1. Preparing for the next Midterm Adventures in production engineering

2. Three easy questions Question #1 Determining delivery routes (Sec 6.6) Question #2 lot sizing with capacity constraints (Sec 7.4) Question #3 Scheduling on multiple machines (Sec. 8.7)

3. Problem #1 Apply the delivery route algorithm given in the textbook in attempting to find a minimum cost (distance in miles) delivery routing where each truck has a capacity of 225 units. customer12345 Warehouse2433121820 Customer 1 45271423 Customer 2 102734 Customer 3 1422 Customer 4 16 Weekly Demands (units)65110726891

4. The n-truck problem Let c 0j = cost of one trip from the depot to customer j c ij = cost of trip from customer i to customer j (assume c ij = c ji ) If separate vehicle assigned to each customer then cost is given by: If go from depot to i to j then back to depot, the saving would be found from:

5. The Algorithm Compute s ij for all possible customer pairs Rank the s ij in decreasing order Consider each of the links in descending order of savings and include link (i,j) in a route if it is feasible if infeasible, go to the next link Once the list is exhausted, eliminate those on the current route and begin a new route

6. The Algorithm in Action compute savings s 12 = c 01 + c 02 – c 12 = 24 + 33 – 45 = 12 s 13 = c 01 + c 03 – c 13 = 24 + 12 – 27 = 9 s 14 = 24 + 18 – 14 = 28 s 25 = c 02 + c 05 – c 25 = 33 + 20 – 34 = 19 s 34 = c 03 + c 04 – c 34 = 12 + 18 – 14 = 16 s 23 = c 02 + c 03 – c 23 = 33 + 12 – 10 = 35 customer12345 Warehouse2433121820 Customer 1 45271423 Customer 2 102734 Customer 3 1422 Customer 4 16 Weekly Demands (units) 65110726891

7. Rank order pairwise savings fromto saving Customer 2 Customer 335 Customer 1Customer 4 28 Customer 2 Customer 424 Customer 4 Customer 522 Customer 1 Customer 521 Customer 2 Customer 519 Customer 3Customer 4 16 Customer 1Customer 212 Customer 3 Customer 510 Customer 1Customer 3 9

8. Problem #1 – The Final Solution customer12345 Warehouse2433121820 Customer 1 45271423 Customer 2 102734 Customer 3 1422 Customer 4 16 Weekly Demands65110726891 TruckCustomer RouteMiles traveled 1 2,333+10+12=55 2 1,4,524+14+16+20=74 Total miles traveled per week 129 Savings over sending a separate truck to each customer 2 x (24+33+12+18+20) – 129 = 85 fromto saving Customer 2 Customer 335 Customer 1Customer 4 28 Customer 2 Customer 424 Customer 4 Customer 522 Customer 1 Customer 521 Customer 2 Customer 519 Customer 3Customer 4 16 Customer 1Customer 212 Customer 3 Customer 510 Customer 1Customer 3 9 capacity = 225

9. 9 Lot Sizing with Capacity Constraints Known requirements in each period versus capacities r i = requirement in period i c i = production capacity in period i y i = production level in period i

10. 10 Our Very Next Problem Day12345678910 requirement 8090100 7080100708050 capacity 120 9010070 100 The finishing department of the Capacity Constrained Construction Company (CCCC) has just received its daily production requirements for the next two weeks (10 working days). Because of planned downtime on several of the finishing machines, the department capacity will vary over this time period. Each day, the machines must be setup and aligned for that days production run at a cost of $400. Holding cost on inventory held over each day is $3.00.

11. Check for feasibility Day12345678910 requirement8090100 7080100708050 capacity120 9010070 100 cumulative requirement80170270370440520620690770820 cumulative capacity120240360450550620690790890990

12. Check for feasibility Day12345678910 requirement8090100 7080100708050 capacity120 9010070 100 cumulative requirement80170270370440520620690770820 cumulative capacity120240360450550620690790890990 test for feasibility Ok

13. The simple 2-step algorithm 1. Find next period in which demand > capacity 2. Back-shift demands to period(s) having excess capacity

14. The simple 2-step algorithm in action Day12345678910 requirement8090100 7080100708050 capacity120 9010070 100 exceeds cap10 30 reschedule809011090 Day12345678910 requirement8090100 7080100708050 capacity120 9010070 100 exceeds cap1030 reschedule8090110908070 Day12345678910 requirement8090100 7080100708050 capacity120 9010070 100 exceeds cap30 reschedule80901209010070 8050

15. Costing the schedule requirement8090100 7080100708050 totalcost schedule80901209010070 8050104000 inventory00201040300000100300 4300 Setup cost = $400; holding cost = $3.00 Feasible but not necessarily optimal!

16. The Improvement 2-Step algorithm Start at end and work backwards 1. Determine if it is cheaper to shift entire production lot to prior periods having excess capacity 2. Repeat until no further improvement is possible

17. The Improvement 2-Step algorithm in action Setup cost = $400; holding cost = $3.00 requirement8090100 7080100708050 totalcost schedule80901209010070 8050104000 inventory00201040300000100300 4300 capacity120 9010070 100 requirement8090100 7080100708050totalcost reschedule80901209010070 100 93600 inventory00201040300 500180540 4140 capacity120 9010070 100 excess40300000000100

18. Iteration 2 of the Improvement 2-Step algorithm in action requirement8090100 7080100708050 totalcost reschedule80901209010070 100 93600 inventory00201040300 500180540 4140 capacity120 9010070 100 requirement8090100 7080100708050totalcost reschedule120 9010070100 83200 inventory407090801101000305005701710 4910

19. Iteration 3 of the Improvement 2-Step algorithm in action requirement8090100 7080100708050 totalcost reschedule80901209010070 100 93600 inventory00201040300 500180540 4140 capacity120 9010070 100 requirement8090100 7080100708050totalcost reschedule120 9010070100 83200 inventory40709080110300 5005001500 4700

20. Final Solution requirement8090100 7080100708050 totalcost reschedule80901209010070 100 93600 inventory00201040300 500180540 4140

21. Multi-machine job scheduling Skill set Performance Measures Gantt Charts Multi-machine scheduling Parallel machines Johnson’s algorithm (2-machine problem) The m-machine problem CDS and Gupta heuristics 2 –machine flow shop problem (Jackson’s algorithm)

22. 22 Results for Multiple Machines Parallel Machines Use SPT (min flowtime) or LPT (min makespan) Flow Shops The optimal solution for scheduling n jobs on two machines is always a permutation schedule (that is, jobs are done in the same order on both machines). (This is the basis for Johnson’s algorithm.) For three machines, a permutation schedule is still optimal if we restrict attention to total flow time only. Under rare circumstances, the two machine algorithm can be used to solve the three machine case. Use Cambell, Dudek, & Smith (CDS) or Gupta heuristic for n jobs, m machines Job Shops When scheduling n jobs on 2 machines, use Jackson’s algorithm When scheduling two jobs on m machines, the problem can be solved by graphical means.

23. Parallel machines – one more time Professors James, Jones, and Smith are team teaching the new undergraduate course in production engineering. Each student in the course must turn in a class project report consisting of an introduction, a literature review, a data collection and analysis, a methodology and results, and a conclusion. Each student’s report will be assigned to one of the instructors to evaluate and grade. The table is an estimate of the amount of time it will take to grade each student’s report based upon their respective page counts. IDStudentGrading time in minutes AHarry120 BTom43 CDick83 DJane107 ESally38 FJohn57 GMike111 HPete75 IBetty72 JMary47 KMolly80 LNed138 MTed112 NFred84

24. Min Makespan LPT Solution Grades must be turned in now to the registrar. Determine which reports each instructor should grade to minimize the total time required to grade all the reports. The first report to be assigned should go to Professor James and the second to Professor Jones. Professor JamesProfessor JonesProfessor Smith IDtimeIDtimeIDtime L138 A120 M112 N84222D107227G111223 C83305H75302K80303 J47352I72374F57360 B43395E38398 Time to grade all projects (hours): 398 min = 6.63 hr. Average student waiting time for grade: 273.6 min = 4.56 hrs IDStudenttime AHarry120 BTom43 CDick83 DJane107 ESally38 FJohn57 GMike111 HPete75 IBetty72 JMary47 KMolly80 LNed138 MTed112 NFred84

25. LPT Solution with SPT by Instructor Students are to be notified of their grade by email as soon as their report has been evaluated by one of the instructors. For the solution found in a, determine the order in which each instructor should grade the reports assigned in order to reduce the average time (per student) that students have to wait for their grades. Professor JamesProfessor JonesProfessor Smith IDtimeIDtimeIDtime B43 I72 E38 J4790H75147F5795 C83173D107254K80175 N84257A120374G111286 L138395 374M112398 Time to grade all projects (hours): 398 min = 6.63 hr Average student waiting time for grade: 99.78 min. = 3.33 hr. IDStudenttime AHarry120 BTom43 CDick83 DJane107 ESally38 FJohn57 GMike111 HPete75 IBetty72 JMary47 KMolly80 LNed138 MTed112 NFred84

26. SPT Solution Can the average waiting time for grades in b. be reduced further? If so, determine which reports to assign to each instructor along with the order in which they should be graded. The first report to be assigned should go to Professor James and the second to Professor Jones. Professor JamesProfessor JonesProfessor Smith IDtimeIDtimeIDtime E38 B43 J47 F5795I72115H75122 K80175C83198N84206 D107282G111309M112318 A120402L138447 318 Time to grade all projects (hours): 447 min. = 7.45 hr. Average student waiting time for grade: 190.57 min. = 3.176 hr. IDStudenttime AHarry120 BTom43 CDick83 DJane107 ESally38 FJohn57 GMike111 HPete75 IBetty72 JMary47 KMolly80 LNed138 MTed112 NFred84

27. Johnson Algorithm – one more time Professors Johnson and Jackson operate a landscaping business over their summer break in order to pay for their rent and food until the fall term begins. Johnson prepares the beddings and Jackson follows with the planting and mulching. For the coming month, they have lined up 12 customers with the job times given in the table to the right time in hours customerprepare beddings plant and mulch A67 B812 C 20 D46 E 5 F1518 G620 H1211 I227 J3127 K1415 L930

28. Johnson in action SORTED BY PREPARE BEDDINGS SORTED BY PLANTING AND MULCHING customerprepare beddings customerplant and mulch D4 E5 A6 D6 G6 A7 B8 I7 L9 H11 C12 B H K15 K14 F18 F15 C20 E G I22 J27 J31 L30 123456789101112123456789101112 D AGBLCKFJHIE

29. Total time to complete all the jobs 123456789101112 Customer D AGBLCKFJHIE prepare time 4668912141531122220 finish 410162433455974105117139159 plant time 67201230201518271175 start time 4101737497999114132159170177 finish time 101737497999114132159170177182 182 hr. / 8 hr. per day = 22.75 days Average flowtime (hours)102.0833 Days12.76042

30. More than two machines JobMachine 1Machine 2Machine 3Machine 4 J111362 J210121818 J3179134 J4121726 J5113516 JobMachine 1Machine 4 J112 J21018 J3174 J4126 J51116 Johnson sequence: 1 -2 – 5 – 4 – 3 C max = 88

31. More than two machines JobMachine 1Machine 2Machine 3Machine 4 J111362 J210121818 J3179134 J4121726 J5113516 JobMachine 1&2Machine 3&4 J1148 J22236 J32617 J4298 J51421 Johnson sequence: 5 – 2 – 3 -1 -4 C max = 85

32. More than two machines JobMachine 1Machine 2Machine 3Machine 4 J111362 J210121818 J3179134 J4121726 J5113516 JobMachine 1,2&3Machine 2,3&4 J12021 J24048 J33926 J43125 J51924 Johnson sequence: 5 – 1 – 2 – 3 - 4 C max = 85

33. The Gantt Chart job5 job 2 job 3 job 1 job 4 startt5 finis hstartt2 finis hstartt3 finis hstartt1 finis hstartt4finish Machine 1 011 1021 1738 139 1251 Machine 2 1131421123338947 1360 1777 Machine 3 14519331851 1364 67077279 Machine 4 191635511869 473 27579685 5 – 2 – 3 -1 -4 Optimum makespan is 80 Gupta’s algorithm will find it! Or try all 5! (120) permutations An exercise for the student

34. The 2-machine Job Shop Mr. Otto Mo Beale operates an automobile repair service. Otto has two servicing bays in his shop. The “A” bay has a hydraulic lift and is used for tire rotations, brake replacements, oil and filter changes, transmission repairs, etc. The “B: bay has wheel alignment equipment and an engine diagnostic computer. It is also used for tune-ups, carburetor repair, battery and belt replacements, etc. There are 12 automobiles scheduled for servicing today with the following service times in minutes required on each bay. 123456789101112 routeA,BBA B,AB, A,A,B B,A A time403025358045403034 B time35254025607535557560

35. The Jackson algorithm in action 123456789101112 routeA,BBA B,AB, A,A,B B,A A time403025358045 3034 B time55254025607535557560 AB = {1,4,9,10}BA = {5,11,12}A = {3,8}B = {2,6,7} A Bay: {AB} Job14910 A 40254545 B55403555 Johnson’s algorithm: 4 – 1 – 10 – 9

36. The Jackson algorithm in action 123456789101112 routeA,BBA B,AB, A,A,B B,A A time403025358045 3035 B time55254025607535557560 AB = {1,4,9,10}BA = {5,11,12}A = {3,8}B = {2,6,7} B Bay: {BA} Job51112 A 353035 B257560 Johnson’s algorithm: 11 – 12 – 5 Reverse: 5 – 12 – 11

37. 37 Jackson’s Algorithm to determine the order within a set: Machine A: {AB} jobs ordered by Johnson’s algorithm, then {A} in any order, followed by {BA} jobs in reverse Johnson’s order. Machine B: {BA} jobs ordered by reverse Johnson’s algorithm, then {B} in any order, followed by {AB} jobs in Johnson’s order.

38. The Jackson algorithm in action B Bay: {BA} Job51112 A 353035 B257560 Johnson’s algorithm: 11 – 12 – 5 Reverse: 5 – 12 – 11 A Bay: {AB} Job14910 A 40254545 B55403555 Johnson’s algorithm: 4 – 1 – 10 – 9 A : 4 – 1 – 10 – 9 – 3 – 8 - 5 – 12 – 11 B: 5 – 12 – 11 – 2 – 6 – 7 – 4 – 1 – 10 – 9 AB = {1,4,9,10}BA = {5,11,12}A = {3,8}B = {2,6,7}

39. How long must the Otto work today? 12345678910 A Bay 4 11093851211 A Time 254045 308035 30 start time 02565110155185265300335 finish 2565110155185265300335365 B Bay 5121126741109 B time 2560752560754055 35 start time 02585160185245320360415470 finish time 2585160185245320360415470505 505 minutes / 60 = 8.42 hours

40. REGISTER TODAY! Midterm #2