Three groups of Lot-sizing procedures are available:

- Static lot-sizing procedures
- Period lot-sizing procedures
- Optimum lot-sizing procedures

I will explain here about one of the Optimum lot-sizing procedures: Least unit cost procedure (WI).

**Feature**

Starting from the shortage date, successive requirements are grouped together to form lots until total costs per unit reach a minumum level. The total cost are equal to the sum of the lot size independent costs plus the total storage costs.

**Prerequisite**

– Lot size field (MRP 1 view) is set with WI for Least unit cost procedure.

– Lot size independent costs (MRP 1 view) and storage costs indicator (MRP 1 view) is set.

Ordering costs is 100 in this sample.

Storage costs percentage for storage cost indicator 1 of plant 1000 is 10% in this sample. It is maintained in customizing OMI4.

– Price is set. It depends on price control. If price control is V, the price is “Moving price”; if price control is S, the price is “Standard price”.

Price is 20 in this sample.

– Requirements exist. In this sample, there are 4 planned independent requirements (PIRs). They are with date 2015/2/6, 2015/2/13, 2015/2/20 and 2015/2/27.

**Calculation Logic**

**Calculation result of this sample.**

Requirement Date |
Requirement qty |
Lot Size |
Lot Size Ind. Cost | Storage Cost |
Total Cost |
Unit Cost |
---|---|---|---|---|---|---|

2015/2/6 | 1000 | 1000 | 100 | 0 | 100 | 0.100 |

2015/2/13 | 1000 | 2000 | 38.36 | 138.36 | 0.069 | |

2015/2/20 | 1000 | 3000 | 76.71 | 215.07 | 0.072 | |

2015/2/27 | 1000 | 4000 | 115.07 | 330.14 |
0.083 |

Since the Unit cost for Lot Size 2000 is minimum, requirements are grouped together to form lots with qty 2000.

**Formula**

– Lot size = current Lot Size + new requirement qty.

– Storage Cost = ( Requirement * Price * Storage Cost Percentage * Time in shortage ) / ( 100 * 365)

– Total Cost = current Total Cost + Newly calculated Storage Cost

– Unit Cost = Total Cost / Lot Size

**Explaination**

On 2015/2/6, shortage occurs.

Time in shortage is 0 days.

Storage cost = 0.

Total cost = storage cost 0 + Lot size ind. cost 100 = 100

Unit cost = Total cost 100 / Lot size 1000 = 0.100

On 2015/2/13

Time in shortage is 7 days.

Storage cost = (requirement qty 1000 * price 20 * percentage 10 * Time in shortage 7 ) / (100 * 365) = 38.36

Total cost = storage cost 38.36 + current total cost 100 = 138.36

Unit cost = Total cost 138.36 / Lot size 2000 = 0.069

On 2015/2/20

Time in shortage is 14 days.

Storage cost = (requirement qty 1000 * price 20 * percentage 10 * Time in shortage 14 ) / (100 * 365) = 76.71

Total cost = storage cost 76.71 + current total cost 138.36 = 215.07

Unit cost = Total cost 215.07 / Lot size 3000 = 0.072

On 2015/2/27

Time in shortage is 21 days.

Storage cost = (requirement qty 100 * price 20 * percentage 10 * Time in shortage 21 ) / (100 * 365) = 115.07

Total cost = strorage cost 115.07 + current total cost 215.07 = 330.14

Unit cost = Total cost 330.14 / Lot size 4000 = 0.083

**Source Code**

See attached text file.

Thanks Jessica,

Good document.

Kind Regards,

Mariano

Hi Jessica,

Thanks for sharing such a good document

and one more clarity i want to know on which type of business scenarios we will use the optimum lot size ?

Regards,

Venkatesh

Hi Venkatesh,

Sorry for the late reply.

Static and period lot-sizing procedure does not consider the cost caused from storage, etc. For the business scenarios in which minimized cost are important, consultants can consider using optimize lot size.

best regards,

Jessica

Hi Jessica,

thank you so much for this outstanding document. It will be a great help for me.

Good Job.

Thanks.

Regards

Christian

Hi Jessica Li

Do you have any experience with Least unit cost procedure in APO PP/DS?

Thanks for your feedback and regards

THANKS A LOT Jessica Li