Push Vs. Pull
Part Two – Focus on MRP
IE 3265 POM
R. R. Lindeke, Ph. D.
Components of a Pull System

Master Production Schedule – a breakout of
the Aggregate Plan


Actual MRP System




Uses forecast demands/Firm Customer Orders/Safety
Stock Levels/Internal Orders
Structured Bill of Materials
Production Times
Detailed Job Shop Schedules
Detailed Inventory Records
Defining Explosion Calculus


This is the crux of MRP
MRP converts “Dependent Demands” derived from
the Zero Level of an ordered product into a series of
Job Orders for:




Customer Orders
Internal Orders
Safety Stocks to meet unexpected demands
The MRP explodes each order of “0’s” into the
required series of Children orders needed internally
to build the required products
Some Other Important Terms

Gross Requirements:


Scheduled Receipts:


Total needs of a part for ALL parents in a MPS & safety stock
along with internal needs – like for maintenance parts,
replacement parts, etc
These are open orders placed but not yet received
Planned Receipts:

Planning done over time horizon to keep on hand inventories
from dropping below zero (safety stock level) which are
backed up into orders to arrive when the shortage would
have occurred
Some Other Important Terms, cont.

Planned Order Releases:





A release when an order for a specified quantity (EOQ,
L4L, etc) is done
This should happen in the correct “Time Bucket” – a time
bucket is the time stage used during projection
Order release considers: Setup; Processing time;
Handling time; Delays, curing, waiting times
These apply to each item in the product string
We will consider a Seat Assembly for a Ladder Back
Chair
Looking at a Ladder Back Chair:
Structured Bill of Materials for Ladder
Back Chair
Focusing on the Seat Assembly

We can use several Order Schemes:



EOQ or other fixed order quantity (FOQ) schemes
POQ periodic order quantities which will order a
computed amount needed to cover orders over an
extended (specified) period of time. The orders are
released at fixed time intervals corresponding to the time
buckets
Lot-for-Lot (L4L) – a simple plan that puts out a “time
bucket” order to match lot demand – simple but not
necessarily the most economical
Planning for the Seat Ass’bly


Used in two parent products: Ladder back and Kitchen Chairs
Requirements:
–
–
–
–



Wk 1:
Wk 4:
Wk 6:
Wk 7:
150 LBC
120 KC
150 LBC
120 KC
Lets try it as a L4L plan – zero on hand inventory, no safety
stocks
Lead time for Chair assembly is 2 weeks
Assume we have 37 in inventory from previous plans and
expect a delivery of 230 as we enter this period
L4L Plan (manual MRP!)
1
Gross
Re’ment:
150
Sch.
Receipts
230
Proj.
Inventory
37 +
80 =
117
2
4
5
120
117
Planned
Receipts
Planned
O.
Release
3
117
0
0
3
3
150
120
6
7
150
120
0
0
150
120
8
9
0
0
Using a (3 period) Periodic O.Q.
1
Gross
Re’ment:
150
Sch.
Receipts
230
Proj.
Inventory
37+80
= 117
2
4
5
120
153
153
6
7
8
9
0
0
150 120
117 117 150 150
Planned
Receipts
Planned
O.
Release
3
0
0
120
120
Focusing on the “Periods”

Period 1 (1-2-3):




needs are 150
Receipts/onhands are 230 + 37
Xs is 117 which enters inventory
Period 2 (4-5-6):



needs are (120 + 150) = 270 units
Less onhand (117) means we need 153 units for period
to arrive at start of period
Release order in period two to arrive by period 4 – start
of second period
Final Period:

Period 3 (7-8-9):



Needs are: 120 units
No on-hand inventory is carried forward so must release
an order in period 5 (for 120) to arrive in period 7 when
needed
Next, lets look at a deeper need:
–
The frame parts of the seat Ass’bly which has a 1
week lead time (each seat ass’bly requires 4
frame pieces)
Try it with L4L
1
Gross
Re’ment:
150
Sch.
Receipts
230
Proj.
Inventory
37 +
80 =
117
2
3
5
120
117
117
Planned
Receipts
0
0
3
Planned
O. Rel
(SA)
POR
Frames
4
3
12
150
600
480
120
6
7
150
120
0
0
150
120
8
9
0
0
Effect on POQ method
1
Gross
Req’men:
150
Sch.
Receipts
230
Proj.
Inventory
37+80
= 117
2
4
5
120
153
Planned
O. Rel
(SA)
153
612
7
0
0
120
120
480
6
8
9
0
0
150 120
117 117 150 150
Planned
Receipts
P. O. R.
frames
3
Costing the plans -- Basis is Setup vs.
Holding (as expected!)

Given: $0.40/ass’bly holding; $120 Setup
–
L4L on SA:

–
POQ:


.4*(117*3) + 3*120 = 140.4 + 360 = $500.40
.4*(3*117 + 2*150) + 2*120 = 260.4 + 120 = $500.4
Given: $0.10/ass’bly holding; same setup
–
L4L on SA:

–
35.10 + 360 = $395.10
POQ:

65.10 + 240 = $305.10
Continuing but adding in Frame
Ass’blies


For Frames: K = $100; h = $0.02/piece
L4L:


SA Costs + (1092*.02) + 3*100 = 395.10 + 21.84
+ 300 = $716.94
POQ:

SA Costs + .02(612 + 480) + 2*100 = 305.10 +
21.84 + 200 = $526.94
Using EOQ (previously found to be
Recognizing
230)
shortfalls in P4 and
P7 we release EOQ
orders to fill needs
1
Gross
Req’men:
150
Sch.
Receipts
230
Proj.
Inventory
37+80
= 117
2
4
5
120
230
230
6
7
8
9
150 120
117 117 227 227
Planned
Receipts
Planned
O.
Release
3
77
187 187 187
230
230
Comparing Methods

Compute Average Inventory (period)
–
Here:




L4L is (3*117)/9 = 29
POQ is (3*117 + 2*150)/9 = 72.3
EOQ is (3*117 + 2*227 + 77 +3*187)/9 = 160.3
The choice is based on comparing setup
costs vs. the average inventory costs
Statement of the Lot Sizing Problem –
looking for an optimal


Assume there is a known set of requirements (r1,
r2, . . . rn) over an n period planning horizon. Both
the set up cost, K, and the holding cost, h, are
given.
The objective is to determine production quantities
(y1, y2, . . ., yn) to meet the requirements at
minimum cost. The feasibility condition to assure
there are no stockouts in any period is:
j
j
 y  r
i 1
i
i 1
i
for 1  j  n
Methods

Property of the optimal solution: every optimal solution orders
exact requirements: that is,
y1  r1 or y1  r1  r2 ,. . ., or y1  r1  r2  ...  rn

One method that utilizes this property is the Silver Meal
Heuristic. The method requires computing the average cost for
an order horizon of j periods for j = 1, 2, 3, etc. and stopping at
the first instance when the average cost function increases. The
average cost for a production quantity spanning j periods, C(j),
is given by:
C ( j )  ( K  hr2  2hr3  ...  ( j  1)hrj ) / j
Methods (continued)


Another method that is popular in practice is part
period balancing. Here one chooses the order
horizon to most closely balance the total holding cost
with the set-up cost.
Finally, a third heuristic is known as the least unit
cost heuristic. Here one minimizes the average cost
per unit of demand (as opposed to the average cost
per period as is done in the Silver Meal heuristic.)
The average cost per unit of demand over j periods
is given by:
C ( j )  ( K  hr2  2hr3  ...  ( j  1)hrj ) /(r1  r2  ...  rj )
Methods (concluded)



Experimental evidence seems to favor the
Silver Meal Heuristic as the most cost
efficient among the four discussed in the text.
Optimal lot sizes can be found by using
backwards dynamic programming.
The (best?) heuristic method for lot sizing
subject to capacity constraints is shown next
Lets Try Silver-Meal (assumes K is
$140; holding/unit/period is $0.10)

Over our 9 period time horizon (looking at Seat Ass’bly
Demands)








R: (120, 0, 0, 120, 0, 150, 120, 0, 0)
C(1) = K = 140
C(2) = (K + .1*0)/2 = 70
C(3) = (K + .1*0 + 2*.1*0)/3 = 46.67
C(4) = (K + .1*0 + 2*.1*0 + 3*.1*120)/4 = 44
C(5) = (K + .1*0 + 2*.1*0 + 3*.1*120 + 4*.1*0)/5 = 35.20
C(6) = (K + .1*0 + 2*.1*0 + 3*.1*120 + 4*.1*0 + 5*.1*150) =
41.83
Since C(6) > C(5) Stop. 1st order release is (R1 .. R5) =
240 units but due to lead time we must release it in period
-2 (before the current time horizon)!!!
Continuing

Reset clock to Period 6 as new Start Point:






C(1)6 = K = 140
C(2)7 = (K + .1*120)/2 = 76
C(3)8 = (K + .1*120 + 2*.1*0)/3 = 50.67
C(4)9 = (K + .1*120 + 2*.1*0 + 3*.1*0)/4 = 38
Done with plan – so make a lot (in period 4 due to lead
time) for arrive in period 6 for 270 units.
Notice: each time the cost inverts:

we reset the clock and trigger an order release of the
amount required by the sum of requirements identified
before the cost inverted
One Last Topic of Interest

Capacity Constraints
–
Built from a comparison of:




R the requirements vector
C the Capacity vector
Model is feasible only if: (Ci) >= (Ri)
Lets examine a simple implication:




R: (17, 37, 92, 55, 80): (Ri) = 281
C: (60, 60, 60, 60, 60): (Ci) = 300
This is a feasible plan – BUT –
It is obvious that we could be short in period 3 and 5 without
some smart planning!
Lets Try to Be Smart! (Initial Thought –
lets try L4L)
1
2
3
4
5
Gross
Req’r
17
37
92
55
80
Planned
Rec’pt
17
37
60 
55
60 
Xess
Cap
43
23
-32 
5
-20 
Fixing the Shortfalls – push back
needs and build inventory!

Move P3’s shortfall back in time
–
–

23 to P2 (which fills capacity of P2)
9 additional ones back to P1
Move P5’s shortfall back too
–
–
–
5 to P4 (filling capacity of P4)
the other 15 all the way back to P1
After redistribution: P1 Production requirements are
9 + 15 + 17 = 41
Revised ‘Schedule’
1
2
3
4
5
Gross
Req’r
17
37
92
55
80
Planned
Rec’pt
41
60
60
60
60
Inventory
24
47
15 
20
0
Shortcomings of MRP

Uncertainty. MRP ignores demand uncertainty,
supply uncertainty, and internal uncertainties that arise
in the manufacturing process.

Capacity Planning. Basic MRP does not take
capacity constraints into account.
 Rolling Horizons. MRP is treated as a static system
with a fixed horizon of n periods. The choice of n is
arbitrary and can affect the results.

Lead Times Dependent on Lot Sizes. In MRP
lead times are assumed fixed, but they clearly depend
on the size of the lot required.
Shortcomings of MRP, cont.

Quality Problems. Defective items can destroy the
linking of the levels in an MRP system.

Data Integrity. Real MRP systems are big
(perhaps more than 20 levels deep) and the integrity
of the data can be a serious problem.

Order Pegging. A single component may be used
in multiple end items, and each lot must then be
pegged to the appropriate item.