ABWL-EK Kapitel 4

advertisement
Chapter 4
Planning the Production Program
Chapter 4/1
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
Chapter 4/2
Aggregate Planning
2 extreme scenarios in case of seasonal demand:
• Always produce the demand (forecast)
„Synchronisation“ (zero inventory plan)
 cost of hiring/firing, overtime, subcontracting, idle time, …
• Always produce average yearly demand (high utilization)
„Emancipation“ (level workforce plan)
 inventory holding cost
Goal:
•
Trade-off between these costs  minimize total costs
•
Solution by column minimum procedure
Chapter 4/3
Synchronisation
Synchronisation:
No active planning, just reaction on demand (forecasts)
Always produce the demand (forecast)
overview
Chapter 4/4
Emancipation
Emancipation:
More or less constant demand, constant (high) resource utilization,
fluctuating demand is fulfilled by building up and depleting inventory.
Constant Production
Build up Reduce
Inventory Inventory
overview
Chapter 4/5
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
Chapter 4/6
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
Chapter 4/7
Example I - Table
for period
Period
Normal
1
Extra
Normal
2
Extra
production
in period
1
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
0,0
1,0
2,0
3,0
4,0
100
1,5
2,5
3,5
4,5
5,5
10
0,0
1,0
2,0
3,0
100
1,5
2,5
3,5
4,5
10
0,0
1,0
2,0
100
1,5
2,5
3,5
10
0,0
1,0
100
1,5
2,5
10
0,0
100
1,5
10
Normal
3
Extra
Normal
4
Extra
Normal
5
Extra
Normal
6
Extra
Demand
90
110
50
110
100
130
No extra cost
Advance
production:
holding cost
h*(# periods)
h=1
Extra capacity
extra cost u
in 2nd half row
u = 1,5
formula
No shortages
permitted
(otherwise
shortage cost)
Chapter 4/8
Example I – Column Minimum Procedure
Prod
90
10
10
Prod
100
100
0
Column
Minimum
Procedure
100
100
100
0
50
10
10
30
50 40
70
70
0
100
100
100
0
100
100
total cost
10
10
110
10
10
100
100
10
10
110
30 20 10
Chapter 4/9
Example I – Cost & Production Plan
Total cost
table
=
590 * C + 10 * 1 + 10 * 1 + 10 * 3 + 10 * 2,5 + 10 * 1,5
Production cost
Cost of production
using extra capacity
Holding cost
= 590 * C + 90 GE
Production plan
1. Per.
2. Per.
3. Per.
4. Per.
5. Per.
6. Per.
Normal
100
100
70
100
100
100
Extra
0
0
0
0
10
10
Chapter 4/10
Example II
2 sources of extra capacity
• k = 1 overtime &
• k = 2 subcontracing
table
Chapter 4/11
Example II – Variant 1
• 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 = 2780 - 2550 = 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
Chapter 4/12
Example II – Variant 2
• 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 = 2780 - 2550 = 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
700
data
Chapter 4/13
Example II – Solution
• Column minimum
procedure
50
150
100
• Total cost =
100*0
+(700+700+700)*40
+(50+50)*50
+50*52
+150*70
+50*72
= 105700
Chapter 4/14
Download