1. Transportation, Assignment, and Network Algorithms 8 To accompany Quantitative Analysis for Management, Twelfth Edition, by Render, Stair, Hanna and Hale Power Point slides created by Jeff Heyl Copyright ©2015 Pearson Education, Inc.

2. After completing this module, students will be able to: LEARNING OBJECTIVES Copyright ©2015 Pearson Education, Inc.M8 – 2 1.Use the northwest corner and stepping-stone methods. 2.Solve assignment problems with the Hungarian (matrix reduction) method. 3.Use the shortest-route method.

3. M8.1Introduction M8.2The Transportation Algorithm M8.3Special Situations with the Transportation Algorithm M8.4The Assignment Algorithm M8.5Special Situations with the Assignment Algorithm M8.7Shortest-Route Problem MODULE OUTLINE Copyright ©2015 Pearson Education, Inc.M8 – 3

4. Introduction Linear programming can be used to solve distribution and network models There are some specialized algorithms that can be used to find the best solutions more quickly Often used instead of linear programming Copyright ©2015 Pearson Education, Inc.M8 – 4

5. The Transportation Algorithm An iterative procedure to find the optimal solution to a transportation problem Copyright ©2015 Pearson Education, Inc.M8 – 5 TO FROM WAREHOUSE AT ALBUQUERQUE WAREHOUSE AT BOSTON WAREHOUSE AT CLEVELAND FACTORY CAPACITY DES MOINES FACTORY $5$4$3 100 EVANSVILLE FACTORY $8$4$3 300 FORT LAUDERDALE FACTORY $9$7$5 300 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.1 – Transportation Table for Executive Furniture Corporation Des Moines capacity constraint Cell representing a source-to- destination (Evansville to Cleveland) shipping assignment that could be made Total demand and total supply Cleveland warehouse demand Cost of shipping 1 unit from Fort Lauderdale factory to Boston warehouse

6. The Transportation Algorithm Developing an Initial Solution: Northwest Corner Rule 1.Exhaust the supply (factory capacity) at each row before moving down to the next row 2.Exhaust the (warehouse) requirements of each column before moving to the right to the next column 3.Check that all supply and demands are met Copyright ©2015 Pearson Education, Inc.M8 – 6

7. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 7 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 200100 FORT LAUDERDALE (F) $9$7$5 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.2 – Initial Solution to Executive Furniture Problem Using the Northwest Corner Method Means that the firm is shipping 100 units along the Fort Lauderdale-Boston route

8. The Transportation Algorithm The cost of this shipping assignment Copyright ©2015 Pearson Education, Inc.M8 – 8 ROUTE UNITS SHIPPED PER-UNIT COST ($) TOTAL COST ($) FROMTOx= DA1005500 EA20081,600 EB1004400 FB1007700 FC20051,000 Total4,200

9. The Transportation Algorithm Stepping-Stone Method: Finding a Least- Cost Solution –Iterative technique for moving from an initial feasible solution to an optimal feasible solution –Two distinct parts Testing current solution to determine if improvement is possible Making changes to the current solution to obtain an improved solution Copyright ©2015 Pearson Education, Inc.M8 – 9

10. The Transportation Algorithm One rule must first be observed The number of occupied routes (or squares) must always be equal to one less than the sum of the number of rows plus the number of columns Copyright ©2015 Pearson Education, Inc.M8 – 10 Occupied shipping routes (squares) = Number of rows + Number of columns – 1 5 = 3 + 3 – 1 When the number of occupied routes is less than this, the solution is called degenerate

11. The Transportation Algorithm Testing the solution for possible improvement –Test each unused shipping route (or square) in the transportation table by asking “What would happen to total shipping costs if one unit of our product were tentatively shipped on an unused route?” Copyright ©2015 Pearson Education, Inc.M8 – 11

12. The Transportation Algorithm Five Steps to Test Unused Squares with the Stepping-Stone Method 1.Select an unused square to be evaluated. 2.Beginning at this square, trace a closed path back to the original square via squares that are currently being used and moving with only horizontal and vertical moves. 3.Beginning with a plus (+) sign at the unused square, place alternate minus (–) signs and plus signs on each corner square of the closed path. Copyright ©2015 Pearson Education, Inc.M8 – 12

13. The Transportation Algorithm 4.Calculate an improvement index by adding together the unit cost figures found in each square containing a plus sign and then subtracting the unit costs in each square containing a minus sign. 5.Repeat steps 1 to 4 until an improvement index has been calculated for all unused squares. If all indices computed are greater than or equal to zero, an optimal solution has been reached. If not, it is possible to improve the current solution and decrease total shipping costs. Copyright ©2015 Pearson Education, Inc.M8 – 13

14. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 14 TO FROM ALBUQUERQUEBOSTONCLEVELAND FACTORY CAPACITY DES MOINES 543 100 EVANSVILLE 843 300 200100 FORT LAUDERDALE 975 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.3 – Evaluating the Unused Des Moines–Boston Shipping Route $5 99 100 $4 1 100 99$8 200 201 – + –+ – + + – Result of Proposed Shift in Allocation= 1 x $4 – 1 x $5 + 1 x $8 – 1 x $4 = + $3 Evaluation of Des Moines-Boston Square

15. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 15 TO FROM ALBUQUERQUEBOSTONCLEVELAND FACTORY CAPACITY DES MOINES 543 100 EVANSVILLE 843 300 200100 FORT LAUDERDALE 975 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.3 – Evaluating the Unused Des Moines–Boston Shipping Route $5 99 100 $4 1 100 99$8 200 201 – + –+ – + + – Result of Proposed Shift in Allocation= 1 x $4 – 1 x $5 + 1 x $8 – 1 x $4 = + $3 Evaluation of Des Moines-Boston Square Improvement index (I ij ) for Des Moines- Boston route Des Moines- Boston index = I DB = +$4 – $5 + $8 – $4 = +$3

16. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 16 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 200100 FORT LAUDERDALE (F) $9$7$5 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.4 – Evaluating the Des Moines–Cleveland (D–C) Shipping Route Start – + + – Closed path is + DC – DA + EA – EB + FB – FC

17. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 17 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 200100 FORT LAUDERDALE (F) $9$7$5 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.4 – Evaluating the Des Moines–Cleveland (D–C) Shipping Route Start – + + – Closed path is + DC – DA + EA – EB + FB – FC Des Moines–Cleveland= I DC improvement index= +$3 – $5 + $8 – $4 + $7 – $5 = +$4 Evanston–Cleveland index= I EC = +$3 – $4 + $7 – $5 = +$1 (closed path: + EC – EB + FB – FC) Fort Lauderdale–Albuquerque index= I FA = +$9 – $7 + $4 – $8 = –$2 (closed path: + FA – FB + EB – EA)

18. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 18 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 200100 FORT LAUDERDALE (F) $9$7$5 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.5 – Stepping-Stone Path Used to Evaluate Route F–A – + + –

19. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 19 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 100200 FORT LAUDERDALE (F) $9$7$5 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.6 – Second Solution to the Executive Furniture Problem

20. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 20 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 100200 FORT LAUDERDALE (F) $9$7$5 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.6 – Second Solution to the Executive Furniture Problem D to B = I DB = +$4 – $5 + $8 – $4 = +$3 (closed path: +DB – CA + EA – EB) D to C = I DC = +$3 – $5 + $9 – $5 = +$2 (closed path: +DC – DA + FA – FC) E to C = I EC = +$3 – $8 + $9 – $5 = –$1 (closed path: +EC – EA + FA – FC) F to B = I FB = +$7 – $4 + $8 – $9 = +$2 (closed path: +FB – EB + EA – FA)

21. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 21 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 100200 FORT LAUDERDALE (F) $9$7$5 300 100200 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.7 – Path to Evaluate the E–C Route – + + – Start

22. The Transportation Algorithm Total cost of third solution Copyright ©2015 Pearson Education, Inc.M8 – 22 ROUTE DESKS SHIPPED PER-UNIT COST ($) TOTAL COST ($) FROMTOx= DA1005500 EB2004800 EC1003300 FA20091,800 FC1005500 Total3,900

23. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 23 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 200100 FORT LAUDERDALE (F) $9$7$5 300 200100 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.8 – Third and Optimal Solution

24. The Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 24 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) FACTORY CAPACITY DES MOINES (D) $5$4$3 100 EVANSVILLE (E) $8$4$3 300 200100 FORT LAUDERDALE (F) $9$7$5 300 200100 WAREHOUSE REQUIREMENTS 300200 700 TABLE M8.8 – Third and Optimal Solution D to B= I DB = +$4 – $5 + $9 – $5 + $3 – $4 = +$2 (path: +DB – DA + FA – FC + EC – EB) D to C= I DC = +$3 – $5 + $9 – $5 = +$2 (path: +DC – DA + FA – FC) E to A= I EA = +$8 – $9 + $5 – $3 = +$1 (path: +EA – FA + FC – EC) F to B= I FB = +$7 – $5 + $3 – $4 = +$1 (path: +FB – FC + EC – EB)

25. The Transportation Algorithm Summary of Steps in Transportation Algorithm (Minimization) 1.Set up a balanced transportation table. 2.Develop initial solution using the northwest corner method. 3.Calculate an improvement index for each empty cell using the stepping-stone method. If improvement indices are all nonnegative, stop; the optimal solution has been found. If any index is negative, continue to step 4. 4.Select the cell with the improvement index indicating the greatest decrease in cost. Fill this cell using a stepping-stone path and go to step 3. Copyright ©2015 Pearson Education, Inc.M8 – 25

26. Special Situations with the Transportation Algorithm Unbalanced Transportation Problems –Total demand is not equal to total supply –Unbalanced problems handled by introducing Dummy sources or Dummy destinations Copyright ©2015 Pearson Education, Inc.M8 – 26

27. Special Situations with the Transportation Algorithm Demand less than supply –Add dummy column representing a fake destination Demand greater than supply –Add dummy row representing a fake source Copyright ©2015 Pearson Education, Inc.M8 – 27

28. Special Situations with the Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 28 TO FROM ALBUQUERQUE (A) BOSTON (B) CLEVELAND (C) DUMMY WAREHOUSE FACTORY CAPACITY DES MOINES (D) 5430 250 EVANSVILLE (E) 8430 300 5020050 FORT LAUDERDALE (F) 9750 300 150 WAREHOUSE REQUIREMENTS 300200 150850 TABLE M8.9 – Initial Solution to an Unbalanced Problem Where Demand Is Less Than Supply New Des Moines capacity Total cost = 250($5) + 50($8) + 200($4) + 50($3) + 150($5) + 150($0) = $3,350

29. Special Situations with the Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 29 TO FROM WAREHOUSE AWAREHOUSE BWAREHOUSE C PLANT SUPPLY PLANT W $6$4$9 200 PLANT X $10$5$8 175 PLANT Y $12$7$6 75 WAREHOUSE DEMAND 250100150 450 500 TABLE M8.10 – Unbalanced Transportation Table for Happy Sound Stereo Company Totals do not balance

30. Special Situations with the Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 30 TO FROM WAREHOUSE AWAREHOUSE BWAREHOUSE C PLANT SUPPLY PLANT W $6$4$9 200 PLANT X $10$5$8 175 5010025 PLANT Y $12$7$6 75 DUMMY PLANT 000 50 WAREHOUSE DEMAND 250100150500 TABLE M8.11 – Initial Solution to an Unbalanced Problem in Which Demand Is Greater Than Supply Total cost of initial solution = 200($6) + 50($10) + 100($5) + 25($8) + 75($6) + 50($0) = $2,850

31. Special Situations with the Transportation Algorithm Degeneracy in an initial solution Copyright ©2015 Pearson Education, Inc.M8 – 31 TO FROM CUSTOMER 1CUSTOMER 2CUSTOMER 3 WAREHOUSE SUPPLY WAREHOUSE 1 826 100 WAREHOUSE 2 1099 120 10020 WAREHOUSE 3 7107 80 CUSTOMER DEMAND 100 300 TABLE M8.12 – Initial Solution of a Degenerate Problem

32. Special Situations with the Transportation Algorithm Degeneracy during later solution stages Copyright ©2015 Pearson Education, Inc.M8 – 32 TO FROM WAREHOUSE 1WAREHOUSE 2WAREHOUSE 3 FACTORY CAPACITY FACTORY A 8516 70 FACTORY B 15107 130 508020 FACTORY C 3910 80 50 WAREHOUSE REQUIREMENT 1508050280 TABLE M8.13 – Bagwell Paint Transportation Table Total shipping cost = $2,700

33. Degeneracy during later solution stages Special Situations with the Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 33 TO FROM WAREHOUSE 1WAREHOUSE 2WAREHOUSE 3 FACTORY CAPACITY FACTORY A 8516 70 FACTORY B 15107 130 508020 FACTORY C 3910 80 50 WAREHOUSE REQUIREMENT 1508050280 TABLE M8.13 – Bagwell Paint Transportation Table Total shipping cost = $2,700 Factory A – Warehouse 2 index = + 2 Factory A – Warehouse 3 index = + 1 Factory B – Warehouse 3 index = – 15 Factory C – Warehouse 2 index = + 11 Only route with a negative index

34. Degeneracy during later solution stages Special Situations with the Transportation Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 34 TO FROM WAREHOUSE 1WAREHOUSE 3 FACTORY B 157 50 FACTORY C 310 30 50 TABLE M8.14 – Tracing a Closed Path for the Factory B–Warehouse 3 Route – + + –

35. Special Situations with the Transportation Algorithm More Than One Optimal Solution –Multiple solutions are possible –May increase flexibility Maximization Transportation Problem –Improvement indices are negative or zero Unacceptable or Prohibited Routes –Assign a very high cost Other Transportation Methods Copyright ©2015 Pearson Education, Inc.M8 – 35

36. The Assignment Algorithm Each assignment problem has associated with it a table or matrix –Rows contain the objects or people to assign –Columns comprise the tasks or things they are assigned to –Numbers in the table are the costs associated with each particular assignment Copyright ©2015 Pearson Education, Inc.M8 – 36

37. The Assignment Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 37 TABLE M8.15 – Estimated Project Repair Costs for the Fix-It Shop Assignment Problem PROJECT PERSON123 Adams$11$14$6 Brown81011 Cooper9127

38. The Assignment Algorithm The Hungarian Method (Flood’s Technique) –Principle of matrix reduction –Matrix of opportunity costs Copyright ©2015 Pearson Education, Inc.M8 – 38 TABLE M8.16 – Summary of Fix-It Shop Assignment Alternatives and Costs PROJECT ASSIGNMENT LABOR COSTS ($) TOTAL COSTS ($) 123 AdamsBrownCooper11 + 10 + 728 AdamsCooperBrown11 + 12 + 1134 BrownAdamsCooper8 + 14 + 729 BrownCooperAdams8 + 12 + 626 CooperAdamsBrown9 + 14 + 1134 CooperBrownAdams9 + 10 + 625

39. The Assignment Algorithm Three Steps of the Assignment Method 1. Find the opportunity cost table by a.Subtracting the smallest number in each row of the original cost table or matrix from every number in that row. b.Then subtracting the smallest number in each column of the table obtained in part (a) from every number in that column. Copyright ©2015 Pearson Education, Inc.M8 – 39

40. The Assignment Algorithm 2.Test the table resulting from step 1 to see whether an optimal assignment can be made. The procedure is to draw the minimum number of vertical and horizontal straight lines necessary to cover all zeros in the table. If the number of lines equals either the number of rows or columns in the table, an optimal assignment can be made. If the number of lines is less than the number of rows or columns, we proceed to step 3. Copyright ©2015 Pearson Education, Inc.M8 – 40

41. The Assignment Algorithm 3.Revise the present opportunity cost table. This is done by subtracting the smallest number not covered by a line from every uncovered number. This same smallest number is also added to any number(s) lying at the intersection of horizontal and vertical lines. We then return to step 2 and continue the cycle until an optimal assignment is possible. Copyright ©2015 Pearson Education, Inc.M8 – 41

42. The Assignment Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 42 FIGURE M8.1 – Steps in the Assignment Method Set up cost table for problem Step 1 Step 2 Step 3 Optimal (No. of Lines = No. of Rows or Columns) Not Optimal (No. of Lines < No. of Rows or Columns) Find opportunity cost. (a)Subtract smallest number in each row from every number in that row, then (b)subtract smallest number in each column from every number in that column. Revise opportunity cost table in two steps: (a)Subtract the smallest number not covered by a line from itself and every other uncovered number. (b)Add this number at every intersection of any two lines. Optimal solution at zero locations. Systematically make final assignments. (a)Check each row and column for a unique zero and make the first assignment in that row or column. (b)Eliminate that row and column and search for another unique zero. Make that assignment and proceed in a like manner. Test opportunity cost table to see if optimal assignments are possible by drawing the minimum possible number of lines on columns and/or rows such that all zeros are covered.

43. The Assignment Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 43 TABLE M8.17 – Cost of Each Person–Project Assignment for the Fix-It Shop Problem TABLE M8.18 – Row Opportunity Cost Table for the Fix-It Shop Step 1, Part (a) PROJECT PERSON123 Adams$11$14$6 Brown81011 Cooper9127 PROJECT PERSON123 Adams$5$8$0 Brown023 Cooper250

44. The Assignment Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 44 TABLE M8.19 – Total Opportunity Cost Table for the Fix-It Shop Step 1, Part (b) TABLE M8.20 – Test for Optimal Solution to Fix-It Shop Problem PROJECT PERSON123 Adams$5$6$0 Brown003 Cooper230 PROJECT PERSON123 Adams$5$6$0 Brown003 Cooper230 Covering line 1 Covering line 2

45. The Assignment Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 45 TABLE M8.21 – Revised Opportunity Cost Table for the Fix-It Shop Problem TABLE M8.22 – Optimality Test on the Revised Fix-It Shop Opportunity Cost Table PROJECT PERSON123 Adams$3$4$0 Brown005 Cooper010 PROJECT PERSON123 Adams$3$4$0 Brown005 Cooper010 Covering line 2 Covering line 3Covering line 1

46. The Assignment Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 46 TABLE M8.23 – Making the Final Fix-It Shop Assignments (A) FIRST ASSIGNMENT PERSON123 Adams340 Brown005 Cooper010 (B) SECOND ASSIGNMENT PERSON123 Adams340 Brown005 Cooper010 (C) THIRD ASSIGNMENT PERSON123 Adams340 Brown005 Cooper010 ASSIGNMENTCOST ($) Adams to project 36 Brown to project 210 Cooper to project 19 Total cost25

47. Special Situations with the Assignment Algorithm Unbalanced Assignment Problems –Balanced assignment problem – number of rows equals number of columns –For an unbalanced problem add dummy rows or dummy columns Copyright ©2015 Pearson Education, Inc.M8 – 47 TABLE M8.24 – Estimated Project Repair Costs for Fix-It Shop with Davis Included PROJECT PERSON123DUMMY Adams$11$14$6$0 Brown810110 Cooper91270 Davis101380

48. Special Situations with the Assignment Algorithm Maximization Assignment Problems Reformat as minimization problem by converting numbers in table to opportunity costs –Subtract every number in the original payoff table from the largest single number in that table Copyright ©2015 Pearson Education, Inc.M8 – 48 ASSIGNMENTEFFICIENCY Ship 1 to sector D55 Ship 1 to sector C80 Ship 1 to sector B100 Ship 1 to sector A65 Total efficiency300

49. Special Situations with the Assignment Algorithm Copyright ©2015 Pearson Education, Inc.M8 – 49 SECTOR SHIPABCD 120605055 260308075 3801009080 465807570 SECTOR SHIPABCD 180405045 240702025 32001020 435202530 SECTOR SHIPABCD 1400105 2205005 32001020 4150510 SECTOR SHIPABCD 1250100 255000 3501015 40055 TABLE M8.25 – Efficiencies of British Ships in Patrol Sectors TABLE M8.26 – Opportunity Costs of British Ships TABLE M8.27 – Row Opportunity Costs for the British Navy Problem TABLE M8.28 – Total Opportunity Costs for the British Navy Problem

50. Shortest-Route Problem Find the shortest distance from one location to another Shortest-Route Technique –Transport furniture from factory to warehouse –Find shortest route Copyright ©2015 Pearson Education, Inc.M8 – 50 Plant Warehouse 1 2 3 4 5 6 100 50 40 150 200 100 200 FIGURE M8.6 – Roads from Ray’s Plant to Warehouse

51. Shortest-Route Problem Steps of the Shortest-route Technique 1.Find the nearest node to the origin (plant). Put the distance in a box by the node. 2.Find the next-nearest node to the origin (plant), and put the distance in a box by the node. In some cases, several paths will have to be checked to find the nearest node. 3.Repeat this process until you have gone through the entire network. The last distance at the ending node will be the distance of the shortest route. You should note that the distance placed in the box by each node is the shortest route to this node. These distances are used as intermediate results in finding the next-nearest node. Copyright ©2015 Pearson Education, Inc.M8 – 51

52. Shortest-Route Problem Copyright ©2015 Pearson Education, Inc.M8 – 52 FIGURE M8.7 – First Iteration for Ray Design FIGURE M8.8 – Second Iteration for Ray Design Plant Warehouse 1 2 3 4 5 6 100 50 40 150 200 100 200 100 1 2 3 4 5 6 50 40 150 200 100 200 100 150

53. Shortest-Route Problem Copyright ©2015 Pearson Education, Inc.M8 – 53 FIGURE M8.9 – Third Iteration for Ray Design FIGURE M8.10 – Fourth and Final Iteration for Ray Design 1 2 3 4 5 6 100 50 40 150 200 100 200 100 150 190 1 2 3 4 5 6 100 50 40 150 200 100 200 100 150 190 290