Presentation is loading. Please wait.

Presentation is loading. Please wait.

MODELING AND ANALYSIS OF MANUFACTURING SYSTEMS Session 12 MACHINE SETUP AND OPERATION SEQUENCING E. Gutierrez-Miravete Spring 2001.

Similar presentations


Presentation on theme: "MODELING AND ANALYSIS OF MANUFACTURING SYSTEMS Session 12 MACHINE SETUP AND OPERATION SEQUENCING E. Gutierrez-Miravete Spring 2001."— Presentation transcript:

1 MODELING AND ANALYSIS OF MANUFACTURING SYSTEMS Session 12 MACHINE SETUP AND OPERATION SEQUENCING E. Gutierrez-Miravete Spring 2001

2 INTRODUCTION WHOLE SYSTEM DESIGN STRATEGIES VS INDIVIDUAL CELL/WORKSTATION DESIGN STRATEGIES QUESTION: HOW TO TOOL THE MACHINE AND THEN SEQUENCE PRODUCTION ACTIVITIES IN IT? GOAL: MAXIMIZE PRODUCTIVITY

3 OPORTUNITIES SEQUENCE BATCHES SO AS TO MINIMIZE TOOLING CHANGEOVERS SEQUENCE ACTIVITIES SO AS TO MINIMIZE IDLE TIMES OPTIMIZE CELL LAYOUT SO AS TO MINIMIZE ASSEMBLY TIME

4 SEQUENCING AND OPTIMIZATION MANY SEQUENCING PROBLEMS IN MANUFACTURING CELL PLANNING ARE CLASSIC OPTIMIZATION PROBLEMS

5 CELL VS SYSTEM BOTH, CLEVER SEQUENCING OF OPERATIONS AND OVERALL SYSTEM DESIGN ARE IMPORTANT FOR SUCCESS IN MANUFACTURING TIME FRAMES –FOR OVERALL SYSTEM DESIGN: WEEKS OR MONTHS (LONG TERM) –FOR SEQUENCE DESIGN: MINUTES/HOURS (SHORT TERM)

6 TASK ASSIGNMENT

7 LINEAR ASSIGNMENT PROBLEM GOAL : –TO DISTRIBUTE N TASKS AMONG N WORKERS SO AS TO MINIMIZE COST CONSTRAINTS : –1 TASK PER WORKER –1 WORKER PER TASK

8 COST MATRIX C ROWS: WORKERS COLUMNS: MACHINES SUMMARIZES THE ASSIGNMENT COSTS

9 LAP: MATHEMATICAL FORMULATION minimize C  i  j c ij x ij subject to:  i x ij = 1 (for all tasks)  j x ij = 1 (for all workers)

10 Facts IF A CONSTANT IS ADDED TO EVERY ELEMENT OF A ROW OR COLUMN OF C, THE OPTIMAL SOLUTION DOES NOT CHANGE BUT ITS VALUE CHANGES BY THE ADDED CONSTANT IF ALL c ij > 0, ANY SOLUTION WITH COST = 0 MUST BE OPTIMAL

11 Hungarian Algorithm Proceeds by adding and substracting constants from rows and columns so as to maintan a non-negative cost matrix. When a feasible solution is found using only 0 cost cells, optimum has been found.

12 HA Steps STEP 1: COST REDUCTION BY CONSTRUCTION OF THE REDUCED COST MATRIX. THE RCM IS OBTAINED BY SUBSTRACTING FIRST THE MINIMUM ELEMENT IN EACH ROW FROM ALL ELEMENTS IN THE ROW THEN DOING LIKEWISE WITH COLUMNS

13 HA Steps cont’d STEP 2: SEARCH FOR A FEASIBLE SOLUTION USING ONLY THE 0’S IN THE RCM. IF THIS SUCCEEDS, OPTIMAL SOLUTION HAS BEEN FOUND. IF ALL 0’S CAN BE COVERED WITH LESS THAN n HORIZONTAL AND VERTICAL LINES, CONTINUE.

14 HA Steps cont’d STEP 3: FURTHER REDUCTION. FIND THE MINIMUM UNCOVERED ELEMENT. SUBSTRACT THIS REDUCED COST FROM EACH UNCOVERED ELEMENT AND ADD IT TO EACH TWICE-COVERED ELEMENT. GO TO 2. Example 8.1, Table 8.1, Fig. 8.1, pp. 263-

15 TASK SEQUENCING

16 TWO CLASSES OF PROBLEMS EVALUATE CHANGEOVER COSTS WHEN COSTS ARE COMPLETELY DETERMINED BY THE CURRENT JOB SETUP AND THE NEXT JOB TO BE LOADED EVALUATE COSTS WHEN THE ENTIRE SEQUENCE OF JOBS MUST BE KNOWN

17 COMPLETE CHANGEOVERS N JOBS ARE TO BE PERFORMED ON ONE MACHINE UNIT PROCESSING TIME AND BATCH SIZE DETERMINE TOTAL PROCESSING TIME CHANGEOVER TIMES DEPEND ONLY ON CURRENT AND NEXT PRODUCT TOTAL SETUP TIME DEPENDS ON JOB SEQUENCE

18 TRAVELING SALESMAN PROBLEM A SALESMAN MUST VISIT EVERY CITY IN HIS/HER TERRITORY THEN RETURN HOME IN SUCH A WAY THAT THE SMALLEST POSSIBLE TOTAL DISTANCE IS TRAVELLED. TSP CAN BE VISUALIZED WITH A GRAPH OF NODES (CITIES) AND ARC LENGTHS (DISTANCES)

19 TRAVELING SALESMAN PROBLEM DISTANCE BETWEEN CITIES i & j = c ij ALL ARCS ARE BIDIRECTIONAL NO SUBTOURS ALLOWED FIVE CITY COMPLETE TSP GRAPH Fig 8.2a FIVE CITY POSSIBLE TOUR GRAPH Fig 8.2b

20 SOLUTION OF THE TSP CLASSICAL OPTIMIZATION TECHNIQUES ARE HARD TO APPLY WHEN N IS LARGE (See Eqn 8.5-) HEURISTIC METHODS ARE MOST FREQUENTLY USED –NO SUBTOUR CONSTRAINT IS TEMPORARILY RELAXED –SOLVE RESULTING OPTIMIZATION PROBLEM

21 TSP BY CLOSEST INSERTION ALGORITHM 1.- SELECT A STARTING CITY 2.- PROCEED THROUGH N-1 STAGES ADDING A NEW CITY AT EACH STAGE. THE NEW CITY IS SELECTED FROM THOSE CURRENTLY UNASSIGNED SUCH THAT IT IS CLOSEST TO ANY CITY IN THE ACCUMULATED PARTIAL SEQUENCE Example 8.2; Fig 8.3, pp. 268-

22 TSP BY MINIMUM SPANNING TREE A MST IS ANY SET OF N-1 ARCS THAT TOUCH EACH NODE AND HAVE THE SMALLEST SUM OF COSTS FOR ANY SUCH SET BY MODIFYING THE MST SO THAT EACH NODE IS CONNECTED EXACTLY BY TWO ARCS IN A CONNECTED TREE OBTAIN TSP SOLUTION

23 TSP BY SUBTOUR INTEGRATION START WITH A SOLUTION TO THE ASSIGNMENT PROBLEM TRY TO CONNECT TWO SUBTOURS AT A TIME BY SWITCHING ARCS ONCE ALL LAP SUBTOURS ARE COMBINED, GOT TSP SOLUTION Fig 8.4

24 PARTIAL CHANGEOVERS N JOBS ARE TO BE PERFORMED ON ONE MACHINE CAPABLE OF HOLDING M TOOLS JOB j REQUIRES TOOLS Aj TOTAL NUMBER OF TOOLS REQUIRED EXCEEDS M (TOOL CHANGES REQUIRED) NO JOB REQUIRES MORE THAN M TOOLS AT EACH JOB COMPLETION SOME TOOLS MAY HAVE TO BE REMOVED AND NEW TOOLS ADDED Fig 8.5

25 OBJECTIVE TO ORDER JOBS AND TOOL CHANGEOVERS TO MINIMIZE THE TOTAL NUMBER OF TOOLS CHANGED ON THE MACHINE

26 NOTES ALWAYS KEEP M TOOLS ON MACHINE THE ORDERING FOR JOBS ON THE MACHINE IS GIVEN KEEP TOOL NEEDED SOONEST (KTNS) RULE IS OPTIMAL REMOVE ONLY “LONGEST UNTIL NEXT USE” TOOLS

27 JOB ORDERING IF JOB r USES ONLY A SUBSET OF TOOLS USED BY ITS PREDECESOR s CHANGEOVER IS NOT INCREASED (i.e. OPTIMAL SEQUENCE) JOB SEQUENCING PROBLEM IS SIMILAR TO GROUP FINDING IN GT THE TOOL-JOB MATRIX

28 JOB ORDERING SOLUTION METHODS –BINARY CLUSTERING Ex 8.3, Tables 8.3, 8.4 –TSP Table 8.5

29 SOLUTION OF THE PARTIAL CHANGEOVER PROBLEM STEP 1: JOB COMBINATION (REDUCTION) STEP 2: JOB ORDERING STEP 3: TOOL SETUP PLANNING BY KTNS

30 INTEGRATED ASSIGNMENT AND SEQUENCING

31 QUESTION WHAT TO DO WHEN CELL SETUP (TOOLING) AND JOB SEQUENCING ARE RELATED BY NEITHER DICTATES THE OTHER? NEED TO: –MAKE A SEQUENCING DECISION and –MAKE A SETUP DECISION

32 NOTES PROBLEM 1: PLAN SETUP AND OPERATION OF ASSEMBLY CELLS PROBLEM 2: SETUP AND SEQUENCE OF A MACHINE (INTERDEPENDENT TOOLS) GOAL: TO MINIMIZE CYCLE TIME SOLUTION TECHNIQUES: HEURISTICS

33 ASSEMBLY CELL LAYOUT AND SEQUENCING ASSEMBLY ENVIRONMENTS –MASS PRODUCED SINGLE PRODUCT –MULTIPLE PRODUCTS PRODUCED IN ALTERNATING LOTS WITH CHANGEOVER REQUIRED –MIX OF PART TYPES ASSEMBLED SIMULTANEOUSLY IN CELL WITHOUT CHANGEOVER

34 REQUIRED FIND HOW TO PERFORM THE SET OF TASKS THAT HAVE BEEN ASSIGNED TO A WORKSTATION

35 SINGLE PART TYPE A SINGLE PRODUCT FRAME PRODUCT FRAME HAS N LOCATIONS WHERE PARTS ARE TO BE ADDED WORKSTATION HAS N BINS WHERE PART FEEDERS ARE PLACED Fig 8.6

36 REQUIRED ASSIGN FEEDERS TO BINS DETERMINE THE ORDER IN WHICH PARTS ARE TO BE ADDED TO FRAME PROBLEM CAN BE MODELED AS A 2N-CITY TSP

37 BIN ASSIGNMENT AND INSERTION SEQUENCING BIN ASSIGNMENT GOAL: MINIMIZE THE LOADED TRAVEL TIME (DISTANCE) FOR THE ASSEMBLER INSERTION SEQUENCING GOAL: MINIMIZE THE UNLOADED TRAVEL TIME (DISTANCE) FOR THE ASSEMBLER Ex 8.4, Tables 8.6, 8.7; Ex 8.5, Table 8.8

38 MIXED PRODUCTS TYPICALLY, SEVERAL PRODUCT TYPES ARE BEING PRODUCED SIMULTANEOUSLY IN THE SAME CELL DEMAND PROPORTIONS DETERMINED BY BILL OF MATERIALS FOR END PRODUCTS

39 MIXED PRODUCT: UNPACED LINE ASSUME –RELATIVE DEMANDS ARE KNOWN –M FRAMES TO BE MADE –pm IS THE PROPORTION OF TYPE m FRAMES –BINS KEPT ON SAME PLACE GOAL: MINIMIZE AVERAGE ASSEMBLY TIME OF PRODUCT

40 MIXED PRODUCT: UNPACED LINE FOR EACH PART TYPE COMPILE TABLE OF TRAVEL TIMES (Table 8.7) COMBINE TABLES BY WEIGHTED AVERAGE c ij =  m p m c ijm –pm proportion of type m frames –cijm total travel time per type m frame if part i is assigned to bin j –Ex. 8.6, Tables 8.9, 8.10

41 CELL LAYOUT AND SEQUENCING: INTERDEPENDENT TOOLS THE NC PUNCH PRESS –36 TOOL TOOL TURRET (Fig 8.7) –PARTS MAY REQUIRE UP TO 200 HITS –HIT SEQUENCE MAY BE SUBJECT TO PRECEDENCE CONSTRAINTS GOAL: LOAD TOOLS AND SEQUENCE HITS TO MINIMIZE PRODUCTION CYCLE Ex 8.7, Fig. 8.8, Tables 8.11, 8.12


Download ppt "MODELING AND ANALYSIS OF MANUFACTURING SYSTEMS Session 12 MACHINE SETUP AND OPERATION SEQUENCING E. Gutierrez-Miravete Spring 2001."

Similar presentations


Ads by Google