Sequencing
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Sequencing Presentation Transcript

  • 1. Sequencing Pavan Karva Rashmi Navaghane
  • 2. PROBLEM 1
    • There are six job which must go through two machine A and B in the order A-B. Processing time (in hours) is given here:
    • Determine the optimum sequence and the total elapsed time?
    Job 1 2 3 4 5 6 Machine A 8 10 11 12 16 20 Machine B 7 15 10 14 13 9
  • 3. Solution
    • Given
    • Sequence
    • MACHINE A MACHINE B
    Job 1 2 3 4 5 6 Machine A 8 10 11 12 16 20 Machine B 7 15 10 14 13 9 Job 2 1 6 5 4 3
  • 4. Continue…
    • Total Elapsed time
    • Total Elapsed time=85hrs
    Machine A Machine B Sequence Start time Add Time taken End time Start time Add Time taken End time 2 0 + 10 10 10 + 15 25 3 10 + 11 21 25 + 10 35 4 21 + 12 33 35 + 14 49 5 33 + 16 49 49 + 13 62 6 49 + 20 69 69 + 9 78 1 69 + 8 77 78 + 7 85 Total Time 77 68
  • 5. Continue…
    • Idle Time for M/c A=
    • Total Elapsed Time- Total time of M/c A
    • 85-77=8 hrs.
    • Idle Time for M/c B=
    • Total Elapsed Time- Total time of M/c B
    • 85-68=17 hrs.
  • 6. PROBLEM 2
    • There are five jobs, each of which must go through machines A, B and C in the order ABC. Processing time (in hours) are given below:
    • Determine the optimal processing sequence and total elapsed time?
    Job 1 2 3 4 5 Machine A 10 11 8 7 6 Machine B 6 4 5 3 2 Machine C 9 5 4 6 8
  • 7. Solution
    • Given
    • Step 1- 1 st we have to convert this problem into two machine problem. For that we have to check following condition:
    • Min A or Min C >= Max B
    • here Min A=6, Min C=4, Max B=6.
    • therefore 6=6
    • Min A=Max B
    Job 1 2 3 4 5 Machine A 10 11 8 7 6 Machine B 6 4 5 3 2 Machine C 9 5 4 6 8
  • 8. Continue….
    • Consolidation or Conversion Table
    • New job timing According to Consolidation Table
    Job New M/c 1 New M/c 2 P(A+B) New time P(B+C) New time 1 10+6= 16 6+9= 15 2 11+4= 15 4+5= 9 3 8+5= 13 5+4= 9 4 7+3= 10 3+6= 9 5 6+2= 8 2+8= 10 Job 1 2 3 4 5 New M/c 1 16 15 13 10 8 New M/c 2 15 9 9 9 10
  • 9. Continue…..
    • Sequencing According to consolidation Table:
    • Consolidated table:
    • Job sequence:
    Job 1 2 3 4 5 New M/c 1 16 15 13 10 8 New M/c 2 15 9 9 9 10 Job 3 4 2 1 5
  • 10. Other Possible Sequences Job 5 4 1 2 3 Job 5 4 1 3 2
  • 11. Continue…
    • Total Elapsed time
    Machine A Machine B Machine C Sequence Start time Total time End time Start time Total time End time Start time Total time End Time 5 0 6 6 6 2 8 8 8 16 3 6 8 14 14 5 19 19 4 23 1 14 10 24 24 6 30 30 9 39 2 24 11 35 35 4 39 39 5 44 4 35 7 42 42 3 45 45 6 51 TOTAL TIME 42 20 29
  • 12. Continue…
    • Idle Time for M/c A=
    • Total Elapsed Time- Total time of M/c A
    • 51-42=9 hrs.
    • Idle Time for M/c B=
    • Total Elapsed Time- Total time of M/c B
    • 51-20=31hrs.
    • Idle Time for M/c C=
    • Total Elapsed Time- Total time of M/c B
    • 51-29=22hrs.
    • Total Elapsed time=51 hrs
  • 13. PROBLEM 3
    • There are 4 job ABCD are required to be processed on four machine M1, M2, M3, M4 in that order. Determine optimal sequence and total elapsed time.
    Job M1 M2 M3 M4 A 13 8 7 14 B 12 6 8 19 C 9 7 5 15 D 8 5 6 15 MACHINES
  • 14. Continue…
    • Given
    • Step 1- 1 st we have to convert this problem into two machine problem. For that we have to check following condition:
    • Min M1 or Min M4 >= Max M2 or Max M3
    • here Min M1=8, Min M4=14, Max M2=7, Max M3=8.
    • therefore 8=8
    • Min M1=Max M3
    Job M1 M2 M3 M4 A 13 8 7 14 B 12 6 8 19 C 9 7 5 15 D 8 5 6 15
  • 15. Continue…
    • Consolidation or Conversion Table
    • New job timing According to Consolidation Table
    JOB MACHINES 5 MACHINES 6 P(M1+M2+M3) NEW TIME P(M2+M3+M4) NEW TIME A 13+8+7 28 8+7+14 29 B 12+6+8 26 6+8+19 33 C 9+7+5 21 7+5+15 27 D 8+5+6 19 5+6+15 26 Job A B C D New M/c 5 28 26 21 19 New M/c 6 29 33 27 26
  • 16. Continue…
    • Sequencing According to consolidation Table:
    • Consolidated table:
    • Job sequence:
    Job A B C D New M/c 5 28 26 21 19 New M/c 6 29 33 27 26 C A B D JOB
  • 17. Elapsed time calculation
    • START TIME – ST,TIME TAKEN –TT,END TIME -ET
    Machine 1 Machine 2 Machine 3 Machine 4 S.Q ST TT ET ST TT ET ST TT ET ST TT ET D 0 8 8 8 5 13 13 6 19 19 15 34 B 8 12 20 20 6 26 26 8 34 34 19 53 A 20 13 33 33 8 41 41 7 48 53 14 67 C 33 9 42 42 7 49 49 5 54 67 15 82 T.T 49 26 26 63
  • 18. Continue…
    • Total Elapsed time= 82 hrs
    • Idle Time for M/c 1=
    • Total Elapsed Time- Total time of M/c 1
    • 82-42= 40hrs.
    • Idle Time for M/c 2=
    • Total Elapsed Time- Total time of M/c 2
    • 82-26= 56hrs.
    • Idle Time for M/c 3=
    • Total Elapsed Time- Total time of M/c 3
    • 82-26= 56hrs.
    • Idle Time for M/c 4=
    • Total Elapsed Time- Total time of M/c 4
    • 82-63=19hrs.
  • 19.
    • Thank You