D08540000120114006Session 9_Production Planning and Control

7/28/2019 D08540000120114006Session 9_Production Planning and Control

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Session 11Production Planning and ControlLot sizing, Shop floor Scheduling

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D 0 8 5 4Supply Chain : Manufacturing and Warehousing


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Bina Nusantara University

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Lot Sizing, Shop Floor Schedule

Lot Sizing & Lean Manufacturing Strategy

Why Small Lots?

Small lot production (ideally one piece) is animportant component of many Lean Manufacturing

strategies. Lot size directly affects inventory and

scheduling. Other effects are less obvious but equally

important. Small lots reduce variability in the system

and smooth production. Small lots also enhance

quality in many ways.

Bina Nusantara University

3


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The Lean Manufacturing literature gives little

guidance on lot sizing other than statements such as:

"the lot size should be one" or "lot sizing is irrelevant."

This series of papers examines the lot sizing problem in

Lean Manufacturing. It offers a rational alternative to

the slogans and edicts.

The effects of small lots differ somewhat between Make

To Order (MTO) and Make To Stock (MTS)

environments but they are important in either situation.


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In MTO environments, the ability to make smaller lots economicallymakes it practical to accept smaller orders. This can open newmarket segments or eliminate middlemen from the logisticschain.

One of our former clients restructured the entire vinyl siding

industry when they achieved reliable delivery of small lots directlyto retail outlets.

In an MTS environment, small lots translate directly to smallerinventories. Inventory carrying costs are significant and arediscussed further below. In fast-changing fashion or technologymarkets, obsolete inventory may make the difference between profitand loss. Smaller lots often enable conversion from MTS to MTO.

Many factories that deliver to their customers in MTO operateintermediate processes in MTS. The MTS discussion applies tothose intermediate and upstream operations.


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Scheduling via Lot Sizing

Case: TJ International Parallam Product

Issues:Press is bottleneck

12 hour changeover time to switch widths (12, 14, 16, 19)

Many products made from each width, but less 14/16 than16/19

Currently try to run at least a week between changeovers

Seasonal demand: inventory build up in off-season

Problem: determine run sequence each month

Lathe/Clip Press Saw


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Scheduling via Lot Sizing

Notation:

Problem: choose lot sizes for the month to meetdemand as efficiently as possible.

variable)(decisionproductfor(parts)sizelot

productfor(hr)timesetup

productof(parts/hr)rateproduction

productfor(parts/hr)demandhourly

productforth)(parts/mondemandmonthly

nutilizatiodesired

hours/daydays/monthavailablehours

ix

is

ip

id

iD

u

H

i

i

i

i

i


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Cyclic Schedule

Approach: fix lot sizes for all products and computenumber of cycles that can be run in the month as:

If the horizon is short, Nshould be rounded downto the nearest integer. Otherwise, a non-integerNis ok since you will be rescheduling anyway.

Lot Sizes: The lot sizes will be:

These, of course, need to be rounded or ceilinged.

Problem:You dont necessarily want to make thesame number of runs of a low demand product as of

a high demand product.

i i

i ii

s

pDuHN

N

Dx ii

spare capacity

setup/cycle


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Non-Cyclic Schedules

Fundamental Issue: What is your objective?

Minimize Sum of Run Lengths? If we write out theLagrangian for the nonlinear optimization problem andoptimize assuming continuous lot sizes, we get

where

Problems: Since we solved for lot sizes, we wind up with a non-integer number

of setups for the month. How do we allocate the setup time?

What order do we run the products in the non-cyclic schedule?

Conclusion: We should solve for the number of run cycles for eachproduct. But what objective do we use?

iiii psdx

2

i ii

i iii

pdu

psd


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MiniMax Schedule

Objective: minimize the maximum run length.

We are never too far away from changing over to

another product.

Perhaps a good surrogate for maximizing flexibility.

Notation: Let

producedisproducteach timelengthrun

inrunisproducttimesofnumberthe

ipn

Ds

Hin

ii

ii

i

decision variable


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MiniMax Schedule (cont.)

IP Formulation:

integer

allfor1

constraintminimax,allfor)(

constraintcapacity

:

min

i

i

iiii

i iii ii

n

in

iTpnDs

pDuHsn

st

T


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Example

Data:H= 22 days 18 hrs/day = 396 hrs

u = 0.9

Minimal Run Schedule: Suppose the plant has a policy of never making less than 1500 so

their current lot sizes are 15000, 12000, 1500, 1500. This takes 19.2days, close to the 22 days available, but it means that some productwont be produced until pretty late in the month.

i 1 2 3 4

Di 15000 12000 500 250

pi 100 100 75 50

si 8 8 6 4

Di/ i 150 120 6.7 5Total Prod Time

Req = 281.67 hrs


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Example (cont.)

Cyclic Schedule: The number of cycles per month isN= (396 0.9 - 281.67)/26 = 2.87

yielding lot sizes of

But N= 2.87 means that the first two cycles in the

month use the above lot sizes while the last one uses

reduced lot sizes (87% of base).

i 1 2 3 4xi 5219 4175 174 87


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Example (cont.)

MiniMax Schedule:

We can implement the above formulation as follows:

1. Start with ni= 1 for all i. Note that the product with max run length is

product 1 with a run of 158 hours.

2. Add setup for product 1, so n1 = 2, and its max run length becomes83 and now product 2 has the longest run length at 128 hours.

3. Add a setup for product 2, so n2 = 2, and its max run length becomes

68 and now product 1 has longest run length at 83 hours.

4. Continue adding setups to product with longest run until we run out of

capacity. The optimal solution is:

i 1 2 3 4

ni 4 4 1 1

xi 3750 3000 500 250


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Example (cont.)

Results: The schedule goes as follows:

1. We set up for the 4th product and get it out of theway.

2. Likewise for the 3rd.

3. Now we run 3750 of product 1 and 3000 of product 2in succession for 4 times.

The longest anyone would ever have to wait wouldbe for product 2 at the beginning of the month at 3.73

days. After that the longest wait would be 2.53 days.

Conclusion: Minimax schedule results in shorter wait forproduct than either minimal run schedule or cyclicschedule. And it is simple to compute.


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TJ International Case

Non-Cyclic Scheduling?

Fewer runs of 12, 14

But long setups make running anything twice in the

month difficult.Skip 12, 14 some months?

How to Handle Seasonality?

Could run high volume products in off-season (sure to

sell)But running 12, 14 products would help skip setups

and increase capacity in peak season

Need to choose fractile of demand to stock in off-

season (75%?)


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Scheduling in Pull Environments

Simple CONWIP Lines:

Simple sequence sufficient.

No setups EDD sequence.

CONWIP Lines with Shared Resources:

No setups EDD within lines.

Need to augment sequence at shared lines.

FISFO at shared resources. Routing A

Routing B

Routing A

Routing B

FISFO

sequencing


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Scheduling in Pull Environments (cont.)

CONWIP Lines with Setups:

Sequence still reasonable.

Must balance capacity loss with due date

requirements.Exact solution impossible.

Heuristics starting with EDD and rearranging

sequence can work well.

More General Environments:

Many routings make simple sequence insufficient.

Backward scheduling due dates may be infeasible.

Forward scheduling may be suboptimal.


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A Sequencing Example

Problem Description:

16 jobs

Each job takes 1 hour on single machine (bottleneck

resource)4 hour setup to change families

Fixed due dates

Find feasible solution if possible


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A Sequencing Example (cont.) EDD Sequence:

Average Tardiness: 10.375

Job Due CompletionNumber Family Date Time Lateness

1 1 5.00 5.00 0.002 1 6.00 6.00 0.003 1 10.00 7.00 -3.004 2 13.00 12.00 -1.00

5 1 15.00 17.00 2.006 2 15.00 22.00 7.007 1 22.00 27.00 5.008 2 22.00 32.00 10.009 1 23.00 37.00 14.00

10 3 29.00 42.00 13.0011 2 30.00 47.00 17.0012 2 31.00 48.00 17.0013 3 32.00 53.00 21.0014 3 32.00 54.00 22.0015 3 33.00 55.00 22.0016 3 40.00 56.00 16.00


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Sequencing Example (cont.)

Greedy Approach:

Consider all pairwise interchanges

Choose one that reduces average tardiness by

maximum amountContinue until no further improvement is possible


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Sequencing Example (cont.) First Interchange: Exchange jobs 4 and 5.

Average Tardiness: 5.0 (reduction of 5.375!)

Job Due Completion

Number Family Date Time Lateness1 1 5.00 5.00 0.002 1 6.00 6.00 0.003 1 10.00 7.00 -3.00

5 1 15.00 8.00 -7.004 2 13.00 13.00 0.006 2 15.00 14.00 -1.007 1 22.00 19.00 -3.008 2 22.00 24.00 2.009 1 23.00 29.00 6.00

10 3 29.00 34.00 5.0011 2 30.00 39.00 9.0012 2 31.00 40.00 9.0013 3 32.00 45.00 13.0014 3 32.00 46.00 14.0015 3 33.00 47.00 14.0016 3 40.00 48.00 8.00


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Sequencing Example (cont.) Configuration After Greedy Search:

Average Tardiness: 0.5 (9.875 lower than EDD)

Job Due Completion


5 1 15.00 8.00 -7.004 2 13.00 13.00 0.006 2 15.00 14.00 -1.008 2 22.00 15.00 -7.007 1 22.00 20.00 -2.009 1 23.00 21.00 -2.00

11 2 30.00 26.00 -4.0012 2 31.00 27.00 -4.0010 3 29.00 32.00 3.0013 3 32.00 33.00 1.0014 3 32.00 34.00 2.0015 3 33.00 35.00 2.0016 3 40.00 36.00 -4.00


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Sequencing Example (cont.) A Better (Feasible) Sequence:

Average Tardiness: 0

Job Due Completion


5 1 15.00 8.00 -7.004 2 13.00 13.00 0.006 2 15.00 14.00 -1.008 2 22.00 15.00 -7.00

11 2 30.00 16.00 -14.0012 2 31.00 17.00 -14.007 1 22.00 22.00 0.009 1 23.00 23.00 0.00

13 3 32.00 28.00 -4.0010 3 29.00 29.00 0.0016 3 40.00 30.00 -10.0014 3 32.00 31.00 -1.0015 3 33.00 32.00 -1.00


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Diagnostic Scheduling

Goals:

Schedule at appropriate level of detail for environment.

Make use of realistic, obtainable data.

Accommodate intangibles in decision-support mode.

Approach:Deterministic model.

Based on conveyor model.

Release as late as possible subject to due dates.

Provide diagnostics on infeasibilities.

Types of Infeasibility:WIP (must move out due dates).

Capacity (can move due dates or increase capacity).


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Scheduling Infeasibilities Example

Problem Description:

Capacity = 100/day

Minimum Practical Lead Time = 3 days

WIP in system with 1, 2, 3 days to go = 95, 90, 115


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Scheduling Infeasibilities Example (cont.)

* WIP infeasibility

** capacity infeasibility

Days from Amount Cumulative CumulativeStart Due Amount Due Capacity

1 90 90 952 100 190 185*3 90 280 2854 80 360 385

5 70 430 4856 130 560 5857 120 680 6858 110 790 785**9 110 900 885

10 110 1010 98511 100 1110 108512 90 1200 1185

13 90 1290 128514 90 1380 138515 90 1470 1485


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Cumulative Demand vs. Cumulative Capacity

0200

400

600

8001000

1200

1400

1600

0 2 4 6 8 10 12 14

Day from Start

Uints Cum Due

Cum Capacity


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Surplus

-30

-20

-10

0

10

2030

40

50

60

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Days from Start

Surplus

Units

WI P

Infeasibility

Capacity

Infeasibility

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D08540000120114006Session 9_Production Planning and Control

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