Supply Chain Management
Chapter 8
Aggregate Planning
in the Supply Chain
8-1
Production System
 In a broader sense, a production system is anything that takes inputs
and transforms them into outputs.
 Production system is the collection of people, equipment, and
procedures organized to accomplish the operations of a company (or
other organization).
– Manufacturing
– Service
8-2
Production Planning
 Production Planning is the analysis, design and management of
production systems.
Objective: Transform a variety of inputs (such as raw material, labor,
capital, etc.) into outputs (goods and services) in a manner that is
both efficient in using resources and effective in achieving high
customer satisfaction.
 In today’s competitive business environment it is important to know
methods and specific analysis tools for operational decisions to
effectively manage these systems.
8-3
Production Planning Decisions
 Long-term (Strategic) Decisions:
– Top Management Decisions
– 3-10 years
– Decisions: Capacity, Product, Supplier needs, Quality Policy
 Intermediate-term (Tactical) Decisions:
– Middle Management Decisions
– 6 months - 3 years
– Decisions: Work-force levels, processes, production rates, inventory
levels, contracts with suppliers, quality level, quality costs
 Short-term (Operational) Decisions:
– Operational Management Decisions
– 1 week - 6 months
– Decisions: Allocation of jobs to machines, overtime, undertime,
subcontracting, delivery dates for suppliers, product quality
8-4
Production Planning and Control
General Framework
Resources
Planning
Aggregate
Planning
Demand
Management
Rough-cut
Capacity
Planning
Master Production
Scheduling
Detailed
Capacity
Planning
Detailed Material
Planning
Material and
Capacity Plans
Shop Floor
Systems
Purchasing
8-5
Role of Aggregate Planning
in a Supply Chain
Aggregate planning:
– process by which a company determines levels of capacity,
production, subcontracting, inventory, stockouts, and pricing
over a specified time horizon
– goal is to maximize profit
– decisions made at a product family level
– time frame of 3 to 18 months
– how can a firm best use the facilities it has?
8-6
Role of Aggregate Planning
in a Supply Chain
Specify operational parameters over the time horizon:
–
–
–
–
–
–
–
production rate
workforce
overtime
machine capacity level
subcontracting
backlog
inventory on hand
All supply chain stages should work together on an
aggregate plan that will optimize supply chain
performance
8-7
The Aggregate Planning Problem
Given the demand forecast for each period in the
planning horizon, determine the production level,
inventory level, and the capacity level for each period
that maximizes the firm’s (supply chain’s) profit over
the planning horizon
Specify the planning horizon (typically 3-18 months)
Specify the duration of each period
Specify key information required to develop an
aggregate plan
8-8
Information Needed for
an Aggregate Plan
Demand forecast in each period
Production costs
– labor costs, regular time ($/hr) and overtime ($/hr)
– subcontracting costs ($/hr or $/unit)
– cost of changing capacity: hiring or layoff ($/worker) and
cost of adding or reducing machine capacity ($/machine)
Labor/machine hours required per unit
Inventory holding cost ($/unit/period)
Stockout or backlog cost ($/unit/period)
Constraints: limits on overtime, layoffs, capital
available, stockouts and backlogs
8-9
Outputs of Aggregate Plan
Production quantity from regular time, overtime, and
subcontracted time: used to determine number of
workers and supplier purchase levels
Inventory held: used to determine how much warehouse
space and working capital is needed
Backlog/stockout quantity: used to determine what
customer service levels will be
Machine capacity increase/decrease: used to determine
if new production equipment needs to be purchased
A poor aggregate plan can result in lost sales, lost
profits, excess inventory, or excess capacity
8-10
Fundamental Tradeoffs in
Aggregate Planning
Capacity (regular time, overtime, subcontract)
Inventory
Backlog / lost sales
Basic Strategies
Chase strategy
Time flexibility from workforce or capacity
Level strategy
8-11
Aggregate Planning Strategies
Trade-off between capacity, inventory,
backlog/lost sales
Chase strategy – using capacity as the lever
Time flexibility from workforce or capacity
strategy – using utilization as the lever
Level strategy – using inventory as the lever
Mixed strategy – a combination of one or more of
the first three strategies
8-12
Chase Strategy
Production rate is synchronized with demand by
varying machine capacity or hiring and laying off
workers as the demand rate varies
However, in practice, it is often difficult to vary
capacity and workforce on short notice
Expensive if cost of varying capacity is high
Negative effect on workforce morale
Results in low levels of inventory
Should be used when inventory holding costs are high
and costs of changing capacity are low
8-13
Level Strategy
Maintain stable machine capacity and workforce
levels with a constant output rate
Shortages and surpluses result in fluctuations in
inventory levels over time
Inventories that are built up in anticipation of future
demand or backlogs are carried over from high to low
demand periods
Better for worker morale
Large inventories and backlogs may accumulate
Should be used when inventory holding and backlog
costs are relatively low
8-14
Time Flexibility Strategy
Can be used if there is excess machine capacity
Workforce is kept stable, but the number of hours
worked is varied over time to synchronize production
and demand
Can use overtime or a flexible work schedule
Requires flexible workforce, but avoids morale
problems of the chase strategy
Low levels of inventory, lower utilization
Should be used when inventory holding costs are
high and capacity is relatively inexpensive
8-15
Comparison of production planning
strategies
Item
Chase Demand
Level Capacity
Labor skill required
Low
High
Job discretion
Low
High
Working conditions
Sweatshop
Pleasant
Training required
Low
High
Labor turnover
High
Low
Supervision required
High
Low
8-16
Aggregate Planning Problem
Costs in Aggregate Planning:
•Material Cost
•Inventory Holding Cost
•Shortage cost
•Regular Time Costs
•Overtime and Subcontracting Costs
•Hiring and Firing Costs
•Idle Time Costs
•Backlogging costs
•Costs associated with lost sales
•Control system cost.
8-17
Data for the aggregate planning
problem
Item
Demand Forecast
Number of Working Days
January
February
March
April
May
June
Totals
2250
1425
1000
850
1150
1725
8400
22
19
21
21
22
20
125
1,425
1,000
850
1,150
1,725
Costs:
Materials
100
Inventory holding cost
1.5
Marginal cost of stockout
Marginal cost of subcontracting
5
20
Hiring and training cost
200
Layoff cost
250
Labor hours required
5
Straight-line cost (first eight hours each day)
4
Overtime cost (time and a half)
6
Beginning Inventory
Production Requirement (Demand
Forecast - Beginning Inventory)
400
1,850
8-18
Production Plan 1: Exact Production;
Vary Workforce
Item
January
February
March
April
May
June
Production Requirement
1,850
1,425
1,000
850
1,150
1,725
Production Hours Required
(Production Requirement x 5
Hr./Unit)
9,250
7,125
5,000
4,250
5,750
8,625
22
19
21
21
22
20
176
152
168
168
176
160
53
47
30
25
33
54
New Workers Hired (Assuming opening
workforce equal to first month's
requirement of 53 workers.)
0
0
0
0
8
21
Hiring Cost (new Workers Hired x $200)
$0
$0
$0
$0
$1,600
$4,200
0
6
17
5
0
0
Layoff Cost (Workers Laid Off x $250)
$0
$1,500
$4,250
$1,250
$0
$0
$7,000
Straight Time Cost (Production Hours
Required x $4)
$37,000
$28,500
$20,000
$17,000
$23,000
$34,500
$160,000
Total Cost
$172,800
Working Days per Month
Hours per Month per Worker (Working
Days x 8 Hrs/Day)
Workers Required (Production Hours
Required/Hours per Month per
Worker, must round this number
up)
Workers Laid Off
Total
$5,800
8-19
8-20
Production Plan 3: Constant Low
Workforce; Subcontract
January
Production Requirement
February
March
April
May
June
Total
1,850
1,425
1,000
850
1,150
1,725
22
19
21
21
22
20
4,400
3,800
4,200
4,200
4,400
4,000
Actual Production (Production Hours
Available/5 hr.per Unit)
880
760
840
840
880
800
Units Subcontracted (Production
Requirement - Actual Production)
970
665
160
10
270
925
Subcontracting Cost (Units
Subcontracted x $20)
$19,400
$13,300
$3,200
$200
$5,400
$18,500
$60,000
Straight Time Cost (Production Hours
Available x $4)
$17,600
$15,200
$16,800
$16,800
$17,600
$16,000
$100,000
Working Day per Month
Production Hours Available (Working
Day x 8 Hrs./Day x 25 Workers)*
*Minimum production requirement. In
this example, April is minimum of 850
units. Number of workers required for
April is (850 x 5)/(21 x 8) = 25
Total Cost
$160,000
8-21
Production Plan 4: Constant Low
Workforce; Overtime
January
Production Requirement
February
March
April
May
June
Total
1,850
1,425
1,000
850
1,150
1,725
22
19
21
21
22
20
Production Hours Available (Working
Day x 8 Hrs./Day x 25 Workers)*
4,400
3,800
4,200
4,200
4,400
4,000
Regular Production (Production Hours
Available/5 hr.per Unit)
880
760
840
840
880
800
Units Produced with Overtime
(Production Requirement - Actual
Production)
970
665
160
10
270
925
Overtime Cost (Overtime Units
x5hr/unit x $6)
$29,100
$19,950
$4,800
$300
$8,100
$27,750
$90,000
Straight Time Cost (Production Hours
Available x $4)
$17,600
$15,200
$16,800
$16,800
$17,600
$16,000
$100,000
Working Day per Month
*Minimum production requirement. In
this example, April is minimum of 850
units. Number of workers required for
April is (850 x 5)/(21 x 8) = 25
Total Cost
$190,000
8-22
Aggregate Planning at
Red Tomato Tools
Month
January
February
March
April
May
June
Demand Forecast
1,600
3,000
3,200
3,800
2,200
2,200
8-23
Aggregate Planning
Item
Materials
Inventory holding cost
Marginal cost of a stockout
Hiring and training costs
Layoff cost
Labor hours required
Regular time cost
Over time cost
Cost of subcontracting
Cost
$10/unit
$2/unit/month
$5/unit/month
$300/worker
$500/worker
4/unit
$4/hour
$6/hour
$30/unit
8-24
Aggregate Planning
(Define Decision Variables)
Wt = Workforce size for month t, t = 1, ..., 6
Ht = Number of employees hired at the beginning of month t,
t = 1, ..., 6
Lt = Number of employees laid off at the beginning of month t,
t = 1, ..., 6
Pt = Production in month t, t = 1, ..., 6
It = Inventory at the end of month t, t = 1, ..., 6
St = Number of units stocked out at the end of month t,
t = 1, ..., 6
Ct = Number of units subcontracted for month t, t = 1, ..., 6
Ot = Number of overtime hours worked in month t, t = 1, ..., 6
8-25
Aggregate Planning
(Define Objective Function)
6
6
t 1
t 1
Min 640 W t   300 H t
6
6
6
t 1
t 1
t 1
  500 Lt   6 Ot   2 I t
6
6
6
t 1
t 1
t 1
  5 S t  10 Pt   30 C t
8-26
Aggregate Planning (Define
Constraints Linking Variables)
Workforce size for each month is based on hiring
and layoffs
W t  W t 1  H t  Lt, or
W t  W t 1  H t  Lt  0
for t  1,...,6, where W 0  80.
8-27
Aggregate Planning (Constraints)
Production for each month cannot exceed capacity
,
4

40

P t W t Ot
40W t  Ot 4  Pt  0,
for t  1,...,6.
8-28
Aggregate Planning (Constraints)
Inventory balance for each month
I t 1  Pt  C t  Dt  S t 1  I t  S t ,
I t 1  Pt  C t  Dt  S t 1  I t  S t  0,
for t  1,...,6,where I 0  1,000 ,
S 0  0,and I 6  500 .
8-29
Aggregate Planning (Constraints)
Over time for each month
Ot  10 W t,
10 W t Ot  0,
for t  1,...,6.
8-30
Scenarios
Increase in holding cost (from $2 to $6)
Overtime cost drops to $4.1 per hour
Increased demand fluctuation
8-31
Increased Demand Fluctuation
Month
January
February
March
April
May
June
Demand Forecast
1,000
3,000
3,800
4,800
2,000
1,400
8-32
Aggregate Planning in Practice
Think beyond the enterprise to the entire supply chain
Make plans flexible because forecasts are always
wrong
Rerun the aggregate plan as new information emerges
Use aggregate planning as capacity utilization
increases
8-33
Production Planning and Control
General Framework
Resources
Planning
Aggregate
Planning
Demand
Management
Rough-cut
Capacity
Planning
Master Production
Scheduling
Detailed
Capacity
Planning
Detailed Material
Planning
Material and
Capacity Plans
Shop Floor
Systems
Purchasing
8-34
Aggregate Production Plan and
Master Production Schedule
Actual production plan should consider individual
products, smaller time units, production sequence etc.
Aggregate production plan is disaggregated to form
the master production schedule(MPS).
8-35
Aggregate Production Plan and
Master Production Schedule
Production Plan
Months
1
2
Marker Family 100000 120000
Master Production Schedule
Products
Weeks
1 2 3 4
1 2 3
Red 10 10 10 10 15 15 15
Blue 10 10 10 10 10 10 10
Green 5 5 5 5
5 5 5
4
15
10
5
8-36
Capacity Planning and
Material Requirements Planning
 Rough-cut Capacity Planning: To check feasibility of MPS.
– Quick check on capacity of key resources
– Use Bill of Resource (BOR) for each item in MPS
– Infeasibilities addressed by altering MPS or adding capacity (e.g., overtime)
 Material Requirements Planning(MRP): Breaking the MPS into a
production schedule for each component of an end-item. Determines the
material requirements and timings for each phase of production.
 Detailed Capacity Planning: To check feasibility of MRP. Supplements the
process of checking material shortage.
–
–
–
–
Uses routing data (work centers and times) for all items
Generates usage profile of all work centers
Identifies overload conditions
More detailed than RCCP
8-37
Materials Requirements Planning
(MRP)
 Materials Requirements Planning (MRP) determines time-phased
requirements (period-by-period) for all purchased and manufactured
parts such as raw materials, components, parts, subassemblies, etc.
 Three major inputs are MPS, Inventory Status and Bill of Materials
(BOM), also called the Product Structure.
 The major output of MRP is planned-order releases: Purchase orders
and work orders(production plan).
8-38
Product Structure Example
8-39
Product and Part Complexity
Product
Approximate number of
components
Mechanical pencil (modern)
10
Ball bearing (modern)
20
Rifle (1800)
50
Sewing machine (1875)
150
Bicycle chain
300
Bicycle (modern)
750
Early automobile (1910)
Automobile (modern)
Commercial airplane (1930)
Commercial airplane (modern)
Space shuttle (modern)
2000
20000
100000
1000000
10000000
8-40
Steps in MRP
 Explosion: Evaluating the gross requirements of each component
 Netting: Adjusting gross requirements to account for on-hand
inventory or quantity on order.
 Offsetting:Determining the timing of order releases
 Lot Sizing: Determining the batch size to be purchased or produced
8-41
MRP Example
Trumpet
Bell Assembly (1)
Valve Casing Assembly (1)
Lead Time=2 wks
Lead Time=4 wks
Slide Assemblies (3)
Valves (3)
Lead Time=2 wks
Lead Time=3 wks
Total Lead Time= 7 wks
Week
Trumpet Demand
8
42
9
42
10
32
11
12
12
26
13
112
14
45
15
14
16
76
17
38
8-42
Product Structure Example
Valve Casing Assembly:
Week
Gross Requirement
Net Requirement
Time-Phased Net Requirement
4
5
6
7
42
42
32
12
8
42
42
26
9
42
42
112
10
32
32
45
11
12
12
14
12
26
26
76
13
112
112
38
14
45
45
15
14
14
16
76
76
17
38
38
Valves: (Assume an inventory of 282 at hand at week 3)
Week
Gross Requirements
On Hand Inventory
Net Requirement
Time-Phased Net Requirement
3
282
66
4
126
156
36
5
126
30
78
6
96
7
36
8
78
9
336
10
135
11
42
12
228
13
114
66
336
36
135
78
42
336
228
135
114
42
228
114
8-43
Lot Sizing
Setup/Ordering Costs: Every time a new production is started, a
setup of machines/labor is required and there is an associated cost
with it.
Similarly, every time an order is given for a product, there is an
ordering cost for transportation, managerial issues etc. Thus, every
time we start a new production or give a new order, we want it to be
for high quantities.
However, if we order for high quantities, we have to carry high levels
of inventory. Thus, there is a tradeoff between inventory costs and
setup costs.
Question: What is the optimal time and amount to order?
8-44
Lot Sizing
Parameters:
h(t):unit inventory holding cost in period t.
A(t): setup/ordering cost in period t.
Decision variables:
X(t): Amount ordered in period t.
O(t) = 1 if an order is given in period t
= 0 otherwise
I(t): Inventory level at the end of period t.
Min
∑(h(t)I(t) + A(t)O(t))
s.t.
I(t) = I(t-1)+X(t)-D(t)
X(t) ≤ O(t).M
I(t), X(t)≥0, O(t) = 0 or 1
8-45
Lot Sizing
Ex: Over the next 5 weeks, the net requirement of our company for a
product is 18, 30, 20, 5, 20. The holding cost is $2 per unit per week and
the ordering cost is $80. What are the optimal times and amounts to
order?
•Simple Rules
•Lot for Lot (L4L): Order 1 period of future demand
•Fixed period demand: Order m periods of future demand
•Fixed Order Quantity: Order fixed amounts
•Heuristic Methods
•Silver-Meal Method: Decision based on average cost per period
•Least unit cost: Decision based on average cost per unit
•Part-Period Balancing: Decision based on total variable cost per order
•Exact Methods:
•Wagner-Whitin Algorithm
•Integer Programming
8-46
Lot Sizing Example
Ex: Over the next 5 weeks, the net requirement of our company for a
product is 18, 30, 20, 5, 20. The holding cost is $2 per unit per week
and the ordering cost is $80. What are the optimal times and amounts to
order?
Silver-Meal Heuristic:
C(T)=Average cost per period if the current order is for the next T
periods.
Evaluate C(T) for T=1,2… and stop when C(T)>C(T-1).
C(1)=80/1=80
C(2)=(80+2*30)/2=70
C(3)=(80+2*30+2*2*20)/3=73.33 Stop
Order 48 units at week 1.
8-47
Lot Sizing Example
Now go to week 3 and start over.
C(1)=80/1=80
C(2)=(80+2*5)/2=45
C(3)=(80+2*5+2*2*20)/3=56.67 Stop
Order 25 units at week 3.
Go to week 5. Since week 5 is the final week, order
20 units at week 5.
Total Cost=80+2*30+80+2*5+80=310
Is it optimal? No, because
If we consider ordering 18 at week 1, 55 at week 2 and
20 at week 5. Then,
Total Cost=80+80+2*20+2*2*5+80=300<310
8-48
Lot Sizing Example
Least Unit Cost Heuristic:
C(T)=Average cost per unit if the current order is for the next T periods.
Evaluate C(T) for T=1,2… and stop when C(T)>C(T-1).
C(1)=80/18=4.4
C(2)=(80+2*30)/48=2.9
C(3)=(80+2*30+2*2*20)/68=3.23 Stop
Order 48 units at week 1.
Now go to week 3 and start over.
C(1)=80/20=4
C(2)=(80+2*5)/25=3.6
C(3)=(80+2*5+2*2*20)/45=3.77 Stop
Order 25 units at week 3.
Go to week 5. Since week 5 is the final week, order 20 units at week 5.
Total Cost=80+2*30+80+2*5+80=310
8-49
Lot Sizing Example
Part Period Balancing
C(T)=Total inventory cost for the current order if the order is for the next T periods.
Evaluate C(T) for T=1,2… and stop when C(T)>A=80.
C(1)=0
C(2)=2*30<80
C(3)=(2*30+2*2*20)>80 Stop
Order 48 units at week 1.
Now go to week 3 and start over.
C(1)=0
C(2)=2*5<80
C(3)=(2*5+2*2*20)>80 Stop
Order 25 units at week 3.
Go to week 5. Since week 5 is the final week, order 20 units at week 5.
Total Cost=80+2*30+80+2*5+80=310
8-50
Dynamic Lot Sizing Notation
t a period (e.g., day, week, month); we will consider t = 1, … ,T, where T
represents the planning horizon.
Dt demand in period t (in units)
ct unit production cost (in dollars per unit), not counting setup or inventory
costs in period t
At fixed or setup cost (in dollars) to place an order in period t
ht holding cost (in dollars) to carry a unit of inventory from period t to period t
+1
Qt the unknown size of the order or lot size in period t
decision variables
8-51
Wagner-Whitin Example
Data
t
Dt
ct
At
ht
1
2
3
4
5
6
7
8
9
10
20 50 10 50 50 10 20 40 20 30
10 10 10 10 10 10 10 10 10 10
100 100 100 100 100 100 100 100 100 100
1
1
1
1
1
1
1
1
1
1
Lot-for-Lot Solution
t
Dt
Qt
It
Setup cost
Holding cost
Total cost
1
20
20
0
100
0
100
2
50
50
0
100
0
100
3
10
10
0
100
0
100
4
50
50
0
100
0
100
5
50
50
0
100
0
100
6
10
10
0
100
0
100
7
20
20
0
100
0
100
8
40
40
0
100
0
100
9
20
20
0
100
0
100
10
30
30
0
100
0
100
Total
300
300
0
1000
0
1000
Since production cost c is constant, it can be ignored.
8-52
Wagner-Whitin Example (cont.)
Data
t
Dt
ct
At
ht
1
2
3
4
5
6
7
8
9
10
20 50
10
50
50
10
20
40
20
30
10 10
10
10
10
10
10
10
10
10
100 100 100 100 100 100 100 100 100 100
1
1
1
1
1
1
1
1
1
1
Fixed Order Quantity Solution
t
Dt
Qt
It
Setup cost
Holding cost
Total cost
1
20
100
80
100
80
180
2
50
0
30
0
30
30
3
4
5
10 50 50
0 100 0
20 70 20
0 100 0
20 70 20
20 170 20
6
7
8
10 20 40
0 100 0
10 90 50
0 100 0
10 90 50
10 190 50
9
20
0
30
0
30
30
10 Total
30 300
0 300
0
0
0 300
0 400
0 700
8-53
Wagner-Whitin Property
A key observation
If we produce items in t (incur a setup cost) for use to satisfy demand in t+1,
then it cannot possibly be economical to produce in t+1 (incur another setup
cost) .
Either it is cheaper to produce all of period t+1’s demand in period t, or all
of it in t+1; it is never cheaper to produce some in each.
Under an optimal lot-sizing policy
(1) either the inventory carried to period t+1 from a previous period will
be zero (there is a production in t+1)
(2) or the production quantity in period t+1 will be zero (there is no
production in t+1)
Does fixed order quantity solution violate this property? Why?
8-54
Basic Idea of Wagner-Whitin
Algorithm
By WW Property, either Qt=0 or Qt=D1+…+Dk for some k.
If jk* = last period of production in a k period problem,
then we will produce exactly Dk+…DT in period jk*. Why?
We can then consider periods 1, … , jk*-1 as if they are an
independent jk*-1 period problem.
8-55
Wagner-Whitin Example
t
Dt
ct
At
ht
1
2
3
4
5
6
7
8
9
10
20 50 10 50 50 10 20 40 20 30
10 10 10 10 10 10 10 10 10 10
100 100 100 100 100 100 100 100 100 100
1
1
1
1
1
1
1
1
1
1
 Step 1: Obviously, just satisfy D1 (note we are neglecting production
cost, since it is fixed).
Z1*  A1  100
j1*  1
 Step 2: Two choices, either j2* = 1 or j2* = 2.
Z 2*
 A1  h1 D2 , produce in 1
 min  *
Z1  A2 , produce in 2
100  1(50)  150
 min 
 100  100  200
 150
j2*  1
8-56
Wagner-Whitin Example (cont.)
 Step3: Three choices, j3* = 1, 2, 3.
 A1  h1 D2  (h1  h2 ) D3 , produce in 1

Z 3*  min Z1*  A2  h2 D3 ,
produce in 2
 Z*2  A3 ,
produce in 3
100  1(50 )  (1  1)10  170

 min 100  100  (1)10
 210
150  100
 250
 170
j3*  1
8-57
Wagner-Whitin Example (cont.)
 Step 4: Four choices, j4* = 1, 2, 3, 4.
 A1  h1 D2  (h1  h2 ) D3  (h1  h2  h3 ) D4 , produce
Z*  A  h D  (h  h ) D ,
produce
 1
2
2 3
2
3
4
*
Z 4  min  *
produce
 Z 2  A3  h3 D4 ,
 Z*3  A4 ,
produce
100  1(50)  (1  1)10  (1  1  1)50  320
100  100  (1)10  (1  1)50
 310

 min 
 300
150  100  (1)50
170  100
 270
in 1
in 2
in 3
in 4
 270
j4*  4
8-58
Planning Horizon Property
 In the Example:
– Given fact: we produce in period 4 for period 4 of a 4 period
problem.
– Question: will we produce in period 3 for period 5 in a 5 period
problem?
– Answer: We would never produce in period 3 for period 5 in a 5
period problem.
If jt*=t, then the last period in which production occurs in
an optimal t+1 period policy must be in the set t, t+1,…t+1.
(this means that it CANNOT be t-1, t-2……)
8-59
Wagner-Whitin Example (cont.)
 Step 5: Only two choices, j5* = 4, 5.
*

Z
*
3  A4  h4 D5 , produce in 4
Z 5  min  *
produce in 5
 Z 4  A5 ,
170  100  1(50)  320
 min 
 370
270  100
 320
j5*  4
 Step 6: Three choices, j6* = 4, 5, 6.
– And so on.
8-60
Wagner-Whitin Example Solution
Last Period
with Production
1
2
3
4
5
6
7
8
9
10
Zt
jt
Planning Horizon (t)
1
2
3
4
5
6
7
8
100 150 170 320
200 210 310
250 300
270 320 340 400 560
370 380 420 540
420 440 520
440 480
500
100 150 170 270 320 340 400 480
1
1
1
Produce in period 1
for 1, 2, 3 (20 + 50 +
10 = 80 units)
4
4
4
4
Produce in period 4
for 4, 5, 6, 7 (50 + 50 +
10 + 20 = 130 units)
7
9
10
520
520
580
520
610
580
610
620
580
7 or 8
8
Produce in period 8
for 8, 9, 10 (40 + 20 +
30 = 90 units
8-61
Wagner-Whitin Example Solution
(cont.)
 Optimal Policy:
– Produce in period 8 for 8, 9, 10 (40 + 20 + 30 = 90 units)
– Produce in period 4 for 4, 5, 6, 7 (50 + 50 + 10 + 20 = 130 units)
– Produce in period 1 for 1, 2, 3 (20 + 50 + 10 = 80 units)
t
Dt
Qt
It
Setup cost
Holding cost
Total cost
1
20
80
60
100
60
160
2
50
0
10
0
10
10
3
4
5
10 50 50
0 130 0
0 80 30
0 100 0
0 80 30
0 180 30
6
10
0
20
0
20
20
7
8
9
20 40 20
0 90 0
0 50 30
0 100 0
0 50 30
0 150 30
10 Total
30 300
0 300
0
0
0 300
0 280
0 580
8-62