preparing for the next midterm adventures in production engineering

Preparing for the next Midterm

Adventures in production engineering

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)

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.

customer 1 2 3 4 5

Warehouse 24 33 12 18 20

Customer 1 45 27 14 23

Customer 2 10 27 34

Customer 3 14 22

Customer 4 16

Weekly Demands (units) 65 110 72 68 91

The n-truck problem Let c0j = cost of one trip from the depot to customer j cij = cost of trip from customer i to customer j

(assume cij = cji) 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:

01

2n

jj

c

0 0ij i j ijs c c c

The Algorithm

Compute sij for all possible customer pairs Rank the sij 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

The Algorithm in Action

compute savingss12 = c01 + c02 – c12 = 24 + 33 – 45 = 12

s13 = c01 + c03 – c13 = 24 + 12 – 27 = 9

s14 = 24 + 18 – 14 = 28

s25 = c02 + c05 – c25 = 33 + 20 – 34 = 19

s34 = c03 + c04 – c34 = 12 + 18 – 14 = 16

s23 = c02 + c03 – c23 = 33 + 12 – 10 = 35

customer 1 2 3 4 5

Warehouse 24 33 12 18 20

Customer 1 45 27 14 23

Customer 2 10 27 34

Customer 3 14 22

Customer 4 16

Weekly Demands (units)

65 110 72 68 91

Rank order pairwise savings

from to savingCustomer 2 Customer 3 35Customer 1 Customer 4 28Customer 2 Customer 4 24Customer 4 Customer 5 22Customer 1 Customer 5 21Customer 2 Customer 5 19Customer 3 Customer 4 16Customer 1 Customer 2 12Customer 3 Customer 5 10Customer 1 Customer 3 9

Problem #1 – The Final Solution

customer 1 2 3 4 5

Warehouse 24 33 12 18 20

Customer 1 45 27 14 23

Customer 2 10 27 34

Customer 3 14 22

Customer 4 16

Weekly Demands 65 110 72 68 91

Truck Customer Route Miles traveled

1 2,3 33+10+12=55

2 1,4,5 24+14+16+20=74Total miles traveled per week 129Savings over sending a separate truck to each customer

2 x (24+33+12+18+20) – 129 = 85

from to savingCustomer 2 Customer 3 35Customer 1 Customer 4 28Customer 2 Customer 4 24Customer 4 Customer 5 22Customer 1 Customer 5 21Customer 2 Customer 5 19Customer 3 Customer 4 16Customer 1 Customer 2 12Customer 3 Customer 5 10Customer 1 Customer 3 9

capacity = 225

9

Lot Sizing with Capacity Constraints

Known requirements in each period versus capacitiesri = requirement in period ici = production capacity in period iyi = production level in period i

1 2

1 2

1 2

, ,...,

, ,...,

, ,..., , 1

n

n

n i i

r r r

c c c

Find y y y where y c for i n

1 1

1,...,j j

i ii i

c r for j n

10

Our Very Next Problem

Day 1 2 3 4 5 6 7 8 9 10requirement 80 90 100 100 70 80 100 70 80 50capacity 120 120 120 90 100 70 70 100 100 100

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.

Check for feasibility

Day 1 2 3 4 5 6 7 8 9 10requirement 80 90 100 100 70 80 100 70 80 50capacity 120 120 120 90 100 70 70 100 100 100cumulative requirement 80 170 270 370 440 520 620 690 770 820cumulative capacity 120 240 360 450 550 620 690 790 890 990

Check for feasibility

Day 1 2 3 4 5 6 7 8 9 10requirement 80 90 100 100 70 80 100 70 80 50capacity 120 120 120 90 100 70 70 100 100 100cumulative requirement 80 170 270 370 440 520 620 690 770 820cumulative capacity 120 240 360 450 550 620 690 790 890 990test for feasibility Ok Ok Ok Ok Ok Ok Ok Ok Ok Ok

The simple 2-step algorithm

1. Find next period in which demand > capacity

2. Back-shift demands to period(s) having excess capacity

The simple 2-step algorithm in action

Day 1 2 3 4 5 6 7 8 9 10requirement 80 90 100 100 70 80 100 70 80 50capacity 120 120 120 90 100 70 70 100 100 100exceeds cap 10 10 30reschedule 80 90 110 90

Day 1 2 3 4 5 6 7 8 9 10requirement 80 90 100 100 70 80 100 70 80 50capacity 120 120 120 90 100 70 70 100 100 100exceeds cap 10 30reschedule 80 90 110 90 80 70

Day 1 2 3 4 5 6 7 8 9 10requirement 80 90 100 100 70 80 100 70 80 50capacity 120 120 120 90 100 70 70 100 100 100exceeds cap 30reschedule 80 90 120 90 100 70 70 70 80 50

Costing the schedule

requirement 80 90 100 100 70 80 100 70 80 50total costschedule 80 90 120 90 100 70 70 70 80 50 10 4000inventory 0 0 20 10 40 30 0 0 0 0 100 300 4300

Setup cost = $400; holding cost = $3.00

Feasible but not necessarily optimal!

The Improvement 2-Step algorithm

Start at end and work backwards1. Determine if it is cheaper to shift entire

production lot to prior periods having excess capacity

2. Repeat until no further improvement is possible

The Improvement 2-Step algorithm in action

Setup cost = $400; holding cost = $3.00

requirement 80 90 100 100 70 80 100 70 80 50total costschedule 80 90 120 90 100 70 70 70 80 50 10 4000inventory 0 0 20 10 40 30 0 0 0 0 100 300 4300

capacity 120 120 120 90 100 70 70 100 100 100requirement 80 90 100 100 70 80 100 70 80 50total costreschedule 80 90 120 90 100 70 70 100 100 9 3600inventory 0 0 20 10 40 30 0 30 50 0 180 540

4140

capacity 120 120 120 90 100 70 70 100 100 100excess 40 30 0 0 0 0 0 0 0 100

Iteration 2 of the Improvement 2-Step algorithm in action

requirement 80 90 100 100 70 80 100 70 80 50total costreschedule 80 90 120 90 100 70 70 100 100 9 3600inventory 0 0 20 10 40 30 0 30 50 0 180 540

4140

capacity 120 120 120 90 100 70 70 100 100 100requirement 80 90 100 100 70 80 100 70 80 50total costreschedule 120 120 120 90 100 70 100 100 8 3200inventory 40 70 90 80 110 100 0 30 50 0 570 1710

4910

Iteration 3 of the Improvement 2-Step algorithm in action


4140

capacity 120 120 120 90 100 70 70 100 100 100requirement 80 90 100 100 70 80 100 70 80 50total costreschedule 120 120 120 90 100 70 100 100 8 3200inventory 40 70 90 80 110 30 0 30 50 0 500 1500

4700

Final Solution


4140

Multi-machine job schedulingSkill 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

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.

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.

ID Student Grading time in minutes

A Harry 120B Tom 43C Dick 83D Jane 107E Sally 38F John 57G Mike 111H Pete 75I Betty 72J Mary 47K Molly 80L Ned 138M Ted 112N Fred 84

Min MakespanLPT 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 James Professor Jones Professor SmithID time ID time ID timeL 138 138 A 120 120 M 112 112N 84 222 D 107 227 G 111 223C 83 305 H 75 302 K 80 303J 47 352 I 72 374 F 57 360B 43 395 E 38 398

Time to grade all projects (hours): 398 min = 6.63 hr.Average student waiting time for grade: 273.6 min = 4.56 hrs

ID Student timeA Harry 120B Tom 43C Dick 83D Jane 107E Sally 38F John 57G Mike 111H Pete 75I Betty 72J Mary 47K Molly 80L Ned 138M Ted 112N Fred 84

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 James Professor Jones Professor SmithID time ID time ID timeB 43 43 I 72 72 E 38 38J 47 90 H 75 147 F 57 95C 83 173 D 107 254 K 80 175N 84 257 A 120 374 G 111 286L 138 395 374 M 112 398

Time to grade all projects (hours): 398 min = 6.63 hr Average student waiting time for grade: 99.78 min. = 3.33 hr.


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 James Professor Jones Professor SmithID time ID time ID timeE 38 38 B 43 43 J 47 47F 57 95 I 72 115 H 75 122K 80 175 C 83 198 N 84 206D 107 282 G 111 309 M 112 318A 120 402 L 138 447 318

Time to grade all projects (hours): 447 min. = 7.45 hr.Average student waiting time for grade: 190.57 min. = 3.176 hr.


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 hourscustomer prepare

beddingsplant and mulch

A 6 7B 8 12C 12 20D 4 6E 20 5F 15 18G 6 20H 12 11I 22 7J 31 27K 14 15L 9 30

Johnson in action

SORTED BY PREPARE BEDDINGS

SORTED BY PLANTING AND MULCHING

customer prepare beddings

customer plant and mulch

D 4 E 5A 6 D 6G 6 A 7B 8 I 7L 9 H 11C 12 B 12H 12 K 15K 14 F 18F 15 C 20E 20 G 20I 22 J 27J 31 L 30

1 2 3 4 5 6 7 8 9 10 11 121 2 3 4 5 6 7 8 9 10 11 12D A G B L C K F J H I E

Total time to complete all the jobs

1 2 3 4 5 6 7 8 9 10 11 12Customer D A G B L C K F J H I Eprepare time 4 6 6 8 9 12 14 15 31 12 22 20finish 4 10 16 24 33 45 59 74 105 117 139 159plant time 6 7 20 12 30 20 15 18 27 11 7 5start time 4 10 17 37 49 79 99 114 132 159 170 177finish time 10 17 37 49 79 99 114 132 159 170 177 182

182 hr. / 8 hr. per day = 22.75 days

Average flowtime (hours) 102.0833Days 12.76042

More than two machines

Job Machine 1 Machine 2 Machine 3 Machine 4J1 1 13 6 2J2 10 12 18 18J3 17 9 13 4J4 12 17 2 6J5 11 3 5 16

Job Machine 1 Machine 4J1 1 2J2 10 18J3 17 4J4 12 6J5 11 16

Johnson sequence:1 -2 – 5 – 4 – 3Cmax = 88

More than two machinesJob Machine 1 Machine 2 Machine 3 Machine 4J1 1 13 6 2J2 10 12 18 18J3 17 9 13 4J4 12 17 2 6J5 11 3 5 16

Job Machine 1&2 Machine 3&4J1 14 8J2 22 36J3 26 17J4 29 8J5 14 21

Johnson sequence:5 – 2 – 3 -1 -4 Cmax = 85

More than two machinesJob Machine 1 Machine 2 Machine 3 Machine 4J1 1 13 6 2J2 10 12 18 18J3 17 9 13 4J4 12 17 2 6J5 11 3 5 16

Job Machine 1,2&3 Machine 2,3&4J1 20 21J2 40 48J3 39 26J4 31 25J5 19 24

Johnson sequence:5 – 1 – 2 – 3 - 4Cmax = 85

The Gantt Chart

job5 job 2 job 3 job 1 job 4

start t5finish start t2

finish start t3

finish start t1

finish start t4 finish

Machine 1 0 11 11 11 10 21 21 17 38 38 1 39 39 12 51Machine 2 11 3 14 21 12 33 38 9 47 47 13 60 60 17 77Machine 3 14 5 19 33 18 51 51 13 64 64 6 70 77 2 79Machine 4 19 16 35 51 18 69 69 4 73 73 2 75 79 6 85

5 – 2 – 3 -1 -4

Optimum makespan is 80Gupta’s algorithm will find it!Or try all 5! (120) permutationsAn exercise for the student

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.

1 2 3 4 5 6 7 8 9 10 11 12

route A,B B A A,B B,A B, B, A, A,B A,B B,A B,A

A time

40 30 25 35 80 45 40 30 34

B time

35 25 40 25 60 75 35 55 75 60

The Jackson algorithm in action

1 2 3 4 5 6 7 8 9 10 11 12


A time

40 30 25 35 80 45 45 30 34

B time

55 25 40 25 60 75 35 55 75 60

AB = {1,4,9,10} BA = {5,11,12} A = {3,8} B = {2,6,7}

A Bay: {AB}Job 1 4 9 10A 40 25 45 45

B 55 40 35 55

Johnson’s algorithm:4 – 1 – 10 – 9


1 2 3 4 5 6 7 8 9 10 11 12


A time

40 30 25 35 80 45 45 30 35

B time

55 25 40 25 60 75 35 55 75 60

AB = {1,4,9,10} BA = {5,11,12} A = {3,8} B = {2,6,7}

B Bay: {BA}Job 5 11 12A 35 30 35

B 25 75 60

Johnson’s algorithm: 11 – 12 – 5Reverse:5 – 12 – 11

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 reverseJohnson’s order.

Machine B: {BA} jobs ordered by reverse Johnson’s algorithm,then {B} in any order, followed by {AB} jobs inJohnson’s order.


B Bay: {BA}Job 5 11 12A 35 30 35

B 25 75 60

Johnson’s algorithm: 11 – 12 – 5Reverse:5 – 12 – 11

A Bay: {AB}Job 1 4 9 10A 40 25 45 45

B 55 40 35 55

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}

How long must the Otto work today?

1 2 3 4 5 6 7 8 9 10A Bay 4 1 10 9 3 8 5 12 11A Time 25 40 45 45 30 80 35 35 30start time 0 25 65 110 155 185 265 300 335finish 25 65 110 155 185 265 300 335 365B Bay 5 12 11 2 6 7 4 1 10 9B time 25 60 75 25 60 75 40 55 55 35start time 0 25 85 160 185 245 320 360 415 470finish time 25 85 160 185 245 320 360 415 470 505

505 minutes / 60 = 8.42 hours

REGISTER TODAY!

Midterm #2

preparing for the next midterm adventures in production engineering

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