Alternative lot sizing schemes
Anadolu University
Industrial Engineering Department
Source: Nahmias, S., Production And Operations Analysis, McGraw-Hill /Irwin, Fifth
Edition, 2005, ISBN: 0073018651
The three major control phases of the
productive system
Information requirements for each
end item over the planning horizon
Phase 1
Master Production
Schedule
Lot sizing rules and
capacity planning
Materials Requirements
Planning
Requirements
for raw material
Detailed shop
floor schedule
Phase 2
Phase 3
The explosion calculus
• Explosion calculus is as term that refers to the set of rules by which gross
requirements at one level of product structure are translated into a
production schedule at that level and requirements at lower level.
End item level
Parent level
(level 1)
Child level
(level2)
End item
A(2)
1 week
C(1)
2 weeks
B(1)
2 weeks
D(2)
1 week
C(2)
2 weeks
E(3)
2 weeks
Example 7.1
• The Harmon Music Company produces a variety of wind instruments at its
plant in Joliet, Illinois. Because the company relatively small, it would like
to minimize the amount of money tied to inventory. For that reason
production levels are set to match predicted demand as closely as
possible. In order to achieve this goal, the company has adopted an MRP
system to determine production quantities.
• One of the instruments produced is the model 85C trumpet. The trumpet
retails for $800 and has been a reasonably profitable item for the
company.
• Figure gives the product
structure diagram for the
construction of the trumpet.
Trumpet
(end item)
Bell Assembly (1)
Lead Time: 2 weeks
Valve casing (1)
Lead times 4 weeks
Slide assemblies (3)
Lead Time : 2 weeks
Valves (3)
Lead times : 3
weeks
Explosion of trumpet BOM
Week
Demand
2
3
4
5
6
7
8 9 10 11 12 13 14 15 16 17
77 42 38 21 26 112 45 14 76 38
Week
Scheduled receipts
8
12
week
net predicted demand
8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
Week
Gross requirements
Net requirements
Time-phased net requirements
planned order relase (lot-for-lot)
6
8
42
42
42 42 32
42 42 32
Time-phased net requirements
planned order relase (lot-for-lot)
Week
Gross requirements
Scheduled receipts
On-han inventory
Net requirements
Time-phased net requirements
planned order relase (lot-for-lot)
7
9 10 11
6 9
9
42
42
12
12
10 11
32 12
32 12
26 112
26 112
23 trumpet at the end of the week 7
12 13
26 112
26 112
45 14
45 14
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
2
3
4 5 6 7
126 126 96 36
96
186 60 30
0 0 66 36
66 36 78 336 135
66 36 78 336 135
Trumpet
14
45
45
76
76
15 16 17 Bell Assembly
14 76 38
14 76 38
38
38
Valve Casing
Assembly
8 9 10 11 12 13 14 15 16 17 valves
78 336 135 42 228 114
78 336 135 42 228 114
42 228 114
42 228 114
Alternative Lot-Sizing Schemes
•
•
•
•
EOQ lot sizing
The silver-meal heuristic
Least Unit Cost
Part Period Balancing
EOQ Lot Sizing
• To apply the EOQ formula, we need three
inputs:
– The average demand rate, 
– The holding cost rate, h
– Setup cost, K
2 K
Q
h
EOQ Lot Sizing- Valve casing assembly in Ex.7.1
• Suppose that the setup operation for the machinery used in
this assembly operation takes two workers about three hours.
– The worker average cost: $22 per hour
– K= (22)(2)(3) = $132
• The company uses a holding cost on a 22 percent annual
interest rate.
– Each valve casing assembly costs $141.82 in materials and value added
for labor,
– Holding cost, h = (141.82)(0.22) / 52 = 0.60
• Lot-for-lot policy requires
– Total holding cost for
– Total setup cost
=0
= (132)(10) = $1320
EOQ Lot Sizing- Valve casing assembly in Ex.7.1
2 K
(2)(132)(43.9)
Q

 139
h
0.6
Week
Gross requirements
Net requirements
Time-phased net requirements
Planned order relase (EOQ)
Planned deliveries
Ending inventory
4
5
6
7
8 9 10
42 42 32
42 42 32
42 42 32 12 26 112 45
139 0 0 0 139 0 139
139 0 0
97 55 23
11 12 13 14
12 26 112 45
12 26 112 45
14 76 38
0 0 139
0 139 0 139
11 124 12 106
15 16 17
14 76 38
14 76 38
0 0 139
92 16 117
Lot-for-lot policy requires
Cumulative ending inventory
97
+55
+23
+11 +124 +12 +106 +92
+16 +117
= 653
Total holding cost for = (0.6)(653) = $391.8
Total setup cost
= (132)(4) )= $528
The total cost of EOQ policy )= $528 + $391.8 = $ 919.8
The silver meal heuristic
• The silver meal heuristic
– named for Harlan Meal and Edward Silver
– is a forward method that requires determining the average cost per
period as a function of the number of periods the current order is to
span,
– and stopping the computation when this function first increases.
• C(T) as the average holding and setup cost per period if the
current orders spans the next T periods.
The silver meal heuristic
• Let (r1,…,rn) be the requirements over n-period horizon.
• Consider period 1.
– If we produce just enough in period 1 to meet the demand in period 1, then
just incur the order cost K.
C(1) = K
– If we order enough in period 1 to satisfy the demand both periods 1 and 2
then we must hold r2 for one period.
C(2)= (K + hr2)/2
– Similarly
C(3)= (K + hr2 + 2hr3)/3
– and in general,
C(j)= (K + hr2 + 2hr3 + … + (j-1)hr3) /j
– Once C(j)>C(j-1), we stop and set y1 = r1 + r2 + … + rn , and begin the process
again starting at period j.
The silver meal heuristic – Example 7.2
• A machine shop uses the Silver-Meal heuristic to schedule production lot
sizes for computer casing. Over the next five weeks the demands for the
casing are r = (18, 30, 42, 5, 20). The holding cost is $2 per case per week,
and the production setup cost is $80. Find the recommended lot sizing.
• Starting in period 1
–
–
–
–
C(1) = 80
C(2) =[80 + (2)(30)] /2 = 70
C(3) = [80 + (2)(30)+ (2)(2)(42)] /3 = 102.67 stop because C(3)>C(2).
Set y1 = r1 + r2 = 18 + 30 = 48
• Starting in period 3
–
–
–
–
C(1) = 80
C(2) =[80 + (2)(5)] /2 = 45
C(3) = [80 + (2)(5)+ (2)(2)(20)] /3 = 56.67 stop
Set y3 = r3 + r4 = 42 + 5 = 47
• Because period 5 is final period in the horizon, we do not need to start the
process again.
Set y5 = r5 = 20
y = (48, 0, 47, 0, 20)
Least unit cost
• The least unit cost (LUC) heurist
– Similar to Silver – Meal heuristic except instead of dividing the cost over j
periods by the number of periods, j, we divide it by the total number of units
demanded through period j, r1 + r2 + … + rj .
• Define C(T) as the average holding and setup cost per unit for a Tperiod order horizon. Then,
C(1) = K/r1
C(2)= (K + hr2)/(r1 + r2)
…
C(j)= [K + hr2 + 2hr3 + … + (j-1)hr3] /(r1 + r2 + … + rj )
– As with the Silver-Meal heuristic, this computation is stopped when C(j)>C(j-1),
and production level is set to r1 + r2 + … + rj-1 .
– The process is then repeated, starting at period j and continuing until the end
of the planning horizon is reached.
Least Unit Cost – Example 7.4
• Assume the same requirements schedule and costs as given in
Ex.7.2
• Starting in period 1
–
–
–
–
C(1) = 80/18 = 4.44
C(2) =[80 + (2)(30)] /(18+30) = 2.92
C(3) = [80 + (2)(30)+ (2)(2)(42)] /(18+30+42) = 3.42 stop because C(3)>C(2).
Set y1 = r1 + r2 = 18 + 30 = 48
• Starting in period 3
– C(1) = 80 /42 = 1.90
– C(2) =[80 + (2)(5)] /(42+5) = 1.92 stop
– Set y3 = r3 = 42
• Starting in period 4
– C(1) = 80/5 = 16
– C(2) =[80 + (2)(20)] /(5+20) = 4.8
• As we reached the en of the horizon, we set y4 = r4 + r5 = 5 + 20 =25. The
solution obtain by LUC heuristic y = (48, 0, 42, 25, 0)
Part Period Balancing
• Although Silver-Meal heuristic seems to give better results in a greater
number of cases, part period balancing seems to be more popular in
practice.
• The method is set the order horizon equal to the number of periods that
most closely matches the total holding cost with the setup cost over that
period.
• The order horizon that exactly equates holding and setup costs will rarely
be an integer number of periods.
Part Period Balancing- Example 7.5
• Consider Example 7.2.
• Starting in period 1
Order horizon
1
2
3
Total holding cost
0
60
(30)(2)
228
(30)(2)+(2)(2)(42)
– Because 228 exceeds the setup cost of 80, we stop. As 80 is closer to 60 than to 228, the
first order horizon is two periods. y1 = r1 + r2 = 18 + 30 = 48
• We start again in period 3
Order horizon
1
2
3
–
–
–
–
Total holding cost
0
10
90
(5)(2)
(5)(2)+(2)(2)(20)
We have exceeded the setup cost of 80, so we stop.
Because 90 is closer to 80 than 10, the order horizon is three periods.
y3 = r3 + r4 + r5 = 42+ 5 + 20= 67
y = (48,0,67,0,0)
Comparison of results
Silver-Meal
Demads
LUC
Part Period
Balancing
r = (18, 30, 42, 5, 20).
Solution
y = (48, 0, 47, 0, 20)
y = (48, 0, 42, 25, 0)
y = (48,0,67,0,0)
Holding inventory
30 + 5 = 35
30+20 = 50
30 + 5+(2)(20)=75
Holding cost
(35)(2) = 70
(50)(2) =100
(75)(2)=150
Setup cost
(3) (80) = 240
(3)(80) = 240
(2)(80) = 160
Total cost
310
340
310
Incorporating lot-sizing algorithms into the explosion
calculus
• Consider valve casing assembly in the Ex.7.1
• Time phased net requirements for valve casing are
Time-phased net requirements
planned order relase (lot-for-lot)
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
Valve Casing
Assembly
• The setup cost the valve casing is K=$132, and holding cost h=$0.60 per
assembly per week.
Silver Meal heuristic.
• Starting in week 4
–
–
–
–
–
–
C(1) = 132
C(2) =[132 + (0.6)(42)] /2 = 78.6
C(3) = [132 + (0.6)[42+ (2)(32)]] /3 = 65.2
C(4) = [132 + (0.6)[42+ (2)(32)+(3)(12)]] /4 = 54.3
C(5) = [132 + (0.6)[42+ (2)(32) +(3)(12) +(4)(26)]] /5 = 55.92 , stop.
Set y4 = 42+ 42+ 32 + 12= 128
Incorporating lot-sizing algorithms into
the explosion calculus
• Starting in week 8
–
–
–
–
–
–
C(1) = 132
C(2) =[132 + (0.6)(112)] /2 = 99.6
C(3) = [132 + (0.6)[112+ (2)(45)]] /3 = 84.4
C(4) = [132 + (0.6)[112+ (2)(45)+(3)(14)]] /4 = 69.6
C(5) = [132 + (0.6)[112+ (2)(45)+(3)(14)+(4)(76)]] /5 = 92.12 , stop.
Set y8 = 26+ 112+ 45 + 14= 197
• y12 = 76 + 38
Week
Gross requirements
Net requirements
Time-phased net requirements
Planned order relase (S-M)
Planned deliveries
Ending inventory
4
5
6
7
8
42
42
42 42 32 12 26
128 0 0 0 197
128
86
9
42
42
112
0
0
44
10
32
32
45
0
0
12
11
12
12
14
0
0
0
12
26
26
76
114
197
171
13
112
112
38
0
0
59
14 15 16 17
45 14 76 38
45 14 76 38
0
14
0 114
0 38
0
0
• S-M policy requires
– Total holding cost for = (0.6)(424) = $254.4
– Total setup cost
= (132)(3)= $396
– The total cost of EOQ policy )= $396 + $254.4 = $ 650.50
• Lot for lot : TCL4L=$1320
• EOQ :
TCEOQ=$919.8
Lot sizing with capacity constraints
• Capacity constraint clearly makes the problem far more realistic.
• However it also makes the problem more complex.
• Even finding a feasible solution may not be obvious. Consider the
Example: r=(52, 87, 23, 56) and capacity for each period c=(60, 60, 60, 60).
• First we must determine if the problem is feasible
• On the surface the problem looks solvable, 4*60 =240>218
• But problem is infeasible. ! 60+60=120<139
• Feasibility condition
j
j
 c  r
i 1
i
i 1
i
for j  1,, n.
• Even feasibility condition is satisfied, it is not obvious how to find feasible
solution.
Lot sizing with capacity constraints-Example 7.7
r= (20, 40, 100, 35, 80, 75, 25)
c= (60, 60, 60, 60, 60, 60, 60)
Checking for feasibility:
r1 = 20
r1 + r2 = 60,
r1 + r2 + r3 = 160,
r1 + r2 + r3 + r4 = 195,
r1 + r2 + r3 + r4 + r5 = 275,
r1 + r2 + r3 + r4 + r5 + r6 = 350,
r1 + r2 + r3 + r4 + r5 + r6 + r7 = 375,
c1 = 60
c1 + c2 = 120,
c1 + c2 + c3 = 180,
c1 + c2 + c3 + c4 = 240,
c1 + c2 + c3 + c4 + c5 = 300,
c1 + c2 + c3 + c4 + c5 + c6 = 360,
c1 + c2 + c3 + c4 + c5 + c6 + c7 = 420,
50
40
60
60
60
55
60
r‘ = ( 20, 40,
100,
35,
80,
75,
25
)
c =(
60,
60,
60,
60,
60
)
60, 60,
Setting y=r’ gives a feasible solution.
60
Improvement step
•
•
•
•
Modified requirements schedule r’ is feasible.
Is there another feasible policy that has lower cost.
Variety of reasonable improvement rules can be used
For each lot that is scheduled
– Start from the last and work backward to the beginning
– Determine whether it is cheaper to produce the units composing that lot by
shifting production to prior periods of excess capacity
– By eliminating a lot, one reduces setup const in that period to zero, but
shifting production to prior periods increases the holding cost.
Example 7.8
•
•
•
•
Assume that K=$450 and h=$2
r=(100, 79 , 230, 105, 3, 10, 99, 126, 40)
c=(120, 200, 200, 400, 300, 50, 120, 50 , 30)
Feasibility check cum. c 100 179 409 514 517 527 626
cum. r
752 792
120 320 520 920 1220 1270 1390 1440 1470
feasible.
• r'=(100, 109, 200, 105, 28, 50, 120, 50, 30)
r'
c
y
Excess capacity
1
100
120
100
20
2
109
200
109
91
3
200
200
200
0
4
105
400
105
295
5
28
300
28
272
6
50
50
50
0
7
120
120
120
0
8
50
50
50
0
9
30
30
30
0
Example 7.8
r'
c
y
Excess capacity
1
100
120
100
20
2
109
200
109
91
3
200
200
4
105
400
200
263
105
0
137
295
5
6
7
28
50
120
300
50
120
0
158 4
3
2
108
58
0
28
50
120
300
142
192
242
272
0
0
8
50
50
9
30
30
1
0
50
0
30
0
0
(4 period) ($2 per unit per period) (30 unit) =$240 <K=$450
•
•
•
•
•
y=(100, 109, 200, 263, 0, 0 , 120, 0 , 0)
r=(100, 79 , 230, 105, 3, 10, 99, 126, 40)
Setup cost = 5 x 450 = 2250
Holding cost = 2 x (0+30+0+158+155+145+166+40+0) = 2 x 694 = $ 1388
Total cost of this policy $3638, ($4482 for initial feasible policy).
Optimal Lot-Sizing for the time Varying
Demand
• Heuristic techniques (EOQ lot sizing, The silver-meal heuristic, Least Unit
Cost, Part Period Balancing) are easy to use and give lot sizes with costs
that are near the true optimum.
• Lot size problem can be expressed as a shortest path problem.
• The dynamic programming can be used to find shortest path.
Assume that|
1. Forecasted demands over the next n periods are known and given by
vector r=(r1,r2,…,rn).
2. Costs are charged against holding at $h per unit per period and $K per
setup.
Example 7A.1
• The forecast demand for an electronic assembly produced at Hi-tech, a
local semiconductor fabrication shop, over the next four weeks is 52, 87,
23, 56. There is only one setup each week for production of these
assemblies, and there is no back-ordering of excess demand. Assume that
the shop has the capacity to produce any number of assemblies in a week.
• Let y1, y2, y3, y4 be the order quantities in each week.
• Clearly y152 , if we assume that ending inventory in week 4 is zero, then
y1 218 (52+87+23+56)
• y1 can take any one of 167 possible values
• It is thus clear that for even moderately sized problems the number of
feasible solution is enormous.
Wagner-Whitin Algorithm
• The Wagner – Whitin algorithm is based on the following
observation:
– Result. An optimality policy has the property that each value of y is
exactly the sum of a set of future demands(exact requirement policy).
y1 =r1,
or y1=r1+r2+…,
or y1=r1+r2+…+rn
y2 =0,
or y2=r2
or y2=r2+r3+…,
or y2=r2+r3+…+rn
…
yn =0,
or yn =rn.
– The number of exact requirement policy is much smaller than the total
number of feasible policies.s
Wagner-Whitin Algorithm
• Because y1 must satisfy exact requirements, it can take only values : 52,
139, 162, 218
• Ignoring value of y1, y2 can take 0, 87, 110, 166.
• Every exact policy is the form (i1,i2,…,in) , where ij are either 0 or 1.
• For example the policy (1,0,1,0) means that production occurs in period 1
and 3 only, that is y= (139, 0, 79, 0).
• For this example, there are exactly 23 = 8 distinct exact requirement
policies.
Wagner-Whitin Algorithm
•
•
A convenient way to look at the problem is as a one-way network with the
number of nodes equal to exactly one more than the number of periods.
Every path through to network corresponds to a specific exact requirements
c15
policy.
c
14
c13
1
c12
2
c23
3
c34
c24
c45
4
5
c35
c25
•
•
•
•
For any pair (i,j) with i<j, if the arc (i,j) is one path, it means that ordering takes
place in period i and the order size is equal to the sum of requirements in
periods i, i+1, i+2 ,…, j-1.
Period j is the next period of ordering.
All paths end at period n+1,
The value of each arc, cij, is defined as setup and holding cost of ordering in
period i to meet requirements through j-1.
Example 7A.2
• r = (52, 87, 23, 56), h=$1, K=$75
• The first step compute cij for 1i4 and 2j5
–
–
–
–
–
–
–
–
–
–
–
c12 =75 (setup cost)
c13 =75 + 87 = 162
c14 =75 + (23 x 2) + 87 = 208
c14 =75 + (56 x 3) + (23 x 2) + 87 = 376
c23 =75
c24 =75 + 23 = 98
c25 =75 + (56 x 2) + 23 = 210
c34 =75
c35 =75 + 56= 131
i \ j 1
c23 =75 (setup cost)
1
c45 =75
2
3
4
Path
Cost
1-2-3-4-5
300
1-2-4-5
248
1-2-5
285
2
3
4
5
1-2-3-5
281
75
162
208
376
1-3-4-5
312
75
98
210
1-3-5
293
75
131
1-4-5
283
75
1-5
376
Solution by Dynamic Programming
• The total number of exact policies for a problem of n periods is 2n-1.
• As n gets large, total enumeration is not efficient.
• Dynamic programming is a recursive solution technique that can
significantly reduce the number of computations required.
• Dynamic programming is based on the principle of optimality.
• Define fk as the minimum cost starting at node k.
f k  min (ckj  f j )
j k
• Initial condition is fn+1 =0
for k  1,2,, n
Solution by Dynamic Programming
f5  0
f 4  min (c4 j  f j )
j 4
 75
at j  5
c34  f 4 
75  75
150
f3  min (c3 j  f j )  min 
  min 
  min  
j 3
131  0 
131
c35  f 5 
 131 at j  5
c23  f 3 
75  131
206






f 2  min (c2 j  f j )  min c24  f 4   min 98  75   min 173 
j 2
c  f 
210  0 
210
5




 25
 173 at j  4
c12 
c 
 13
f1  min (c1 j  f j )  min 
j 1
c14 
c15 
 248 at j  2
f2 
75  173 
248
162  131
293
f 3 





min

min





f4 
208

75
283




376  0 
376
f 5 