Planning the Production Program Chapter 4 Planning the Production Program

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

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

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

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

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

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

Example I - Table for period production in period No extra cost 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 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)

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

Example I – Cost & Production Plan Total cost = 590 * C + 10 * 1 + 10 * 1 + 10 * 3 + 10 * 2,5 + 10 * 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

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

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 = 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

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 = 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 700

Example II – Solution Column minimum procedure Total cost = 100*0 +(700+700+700)*40 +(50+50)*50 +50*52 +150*70 +50*72 = 105700 50 100 150