Capacity Planning

Product/Operations Management

Chapter 3.0 – Capacity Planning

Capacity –  an upper value or limit on the load that an operating unit can handle.

Decisions involved in Capacity Planning:

1.      Machine requirements

a.      Purchase new equipment

b.      Lease or buy second hand machines

2.      Number of workers to be maintained

Importance of Capacity Decisions

1. Impacts ability to meet future demands

2. Affects operating costs

3. Major determinant of initial costs

4. Involves long-term commitment

5. Affects competitiveness

6. Affects ease of management

7. Globalization adds complexity

8. Impacts long range planning

Alternatives to consider in Capacity Planning:

1.      Produce to stock during the slack periods

2.      Promise later delivery dates on peak periods

3.      Introduce new product that would complement seasonal products

4.      Subcontract during peak periods

5.      Allocate resources

Determinants of Effective Capacity

·         Facilities factors – design, size, expansion, location, labor supply, energy sources

·         Product/service factors – product or service design

·         Process factors – quantity capability of a process

·         Human factors – the human tasks involve, training and skill

·         Operation factors – scheduling, inventory stocking, etc.

·         External factors – product standards and quality & performance standards

·         Supply chain factors-  sources and quality materials

Capacity Types

1.      Design Capacity – maximum output that can possibly be attained; output based on design specifications and allowances are not considered.

2.       Effective (Peak) Capacity – maximum possible output given plant operation characteristics: plant maintenance requirements, product mix, working schedules, etc.

3.       Actual Capacity – rate of actual output actually achieved by the operations.

Most common measures of capacity performance being used are Efficiency and Utilization

                                               Efficiency    =    Actual Capacity

                                                                        Effective Capacity

                                              Utilization    =   Actual Capacity

                                                                         Design Capacity  

                                  Both measures expressed as percentages

Example3-1:  Compute the efficiency and the utilization of the vehicle repair department.

Design Capacity – 50 trucks per day

Effective Capacity – 40 trucks per day

Actual output – 36 trucks per day

Efficiency =  AC   =    36/40 = 90% (very good)    

                      EC

Utilization = AC   =   36/50 = 72% (good)

                      DC

Example 3-2 : A department works on 8 –hour shift, 250 days a year and has these figures:

Product	Annual Demand	Standard  Processing Time per unit (hr.)	Processing time      needed       (hr.)
# 1    # 2    #3	400   300  700	5.0       8.0      2.0	2,000     2,400    1,400    5,800

                     How many machines are needed?

Formula:

                 N(no. of machines) = Total Process time(annual)

                                                         Total capacity time

                                                   =              5,800 hrs                   

                                                       (8 x 250)2000 hrs/machine         = 2.90 machines = 3 machines

Example 3-3: A plant engineer finds in a particular month a tool will be required to process 1390 units of a part & that the actual time per unit of output will probably be 27 minutes. If the plan were to operate on a single shift, each tool would be available for 173 hours/month. However, current martketi9ng forecasts dictate the need to work for 2 shifts. How many of this tool is required?

Solution:

Given: Demand = 1390 units/month

            Drill Capacity = 173 hrs/month-shift/tool

           Processing time = 27 min/unit

     27 min/unit x 1 hr/60 min = 0.45 hrs/unit

                 N(no. of tools) = 1390 x 0.45

                                             173 x 2 shifts        = 1.81 or 2 tools

Example 3-4: Suppose that a department has two machines stuffed for one 8-hour shift for five days a week per machine. During the last month, records revealed that a weekly average of 8 hours of downtime occurred, reducing total hours available for production. Records also showed that during last month, the department produced a weekly average of 90 standard hours of output. What are the department’s efficiency and utilization index values?

(Solution)

Given: Design Capacity = 8hrs/day-mach. X 5 days x 2 mach. = 80 hrs.

             Effective Capacity = DC – downtime = 80 hrs. – 8 hrs. = 72 hrs.

             Actual Output = 90 hrs.

Efficiency = 90/72 x 100% = 125.0 %

Utilization = 72/80 x 100% = 90.0 %

Example 3-5: A metal processing firm wishes to install enough automatic equipment to produce 250,000 good castings per year. The molding operation takes 1.5 minutes per casting, but its output is typically about 3 percent defective. How many molding machines will be required if each one is available for 2000 hours of capacity?

(Solution)

 System Capacity = Actual Output       

                                     Efficiency

                               = 250,000

                                    0.97   ( 100% – 3% )

                               = 257,732 units/year

                               = 257,732 units/yr

                                     2000 hrs/yr

                               =  129 units/hr

Given: Molding Capacity = 1.5 mins/unit x 1 hr/60 mins = 40 units/mach-hr.

Number of machine, N = 129 units/hr

                                              40 units/hr

                                          = 3.2 molding eqpt.

Example 3-6: A service business finds it necessary to type approximately 4,500 correct labels during each eight-hour day operation. An Average of 0.75 minute is actually required to type and verify each address label. Any label error is discarded or scrapped. Past experience suggests that ten percent of the total labels contain errors.

a.      Compute for the total number of labels to be scheduled for typing.

b.      Calculate the number of typewriters/typists needed.

(Solution)

a.        Required Output = Required output                   b.    Number, N = 5000 x 0.75 min

                                                 100% - Sc%                                                           8 x 60 mins/hr

                                     = 4,500 labels

                                        100% - 10%                                                                = 7.8 typewriters/typists

                                     = 5,000 labels

Example 3-7: A specified production schedule shows that 200 good units of a specific assembly part are to be manufactured in a certain week. The routing sheet for the past indicates that one of the operation calls for a power hack-saw of which the normal operating time is 0.072 hr/unit. Calculate the number of machine of this type will be required to process the part in the week under consideration if:

a.      No scrap is expected; labor efficiency is 100% and each saw mill will be operated 40 hrs/week.

b.      5% scrap

c.       3% scrap, 105% labor efficiency

d.      40% scrap, 70% labor efficiency, 50 hrs/week

(Solution)

a.      N = 200 units/week x 0.072 mach-hrs/unit

                                   40 hrs/week

                   = 0.36

b.      N = 200/(100% - 5%) x 0.072

                       40

    = 0.3789

c.       N = 200/(100% - 3%) x 0.072/1.05

                       40

    = 0.353

d.      N = 200/(100% - 40%) x 0.072/.70

                       50

    = 0.685

Cost – Volume Analysis

·         Focuses on relationships between cost, revenue and volumes of output

·         To estimate the income under different operating conditions

Fixed Costs – constant costs regardless of volume of output ( light, rent, administrative costs, taxes)

Variable Costs – vary directly with the volume of output ( labor & materials)

Cost – Volume Symbols

 FC – Fixed Cost                                                VC – Variable Cost

TC – Total Cost                                                  TR – Total Revenue

R – Revenue per Unit                                     Q – Quantity/Volume of Output

BEP – Break Even Point                                                 Qbep – Break Even Quantity

P= Profit                                                              SP – Specified Profit

Formula:

BEP = Total Cost = Total Revenue                              Under Specified Profit

TC = FC + VC x Q                                                                               Q (Volume) = SP + FC

                                                                                                                                        R-VC

TR = R x Q

                                                                                                                        Qbep =     FC

P = TR –TC                                                                                                                     R-VC

   = R x Q – (FC + VC + Q)

Example 3-8:  The Owner Of the ICST Pies is thinking of adding newline of pies which will require leasing new equipment for a monthly payment of P 6,000. Variable cost would be P 2.00 per pie & pies would retail at P 7.00 each.

a.       How many pies must be sold to break even?

b.      What would be the profit(loss) if 1,000 pies are made & sold in a month?

c.       How many pies must be sold to realize a profit of P 4,000

Given: FC = P6,000; VC = P 2.00/pie; R = P 7.00/pie

a.       Qbep =    FC          = P 6,000         = 1,200/pies

              R – VC            7 -2

b.      For Q = 1,000

P = R x Q – ( FC + VC x Q )

= 7 x 1,000 – ( 6,000 + 2 x 1,000)       =  - P 1,000

c.       P = P 4,000 solve for Q

              Q = SP + FC         = P 4,000 + 6,000      = 10,000       =  2,000 pies

                      R – VC                     7-2                            5

              OR

             P 4,000 = 7Q – ( 6,000 + 2Q)

   

           Rearrange:

                7Q -2Q = 4,000 + 6,000                    =  5Q = 10,000                          =   Q = 2,000

Example 3-9: A manager has the option of purchasing 1, 2 or 3 machines, below are the fixed costs and potential volumes:

Number of Machines	Total Annual Fixed Costs	Corresponding Range of Output
1	P 9,600	0 to 300
2	15,000	301 to 600
3	20,000	601 to 900

Variable cost is P 10/unit; & revenue is P 40/unit

a.       Determine the break- even point for each range.

b.      If projected annual demand is between 580 to 660 units, how many machines should  the manager purchase.

(Solution)

a.       Qbep =  FC

                           R-VC

               One Machine = 9,600             =  320 units

                            40 -10

Two Machines  = 15,000        =  500 units

                                 40-10

Three Machines = 20,000       = 666.67 units

                                  40-10

b.      Two (2) Machines

Financial Analysis – used in capacity planning, a common approach to rank investment proposals.

Cash Flow – refers to the difference between cash received from sales and other sources and the cash outflow for labor, materials, overheard and taxes.

Present Value – expresses in current value the sum of all future cash flows on an investment proposal

3 common methods of financial analysis:

1.       Payback – focuses on the length of time it will take for an investment to return its original cost.

Example: Original Cost of P 6,000 and a monthly net cash flow of P 1,000. The payback period is 6 months. Payback ignores the time value of money.

2.       Present Value (PV)- summarizes the initial cost of an investment; its estimated annual cash flows, and any expected salvage value in a single value called ‘ equivalent current value ‘, taking into account the time value of money ( ex. Interest rate).

3.       Internal Rate of Return (IRR) – summarizes the initial cost, expected annual cash flows & estimated future salvage value of an investment proposal in an equivalent interest rate. This method identifies the rate of return that equates the estimated future returns & initial cost.