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15/04/2024

Lean Thinking & Lean Tools

- Flow without Interruption: Session 2

Professor Dongping Song

Liverpool University Management School

Email: [email protected]

Learning Outcomes

• Techniques to Solve Line Balancing

• Longest-Task-Time (LTT) Heuristic; STT

• Kilbridge and Wester Method (KWM)

• Ranked Positional Weights (RPW) Method

• Incremental Utilisation Heuristic

• Other Ways To Improve Line Balance Computer Method

Mathematical Methods

• Mixed-model production leveling

• Weighted Average Time method/rule

• Production levelling in logistics

15/04/2024

Review Terminologies

• Task –the elements of work; operation

• Task precedence

• Task time (also operation cycle time)

• Workstation

• Work center

• Production rate

• Cycle time (line cycle time)

• Stacked time

• Takt time

Techniques to Solve Line Balancing &

Production Levelling

• Heuristic Methods -- based on simple rules:

– Longest-Task-Time Heuristic

– Shortest-Task-Time Heuristic

– KWM --Kilbridgeand WesterMethod (1961)

– Ranked Positional Weights Method

– Incremental Utilisation Heuristic

• Computerised Tools, e.g. COMSOAL (1996)

• Mathematical Methods

– Linear/Integer Programming

– Dynamic Programming

15/04/2024

Line Balancing - Longest-Task-Time

Heuristic

• LTTis also called Largest-Candidate Rule (LCR).

• Step 1. Construct the precedence diagram. List all elements

(tasks) in descending order of Te(= task time) value.

• Step 2. To assign elements to the first workstation, start at the

top of the list for the available elements, selecting the first

feasible elementfor placement at the station. A feasible

elementis one that satisfies the precedence requirements

and does not cause the sum of the Tevalue at workstation to

exceed the cycle time CT. Open a new workstation as needed.

• Step 3. Repeat step 2.

Select the available and feasible task with the largest Te.

Line Balancing – Kilbridge and Wester

Method

• Step 1. Construct the precedence diagram so those nodes

representing work elements of identical precedence are

arranged verticallyin columns.

• Step 2. List the elements in order of their columns, column I

at the top of the list. If an element can be located in more

than one column, list all columns by the element to show the

transferability of the element.

• Step 3. To assign elements to workstations, start with the

column I elements. Continue the assignment procedure in

order of column number until the cycle time is reached (CT).

Select tasks according to their position in the precedence diagram

15/04/2024

Line Balancing – Ranked Positional

Weights Method

• Introduced by Helgeson and Birnie in 1961.

• Combined the LTT and KWM methods.

• The RPW takes account of both the Te value of the

element and its position in the precedence diagram.

• Need to calculate RPW for all elements;

• Then, the elements are assigned to workstations in

the general order of their RPW values.

Procedure of RPW

• Step 1. Construct the precedence diagram. Then calculate the

RPW for each element by summing the elements Tetogether

with the Tevalues for all the elements that follow it in the

arrow chain of the precedence diagram.

• Step 2. List the elements in the order of their RPW, largest

RPW at the top of the list. For convenience, include the Te

value and immediate predecessors for each element.

• Step 3. Assign elements to stations according to RPW priority

rule, avoiding precedence constraint and cycle time violations.

Select tasks considering both their position and time value.

15/04/2024

Example: Small electrical appliance

No. Task description Time (min) Precedence by

1 Place frame on work holder and clamp 0.2 --

2 Assemble plug, grommet to power cord 0.4 --

3 Assemblebrackets to frame 0.7 1

4 Wire power cord to motor 0.1 1,2

5 Wire power cord to switch 0.3 2

6 Assemblemechanism plate to bracket 0.11 3

7 Assemble blade to bracket 0.32 3

8 Assemble motor to bracket 0.6 3,4

9 Align blade and attach to motor 0.27 6,7,8

10 Assemble switchto motor bracket 0.38 5,8

11 Attach cover, inspect, and text 0.5 9,10

12 Place in tote pan for packing 0.12 11

Assume cycle time is 1 minute. To determine number of WS and assignment of tasks.

Example: Precedence Diagram

0.11

0.7

0.32 0.27

0.2

7 9

0.1 0.5 0.12

4 11 12

0.6 0.38

0.4

8 10

0.3

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Example: LTT – Step 1

Task Time Immediate

(Te) predecessor

3 0.7 1

8 0.6 3,4

11 0.5 9,10

2 0.4 ---

10 0.38 5,8

7 0.32 3

5 0.3 2

9 0.27 6,7,8

1 0.2 ---

12 0.12 11

6 0.11 3

4 0.1 1,2

List all tasks in descending order of task time

Example: LTT – Step 2,3

Task Time Immediate

WS Task Te sum(Te) Available

(Te) predecessor

3 0.7 1 1 2 0.4 (2, 1)

8 0.6 3,4 5 0.3 (5, 1)

11 0.5 9,10 1 0.2 (1)

2 0.4 --- 4 0.1 1.00 (3, 4)

10 0.38 5,8 2 3 0.7 (3)

7 0.32 3 6 0.11 0.81 (8, 7, 6)

5 0.3 2 3 8 0.6 (8, 7)

9 0.27 6,7,8 10 0.38 0.98 (7, 10)

1 0.2 --- 4 7 0.32 (7)

12 0.12 11 9 0.27 0.59 (9)

6 0.11 3 5 11 0.5 (11)

4 0.1 1,2 12 0.11 0.62 (12)

7 9

1 1

1 1 2

2 8

15/04/2024

Example: KWM – Step 1

I II III IV V VI

0.11

0.7

0.32 0.27

0.2

7 9

0.1 0.5 0.12

4 11 12

0.6 0.38

0.4

8 10

0.3

Arrange the tasks in columns vertically

Example: KWM – Step 2

Task Column Time Sum of

(Te) Column Te

1 I 0.2

2 I 0.4 0.6

3 II 0.7

4 II 0.1

5 II,III 0.3 1.1

6 III 0.11

7 III 0.32

8 III 0.6 1.03

9 IV 0.27

10 IV 0.38 0.65

11 V 0.5

12 VI 0.12 0.62

List the tasks in order of their columns

15/04/2024

Example: KWM – Step 3

Task Column Time Sum of

WS Task Te sum(Te)

(Te) Column Te

1 I 0.2 1 1 0.2

2 I 0.4 0.6 2 0.4

3 II 0.7 4 0.1

4 II 0.1 5 0.3 1.00

5 II,III 0.3 1.1 2 3 0.7

6 III 0.11 6 0.11 0.81

7 III 0.32 3 7 0.32

8 III 0.6 1.03 8 0.6 0.92

9 IV 0.27 4 9 0.27

10 IV 0.38 0.65 10 0.38 0.65

11 V 0.5 5 11 0.5

12 VI 0.12 0.62 12 0.11 0.62

Assign the tasks into workstations

Example: RPW – Step 1

• For Task 1, the tasks that follow it in the arrow chain are

3,4,6,7,8,9,10,11,12.

• The Ranked Positional Weight (RPW) for Task 1 would be the sum

of the Tefor all these tasks, plus Tefor Task 1, which is 3.3.

• The same can be done for other Tasks.

0.11 Calculate RPWs for all tasks

0.7

3 0.32 0.27

0.2

7 9

0.1 0.5 0.12

4 11 12

0.6 0.38

0.4

8 10

0.3

15/04/2024

Example: RPW – Step 2

Task RPW Te Immediate

predecessor

1 3.3 0.2 ---

3 3 0.7 1

2 2.67 0.4 ---

4 1.97 0.1 1,2

8 1.87 0.6 3,4

5 1.3 0.3 2

7 1.21 0.32 3

6 1.00 0.11 3

10 1.00 0.38 5,8

9 0.89 0.27 6,7,8

11 0.62 0.5 9,10

12 0.12 0.12 11

List all tasks in descending order of RPW

Example: RPW – Step 3

Task RPW Te Immediate

WS Task Te sum(Te) Available

predecessor

1 3.3 0.2 --- 1 1 0.2 (2, 1)

3 3 0.7 1 3 0.7 0.90 (2, 3)

2 2.67 0.4 --- 2 2 0.4 (2,6,7)

4 1.97 0.1 1,2 4 0.1 (4,5,6,7)

8 1.87 0.6 3,4 5 0.3 (5,6,7,8)

5 1.3 0.3 2 6 0.11 0.91 (6,7,8)

7 1.21 0.32 3 3 8 0.6 (7,8)

6 1.00 0.11 3 7 0.32 0.92 (7,10)

10 1.00 0.38 5,8 4 10 0.38 (10,9)

9 0.89 0.27 6,7,8 9 0.27 0.65 (9)

11 0.62 0.5 9,10 5 11 0.5 (11)

12 0.12 0.12 11 12 0.11 0.62 (12)

7 9

1 1

1 1 2

2 8

15/04/2024

Example: STT

Task Time Immediate

(Te) predecessor WS Task Te sum(Te) Available

3 0.7 1 1 1 0.2 (2, 1)

8 0.6 3,4 2 0.4 (2, 3)

11 0.5 9,10 4 0.1 (3,4,5)

2 0.4 --- 5 0.3 1.00 (3,5)

10 0.38 5,8 2 3 0.7 (3)

7 0.32 3 6 0.11 0.81 (6,7,8)

5 0.3 2 3 7 0.32 (7,8)

9 0.27 6,7,8 8 0.6 0.92 (8)

1 0.2 --- 4 9 0.27 (9,10)

12 0.12 11 10 0.38 0.65 (10)

6 0.11 3 5 11 0.5 (11)

4 0.1 1,2 12 0.11 0.62 (12)

7 9

1 1

1 1 2

2 8

Example – LTT, KWM, RPW, STT

WS LTT KWM RPW STT

1 1.00 1.00 0.90 1.00

2 0.81 0.81 0.91 0.81

0.59 0.92 0.92 0.92

4 0.98 0.65 0.65 0.65

0.62 0.62 0.62 0.62

15/04/2024

Result Analysis (RPW)

Line Balance Ratio = Stacked Times . X 100%

Largest Op. x No. of Operators

Line Balance Ratio = 4 . X 100%

0.92 x 5

= 87 %

Result Analysis (RPW)

Line Balance Eff = Stacked Times . X 100%

TaktTime. x No. of Operators

Assume Takt time = CT

Line Balance Eff. = 4 . X 100%

1 x 5

= 80 %

15/04/2024

Result Analysis (LTT vs. RPW)

Rules LTT RPW

Takttime 1 1

No of WSs 5 5

Maximum WS time 1 0.92

Balance loss 1 1

Line balance ratio 80% 87%

Line balance efficiency 80% 80%

Learning Outcomes

• Longest-Task-time Heuristic

• Kilbridgeand WesterMethod

• Ranked Positional Weights Method

• Compare LTT, STT, KWM, and RPW

• Incremental Utilisation Heuristic

• Other Ways To Improve Line Balance

• Product leveling

• Mixed-model production leveling

• Weighted Average Time method/rule

• Remarks on production levelling

15/04/2024

Line Balancing – Incremental

Utilization Heuristic

• What if task times are greater than the cycle time?

• Assign tasks to a number of workstations (or

operators) to form a number of work centres

• Using Incremental Utilization Heuristic

– Tasks are added to each WS in order of task precedence

one at a time until utilization is 100% or falling.

– To determine: number of work centres, number of

workstations at each work centre, assignment of tasks to

each work centre.

Utilisation = sum of tasks assigned / (CT * NoOfWS)

Example

Textech, a large electronics manufacturer, assembles model AT7S handheld

calculators at its Midland , Texas, plant. The assembly tasks that must be

performed on each calculator are shown in the Table below.

• The parts used in this assembly line are supplied by materials-handling

personnel to parts bins used in each task. The assemblies are moved along

by belt conveyors between workstations. The productive time is 54

minutes per hour.

• Textechwants this assembly line to produce 540 calculators per hour.

Assuming 100% production line efficiency.

1. Compute the cycle time per calculator in minutes.

2. Compute the minimum number of workstations.

3. How would you combine the tasks into workstations and work

centres to minimize idle time? Evaluate your proposal.

15/04/2024

Task Description Immed. Preced. Task time

A place circuit frame on jig. none 0.18

B place circuit #1 into frame. A 0.12

C place circuit #2 into frame. A 0.32

D place circuit #3 into frame. A 0.45

E Attach circuits to frame. B,C,D 0.51

F Solder circuit connections to central circuit control. E 0.55

G Place circuit assembly in calculator inner frame. F 0.38

H Attach circuit assembly to calculator inner frame. G 0.42

I Place and attach display to inner frame. H 0.30

J Place an attach keyboard to inner frame. I 0.18

K Place and atach top body of calculator to inner frame. J 0.36

L Place and attach power assembly to inner frame. J 0.42

M Place and attach bottom assembly to inner frame. K,L 0.48

N Test circuit integrity. M 0.30

O Place calculator and printed matter in box. N 0.39

Precedence Diagram

B K

/ \ / \

A – C – E – F – G – H – I – J M – N – O

\ / \ /

D L

15/04/2024

Example

Compute the target cycle time per unit:

A=.18

Takt Time= productive time per hour/demand per hour B=.12

C=.32

= 54/540 = 0.10 minute

D=.45

Cycle Time = TT= 0.10 minute E=.51

F=.55

G=.38

Compute the minimum number of WS: H=.42

I=.30

minNumWS= Sum of task times / CT J=.18

K=.36

= 5.36 / 0.10

L=.42

= 53.60 WS M=.48

N=.30

:= 54 WS

O=.39

A=.18

B=.12

C=.32

D=.45

E=.51

F=.55

G=.38

H=.42

I=.30

J=.18

B K K=.36

/ \ / \ L=.42

A – C – E – F – G – H – I – J M – N – O M=.48

N=.30

\ / \ /

O=.39

D L

CT = 0.1; Utilisation = sum of tasks assigned / (CT * NoOfWS);

15/04/2024

B K

/ \ / \ A=.18

A – C – E – F – G – H – I – J M – N – O B=.12

\ / \ / C=.32

D L D=.45

E=.51

F=.55

G=.38

H=.42

I=.30

J=.18

K=.36

L=.42

M=.48

N=.30

O=.39

CT = 0.1; Utilisation = sum of tasks assigned / (CT * NoOfWS);

Summarize the assignment of tasks to

Tasks in work A,B C,D,E F,G,H,J J,K,L,M,O

centers

Work centers 1 2 3 4

Actual no of WS 3 13 17 22

Total number of WS = 3+13+17+22 = 55.

Average Utilisation = minNumOfWS / actualNumOfWS

= 53.6/55

= 97.5%

15/04/2024

Computerized Line Balancing

• Line balancing by hand becomes unwieldy as the problems

grow in size.

• There are software packages that will balance large lines

quickly. IBM'sCOMSOAL(Computer Method for Sequencing

Operations for Assembly Lines) and GE's ASYBL (Assembly

Line Configuration Program) can assign hundreds of work

elements to workstations on an assembly line.

• Random trial and search for better solutions.

• Some software lets the user select specific heuristic, e.g.

ranked positional weight (RPW), longest operation time (LTT),

shortest operation time (STT), most number of following

tasks, and least number of following tasks.

COMSOAL: Computer Method

 Operation sequences are generated by randomly picking a task and

constructing subsequent tasks

 New stations are opened when needed

 New solution that exceeds the best solution so far is discarded

 Better solutions become upper bounds

• NIP(i) = Set of Immediate Predecessors for each task i;

• WIP(i) = Set of tasks for which i is an immediate predecessor

• TK = Set of N tasks

• A = Set of unassigned tasks

• B = Set of tasks from A with all immediate predecessors assigned

• F = Set of tasks from B with tasks times not exceeding remaining cycle

time in the current workstation

15/04/2024

COMSOAL: Algorithm

For generating X trial solutions to minimise the total idle time:

1. Initialisation of algorithm

1.-SET x=0, UB=inf, c=CT 2. Initialisation at each iteration

2.-START NEW SOLUTION: SET x=x+1, A=TK, NIPW(i) = NIP(i)

3.-PRECEDENCE FEASIBILITY: FOR i IN A, IF NIPW(i) = 0 , ADD i TO B

4.-TIME FEASIBILITY: 3. Update available tasks in B

FOR i IN B, IF ti< c ADD i TO F.

4. Update feasible tasks in F

If F empty , go to 5 , otherwise go to 6

5.-OPEN NEW STATION:

IDLE=IDLE + c , c = CT.

5. Discard a bad solution

If IDLE > UB , go to 2, otherwise go to 3

A = Set of unassigned tasks;

B = Set of available tasks from A;

F = Set of feasible tasks from B; c = the remaining capacity of a WS

COMSOAL: Algorithm

6.-SELECT TASK: SET m = card{F} = size of set F;

6. Randomly select tasks in F

RANDOM GENERATE RN in U(0,1);

LET i* = [m*RN]thTASK from F;

6. Task i* has been assigned,

REMOVE i* from A, B, F;

update A, B, F.

c = c –ti;

FOR ALL j in WIP(i*), NIPW(j) = NIPW(j) –{i*};

IF A EMPTY --> 7, OTHERWISE --> 3

6. Update preceding tasks

7.-SCHEDULE COMPLETION

IDLE = IDLE + c

IF IDLE < UB , UB = IDLE --> STORE SCHEDULE

IF x = X , STOP, OTHERWISE --> 2 7. Output the solution of

the assigned tasks

A = Set of unassigned tasks;

B = Set of available tasks from A;

F = Set of feasible tasks from B; c = the remaining capacity of a WS

15/04/2024

Mathematical Methods (IP)

Notation:

• t = task time;

• NIP = set of Immediate Predecessors for task j;

Decisions:

• F n= 1 if WS n is used/opened; 0 otherwise. • Number of WSs;

• X = 1 if task j is assigned to WS n; 0 otherwise. • Assignment of tasks;

Objective: MIN F minimum number of WS to be used.

1. Each task has to be assigned to one and

Constraints:

only one WS;

1.  X = 1, for any j; 2. Tasks on a WS should not exceed CT;

n jn

2.  X * t <= CT*F , for any n; 3. Lower number WSs fill up first;

j jn j n

4. Precedence relationship constraints.

3. F >= F , for any n;

n n+1

4. X <= (X + X + … + X ),for any j, n; and any k in NIP;

jn k1 k2 kn j

More Approaches To Improve Line

Balance

• Dividing work elements

• Changing speed/feed at stations

• Activity/process analysis

– Better workplace layout, redesign tooling & fixturing

• Pre-assembly of components

• Merge tasks/workers

• Parallel stations

15/04/2024

Example: Line Balancing

In the production line’s current form (assuming each task is performed on

one WS by one worker),

• What is the bottleneck activity?

• What is the production capacity per hour of the system?

• Redesign the product flow diagram to balance the line and increase the

capacityper hour of the system.

Activity Average timein second

1 20

2 30

3 10

4 10

5 15

6 17

1 2 3 4 5 6

Example: Line Balancing

1 2 3 4 5 6

(CT, Cap) = (20, 180) (30, 120) (10, 360) (10, 360) (15, 240) (17, 212)

The bottleneck is: WS2

The current capacity is: 3600/30=120

Total task time = stacked time = 20+30+10+10+15+17 = 102 secs.

We aim for 102/6 = 17 secs.

How can we achieve the cycle time approximately at 17 seconds?

15/04/2024

Example: Line Balancing

1 2 3 4 5 6

(CT, Cap) = (20, 180) (30, 120) (10, 360) (10, 360) (15, 240) (17, 212)

One person one hour

(20, 180)

1 2 5 6

(20, 180)

(CT, Cap): (20, 180) (30, 120) (15, 240) (17, 212)

Example: Line Balancing

(20, 180)

1 2 5 6

(CT, Cap): (20, 180) (30, 120) (15, 240) (17, 212)

(20, 180)

Two persons

one hour

2 3

(30, 120) (20, 180)

1 5 6

2 4

(30, 120) (20, 180)

(20, 180) (15, 240) (20, 180) (15, 240) (17, 212)

15/04/2024

Learning Outcomes

• Longest-Task-time Heuristic

• Kilbridgeand WesterMethod

• Ranked Positional Weights Method

• Compare LTT, STT, KWM, and RPW

• Incremental Utilisation Heuristic

• Other Ways To Improve Line Balance

• Product leveling

• Mixed-model production leveling

• Weighted Average Time method/rule

• Remarks on production levelling

Product Leveling

• If we produce more than one product in production

line, what does production Levelling/smoothing

mean?

• Large batches of the same product may reduce set-

up times and changeovers, but usually result in:

– long lead times;

– swelling inventories;

– greater opportunities for defects;

– excessive idle time and/or overtime.

• Product levelling

15/04/2024

Mixed-Model Production

• Mixed-model production

– Averaging both the volumeand the production sequence

of different model types on a production line.

• Example: Toyota Manufacturing

Toyota makes 3 car models -a convertible, hardtop, and a SUV.

Assume that customers are buying 9 convertibles, 9 hardtops,

and 9 SUVs each day. What is the most-efficient way to make

those cars?

Mixed Model Production Leveling

One solutionwould be for Toyota to make all 9 convertibles

in the morning, all 9 hardtops in the afternoon, and all 9 SUVs

in the evening. That would allow people to concentrate on

one kind of work at a time.

Uneveness of equipment and people that make different cars over

morning, afternoon, and evening.

15/04/2024

Mixed Model Production Leveling

In the staging lot, vehicles would pile up between the plant and

the dealers.

Customersdon't buy 9 convertibles in the morning, 9 hardtops

in the afternoon, and 9 SUVs in the evening. They buy different

kinds of cars through the day and week.

Parts Factory

Car Factory Dealer

Mixed Model Production Leveling

Toyota solved the problem by production levelling.

• If customers are buying 9 convertibles, 9 hardtops, and 9 SUVs each day,

• Toyota assembles 3 of each in the morning, 3 of each in the afternoon,

and 3 of each in the evening  3A,3B,3C; 3A,3B,3C; 3A,3B,3C;

• It also distributes the production of convertibles, hard tops, and SUVs as

evenly as possible through each shift: convertible, hard top, SUV,

convertible, hard top, SUV, and so on. ABC,ABC,ABC; ABC,ABC,ABC;

ABC,ABC,ABC;

15/04/2024

Mixed Model Production Leveling

Leveling production also helps to avoid the problem of excess inventory

of finished vehicles. The vehicle plants make the different types of cars

at about the same pace that customers buy those cars. They can adjust

the pace of production as buying patterns change.

As the result, dealersonly need to maintain a minimal inventory of cars

to show and sell.

Parts Factory Car Factory Dealer

Mixed Model Production: Example

The monthly customer demands are 4800, 2400, 1200 units for three

products A, B, C, respectively (over 20 days). We want to meet customer

demands and arrange production to be as mixed as possible.

Solution steps:

1. Determine daily production requirements

2. Determine repeating sequence through;

i. Determining the largestINTEGERdivides into the daily volume

ii. Calculate the minimum ratio, e.g. 240, 120, 60 => 60 is the integer,

and so ratios are 4, 2, 1 (60 repeats daily)

iii. Define a product ordering within the repeat sequence

Monthly Days Daily

Prod. A 4800 20 240

Prod. B 2400 20 120

Prod. C 1200 20 60

All 8400 20 420

15/04/2024

Mixed Model Production: Example

Determine sequence: 240/120/60 → 4/2/1 → AAAABBC → AABABAC

Product ratios: q =240/420=0.571; q =120/420=0.286; q =60/420=0.143;

A B C

Assume 100% efficiency and the available production time per day is 7

hours. Each product requires four tasks with task times in the table.

TaktTime = ? (minutes) for each product and for overall.

For A, 420/240 =1.75; For B, 420/120=3.5;

For C, 420/60=7.0; For all, 420/420 = 1.

Prod. A Prod. B Prod. C

Task 1 0.7 0.6 0.5

Task 2 0.6 0.5 0.6

Task 3 0.5 0.4 0.4

Task 4 0.6 0.6 0.7

Total 2.4 2.1 2.2

Daily Production 240 120 60 total: 420

Mixed Model Production: Example

• How to balance: By assigning tasks to a sequence of workstations (WS)

such that:

– The cycle time (CT) for each product at each WS satisfies the Takt

Time (TT) or Required CT

– Tasks are assigned as efficient as possible

• Weighted Average Time (WAT) method/rule:

– WAT is the time required on average at a WS to perform the task

– RULE: WAT at each WS should not exceed the TT

WAT = ∑∑ q x T ≤ TT; i.e. Sum of the product_ratios * task_time < Takt

j i j ij

Where q = proportion of product j in the sequence (q =D/∑ D)

j j j j

T = time for performing task ion product j

For all the tasks that are assigned to each WS.

15/04/2024

Mixed Model Production: Example

Prod. A B C

Task 1 0.7 0.6 0.5

Task 2 0.6 0.5 0.6

Task 3 0.5 0.4 0.4

Task 4 0.6 0.6 0.7

Total 2.4 2.1 2.2

Daily Production 240 120 60 total: 420

Product ratios: q =0.571; q =0.286; q =0.143;

A B C

For all products, TT= 420/420 = 1min.

WS Task Weighted Average Time (WAT) in minutes

1 1 0.571(0.7) + 0.286(0.6) + 0.143(0.5) = 0.643

2 2,3 0.571(0.6+0.5) + 0.286(0.5+0.4) + 0.143(0.6+0.4) = 1.029

3 4 0.571(0.6) + 0.286(0.6) + 0.143(0.7) = 0.614

Mixed Model Production: Example

Workstation WS1 WS2 WS3

Task 1 2+3 4

A 0.7 1.1 0.6

B 0.6 0.9 0.6

A 0.7 1.1 0.6

B 0.6 0.9 0.6

A 0.7 1.1 0.6

C 0.5 1.0 0.7

Sum 4.5 7.2 4.3

Average 0.643 1.029 0.614

TT=1min; a sequence = seven products; Every 7 minutes, we produce 4A, 2B and 1C.

This will repeat 60 times per day (assume 7 hours per day).

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Mixed Model Production: Example

WS1 WS2 WS3

Task 1 2 3 4

A 0.7 0.6 0.5 0.6

B 0.6 0.5 0.4 0.6

A 0.7 0.6 0.5 0.6

B 0.6 0.5 0.4 0.6

A 0.7 0.6 0.5 0.6

C 0.5 0.6 0.4 0.7

Sum 4.5 6.8 4.7

Average 0.64 0.97 0.67

Some Remarks

• Production Levelling (Heijunka) points to a knowledge-driven

processfor ensuring stability, flow and pull

• Most lean operations strike a balance between product

levelling and volume levelling

• It also needs to strike a balance between having inventory

and meeting demand

• PL requires a lot of data, and can be tough to deal with

• PL is not necessarily much useful for businesses with level and

dependable demand.

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Production Levelling in Logistics

• Milk Run Distribution versus Production Levelling

• Milk Run is a delivery method for mixed loads from different

suppliers planned as cyclically repeated tours.

• Instead of each of several suppliers (e.g. 5) sending a vehicle

every week to meet the weekly needs of a customer, one

vehicle visits each supplier on daily basis and picks up

deliveries for that customer.

• Get its name from dairy industry where one tanker collects

milk every day from several farmers to a milk processing firm

Meyer, A (2015). Milk Run Design: Definitions, Concepts and Solution Approaches, Scientific Publishing.

Individual vs Milk Run Delivery

www.nipponexpress.com/about/csr/environment/cooperation.html

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Comments on Milk Runs

• In Japan, supplier MR are widely used for connecting regular

suppliers with the corresponding plants;

• In Germany (and Europe), MR for supplier-plant-relations are

rarely taken.

• MR allow to ship parts more frequently and in smaller lots;

• Regular schedules reduce the variability of the processes and

increase the planning ability in the inbound network

• Suppliers: reduce their safety stocks;

• Logistics providers: improve reliability and reduce empty

return trips;

• Society: reduce emissions and congestion

Meyer, A (2015). Milk Run Design: Definitions, Concepts and Solution Approaches, Scientific Publishing.

Discussion: Benefits and Implications

Milk run in logistics is a process for inbound deliveries to warehouses. These deliveries can involve internal or

external supply chains.

Example 1. There are many different growers in one region supplying wheat to a mill that makes flour. Each

farm could send a delivery truck with a load of grain to the mill each week (or day). That’s a quite standard SC

operation. In milk run logistics, a truck would go from the mill to the farms. At each stop on the route, the truck

picks up wheat from a different grower. If each farmer only has a partial load to deliver, this is a more efficient

way to collect the grain.

Example 2. In a heavy equipment manufacturing facility, several sections of the facility fabricate different parts

of the machinery. Another team does the assembly. Someone from each fabrication department could deliver

finished pieces to the assembly section. In milk run, a driver is sent to the different sections to load and

transport all the parts to the assembly area. In this situation, one driver takes the place of many, and each

department’s workforce can focus on fabrication.

A milk run can be a more efficient way to handle logistics. But they do take planning. If the route involves

products from different companies, you need an agreement about cost-sharing and other aspects of the

cooperative delivery arrangement. Once the group settles these issues, this delivery method can save time and

money for everyone by pooling operation costs and resources.

Receiving operations also benefit from milk runs. In the example of the mill, instead of processing a dozen

deliveries from different farms, the mill can get a full wheat delivery from just one truck. In the example of

internal facility, the assembly operations only have one delivery to keep track of rather than running down

parts deliveries from multiple fabrication departments.

efficiency; planning; coordination vertically and horizontally; inventory; time; cost;

utilisation; tracking delivery; competitors.

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