1. Planning the Production Program
Chapter 4
Planning the Production Program

2. Planning the Production Program
Based on demand forecasts and orders plan the production quantities for the (main) products for the „next“ periods
2 variants:
Aggregate Planning (aggregated view, tactical planning, medium run) few product groups for the next (months), quarters, or years capacities can be adjusted (hiring/firing, overtime, holidays, subcontracting ...)
Master Production Scheduling (more detailed view, operational planning, short run) all main end products for next few shifts, days, or weeks (or months) capacities more or less fixed (except for overtime)
Typically solved as an LP model

3. Aggregate Planning 2 extreme scenarios in case of seasonal demand:
Always produce the demand (forecast) „Synchronisation“  cost of hiring/firing, overtime, subcontracting, idle time, …
Always produce average yearly demand (high utilization) „Emancipation“  inventory holding cost
Goal:
Trade-off between these costs  minimize total costs
Solution by column minimum procedure

4. Synchronisation Synchronisation:
No active planning, just reaction on demand (forecasts) Always produce the demand (forecast)
overview

5. Emancipation Emancipation:
More or less constant demand, constant (high) resource utilization, fluctuating demand is fulfilled by building up and depleting inventory.
overview
Constant Production
Build up Inventory
Reduce Inventory

6. Column Minimum Procedure
In each period regular capacity can be extended at extra cost (overtime, subcontracting, …)
Cope with fluctuating demand (capacity shortages):
Produce more than demand – build up inventory, OR
Use extra capacity
Solution a special case (just one product group) as a TP
In each cell (row t … production period, half row k … capacity type, and
column  … demand period) the unit extra cost are:
ctk = uk + h( - t)
where: uk ... Extra cost (per unit) of production using extra capacity k (e.g. overtime)
h ... Inventory holding cost per unit and per period,
h( - t) ... Inventory holding per unit if produced  - t periods early
Solve as transportation problem using Column Minimum Procedure
table

7. Example I Given 6 Periods Normal capacity in each Period: 100 units
Just 1 type of extra capacity: k = 1 max. possible extra capacity: 10 units
Cost:
Holding cost: h = 1 € per unit and period
Cost of extra capacity: u1 = 1,5 € for each unit produced in overtime k = 1
Determine optimal production plan

8. Example I - Table for period production in period No extra cost1 2 3 4 5 6
capacity
0,0
1,0
2,0
3,0
4,0
5,0
100
1,5
2,5
3,5
4,5
5,5
6,5 10
Demand 90
110 50
130
No extra cost
Normal
Extra
Advance production: holding cost h*(# periods) h = 1
production in period
Extra capacity extra cost u in 2nd half row u = 1,5
formula
No shortages permitted (otherwise shortage cost)

9. Example I – Column Minimum Procedure
Prod
100 70
110 10 90 10 10
Column Minimum Procedure
100 30 50 10 10 50 40
100
100 10
100
total cost 10 10 10 30 20 10

10. Example I – Cost & Production Plan
Total cost =
590 * C
+ 10 * * * 3
+ 10 * 2, * 1,5
Production cost
Holding cost
Cost of production using extra capacity
table
= 590 * C + 90 GE
Production plan
1. Per.
2. Per.
3. Per.
4. Per.
5. Per.
6. Per.
Normal
100 70
Extra 10

11. Example II 2 sources of extra capacity k = 1 overtime &
k = 2 subcontracing
table

12. Example II – Variant 1 data
Each row now has 3 sub-rows for 3 sources of capayity (normal, overtime, subcontracting)
Make it completely equivalent to TP by adding Dummy Column for unused capacity
Total capacity = 2780 Total demand = 2550 unused capacity = = 230
Initial inventory can be treated in 2 ways: Variant 1: treat as additional (artificial) production row 0 oder Variant 2: subtract from demand of first period
data

13. Example II – Variant 2 data
Each row now has 3 sub-rows for 3 sources of capayity (normal, overtime, subcontracting)
Make it completely equivalent to TP by adding Dummy Column for unused capacity
Total capacity = 2780 Total demand = 2550 unused capacity = = 230
Initial inventory can be treated in 2 ways: Variant 1: treat as additional (artificial) production row 0 oder Variant 2: subtract from demand of first period
data
700

14. Example II – Solution Column minimum procedure
Total cost = 100*0 +( )*40 +(50+50)*50 +50* *70 +50*72 = 50
100
150