Procedure by Groff

The method of Groff is a heuristic method of dynamic determination batches . In this way, lot sizes are determined for production , which cover the requirements and at the same time minimize the sum of setup and storage costs. The procedure was published in 1979.

The method makes use of the fact that in the minimum of the cost function, the marginal setup costs are equal to the marginal storage costs.

Prerequisite and question

To calculate an order or production plan that minimizes costs as much as possible, the setup costs K [euros / lot], the storage costs h [euros / piece / period], and the requirements for the next periods (days, weeks) must be known.

Conflict of goals

Cost history for batch production

You made with fewer, larger lots, the reduced set-up costs , the increase and there is storage costs . Conversely, more lots lead to lower storage costs but higher set-up costs.

Heuristic

Groff's heuristic increases the lot until the following condition is no longer met.

${\ displaystyle B_ {t + n} \ cdot n \ cdot (n + 1) \ leqslant {\ frac {2 \ cdot k} {h}}}$

{\ displaystyle B_ {t + n}:}

Requirement for period t + n

{\ displaystyle t:}

Period in which the order is carried out

{\ displaystyle n + 1:}

Number of periods in the lot

{\ displaystyle k:}

Setup costs for one lot [euros / lot]

{\ displaystyle h:}

Storage costs in euros per piece and period [euros / piece / period]

Calculation example

period	1	2	3	4th	5
requirement	20th	30th	25th	40	15th

Order costs = 200
storage costs = 4

Calculate the upper limit:

${\ displaystyle {\ frac {2 \ cdot 200} {4}} = 100}$

Order 1

Check whether you want to order in the first period for the second period:

${\ displaystyle 30 \ cdot 1 \ cdot (1 + 1) = 60 \ leqslant 100}$ Conditions met!

Check whether you should also order for the third period in the first period:

${\ displaystyle 25 \ cdot 2 \ cdot (2 + 1) = 150 \ geqslant 100}$ Condition not met! Cancellation.

A lot size 50 (= 20 + 30) is ordered in period one. This lot covers the needs for the first and second periods. In the third period, orders will be placed again. The lot size of the order in period three is calculated below.

Order 2

Check whether you want to order for the fourth period in the third period:

${\ displaystyle 40 \ cdot 1 \ cdot (1 + 1) = 80 \ leqslant 100}$ Conditions met!

Check whether the third period should also be ordered for the fifth period:

${\ displaystyle 15 \ cdot 2 \ cdot (2 + 1) = 90 \ leqslant 100}$ Conditions met!

A lot size 80 (= 25 + 40 + 15) will be ordered in period three. This lot covers the needs for periods three, four and five.

literature

Gene K. Groff: A Lot-Sizing Rule for Time-Phased Component Demand. In: Production and Inventory Management. Vol. 20, No. 1, 1979, ISSN 0897-8336 , pp. 47-53.