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Material Requirements Planning



The material requirements planning model is used to determine production requirements for items which are dependent.

The Data



Consider the following example.

The numbers on the left of each item indicate the number of subcomponents which must be used in the parent component. Lead times are 1 week except for item A which has a 2 week lead time.

The framework for MRP is given by the number of lines in the indented bill of materials and the number of time periods. In our screen below representing the example above we show a problem with 7 BOM lines and 8 periods as seen at the top of the screen.

Item names. The item names are entered in this column. The same name will appear in more than one row if the item is used by two parent items such as item e. Note that as a rule, names are unimportant but in MRP names are extremely important. Case (upper/lower) does not matter but spaces matter very much.

Item level. The level in the indented BOM must be given here. The item can not be placed at a level more than one below the item immediately above.

Lead time. The lead time in order to get the item is entered here. The default is one week. This is also the default for the end item.

Number(#) per parent. The number of units of this sub-assembly needed for its parent is entered here. The default is one.

On-hand. The current inventory on hand at the beginning of the problem is listed here. If a sub-assembly is listed twice then it makes sense for the current inventory to appear only one time. However if it appears twice then the starting inventory will be the sum of the listed amounts (see example 2).

Lot size. The lot size can be specified here. A 0 or 1 will perform lot for lot ordering. If another number is placed here then all orders for that item will be in lots which are integer multiples of that number (see example 2).

Demands(entered under pd1 through pd8). The demands are entered for any level 0 item, in the week in which the items are demanded.

Scheduled receipts.. If units are scheduled to be delivered in the future they should be listed in the appropriate time period(column) and item(row) (see example 2).

The Solution Screen



Example 1 - A Simple MRP example

A sample data screen which expresses the problem appears above. The levels indicate that we have an item termed a which has two (level 1) subcomponents named b and c. Subcomponent b has two (level 2) subcomponents named e and f. Subcomponent c has two subcomponents named d and e. Notice that e is a subcomponent of both b and c.

The demand for the end item, a, is 120 units in week 7 and 140 units in week 8. The number of subcomponents used is given in the number column. For example, end item a consists of two subcomponent b which in turn consists of 1 e and 2 fs. At the beginning of the problem, there are no units of any kind of inventory on hand.

Results



A portion of the results is displayed in the screen below.

Total Required. The total number of units required in each week is listed in the first row. For the end item, this is the demand schedule which was input on the data screen. For other items this is computed.

On-hand. The number on hand is listed here. This starts as given on the data screen and is reduced according to needs. A later example will demonstrate on hand inventory.

Scheduled receipt. This is the amount that was scheduled in the original data screen (see example 2)

Net required. The net amount required is the amount needed after the on hand inventory is used. Again, component c illustrates the subtraction (see example 2).

Order release. This is the net required but offset by the lead time.

Printing



We show a printout for the problem for one main reason. The printout of the input is in slightly different form than the screen display. Notice that the software prints an indented bill of materials.

Howard J. Weiss Statistics 802 EMBA
C:\ABPOM\EXAMPLES\EXAMPLE1.MAT 1/27/97 3:42 PM
Module/submodule: Material Requirements Planning
Problem title: Example 1
Indented BOM and Results ----------
Indented Bill of Materials
Item Number per On hand Lot Size
ID Leadtime parent Inventory (if not lot for lot)
a 2 1
b 1 2
e 1 1
f 1 2
c 1 3
d 1 5
e 1 4
Demands for level 0 items
Item Id = a
Period Demand
1 0
2 0
3 0
4 0
5 0
6 0
7 120
8 140
9 0
Scheduled receipts for all items which are not end (level 0) items (if any)
a(low level = 0)
pd1 pd2 pd3 pd4 pd5 pd6 pd7 pd8
TOT.REQ. 0 0 0 0 0 0 120 140
ON HAND 0 0 0 0 0 0 0 0
SchdREC. 0 0 0 0 0 0 0 0
NET REQ 0 0 0 0 0 0 120 140
ORD REL. 0 0 0 0 120 140 0 0

Options



The next example will demonstrate some of the features of our MRP module. We modify our previous example as shown below.

First notice that we have changed the level for product d and the subcomponent e below it. Thus e is at level 2 for A but level 1 for D. D is an end item with a demand of 65 in period 8. Second, we have given e an initial inventory of both 10 and 20. (We really should not have both but we want to demonstrate what happens.) Third, we have a scheduled receipt of 800 units in period 2 for item c. Finally, item f must be bought or made in lots that are multiples of 24 units.

The results can be seen below. (We have edited some of the input data by eliminating some 0's)

Module/submodule: Material Requirements Planning

Problem title: Example 2

Indented BOM and Results ----------

Indented Bill of Materials

Item Number per On hand Lot Size

ID Leadtime parent Inventory (if not lot for lot)

a 2 1

b 1 2

e 1 1 10

f 1 2 24

c 1 3

d 1 5

e 1 4 20

Demands for level 0 items

Item Id = a

Period Demand

7 120

8 140

Item Id = d

Period Demand

8 65

Scheduled receipts for all items which are not end (level 0) items (if any)

Item Id = c

Period Receipt

1 0

2 800



a(low level = 0)

pd1 pd2 pd3 pd4 pd5 pd6 pd7 pd8 pd9

TOT.REQ. 0 0 0 0 0 0 120 140 0

ON HAND 0 0 0 0 0 0 0 0 0

SchdREC. 0 0 0 0 0 0 0 0 0

NET REQ 0 0 0 0 0 0 120 140 0

ORD REL. 0 0 0 0 120 140 0 0 0

d(low level = 0)

pd1 pd2 pd3 pd4 pd5 pd6 pd7 pd8 pd9

TOT.REQ. 0 0 0 0 0 0 0 65 0

ON HAND 0 0 0 0 0 0 0 0 0

SchdREC. 0 0 0 0 0 0 0 0 0

NET REQ 0 0 0 0 0 0 0 65 0

ORD REL. 0 0 0 0 0 0 65 0 0

b(low level = 1)

pd1 pd2 pd3 pd4 pd5 pd6 pd7 pd8 pd9

TOT.REQ. 0 0 0 0 240 280 0 0 0

ON HAND 0 0 0 0 0 0 0 0 0

SchdREC. 0 0 0 0 0 0 0 0 0

NET REQ 0 0 0 0 240 280 0 0 0

ORD REL. 0 0 0 240 280 0 0 0 0

c(low level = 1)

pd1 pd2 pd3 pd4 pd5 pd6 pd7 pd8 pd9

TOT.REQ. 0 0 0 0 360 420 0 0 0

ON HAND 0 0 800 800 800 440 20 20 20

SchdREC. 0 800 0 0 0 0 0 0 0

NET REQ 0 0 0 0 0 0 0 0 0

ORD REL. 0 0 0 0 0 0 0 0 0

e(low level = 2)

pd1 pd2 pd3 pd4 pd5 pd6 pd7 pd8 pd9

TOT.REQ. 0 0 0 240 280 0 260 0 0

ON HAND 30 30 30 30 0 0 0 0 0

SchdREC. 0 0 0 0 0 0 0 0 0

NET REQ 0 0 0 210 280 0 260 0 0

ORD REL. 0 0 210 280 0 260 0 0 0

f(low level = 2)

pd1 pd2 pd3 pd4 pd5 pd6 pd7 pd8 pd9

TOT.REQ. 0 0 0 480 560 0 0 0 0

ON HAND 0 0 0 0 16 0 0 0 0

SchdREC. 0 0 0 0 0 0 0 0 0

NET REQ 0 0 0 480 560 0 0 0 0

ORD REL. 0 0 480 576

Notice the on hand inventory for item e. It begins at 30 and remains until it is needed. Notice that for item C a scheduled delivery arrives in period 2 and then goes into inventory until it is needed. Notice that for item F in period 5, 560 units are needed but the order is placed for 576 units because it must be a multiple of 24.
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