Capacity Planning

Capacity Capacity (I): is the upper limit on the load that an operating unit can handle. Capacity (II): the upper limit of the quantity of a product (or product group) that an operating unit can produce (= the maximum level of output) Capacity (III): the amount of resource inputs available relative to output requirements at a particular time

Examples of Capacity Measures

Capacity Design capacity Effective capacity maximum output rate or service capacity an operation, process, or facility is designed for = maximum obtainable output = best operating level Effective capacity Design capacity minus allowances such as personal time, maintenance, and scrap (leftover) Actual output = Capacity used rate of output actually achieved. It cannot exceed effective capacity.

Capacity Efficiency and Capacity Utilization Actual output Efficiency = Effective capacity Utilization = Design capacity

Determinants of Effective Capacity Facilities, layout Product and service factors Process factors Human factors Policy (shifts, overtime) Operational factors Supply chain factors External factors

Capacity Cushion level of capacity in excess of the average utilization rate or level of capacity in excess of the expected demand. Capacity cushion = (designed capacity / capacity used) – 1 Sources of Uncertainty Manufacturing Customer delivery Supplier performance Changes in demand

The „Make or Buy” problem Available capacity Expertise Quality considerations Nature of demand Cost Risk

Developing Capacity Alternatives Design flexibility into systems Take stage of life cycle into account (complementary product) Take a “big picture” approach to capacity changes Prepare to deal with capacity “chunks” Attempt to smooth out capacity requirements Identify the optimal operating level

Adapting capacity to demand through changes in workforce PRODUCTION RATE (CAPACITY)

Adaptation with inventory DEMAND Inventory reduction Inventory accumulation CAPACITY

Adaptation with subcontracting DEMAND SUBCONTRACTING PRODUCTION (CAPACITY)

Adaptation with complementary product PRODUCTION (CAPACITY) DEMAND DEMAND PRODUCTION (CAPACITY)

Production units have an optimal rate of output for minimal cost. Economies of Scale Economies of scale If the output rate is less than the optimal level, increasing output rate results in decreasing average unit costs Diseconomies of scale If the output rate is more than the optimal level, increasing the output rate results in increasing average unit costs Production units have an optimal rate of output for minimal cost. Minimum cost Average cost per unit Rate of output Minimum average cost per unit

Evaluating Alternatives II. Minimum cost & optimal operating rate are functions of size of production unit. Small plant Medium plant Average cost per unit Large plant Output rate

Seminar exercises

Designed capacity in calendar time CD= N ∙ sn ∙ sh ∙ mn ∙ 60 (mins / planning period) CD= designed capacity (mins / planning period) N = number of calendar days in the planning period (365) sn= maximum number of shifts in a day (= 3 if dayshift + swing shift + nightshift) sh= number of hours in a shift (in a 3 shifts system, it is 8 per shift) mn= number of homogenous machine groups

Designed capacity, with given work schedule C’D= N ∙ sn ∙ sh ∙ mn ∙ 60 (mins / planning period) C’D= designed capacity (mins / planning period) N = number of working days in the planning period (≈ 250 wdays/yr) sn= number of shifts in a day (= 3 if dayshift + swing shift + nightshift) sh= number of hours in a shift (in a 3 shifts system, it is 8) mn= number of homogenous machine groups

Effective capacity in working minutes CE = CD - tallowances (mins / planning period) CD= designed capacity tallowances = allowances such as personal time, maintenance, and scrap (mins / planning period) b =  ∙ CE b = expected capacity (that we use in product mix decisions) CE = effective capacity  = performance percentage

Exercise 1. A company produces a product in 2 shifts pattern and 250 days in a given year. Two machines are available in the workstation. The maintenance needs 8 hours in a month. The performance rate in the first shift is 90%, in the second shift 80%. Each product needs 25 minutes to be finished. a) Determine the designed capacity for one year. b) Determine the effective capacity for one year. c) Determine the expected capacity for one year. d) Determine the number of products we can produce. e) Determine the utilization and efficiency if the actual output is 15 800 pieces. d) Determine capacity cushion.

Exercise 2. Set up the product-resource matrix using the following data! RP coefficients: a11: 10, a22: 20, a23: 30, a34: 10 The planning period is 4 weeks (there are no holidays in it, and no work on weekends) Work schedule: R1 and R2: 2 shifts, each is 8 hour long R3: 3 shifts Homogenous machines: 1 for R1 2 for R2 1 for R3 Maintenance time: only for R3: 5 hrs/week Performance rate: 90% for R1 and R3 80% for R2

Solution (bi) Ri = N ∙ sn ∙ sh ∙ mn ∙ 60 ∙  N=(number of weeks) ∙ (working days per week) R1 = 4 weeks ∙ 5 working days ∙ 2 shifts ∙ 8 hours per shift ∙ 60 minutes per hour ∙ 1 homogenous machine ∙ 0,9 performance = = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9 = 17 280 minutes per planning period R2 = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 2 ∙ 0,8 = 30 720 mins R3 = (4 ∙ 5 ∙ 3 ∙ 8 ∙ 60 ∙ 1 – 5 hrs per week ∙ 60 minutes ∙ 4 weeks) ∙ 0,9= (28 800 – 1200 ) ∙ 0,9= 24 840 mins

Solution (RP matrix) P1 P2 P3 P4 10 20 30 b (mins/y) R1 17 280 R2   P1 P2 P3 P4 b (mins/y) R1 10 17 280 R2 20 30 30 720 R3 24 840

Exercise 1.2 Complete the corporate system matrix with the following marketing data: There are long term contract to produce at least: 50 P1 100 P2 120 P3 50 P4 Forecasts says the upper limit of the market is: 10 000 units for P1 1 200 for P2 1 000 for P3 2 000 for P4 Unit prices: P1=100, P2=200, P3=330, P4=100 Variable costs: R1=5/min, R2=8/min, R3=11/min

Solution (CS matrix) 10 20 30 P1 P2 P3 P4 b (mins/y) R1 17 280 R2   P1 P2 P3 P4 b (mins/y) R1 10 17 280 R2 20 30 30 720 R3 24 840 MIN (pcs/y) 50 100 120 MAX (pcs/y) 10 000 1 200 1 000 2 000 price 200 330 Contr. Marg. 40 90 -10

What is the optimal product mix to maximize revenues? P3= (30 720-100∙20-)/30= 957<MAX P4=24 840/10=2484>2000

What if we want to maximize profit? The only difference is in P4 because of its negative contribution margin. P4=50

Exercise 2 P1 P2 P3 P4 P5 P6 6 3 2 4 1 b (hrs/y) R1 2 000 R2 R3 1 000   P1 P2 P3 P4 P5 P6 b (hrs/y) R1 6 2 000 R2 3 2 R3 4 1 000 R4 6 000 R5 1 5 000 MIN (pcs/y) 200 100 400 MAX (pcs/y) 20000 500 2000 p (HUF/pcs) 50 cm (HUF/pcs) 80 40 -30 20 -10

Solution Revenue max. P1=333 P2=400 P3=400 P4=200 P5=1000 Contribution max. P1=333 P2=500 P3=250 P4=100 P5=1000