SAS Enterprise Guide 8.3 is not displaying all Oracle library tables in the Server List. The library appears with a 'Browse' node instead of fully expanding, even after increasing MaxObjectsToDisplay in SEGuide.exe.config . ... View more

Overview of the article series of the Monopoly board game simulation in SAS Viya   Monopoly #1 - Basic Example for the Visit Frequency of the Fields: Gives a basic introduction into the simulation architecture using a SAS datastep and shows how you can generate first simulations and visualization of the visit distributions. Monopoly #2 - Consider "Go-to-Jail" for the Visit Frequency of the Fields: Extends the simulation by considering the Go-to-Jail directive in your simulation Monopoly #3 - Create a visualization dashboard in SAS Visual Analytics: Shows how you can  store the simulation results in a repository to compare the results from different simulations settings and how to create a visualization dashboard in SAS Visual Analytics Monopoly #4 - Consider Chance Cards and Community Chest Cards for the Visit Frequency: Explains how you can additionally consider the instructions from chance and community chest cards for the simulation by extending the SAS datastep This article: Monopoly #5 - Using Stationary Probabilities and SAS/IML to Calculate the Fields' Visit Frequency: Shows how you can use SAS IML software to calculate stationary probabilities for the visit distribution Monopoly #6 - Calculate the Profit per Field and per Player: Extends the simulation by considering purchase options of properties and houses, as well as rent payment obligations. This article also explains how you can extend your vizualisation dashboard in SAS Visual Analytics to show the profitability per field and per player number.   Introduction and Basic Idea If the Monopoly game continues for a very long time, what proportion of visits will occur on each Monopoly field? The answer can be obtained by Monte Carlo Simulation (as shown it the other articles of this series) or by finding the stationary distribution of the Markov chain. In this article you seen an implementation using SAS IML software.  You can download the SAS programs for this article from my GIT repo.   Johannes Palmetshofer, who was my students at Johannes Kepler University in Linz, Austria in 2025, deserves credit for suggesting the idea to perform the calculation using stationary probabilities.  He created a first version in R as a voluntary additional work in the class. I translated and augmented his program in SAS IML.   The transition matrix   A Monopoly board contains 40 fields. You can represent each field as a state in a stochastic process. After rolling the dice, the player moves to another state. Consequently, the movement from one field to another can be represented using a transition probability matrix.   An example for such a 40x40 matrix, the upper left 9x9 cells, is shown below. The from-to matrix reads as: Rows= FROM, Columns = TO.     Because the player must move to one of the possible fields after every dice roll, the probabilities associated with each starting field sum to 1, consequently the row sums must all be 1. To receive stationary probabilities, the instructions and the logic of the Monopoly Board game must be packed in the definition of the transition matrix.   In its 3 subsections, this article shows how the transition matrix is defined for 3 different scenarios Basic Example just by considering the 2 dice Considering the go-to-jail directive Consider the Chance and Community Chest cards   Calculating the stationary probabilities   The calculation of the stationary probabilities involves the following steps: Calculate the eigenvalues and eigenvectors of the transition matrix. Find eigenvalue closest to 1 Extracting the stationary distribution (normalize)   1. Basic Example for Stationary Probabilities (just consider the 2 dice)   1.1 Start IML and Initialize the Transition Matrix   The PROC IML statements starts SAS IML and the j-function in IML create a 40x40 matrix with 0.   proc iml; /* Initialize a 40x40 zero matrix */ mat = j(40, 40, 0);   1.2 Defining the probabilities   The sum two dice can contain values from 2 to 12. The probabilities follow a triangle distribution (2 and 12 being least frequent, 7 being most frequent). Note that the vector starts with probabilities = 0 as you cannot roll a "1" with two dice.   /* Define dice probabilities */ probs = {0 1 2 3 4 5 6 5 4 3 2 1} / 36;   1.3 Filling the transition probabilities matrix   The PROBS vector is used to fill the matrix with the transition probabilities. Note that there is an OUTER loop for the 40 rows where the player is actually located and an inner loop for the 12 fields that can be reached from each position after rolling the dice once.   /* Fill transition probabilities */ do row = 1 to 40; do i = 1 to 12; mat[row, mod(row + i - 1, 40)+1] = t(probs[i]); end; end;   The values of the transition matrix can be shows using the PRINT statement   print mat;     or by created a heatmap using the CALL HEATMAP function.   call HeatmapCont(mat) colorramp="ThreeColor" range={-0.3 0.3}   In the resulting heatmap you see the distribution of the transition probabilities in the matrix.  If you are located at field #1 you will land on fields 6,7,8,9,10 with higher probability. if you are located at field #34 your will land on fields 39,40,1,2,3 with higher probability.     Note that for better illustration of "FROM in rows, TO in columns" nature of the transition matrix, some of the labels of the rows and columns in the heatmap have been overwritten using the following statements. Vectors of numbers and FROM/TO-labels are concatenated in XVAL and YVAL and finally used in the heatmap.   xval = char(T(1:18)) // "," // "T" // "O" // "." // char(t(23:40)); yval = char(T(1:17)) // "," // "F" // "R" // "O" // "M" // "." // char(t(24:40)); call HeatmapCont(mat) colorramp="ThreeColor" range={-0.3 0.3} xvalues = xval yvalues = yval;   1.5 Calculate the eigenvalues and eigenvectors of the transition matrix   You calculate the eigenvalues and eigenvectors of the transition matrix using the following statement:   call eigen(evals, evecs, t(mat)); EVALS contains the Eigenvalues of matrix MAT. Next you find the eigenvalue closest to 1. You calculate the distance from 1 using the following expression.   /* Find eigenvalue closest to 1 */ diffs = abs(evals - 1);   And you locate the position of the eigenvalue which has the minimum distance using the LOC function.   idx = loc(diffs = min(diffs));   After you have identified the position you extract the eigenvector from the matrix of eigenvectors at this position.   /* Get stationary distribution (normalize) */ stationary = abs(evecs[, idx]); stationary = stationary / sum(stationary);   You use the abs() function to make all elements non-negative. This is useful because a stationary probability distribution must contain non-negative values. Eigenvectors themselves can be returned with an arbitrary sign, meaning that the same eigenvector might be represented with either positive or negative values. Finally you normalize the vector. The resulting vector sums to 1 and can therefore be interpreted as a probability distribution.      stationary = stationary / sum(stationary);   The resulting stationary vector contains the long-run probability of visiting each field at the Monopoly board game.   Finally you add an index vector 1:40 as the first column.   /* Add index (field number) as first column */ index = t(1:40); result = index || stationary;   The first 10 rows of this result matrix are shown below     You can finally export this matrix to a SAS dataset using the following statement.   /* Create dataset with index and probability */ create stationary_prob from result[colname={"Field" "Probability"}]; append from result; close stationary_prob;   And use the SGLOT procedure to create a barchart for each field.   %let ScenarioName = "5. Stationary Probabilities: Dice Only"; proc sgplot data=stationary_prob; title Scenario: &scenarioname.; vbar field / response=Probability; xaxis label="Field"; yaxis label="Probability"; run;   The full code can be found in the attached program 10.5.1.     You see that you get equal probabilities for each field. This is not surprising as you have just used the two dice and not considered any relocation instruction (jail, chance cards, ...) so far. The next section explains how this can be done.   2. Considering the Go-To-Jail Directive in the Calculation of the Stationary Probabilities   In this section you consider the fact that field 31 relocates the token to field 11 (go-to-jail). You consider this fact in the definition of the transition matrix after filling the transition probabilities of the 2 dice. You do this with the statements    /* Fill transition probabilities */ do row = 1 to 40; do i = 1 to 12; mat[row, mod(row + i - 1, 40)+1] = t(probs[i]); end; end; ** ADDED: ; /* Go to jail: add probabilities from 31 to 11, zero out 31 */ mat[,11 ] = mat[,11 ] + mat[,31 ]; mat[,31 ] = 0; /* Go to jail: add probabilities from 31 to 11, zero out 31 */ mat[,11 ] = mat[,11 ] + mat[,31 ]; mat[,31 ] = 0;   The first line adds the probability of ending up at field 31 to field 11. The second line sets the probability of remaining on field 31 to 0 (as you are always sent to jail from this field).   In the heatmap you see that the probability of ending up at field 11 (column 11) not only comes from rows where the token starts from fields 1-9.  You see that there are also transition probabilities coming from fields 19-29, where the token lands at field 31 and is relocated to 11. You also see that the probability of landing at field 31 (column 31) is zero.     Performing the calculation of the stationary probabilities as described in section 1 provides a vector with probabilities which is displayed in the barchart below. You see the higher probability at field 11, and 0 probabilities at field 31. Note that there is also a change in probabilities of other fields fields after: higher probabilities after field 11, and lower probabilities after fields 31.     The full code can be found in the attached program 10.5.2.   3. Considering Chance and Community Chest Fields in the Calculation   In the next step you also consider the chance and community chest fields in the calculation of the visit probability.     The "Chance" fields are located at fields 8, 23, and 37.  10 out of 16 cards relocate the token:   3.1 Cards 1, 2, 3, 4, 11, and 14 relocate the token to a specific field   The target fields of this relocation are 40, 1, 25, 12, 11, and 6. You need to increase the probability of each of these fields with probability that the respective cards is picked. The probability for picking up one of the 16 community chest cards is 1/16. You multiply this with the cumulative visit probabilities of fields 8, 23, and 37 where you have to pick up a community chest card. Note that a vector of the target fields is defined, which you process in a DO LOOP.   /* Increase probability for target fields */ target = {40 1 25 12 11 6}; do t_idx = 1 to ncol(target); t = target[t_idx]; mat[,t] = mat[,t] + (mat[,8] + mat[,23] + mat [,37])* 1/16; end;   3.2 Card 10 relocates you 3 spaces backward.   The target fields of this relocation are 5, 20, 34, because the community chest cards are picked up at fields 8, 23 and 37.  Again need to increase the probability of each of these fields with probability that the respective cards is picked.   space3 = {5 20 34}; do t_idx = 1 to ncol(space3); t = space3[t_idx]; mat[,t] = mat[,t] + (mat[,8] + mat[,23] + mat [,37])/3 * 1/16; end;   3.3 Cards 5,6, and 7 relocate to the "nearest" utility or railroad.   Note that in the simulation "nearest" is interpreted as the next utility (railroad) in the direction how tokens move in the game. Here you need to take into account where the token is currently located to determine the target.    Eg. the first line in the code below considers that the nearest railroad after field 8 is located at field 16.     /* Advance to nearest Railroad */ mat[,16] = mat[,16] + mat[,8] * 2/16 ; mat[,26] = mat[,26] + mat[,23] * 2/16 ; mat[,6] = mat[,6] + mat[,37] * 2/16 ; /* Advance to nearest Utility */ mat[,13] = mat[,13] + (mat[,8] + mat[,37]) * 1/16 ; mat[,29] = mat[,29] + mat[,23] * 1/16 ;   3.4 Decrease the probabilities of the source fields   Finally you need to decrease the probability for the source fields 8, 23, and 37. You loop over the vector of these field numbers and set the probability to 6/16, which is the number of cards that do not contain a relocation.   /* Decrease probability for source fields */ source = {8 23 37}; do s_idx = 1 to ncol(source); s = source[s_idx]; mat[,s] = mat[,s] * 6/16; end;   3.5 Considering the community chest cards   The community chest fields are located at fields 3, 18, and 34. Only 2 of the 16 cards relocate the token: Card 1 relocates you to the start field (field 1) Cards 6 sends you to jail.   You consider this in the same ways as for the chance cards. You increase the probabilities of the target fields.   /* Community Chest: 3,18 and 34 visits are moved to 1 in 1/16 and to 11 in 1/16 of the cases*/ /* Increase probability for target fields */ mat[,1] = mat[,1] + (mat[,3] + mat[,18] + mat[,34])/16; mat[,11] = mat[,11] + (mat[,3] + mat[,18] + mat[,34])/16;   You decrease the probability for the source fields   /* Decrease probability for source fields */ mat[,3] = mat[,3] * 14/16; mat[,18] = mat[,18] * 14/16; mat[,34] = mat[,34] * 14/16;   3.6 Full Code   The full code of considering the chance and community chest cards is shown here:   /* Fill transition probabilities */ do row = 1 to 40; do i = 1 to 12; mat[row, mod(row + i - 1, 40)+1] = t(probs[i]); end; end; /* Go to jail: add probabilities from 31 to 11, zero out 31 */ mat[,11] = mat[,11] + mat[,31]; mat[,31] = 0; *** Added: ; /* Increase probability for target fields */ target = {40 1 25 12 11 6}; do t_idx = 1 to ncol(target); t = target[t_idx]; mat[,t] = mat[,t] + (mat[,8] + mat[,23] + mat [,37])* 1/16; end; space3 = {5 20 34}; do t_idx = 1 to ncol(space3); t = space3[t_idx]; mat[,t] = mat[,t] + (mat[,8] + mat[,23] + mat [,37])/3 * 1/16; end; /* Advance to nearest Railroad */ mat[,16] = mat[,16] + mat[,8] * 2/16 ; mat[,26] = mat[,26] + mat[,23] * 2/16 ; mat[,6] = mat[,6] + mat[,37] * 2/16 ; /* Advance to nearest Utility */ mat[,13] = mat[,13] + (mat[,8] + mat[,37]) * 1/16 ; mat[,29] = mat[,29] + mat[,23] * 1/16 ; /* Decrease probability for source fields */ source = {8 23 37}; do s_idx = 1 to ncol(source); s = source[s_idx]; mat[,s] = mat[,s] * 6/16; end; /* Community Chest: 3,18 and 34 visits are moved to 1 in 1/16 and to 11 in 1/16 of the cases*/ /* Increase probability for target fields */ mat[,1] = mat[,1] + (mat[,3] + mat[,18] + mat[,34])/16; mat[,11] = mat[,11] + (mat[,3] + mat[,18] + mat[,34])/16; /* Decrease probability for source fields */ mat[,3] = mat[,3] * 14/16; mat[,18] = mat[,18] * 14/16; mat[,34] = mat[,34] * 14/16;   3.7 Displaying the Results   You see that the transition matrix now shows increased and decreased probabilities for various fields. The probabilities to remain on fields 8, 23, 37 and 3, 18, 34 is lower and the probabilities of some of the target fields (railroads, utilities, ...) is increased.       Performing the calculation of the stationary probabilities as described in section 1 provides a vector with probabilities which is displayed in the barchart below. You see the higher probability at at the target fields (e.g. 1, 6, 13, 16, 26, 29, 40)  and lower probabilities at the source fields ( e.g. 8, 23, 37).       3.8 Optional: Displaying the Monopoly Board Game   You can also (partially) display the board of the Monopoly Game in a heatmap.  First you initialize the field with a small negative number, which is displayed as blue.   /* Display Board */ board = j(11, 11, -0.08); Next you assign the appropriate values from the stationary distribution to the margins of the board.   board[11,2:11] =t(stationary[10:1]); board[2:11,1]=stationary[20:11]; board[1,1:10] =t(stationary[21:30]); board[1:10,11]=stationary[31:40]; Finally you plot the heat map.   call HeatmapCont(board) colorramp="ThreeColor" range={-0.08 0.08} ; You see the board with field 1 in the lower right corner. Note that the labels are incorrect here and do not represent the true field numbers. However the display gives you some impression about the how/low visit probabilities: high in the lower left corner which is field 11 (jail), low in the upper right corner which is fields 31 (go-to-jail).             Related Articles and Links   You can download the SAS programs for this article from my GIT repo.   DO LOOP Blog https://blogs.sas.com/content/iml/ and SAS IML Software https://www.sas.com/en_us/software/iml.html  Ask-the-Expert On-Demand - Implementing a Digital Twin for the Monopoly Board Game Using SAS® Viya®  More SAS Community articles from Gerhard Case Study 8 (Chapter 26, 27 and 28) in my SAS Press Book Applying Data Science - Business Case Studies Using SAS   Youtube: Studying Complex Systems – Simulating the Monopoly Board Game Analytics mal anders: E-Werk, Bahnhof oder Schlossallee? Der Monopoly-Check | SAS Forum Digital 2021     ... View more

JSF job flow scheduler is awesome that i can define the task dependency  like lsf does     but i would like to know more about the exit code detail .  how can i achieve to have the job continue for below conditions :   1. job completed successfully without warning   2. job completed successfully with warning    thank you very much ..   p/s  it is kind of painful to that  job email notification does not available .      ... View more