Results Only ; you need to label classes 2 and 3 appropriately). Your final map should combine the MCE_Comp layer with the dem50 layer and a hillshaded layer of dem50 (3D Analyst Tools Raster Surface Hillshade; accept all of the defaults) (visually combine the hillshaded image with the DEM by using transparency [Display properties]; also don t forget to set the Resample display properties appropriately [cubic]). As always, include a legend, title, your name, data source, etc. in the layout. [3 marks for the map] You should also hand in a one page report (excluding figures) that briefly summarizes the steps you took in order to produce final map, as well as outlines the differences identified in MCE_Comp (i.e., provide an explanation to the botanist as to how you conducted the MCE analysis and as to what the sensitivity analysis showed). You should also ensure that she knows which areas on your map she should survey. As well, you need to include the image of your AHP results. [4 marks] Conclusion In this lab you have gone through the steps of TIN construction and subsequent DEM production. You then created some derivative surfaces from that DEM (slope and aspect), surfaces that were then used in building a multi-criteria model that identified potential habitat for a rare species. Finally, you conducted a limited sensitivity analysis on the results. In conclusion, models such as the TIN and the DEM, as well as the MCE models are not truth, but simply representations designed to assist in the process of deriving new information, aiding in decision making, and planning; as such, an awareness of their limitations is necessary. References Burrough, P.A Principles of Geographic Information Systems for Land Resources Assessment, Clarendon, Oxford. Chang, K-t. and B-w Tasi The effect of DEM resolution on slope and aspect mapping. Cartography and Geographic Information Systems, 18(1): Turner, M.G Landscape Ecology: The Effect of Pattern on Process. Annual Review Ecology, Evolution and Systematics, 20, Check out their web page at An interesting example of the use of a MCE on assessing fire vulnerability. ESRI provides an overview of Geodatabases here: To learn more about the importance of shapefiles, and the reason why ESRI introduced the File geodatabase: 13