1. Capacity Planning

2. Capacity Capacity (I): is the upper limit on the load that an operating unit can handle. 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 (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 Capacity (III): the amount of resource inputs available relative to output requirements at a particular time The basic questions in capacity planning are: The basic questions in capacity planning are: What kind of capacity is needed? What kind of capacity is needed? How much is needed? How much is needed? When is it needed? When is it needed? How does productivity relate to capacity? How does productivity relate to capacity?

3. 1. Impacts ability to meet future demands 2. Affects operating costs (fixed or variable) 3. Major determinant of initial costs 4. Involves long-term commitment 5. Affects competitiveness 6. Affects ease of management 7. Impacts long range planning Importance of Capacity Decisions

4. Examples of Capacity Measures

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

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

7. Actual output 6 Actual output 6 tons/week Efficiency = = = 75% Effective capacity 8 Effective capacity 8 tons/week Actual output 6 Actual output 6 tons/week Utilization = = = 60% Design capacity 10 Utilization = = = 60% Design capacity 10 tons/week Numeric Example Design capacity = 10 tons/week Effective capacity = 8 tons/week Actual output = 6 tons/week

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

9. Key Decisions of Capacity Planning 1. Amount of capacity needed 2. Timing of changes 3. Need to maintain balance 4. Extent of flexibility of facilities

10. Capacity Cushion level of capacity in excess of the average utilization rate or level of capacity in excess of the expected demand. extra demand intended to offset uncertainty C. cushion = (designed capacity / capacity used) - 1 High cushion is needed:  service industries  high level of uncertainty in demand (in terms of both volume and product-mix)  to permit allowances for vacations, holidays, supply of materials delays, equipment breakdowns, etc.  if subcontracting, overtime, or the cost of missed demand is very high

11. Steps for Capacity Planning 1. Estimate future capacity requirements 2. Evaluate existing and future capacity 3. Identify alternatives 4. Conduct financial analysis 5. Assess key qualitative issues 6. Select one alternative 7. Implement alternative chosen 8. Monitor results (feedback)

12. Sources of Uncertainty Manufacturing Manufacturing Customer delivery Customer delivery Supplier performance Supplier performance Changes in demand Changes in demand

13. The „Make or Buy” problem 1. Available capacity 2. Expertise 3. Quality considerations 4. Nature of demand 5. Cost 6. Risk

14. Developing Capacity Alternatives 1. Design flexibility into systems 2. Take stage of life cycle into account (complementary product) 3. Take a “big picture” approach to capacity changes 4. Prepare to deal with capacity “chunks” 5. Attempt to smooth out capacity requirements 6. Identify the optimal operating level

15. Economies of Scale 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 If the output rate is less than the optimal level, increasing output rate results in decreasing average unit costs Diseconomies of scale Diseconomies of scale If the output rate is more than the optimal level, increasing the output rate results in increasing average unit costs If the output rate is more than the optimal level, increasing the output rate results in increasing average unit costs

16. Evaluating Alternatives Production units have an optimal rate of output for minimal cost. Minimum cost Average cost per unit 0 Rate of output Minimum average cost per unit

17. Evaluating Alternatives II. Minimum cost & optimal operating rate are functions of size of production unit. Average cost per unit 0 Small plant Medium plant Large plant Output rate

18. Need to be near customers Need to be near customers Capacity and location are closely tied Capacity and location are closely tied Inability to store services Inability to store services Capacity must be matched with timing of demand Capacity must be matched with timing of demand Degree of volatility of demand Degree of volatility of demand Peak demand periods Peak demand periods Planning Service Capacity

19. Some examples of managing changes in demand

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

21. Adaptation with inventory DEMAND CAPACITY Inventory accumulation Inventory reduction

22. Adaptation with subcontracting DEMAND PRODUCTION (CAPACITY) SUBCONTRACTING

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

24. Seminar exercises

25. Designed capacity in calendar time C D = N ∙ s n ∙ s h ∙ m n ∙ 60 (mins / planning period ) C D = designed capacity (mins / planning period) C D = designed capacity (mins / planning period) N = number of calendar days in the planning period (365) N = number of calendar days in the planning period (365) s n = maximum number of shifts in a day (= 3 if dayshift + swing shift + nightshift) s n = maximum number of shifts in a day (= 3 if dayshift + swing shift + nightshift) s h = number of hours in a shift (in a 3 shifts system, it is 8 per shift) s h = number of hours in a shift (in a 3 shifts system, it is 8 per shift) m n = number of homogenous machine groups m n = number of homogenous machine groups

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

27. Effective capacity in working minutes C E = C D - t allowances (mins / planning period) C D = designed capacity t allowances = allowances such as personal time, maintenance, and scrap (mins / planning period)

28. The resources we can calculate with in product mix decisions b =  ∙ C E b = expected capacity b = expected capacity C E = effective capacity  = performance percentage How many can we produce of a good, that needs 5 minutes/unit, the preparation needs 30 minutes and  = 90%, C E = 10000 mins? How many can we produce of a good, that needs 5 minutes/unit, the preparation needs 30 minutes and  = 90%, C E = 10000 mins? = (0,9*10000-30)/5= 1794 pieces

29. Exercise 1.1 Set up the product-resource matrix using the following data! RP coefficients: a 11 : 10, a 22 : 20, a 23 : 30, a3 4 : 10 RP coefficients: a 11 : 10, a 22 : 20, a 23 : 30, a3 4 : 10 The planning period is 4 weeks (there are no holidays in it, and no work on weekends) The planning period is 4 weeks (there are no holidays in it, and no work on weekends) Work schedule: Work schedule: R 1 and R 2 : 2 shifts, each is 8 hour long R 1 and R 2 : 2 shifts, each is 8 hour long R 3 : 3 shifts R 3 : 3 shifts Homogenous machines: Homogenous machines: 1 for R 1 1 for R 1 2 for R 2 2 for R 2 1 for R 3 1 for R 3 Maintenance time: only for R 3 : 5 hrs/week Maintenance time: only for R 3 : 5 hrs/week Performance rate: Performance rate: 90% for R 1 and R 3 90% for R 1 and R 3 80% for R 2 80% for R 2

30. Solution (b i ) R i = N ∙ s n ∙ s h ∙ m n ∙ 60 ∙  R i = N ∙ s n ∙ s h ∙ m n ∙ 60 ∙  N=(number of weeks) ∙ (working days per week) N=(number of weeks) ∙ (working days per week) R 1 = 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 R 1 = 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 R 2 = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 2 ∙ 0,8 = 38 720 mins R 2 = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 2 ∙ 0,8 = 38 720 mins R 3 = (4 ∙ 5 ∙ 3 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9) – (5 hrs per week maintenance ∙ 60 minutes per hour ∙ 4 weeks ) = 25 920 – 1200 = 24 720 mins R 3 = (4 ∙ 5 ∙ 3 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9) – (5 hrs per week maintenance ∙ 60 minutes per hour ∙ 4 weeks ) = 25 920 – 1200 = 24 720 mins

31. Solution (RP matrix) Solution (RP matrix) P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 b (mins/y) R1R1R1R110 17 280 R2R2R2R22030 30 720 R3R3R3R310 24 720

32. Exercise 1.2 Complete the corporate system matrix with the following marketing data: Complete the corporate system matrix with the following marketing data: There are long term contract to produce at least: There are long term contract to produce at least: 50 P1 50 P1 100 P2 100 P2 120 P3 120 P3 50 P4 50 P4 Forecasts says the upper limit of the market is: Forecasts says the upper limit of the market is: 10 000 units for P1 10 000 units for P1 1 500 for P2 1 500 for P2 1 000 for P3 1 000 for P3 3 000 for P4 3 000 for P4 Unit prices: P1=100, P2=200, P3=330, P4=100 Unit prices: P1=100, P2=200, P3=330, P4=100 Variable costs: R1=5/min, R2=8/min, R3=11/min Variable costs: R1=5/min, R2=8/min, R3=11/min

33. Solution (CS matrix) P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 b (mins/y) R1R1R1R110 17 280 R2R2R2R22030 30 720 R3R3R3R310 24 720 MIN (pcs/y) 5010012050 MAX (pcs/y) 10 000 1 500 1 000 3 000 price100200330100 Contr. Marg. 504090-10

34. What is the optimal product mix to maximize revenues? P 1 = 17 280 / 10 = 1728 < 10 000 P 1 = 17 280 / 10 = 1728 < 10 000 P 2 : 200/20=10 P 2 : 200/20=10 P 3 : 330/30=11 P 3 : 330/30=11 P 4 =24 720/10=2472<3000 P 4 =24 720/10=2472<3000 P 2 = 100 P 2 = 100 P 3 = (30 720-100∙20-120∙30)/30= 837<MAX P 3 = (30 720-100∙20-120∙30)/30= 837<MAX

35. What if we want to maximize profit? The only difference is in P 4 because of its negative contribution margin. The only difference is in P 4 because of its negative contribution margin. P 4 =50 P 4 =50

36. Exercise 2 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 b (hrs/y) R1R1R1R16 2 000 R2R2R2R2 32 3 000 R3R3R3R3 4 1 000 R4R4R4R4 636 000 R5R5R5R5 145 000 MIN (pcs/y) 0200100250400100 MAX (pcs/y) 2000050040010002000200 p (HUF/pcs) 20010040010050100 cm (HUF/pcs) 5080403020-10

37. Solution Revenue max. P 1 =333 P 1 =333 P 2 =500 P 2 =500 P 3 =400 P 3 =400 P 4 =250 P 4 =250 P 5 =900 P 5 =900 P 6 =200 P 6 =200 Contribution max. P 1 =333 P 2 =500 P 3 =400 P 4 =250 P 5 =966 P 6 =100