INITIAL BASICFEASIBLE SOLUTION
TRANSPORTATION PROBLEM         Transport various        quantities of a single    homogeneous commodity    to different de...
TERMINOLOGY USED IN             TRANSPORTATIONAL MODELFeasible solution: Non negative values of xij where i=1, 2……….m and ...
OPTIMAL SOLUTION OF TRANSPORTATION              PROBLEM                                     Obtain an                     ...
INITIAL BASIC FEASIBLE SOLUTION     Methods Available NORTH WEST     LOWEST COST      VOGEL’S   CORNER          ENTRY     ...
UNBALANCED TRANSPORTATION         PROBLEM             Demand>SupplyAdd dummy column in matrix with zero cost             S...
DEMAND>SUPPLYPRODUCT   P    Q    SUPPLYOFFICEA         10   15   20B         20   30   50DEMAND    30   60             70 ...
SUPPLY>DEMAND    PRODUCT        P         Q            SUPPLY    OFFICE    A              10        15           30    B  ...
METHOD FOR OPTIMAL SOLUTION             Stepping stone                method               Modified              Distribut...
NORTH WEST CORNER METHODMost systematic and easiest method for obtaining initial  feasible basic solution
STEPS TO SOLVE THE PROBLEMStep1: construct an empty m*n matrix, completed with rows & columns.Step2: indicate the rows and...
MINIMIZATION USING NWCM     WAREHOUSE W1   W2   W3   SUPPLYPLANTP1            7     6    9    20P2            5     7    3...
Step1: Check whether problem is                balanced WAREHOUSE W1    W2     W3      SUPPLYPLANTP1        7      6      ...
Allocate P1W1 max quantity such that itshould not exceed supply/requirement      WAREHOUSE W1         W2    W3         SUP...
Check whether the requirements of W1               fulfilled     WAREHOUSE W1         W2   W3         SUPPLYPLANTP1       ...
Allocate 1 to P2W1       WAREHOUSE W1           W2   W3    SUPPLY  PLANT  P1            7             6    9     20/0     ...
No. of allocations=No. of rows + No. of columns-1         (For generating optimal solution)      WAREHOUSE W1           W2...
TOTAL TRANSPORTATION COSTPLANT        WAREHOUSE   COSTP1           W1          7*20=140P2           W2          5*1=5P2   ...
LEAST COST ENTRY METHOD     This method takes into  consideration the lowest cost and therefore takes less time to       s...
STEPS TO SOLVE THE PROBLEM• Step1: select the cell with the lowest transportation cost  among all the rows and columns of ...
MINIMIZATION USING LCEM     WAREHOUSE W1       W2        W3         SUPPLYPLANTP1            7         6         9        ...
Find out the cell having least cost     WAREHOUSE W1      W2          W3          SUPPLYPLANTP1            7        6     ...
In this way allocate next minimum costs     until all requirements are fulfilled      WAREHOUSE W1          W2           W...
Final matrix with all allocations     WAREHOUSE W1          W2           W3          SUPPLYPLANTP1            7           ...
No. of allocations=No. of rows + No. of columns-1         (For generating optimal solution)      WAREHOUSE W1           W2...
TOTAL TRANSPORTATION COSTPLANT        WAREHOUSE   COSTP1           W2          6*20=120P2           W1          5*4=20P2  ...
MAXIMIZATION USING LCEMINVESTMENT   P    Q    R    S    AVAILABLE1            95   80   70   60   702            75   65  ...
Subtract all the elements from the largestelement thereby giving minimization case.  INVESTMENT   P     Q    R       S    ...
This is the matrix after subtracting all the             elements from 95 INVESTMENT   P    Q     R       S       AVAILABL...
TRANSPORTATION PROBLEM  Categorized into two types: Minimization    Maximization   problem        problem
MINIMIZATION PROBLEM  In this transportation cost is given which is to      be minimized.
MAXIMIZATION PROBLEM        In this profit is given which is to be maximized.To solve this problem we convert the problem ...
VOGEL’S APPROXIMATION METHOD• BASIS OF ALLOCATION IS UNIT COST PENALTY• THE SUBSEQUENT ALLOCATIONS IN CELLS ARE  DONE KEEP...
STEPS TO SOLVE THE PROBLEMStep1: for each row of the table identify the lowest and the next lowest cost   cell. Find their...
MINIMIZATION USING VAMWAREHOUSE    W1      W2     W3      SUPPLYPLANTP1           7       6      9       20P2           5 ...
Find out ‘Penalties’(P) for each row & column by   subtracting the smallest element from next      smallest element in the...
The row/column having zero demand/supply left             will be cut by a line     WAREHOUSE    W1      W2     W3        ...
Now find next penalty P2 in same wayWAREHOUSE         W1           W2      W3             SUPPLY   P2PLANTP1              ...
In this way find out all the penalties until all               allocations are not doneWAREHOUSE      W1            W2    ...
Final matrix with all allocationsWAREHOUSE    W1        W2           W3          SUPPLYPLANTP1           7         6      ...
No. of allocations=No. of rows + No. of columns-1         (For generating optimal solution)   WAREHOUSE   W1         W2   ...
TOTAL TRANSPORTATION COSTPLANT        WAREHOUSE   COSTP1           W2          6*20=120P2           W1          5*9=45P2  ...
COMPARISON OF NWCM,LCEM,VAM  WAREHOUSE   W1      W2     W3     SUPPLY  PLANT  P1          7       6      9      20  P2    ...
MAXIMIZATION CASE            I    II    III    IV      CAPACITYA           40   25    22     33      100B           44   3...
To solve this first we will convert this inminimization problem by subtracting the largest           element from all elem...
Now we will balance this matrix    Requirement<Capacity Dummy column will be                       added              I   ...
CHOCOLATE   WININGQUESTION FOR  AUDIENCE
Obtain IBFS for the followingMaximization Problem using VAM:         I    II   III    SUPPLYA        6    4    4      100B...
SOLUTION:         I           II        III        SUPPLYA        6    60     4    40   4          100B        5          ...
Transportation Problem in Operational Research
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Transportation Problem in Operational Research

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  • Dr.H.K.Lakshman Rao Ph.D(Mgmt.),MPhil (Statistics) M.Sc(Stat), M.A(D.Ed.),M.A(Pub.Adm),M.A(R.D) M.Sc(Psy,), M.A(Econ,) , M.A(Eng). OR &amp; SQC (ISI.),AMP (IIMA), DDE. SAS Cert. (Former Gen. Manager MFL &amp; Prof. &amp; Head Dept. Management(CEC) Management, Corporate , Statistical, Hospital , Academic , Placement/Six- Sigma, Project Management Consultant &amp; Mentor- IIMA MentorEdge “ANUGRAHA”, 33, Krishnapuri, R.A.Puram, Chennai – 600 028, M: 09381036989,L-044-24616184, e-mail: hklrao@ mail.com Prof H.K.Lakshman Rao was with Madras Fertilizers Ltd (MFL) for 23 years, in various Depts. (Marketing, Marketing Research, and Distribution &amp; HR) and left as General Manager. While in MFL, he was closely involved in Training &amp; Development of staff &amp; Executives in General Management, Marketing &amp; Logistics functions of MFL. He was also developing Man power in other newly established Fertilizer giants such as IFFCO, MCF, SPIC, and NFCL. He has multiple Master's Degrees in : Statistics, Economics, Public Administration, Psychology, Education, English &amp; Rural Development to his credit, a scholar who has conducted several workshops to varied audience and has written several research papers/ Management books and articles. He is passionate about statistics which is his core area. A Six Sigma and a Project Management enthusiast who specializes in application of quantitative techniques in management.He was invited by Radio Brisbane (Australia) for an hour long on line interview on teaching mathematics for school children, for broadcasting in south East Asia. Prior to MFL, he was with Carborundum Universal Ltd (Murugappa Group) in charge of Marketing Research &amp; MIS, Statistical Analysist.SQC. Later, he was Prof &amp; Head of Management Dept. at Crescent Engineering College, Chennai. Dr. Rao, as Alumni IIMA, has been closely associated with academics as, visiting Professor, examiner at several Universities, Management institutes &amp; Business Schools in India and abroad, in addition to IIT &amp; BITS-Pilani. He has been on the recruitment / selection board of several Universities and Corporate organizations, besides being a guide and examiner for Ph.D. Dr. Rao has travelled widely (UK, Malaysia, Singapore, Australia, Japan, Thailand, Nepal, Pakistan &amp; Srilanka) and has addressed students at several Universities and B-Schools/Corporate organizations. He was one among the three man team on deputation from MFL on a 4and half year International study on fertilizer marketing systems in south East Asia and was at East West centre Honolulu, Hawaii USA/ University of Hawaiias a Fulbright Scholar.He has been a statistical consultant on several World Bank sponsored studies in the areas of Health care &amp; Primary education...... Dr.Rao has been associate with several Universities and Business Schools and corporate organizations including: ITC, TATAChem, Baja Auto,Synergy School of Business Skills, World Bank sponsored projects ***************************************************************** Program on Applied Statistics of:
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  • DISTANCE of IGNOU MBA DESIRING ON LINE TUTORING ON sTATISTICAL METHODS. , OPERATIONS RESEARCH , RESEARCH METHODOLOGY AND PROJERCT GUIDANCE CAN CONTACT;
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Transcript of "Transportation Problem in Operational Research"

  1. 1. INITIAL BASICFEASIBLE SOLUTION
  2. 2. TRANSPORTATION PROBLEM Transport various quantities of a single homogeneous commodity to different destinations in such a way that total transportation cost is minimum.
  3. 3. TERMINOLOGY USED IN TRANSPORTATIONAL MODELFeasible solution: Non negative values of xij where i=1, 2……….m and j=1, 2,…n whichsatisfy the constraints of supply and demand is called feasible solution.Basic feasible solution: If the no of positive allocations are (m+n-1).Optimal solution: A feasible solution is said to be optimal solution if it minimizes thetotal transportation cost.Balanced transportation problem: A transportation problem in which the total supplyfrom all sources is equal to the total demand in all the destinations.Unbalanced transportation problem: Problems which are not balanced are calledunbalanced.Matrix terminology: In the matrix, the squares are called cells and form columnsvertically and rows horizontally.Degenerate basic feasible solution: If the no. of allocation in basic feasible solutions isless than (m+n-1).
  4. 4. OPTIMAL SOLUTION OF TRANSPORTATION PROBLEM Obtain an optimal solution by making Initial Basic successiveSTEP1 Feasible STEP2 improvements in Solution IBFS until no further decrease in transportation cost is possible
  5. 5. INITIAL BASIC FEASIBLE SOLUTION Methods Available NORTH WEST LOWEST COST VOGEL’S CORNER ENTRY APPROXIMATIONMETHOD(NWCM) METHOD(LCEM) METHOD(VAM)
  6. 6. UNBALANCED TRANSPORTATION PROBLEM Demand>SupplyAdd dummy column in matrix with zero cost Supply>Demand Add dummy row in matrix with zero cost
  7. 7. DEMAND>SUPPLYPRODUCT P Q SUPPLYOFFICEA 10 15 20B 20 30 50DEMAND 30 60 70 90PRODUCT P Q SUPPLYOFFICEA 10 15 20B 20 30 50DUMMY 0 0 20DEMAND 30 60 90 90
  8. 8. SUPPLY>DEMAND PRODUCT P Q SUPPLY OFFICE A 10 15 30 B 20 30 60 DEMAND 20 50 90 70PRODUCT P Q DUMMY SUPPLYOFFICEA 10 15 0 20B 20 30 0 50DEMAND 20 50 20 90 90
  9. 9. METHOD FOR OPTIMAL SOLUTION Stepping stone method Modified Distribution Method(MODI)
  10. 10. NORTH WEST CORNER METHODMost systematic and easiest method for obtaining initial feasible basic solution
  11. 11. STEPS TO SOLVE THE PROBLEMStep1: construct an empty m*n matrix, completed with rows & columns.Step2: indicate the rows and column totals at the end.Step3: starting with (1,1)cell at the north west corner of the matrix, allocate maximum possible quantitykeeping in view that allocation can neither be more than the quantity required by the respective warehousesnor more than quantity available at the each supply centre.Step 4: adjust the supply and demand nos. in the rows and columns allocations.Step5: if the supply for the first row is exhausted then move down to the first cell in the second row and firstcolumn and go to the step 4.Step 6: if the demand for the first column is satisfied, then move to the next cell in the second column and firstrow and go to step 4.Step 7: if for any cell, supply equals demand then the next allocation can be made in cell either in the nextrow or column.Step 8: continue the procedure until the total available quantity is fully allocated to the cells as required.
  12. 12. MINIMIZATION USING NWCM WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28P3 4 5 8 17DEMAND 21 25 19
  13. 13. Step1: Check whether problem is balanced WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28P3 4 5 8 17DEMAND 21 25 19 65 65
  14. 14. Allocate P1W1 max quantity such that itshould not exceed supply/requirement WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21/1 25 19 65P1 can supply 20 units at max so W1 can be allocated 20 units only by P1
  15. 15. Check whether the requirements of W1 fulfilled WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20/0 20P2 5 7 3 28P3 4 5 8 17DEMAND 21/1 25 19 65Now allocate the remaining requirement to next north west corner element i.e. P2W1
  16. 16. Allocate 1 to P2W1 WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/27 1 P3 4 5 8 17 DEMAND 21/1/0 25 19 65In this way repeat it for all columns until demand of all warehouses is fulfilled
  17. 17. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/27/0 1 25 2 P3 4 5 8 17/0 17 DEMAND 21/1/0 25/0 19/17/0 65 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
  18. 18. TOTAL TRANSPORTATION COSTPLANT WAREHOUSE COSTP1 W1 7*20=140P2 W2 5*1=5P2 W2 7*25=175P2 W2 3*2=6P3 W3 8*17=136TOTAL COST 462
  19. 19. LEAST COST ENTRY METHOD This method takes into consideration the lowest cost and therefore takes less time to solve the problem
  20. 20. STEPS TO SOLVE THE PROBLEM• Step1: select the cell with the lowest transportation cost among all the rows and columns of the transportation table. If the minimum cost is not unique then select arbitrarily any cell with the lowest cost.• Step2: allocate as many units as possible to the cell determined in step 1 and eliminate that row in which either capacity or requirement is exhausted.• Step3:adjust the capacity and the requirement for the next allocations.• Step4: repeat the steps1to3 for the reduced table until the entire capacities are exhausted to fill the requirements at the different destinations.
  21. 21. MINIMIZATION USING LCEM WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28P3 4 5 8 17DEMAND 21 25 19 65 Step1 :This problem is balanced problem
  22. 22. Find out the cell having least cost WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28/9 19P3 4 5 8 17DEMAND 21 25 19/0 65P2W3 is having least cost i.e. 3 so it is allocated 19 units as per the requirement of W3
  23. 23. In this way allocate next minimum costs until all requirements are fulfilled WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/5/0 4 5 19 P3 4 5 8 17/0 17 DEMAND 21/4/0 25/5/0 19/0 65 Now P2W1 and P3W2 are having same cost, we can allocate any of them but in this case P2W1 will beallocated because in P3W2 P3 cant supply anymore units
  24. 24. Final matrix with all allocations WAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20/0 20P2 5 7 3 28/9/5/0 4 5 19P3 4 5 8 17/0 17DEMAND 21/4/0 25/5/0 19/0 65
  25. 25. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/5/0 4 5 19 P3 4 5 8 17/0 17 DEMAND 21/4/0 25/5/0 19/0 65 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
  26. 26. TOTAL TRANSPORTATION COSTPLANT WAREHOUSE COSTP1 W2 6*20=120P2 W1 5*4=20P2 W2 7*5=35P2 W3 3*19=57P3 W1 4*17=68TOTAL COST 300
  27. 27. MAXIMIZATION USING LCEMINVESTMENT P Q R S AVAILABLE1 95 80 70 60 702 75 65 60 50 403 70 45 50 40 904 60 40 40 30 10DEMAND 40 50 60 60 210Convert maximization into minimization
  28. 28. Subtract all the elements from the largestelement thereby giving minimization case. INVESTMENT P Q R S AVAILABLE 1 95 95 80 70 60 70 2 75 65 60 50 40 3 70 45 50 40 90 4 60 40 40 30 10 DEMAND 40 50 60 60 210Here 95 is maximum element so subtract all elements from 95
  29. 29. This is the matrix after subtracting all the elements from 95 INVESTMENT P Q R S AVAILABLE 1 0 15 25 35 70 2 20 30 35 45 40 3 25 50 45 55 90 4 35 55 55 65 10 DEMAND 40 50 60 60 210This is now minimization case which can be further solved as discussed before
  30. 30. TRANSPORTATION PROBLEM Categorized into two types: Minimization Maximization problem problem
  31. 31. MINIMIZATION PROBLEM In this transportation cost is given which is to be minimized.
  32. 32. MAXIMIZATION PROBLEM In this profit is given which is to be maximized.To solve this problem we convert the problem into minimization.Conversion is done by selecting the largest element from Profit Pay off matrix and then subtracting all elements from largest element including itself. Reduced matrix obtain becomes minimization case and then same steps are taken to solve it as is done in minimization problem
  33. 33. VOGEL’S APPROXIMATION METHOD• BASIS OF ALLOCATION IS UNIT COST PENALTY• THE SUBSEQUENT ALLOCATIONS IN CELLS ARE DONE KEEPING IN VIEW THE HIGHEST UNIT COST• IBFS OBATINED BY THIS METHOD IS EITHER OPTIMAL OR VERY NEAR TO OPTIMAL SOLUTION• SO AMOUNT OF TIME REQUIRED TO CALCULATE THE OPTIMUM SOLUTION IS REDUCED
  34. 34. STEPS TO SOLVE THE PROBLEMStep1: for each row of the table identify the lowest and the next lowest cost cell. Find their least cost than the difference shall be zero.Step 2: similarly find the difference of each column and place it below each column. These differences found in the steps 1 and 2 are also called penalties.Step 3: looking at all the penalties. Identify the highest of them and the row or column relative to that penalty. Allocate the maximum possible units to the least cost cell in the selected row or column. Ties should be broken in this orderMaximum difference least cost cell.Maximum difference tie least cost cellMaximum unit’s allocations tie arbitraryStep 4: adjust the supply and demand and cross the satisfied row or column.Step 5:Recompute the column and row differences ignoring deleted rows/columns and go to the step3. Repeat the procedure until the entire column and row totals are satisfied.
  35. 35. MINIMIZATION USING VAMWAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20P2 5 7 3 28P3 4 5 8 17DEMAND 21 25 19 65 Step1 :This problem is balanced problem
  36. 36. Find out ‘Penalties’(P) for each row & column by subtracting the smallest element from next smallest element in the row/columnWAREHOUSE W1 W2 W3 SUPPLY P1PLANTP1 7 6 9 20 7-6=1P2 5 7 3 28 5-3=2 In W3 3 is min cost 19P3 4 5 8 17 5-4=1DEMAND 21 25 19 65P1 5-4=1 6-5=1 9-3=6 6 is max penaltyFind out the maximum penalty and in that row/column find minimum element and allocate it.
  37. 37. The row/column having zero demand/supply left will be cut by a line WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28/9 19 P3 4 5 8 17 DEMAND 21 25 19/0 65W3 is having zero demand now so it will be cut and wont be considered
  38. 38. Now find next penalty P2 in same wayWAREHOUSE W1 W2 W3 SUPPLY P2PLANTP1 7 6 9 20 7-6=1P2 5 9 7 3 28/9/0 7-5=2 2 is max penaltyP3 4 5 8 17 5-4=1 19DEMAND 21/12 25 19/0P2 5-4=1 6-5=1 ------------ In W1 5 is min cost
  39. 39. In this way find out all the penalties until all allocations are not doneWAREHOUSE W1 W2 W3 SUPPLY P3PLANTP1 7 6 9 20 7-6=1P2 5 9 7 3 28/9/0 ----------P3 4 12 5 8 17/5 5-4=1 19DEMAND 21/12/0 25 19/0 In P3 4 is min costP3 7-4=3 6-5=1 ------------ 3 is max penalty
  40. 40. Final matrix with all allocationsWAREHOUSE W1 W2 W3 SUPPLYPLANTP1 7 6 9 20/0 20P2 5 7 3 28/9/0 9 19P3 4 5 8 17/5/0 12 5DEMAND 21/12/0 25/5/0 19/0
  41. 41. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/0 9 19 P3 4 5 8 17/5/0 12 5 DEMAND 21/12/0 25/5/0 19/0 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
  42. 42. TOTAL TRANSPORTATION COSTPLANT WAREHOUSE COSTP1 W2 6*20=120P2 W1 5*9=45P2 W3 3*19=57P3 W1 12*4=48P3 W2 5*5=25TOTAL COST 295
  43. 43. COMPARISON OF NWCM,LCEM,VAM WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19 65 Minimum Transportation Cost NWCM=462 LCEM=300 VAM=295
  44. 44. MAXIMIZATION CASE I II III IV CAPACITYA 40 25 22 33 100B 44 35 30 30 30C 38 38 28 30 70REQUIREMENT 40 20 60 30 200 150This is an unbalanced maximization problem for finding out maximum profit involved
  45. 45. To solve this first we will convert this inminimization problem by subtracting the largest element from all elements I II III IV CAPACITYA 4 19 22 11 100B 0 9 14 14 30C 6 6 16 14 70REQUIREMENT 40 20 60 30 200 15044 is maximum element in the matrix so subtract all the elements from 44 and re-write in matrix form
  46. 46. Now we will balance this matrix Requirement<Capacity Dummy column will be added I II III IV Dummy CAPACITYA 4 19 22 11 0 100B 0 9 14 14 0 30C 6 6 16 14 0 70REQUIREMENT 40 20 60 30 50 200 200 Now this can be solved by any of the IBFS methods
  47. 47. CHOCOLATE WININGQUESTION FOR AUDIENCE
  48. 48. Obtain IBFS for the followingMaximization Problem using VAM: I II III SUPPLYA 6 4 4 100B 5 6 7 25C 3 4 6 75DEMAND 60 80 85
  49. 49. SOLUTION: I II III SUPPLYA 6 60 4 40 4 100B 5 6 15 7 10 25C 3 4 6 75 75DUMMY 0 0 25 0 25DEMAND 60 80 85 MAXIMUM PROFIT= Rs. 1130

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