Computer aided graphics: determining system size, estimating costs and savings

Computers in Industry

23

Applications

Computer Aided Graphics: Determining System Size, EstimatingCosts and Savings D a n i e l J. S c o t t Honeywell, Inc., Defense System Division, 600 Second St. NE, Hopkins, MN 55343, USA This paper describes a method used to custom configure assess and size- a Computer Aided Graphics system for a specific application. Costs and associated savings are based on this configuration using standard financial evaluation techniques. A hypothetical case is presented to demonstrate these methods. Variables are defined to enable the user to intelligently select values appropriate to his application. Additional iterations are suggested to permit fine tuning the analysis. These iterations are limited only by the users' needs and/or imagination. -

Keywords: Computer Aided Design, Computer Aided Manufacturing, Coml~uter Aided Graphics, Computer Graphics, Cost Calculating, Productivity, Efficiency, CAD, CAM, CAG.

Daniel J. Scott received a B.S. in both Mechanical Engineering (1946) and General Engineering (1947) from Iowa State University. He obtained his MBA in 1966 from the University of Minnesota. He is also a Certified Manufacturing Engineer in the Society of Manufacturing Engineers. He has been employed by Honeywell in Minneapolis, Minnesota f o r the past 24 years with responsibilities in Production Engineering. For the past four years, Mr. Scott has worked on CAD/CAM assignments, first in evaluating and selecting turn key systems and presently in bringing a system on-line.

© North-Holland Publishing Company Computers in Industry 2 (1981) 2 3 - 3 0

1. Introduction The United States' position of lagging in rate of productivity improvement has been well publicized. Our need to improve productivity has received equal publicity. These needs are commonly discussed in government, business and academic circles. Many associations address productivity improvement, e.g., the Air Force Integrated Computer-Aided Manufacturing (ICAM) program, American Productivity Center, Inc. (APC), Computer Aided Manufacturing, International (CAM-I), National Science Foundation (NSF), etc. A department of "Productivity" is becoming more common in private enterprise firms. In short, "productivity" is the center of a great deal of interest and attention today. One technology used to improve productivity is that of Computer Aided Graphics (CAG), sometimes used synonymously for Computer Aided Design (CAD). However, some CAD functions can be accomplished without using CAG, e.g., circuit simulation, structural analysis, etc., so perhaps CAG might be called graphical CAD. At any rate, one only needs to review Computer Aided Design/Computer Aided Manufacturing (CAD]CAM) periodicals or seminar papers to become aware of CAG potential. One " p r o o f of the pudding" is to note the number of companies worldwide which have incorporated CAG, the expanding rate of sales in this field, or talk to users to reinforce this CAG impact on improved productivity. However, improvement is not enough - it must be cost effective, too! CAG system costs must be compared with resulting savings in order to analyze cost effectiveness. An immediate problem is how to configure your system, i.e., how much equipment shouM you have? A determination of the number of Input• Output (I/O) stations needed to satisfy the needs of your designated users is essential. System configura-

24

Applications

Compttters it~ D~dttstrv

tion is the principal determining factor to system effectivity, improved productivity, and cost savings.

CO =

MT

I

CT

Elements A "C 2. Definition of Terms

Note: The time of CT is less than that of MT2, i.e., compressed, because the same elements are performed in either case but CT reflects the speed and power of a computer to accomplish them.

2.1. Manual Operation (MO) Time it takes to perform a complete operation without the aid of CAG. This manual operation can be divided into two time segments, MT1 and MT2. Thus, MO = MTI + MT2. MT, : The time for those elements of a job which will be done manually whether CAG is available or not, e.g., receiving work direction and instructions, planning job approach, attending meetings, etc. MT2: The time for those elements of a job which will be done manually when no CAG is available (but could be accomplished more effectively with (CAG) e.g., obtaining data, making calculations, incorporating changes, etc.

2.2.1. Element Descriptions - Figures 1 and 2 1. MT1 (identical for MO and CO) A Receive instructions B Plan approach C Attend meetings etc. 2. MT2 (for MO) or CT (for CO) D Obtain supplies E Get product drawing F Construct lines G Perform calculations H Incorporate changes etc.

2.2. CAG Operation (CO) Time it takes to perform a complete operation using CAG where possible. This CAG operation can be divided into two time segments, MT1 and CT. Thus, CO = MT1 + CT. MT~ : Identical to MT~ described above. CT: The time it takes to perform certain elements of a job using CAG, specifically those elements identified in MT2 above. A graphical description of these operations may help to clarify intended definitions. Assume that identical functions are to be accomplished under two environments - one without CAG equipment (Fig. 1) and one with CAG equipment (Fig. 2). An example could be initiating a product drawing, tool design, NC tape, etc.

MT 1 [

AElements ------~C

+ I

MT2 DElements It H

Time • Fig. 1. Operations AccomplishedWithout CAG.

I

"H 1

Time Fig. 2. The Same Operations Accomplishedwith CAG.

Terminology used in subsequent sections should be defined in order to avoid ambiguity.

MO =

Elements

D

I

2.3. Computer Productivity Factor (CPF) This is: ratio of time to perform a set of function elements manually compared to the time to perform the same element with CAG. With reference to Figs. 1 and 2: CPF -

MTz CT

This ratio can be quite high in some instances. An example might be a flat part containing the same bolt hole pattern in many positions. Manual generation of the drawing could require the same amount of time to display each pattern position. With CAG, the pattern would need to be generated only once and its display in multiple positions would be nearly automatic. 2.4. Total Productivity Factor ( TPF) This is the ratio of time to perform a total operation entirely manually compared to performing the same total operation using CAG where it is applicable

Computers in Industry and effective. Again referring to Figs. 1 and 2, TPF -

MT1 + MT2 MT1 + CT

Since CT is less than MT2, CPF will always be greater than 1.0. TPF will also be greater than 1.0 but will be less than CPF (because, as MTI becomes larger, i.e., approaches infinity, the TPF ratio approaches 1.0). This explains why the operation of a department might have a TPF equal 2.0, yet some elements of the total job - done with CAG - might have a CPF of 6.0 or 8.0.

2.5. CAG Input/output Station (I/O) This is the interactive work station used by the operator. Generally, this I/O station consists of a cathode ray tube (CRT) as an output display and a function keyboard, menu(s), electronic work tablet and pen as input devices.

3. Methodology The approach taken is to calculate the size of the system based on estimates of CAG use/application/ productivity in your particular situation. Once the system size is calculated, its cost can be determined. Anticipated savings can then be compared to system costs and any of a number of evaluation techniques used to determine feasibility of the investment, e.g., ROI, ROA, PAYBACK PERIOD, etc. A hypothetical example follows in Section 4. Familiarization with CAG capabilities and applications is essential in order to make logical estimates for use in your situation. Fortunately, there is a large source of information available through technical journals, society seminars, vendor salesmen and, perhaps best, conversations with actual users. Good "homework" is mandatory. The sequential steps in evaluation are: 1. Decide which groups will use CAG I/O stations and the headcount [i.e., number of personnel] for each group, e.g., designers, draftsmen, production engineers, etc. 2. Estimate, by group, the per cent of function which must be done manually (MTa) and the anticipated Computer Productivity Factor (CPF). 3. Determine, by group, the resulting Total Productivity Factor (TPF). 4. Calculate the number of I/O stations needed to

D.J. Scott/Economics of Computer Aided Graphics

25

accommodate the anticipated workload - this determines system size. 5. Determine system cost. 6. Estimate anticipated savings in user groups. 7. Compare cost and savings, determine if more rigid analysis is appropriate. Note that the above outline covers the design of an initial system. Yet this procedure can also be used to study a system already in place in such areas as: reassigning I/O stations; planning for system expansion; establishing priority of I/O station use for maximum cost effectiveness; establishing record system to collect ob/ective data regarding MT~, CPF, wait time, etc.

4. Hypothetical Case Example Based on your study of CAG capabilities and knowledge of your unique applications, you have determined the following six sets of information: (1) User groups will consist of Design Engineering, Design Drafting, Numerical Control Programming, Production Engineering, Quality Engineering and Tool Design. (2) Headcounts for these user groups without CAG (Table 1). (3) Estimates of MTI and CPF by user group (given in Table 2). (See Table 3 concerning units of measure). "Resultant CT" is based on a manual operation,.MO% = MT1% + MT2% = 100%, and CPF: = MT2%/CT%, thus CT% = MT2%/CPF = (100-MT~%/CPF. Using the figures for DE, CT% = ( 1 0 0 % - 90%)/3 = 3.33%. (4) Current annual labor costs per person (direct cost plus an allowance for indirect costs) are given in Table 4. (5) Table 5 gives CAG equipment costs (turnkey systems).

Table 1. User Group Headeounts (without CAG) Group

Acronym Headeount

Design Engineering Design Drafting Numerical Control Programming Production Engineering Quality Engineering Tool Design

DE DR NC PE QE TD

150 40 5 150 80 30

26

Applications

Computers h~Industry

Table 2. Estimates of MT1 and CPF

Table 4. Current Annual Labor Costs/Person

Group

MT 1 (% of MO)

CPF

Resultant CT (% of MO)

DE DR NC PE QE TD

90% 30% 30% 90% 95% 40%

3 4 10 2 3 6

3.33% 17.50% 7.00% 5.00% 1.67% 10.00%

Group

Annual K$

DE DR NC PE QE TD

50 30 35 40 40 35

Table 5. CAG Equipment Costs (Turn-key Systems)

(6) Finally, r e c u r r i n g a n n u a l costs (1-shift b a s e ) are d e p i c t e d in Table 6.

Control system (1 for each 8 I/O stations) $ 220 K I/O station (interactive) $ 40 K each Output Plotter (1 for each I/O station) $ 20 K each Application software $ 40 K - 1 time only Training and facilities $ 1 0 0 K - 1 time only

T h e above i n f o r m a t i o n can be used to d e t e r m i n e a T o t a l P r o d u c t i v i t y F a c t o r ( T P F ) for e a c h f u n c t i o n a l g r o u p , g r o u p h e a d c o u n t w i t h C A G available, n u m b e r of stations required to support the new headcount a n d t h e h e a d c o u n t r e d u c t i o n due t o t h e i n t r o d u c t i o n o f CAG. Results o f these c a l c u l a t i o n s are s u m m a r i z e d in Table 7.

Table 6. Recurring Annual Costs Management Software Maintenance Hardware Maintenance

1. D e t e r m i n e T P F b y a p p l y i n g MTa a n d C P F i n p u t s for (3) a b o v e t o T a b l e 8. F o r e x a m p l e , regarding Tool Designers, MTa = 4 0 % a n d C P F = 6. Table 8, t h e

55 K 100 K 125 K

Table 3. Units of Measure - Section 2 Definitions

ManualOperation=MO=

l

t4T1 [

Computer Operation = CO = [

MT1 [

MT2 ¢T

] ]

MT2 b where CT - C---~ y definition -

The units for the above values may be expressed either in hours or as a percentage of MO, but units cannot be intermixed in calculations, e.g., determining TPF. As an example, let MT 1 = 40 hours, MT 2 = 120 hours and CPF = 3 (unitless). Then: units =hours: units=%ofMO:

MO 160hr 100%

= MT 1 + MT2 = 4 0 h r + 120hrs = 25% + 75%

units=hours: units=%ofMO:

CO 80hr 50%

= MT 1 + CTwhereCT =MT 2 + C P F = 4 0 h r + 40 hr (120 h r - 3) = 25% + 25% ( 7 5 % + 3 )

MO To calculate TPF - - - CO units = hours: TPF =

MT1 + MT 2 MT 1 + CT

40hr+120hr 160hrs = = 2.00 (unltless) 40 hr + 40 hr 80 hr

25%+ 75% units = % of MO : TPF = - - - 25% + 25%

100% 50%

2.00 (unitless)

Computers in Industry

D.J. Scott/Economics of Computer Aided Graphics

27

Table 7. The Results from Analyzing the Hypothetical Case by Group Group

No. Operators Without CAG

TPF

=

1. DE DR NC PE QE TD

150 40 5 150 80 30

1.07 2.11 2.70 1.05 1.03 2.00

No. Operators With CAG

No. I/O STA

Headcount Reduction

2.

3.

4.

140.2 19.0 1.9 142.9 77.7 15.0

5.0 7.0 0.4 7.5 1.3 3.0

use use use use use use

5 7 1 8 2 3

10 21 3 7 2 15

Total = 26

i n t e r c e p t p o i n t for t h e s e r o w a n d c o l u m n values is 2.00. T h u s , T P F = 2 . 0 0 . 2. T h e n u m b e r o f o p e r a t o r s w i t h C A G is d e t e r m i n e d b y u s i n g h e a d c o u n t i n p u t s f r o m ( 2 ) a b o v e in c o n j u n c -

30 and TPF = 2.00. Thus, the number of operators w i t h C A G = 3 0 + 2 . 0 0 = 15. 3. T h e n u m b e r o f I / O s t a t i o n s c a n b e c a l c u l a t e d using either of the following methods. Continuing with the TD example:

t i o n w i t h t h e T P F j u s t c a l c u l a t e d By d e f i n i t i o n , TPF = total time without CAG + total time with. C A G ; h e n c e t o t a l t i m e w i t h C A G --- t o t a l t i m e w i t h out CAG + TPF. For the TD example, headcount =

A. Background Referring to Section 2 we have Table 3; and referring to

Table 8: Total Productivity Factory (TPF) * MT1 (% of MO)

CAG Productivity Factor (CPF) * 20

10

8

6

4

3

2

95

1.05

1.05

1.05

1.04

1.04

1.03

1.03

90

1.10

1.10

1.10

1.09

1.08

1.07

1.05

80

1.23

1.22

1.21

1.20

1.18

1.15

1.11

70

1.40

1.37

1.36

1.33

1.29

1.25

1.18

60

1.61

1.56

1.54

1.50

1.43

1.36

1.25

50

1.90

1.82

1.78

1.71

1.60

1.50

1.33

40

2.33

2.17

2.11

2.00

1.82

1.67

1.43

30

2.99

2.70

2.58

2.40

2.11

1.88

1.54

20

4.17

3.57

3.33

3.00

2.50

2.14

1.67

10

6.90

5.26

4.71

4.00

3.08

2.50

1.82

* TPF

=

MT1 + MT2 • MT2 M-T-15,~ ana CPF -~~ (refer to Section 2.2 and 2.3)

To use Table 8, the user must predict MT1 and CPF. The intercept point of the row (MT1) and Column (CPF) is the TPF. Assume MT1 = 40%, CPF = 4, then TPF = 1.82.

Calculation: Refer to Section 2, Figs. 1 and 2 MT1% = 40%, so MT2% = 100% - MTI% = 100% - 40% = 60% CT% = MT2% - 60% _ 15% CPF 4 TPF =

4 0 % + 6 0 % 100% - --= 40%+ 15% 55%

1.82

28

Applications

Computers in Industr;

Tables 1 and 2, the following: Headcount = 30, MT1 = 40%, CPF = 6, and CT = 10%. Then: 1. MO (Hr/wk) = Headcount × 40 hr/wk = 30 × 40 = 1200 hr/wk). 2. MT1 (Hr/wk)=MO (hr/wk) XMTl(%) = 1 2 0 0 x 0 . 4 = 480 hr/wk. 3. MT 2 (Hr/wk) = MO (hr/wk) - MT 1 (hr/wk) = 1 2 0 0 - 4 8 0 = 720 hr/wk. 4. CT (Hr/wk) = MT 2 (hr/wk) + CPF by defn = 720 + 6 = 120 hr/wk. 5. CO (Hr/wk) = MT 1 (hr/wk) + CT (hr/wk) = 480 + 120 = 600 hr/wk. 6. TPF = 2.0 (1. above). 7. Headcount using CAG = MO headcount + TPF = 30 + 2 = 15 people. B. 1. 2. 3.

Method 1 Hr needed on I/O = CT = 120 hr/wk. Hr available on I/O = 40 hr/wk. No. I/O's to handle need = " 1 " + " 2 " = 120 + 40 = 3 I/O stations.

C. Method 2 1. % time operator needs I/O = CT + CO = 120 + 600 = 20%. 2. No. of operators per I/O --- 100% + " 1 " --- 100 + 20 = 5 operators per 1/O (i.e. if 1 operator uses 20% of I/O capacity then how many operators use 100% of I/O capacity?). 3. No. of I/O's = Headcount using I/O + " 2 " = 15 + 5 = 3 1/O stations. 4. H e a d c o u n t reduction = (number of operators w i t h o u t using I / O ) - ( n u m b e r o f o p e r a t o r s w h e n using I/O's). This figure for t h e T D e x a m p l e equals 3 0 - 1 5 -- 15 people. With t h e n u m b e r o f I / O s t a t i o n s d e t e r m i n e d , syst e m costs can be calculated using t h e figures f r o m

Table 10. Annual System Savings Group

Headcount Reduction

Annual Wage (KS)

DE DR NC PE QE TD

10 21 3 7 2 15

50 30 35 40 4O 35

Annual Saving (K $)

= = = = = =

500 630 105 280 8O 525

T = $ 2,120

i t e m (5) ( n o n - r e c u r r i n g , T a b l e 5) a n d i t e m (6) (recurring, Table 6) a b o v e . T h e results are s h o w n in T a b l e 9. C o m p a r i s o n o f these b u d g e t a r y cost a n d saving figures ( T a b l e 10) s h o w t h a t the ratio o f A n n u a l Savings to N o n - r e c u r r i n g Costs is 0 . 8 2 o n a one-shift or 1.59 o n a t w o - s h i f t basis. M a n a g e m e n t can determ i n e i f a m o r e t h o r o u g h analysis is advisable b y comparing c o r r e s p o n d i n g ratios o f o t h e r p r o p o s a l s c o m p e t i n g for available capital dollars. A s s u m e t h a t C A G is viewed w i t h f a v o u r a n d t h a t a m o r e t h o r o u g h analysis is r e q u i r e d . T h e n e x t step w o u l d p r o b a b l y b e to c o n s i d e r cash flow a n d the t i m e value o f m o n e y . A n e s t i m a t e o f a n n u a l costs a n d savings m u s t b e m a d e in o r d e r t o use y o u r c o m p a n y ' s p r e f e r r e d m e t h o d o f financial analysis, be it ROI, ROA, P a y b a c k Period, etc. Table 11 was g e n e r a t e d b a s e d o n t h e costs and savings for t h e o n e - s h i f t o p e r a t i o n s h o w n above. F o r ease o f i l l u s t r a t i o n , n o n - r e c u r r i n g costs were split equally b e t w e e n f o u r systems w i t h o n e s y s t e m proc u r e d yearly. R e c u r r i n g costs were also split equally

Table 9. System Costs

1. Non-recurring a. Control System b. I/O Stations c. Output Plotters d. Application Software e. Training/facilities Total 2. Annual Recurring a. Management b. Software Maintenance c. Hardware Maintenance Total

1-Shift Base

2-Shift Base

4 @ $ 220 K 26 @ 40 K 26 @ 20 K 40 K 100 K

2 @ $ 220 K 13 @ 40 K 13 @ 20 K 40 K 75 K

$ 2,580 K

T = $ 1,335 K

$ 55 K 100 K 125 K

$ 80K 100 K 90 K

$280K

T=

$270K

Table 11.10 Year Cost-Savings Projection Year

Non-recurring Cost (KS)

Recurring Cost (KS)

Annual Cost (KS)

Annual Savings (KS)

1 2 3 4 5 6 7 8 9 10

645 645 645 645 0 0 0 0 0 0

0 70 140 210 280 280 280 280 280 280

645 715 785 855 280 280 280 280 280 280

0 265 795 1,325 1,855 2,120 2,120 2,120 2,120 2,120

Computers in Industry per system and were not applied during the year in which a particular system was obtained. Savings were also equally split between systems with no savings during the year a system is bought, 50% of its savings in the second year and a 100% savings in all subsequent years. The point is that you must make similar assumptions based on your company's experience factors. Using the above figures with an analysis program which: 1. Calculates the discounted rate of return which equates the present value cash flow to zero for a selected project life (10 years in example) 2. Computes payback period on a net cash flow ... produces these results: Discounted rate of return = 54.4% Payback Period = 4 yr 6 mo. Additional iterations of financial analysis can be made imposing more stringent conditions. For example, the projected savings of this example were based on multiple users for individual I/O stations but no allowance was made for operator "wait time" when I/O stations are busy, i.e. a "queue factor" (QF). Cost of this idle time would have to be deduced from the projected labor savings. Other considerations might include an allowance for equipment down time to more closely approach "real world" conditions. This could be reflected in reduced savings. Or each user group might be broken into sub functio~as so that a more exact assignment of CPF and MT1 can be made, e.g., go from a single function "drafting" to functions of "mechanical drafting", "PCB drafting" and "IC drafting". The user will have to determine the type and number of iterations which best fit his or her situation. For example, to include an allowance for equipment down time would require more CAG capacity. Again using the TD example in Table 7 and assuming a 10% downtime, the number of I/O stations required would be 3.33 (instead of 3). 1. Hr needed on I/O = CT = 120 hr/wk 2. Hr available on I/O = 40 x 90% = 36 hr/wk 3. No. I/O's to handle need = "1" + "2" = 3.33 I/O Sta Rounding to the next integer now shows a requirement for four I/O stations instead of three. This increases both non-recurring cost (more equipment) and recurring cost (maintenance) which in turn makes any financial analysis appear less favorable. (An off-

D.J. Scott/Economics of ComputerAided Graphics

29

setting "plus" consideration would be that potential "wait time" would be reduced due to the added station).

5. "What If" Analysis As illustrated in the hypothetical case example, values must be selected for six sets of information in order to determine system cost and savings. Since these values are often "best judgment" figures, they are subject to the game of "what if", i.e., what if some or all of the selected values were changed. A computer program can be written so that all selected values are stored in assigned memory locations and recalled during program execution. Changing a value in memory allows the program to be rerun with nearly instantaneous results based on these "what if' speculations. Such a program has been written for the TI59 programmable calculator. Once selected values are entered into memory, execution of the program results in calculations of system "non-recurring costs", "annual recurring costs" and "annual savings". The results of program execution as shown are: non-recurring cost = $ 2,580,000 annual recurring cost = 280,000 annual savings = 2,120,000 i.e., the same values as shown for the hypothetical case example in Section 4. "What ip' the number of Tool Designers is 22 instead of 30, the manual percentage (MT1) of a Draftsman is 0.45 instead of 0.3 and the cost of a I/O station is $ 47,200 instead of $ 40,000? How do these changes impact cost and savings? These three values are substituted in their respective data memory locations and ,the program rerun. Resulting calculations are: non-recurring cost = $ 2,700,000 annual recurring cost = 280,000 annual savings = 1,830,000 This program could be expanded to include revised payback and rate of return calculations. It illustrates the use and value of a computer program which permits rapid substitution of the value of variables to analyze extremes, i.e., minimum and maximum expectations.

30

Applications

6. Method Considerations Any analysis which relies in part on subjective data is subject to criticism. A risk analysis should be made based on real constructive criticism and "other" potential savings. The following "pro" and "con" statements list a few of the more obvious considerations regarding confidence in results of this proposed analysis.

6.1. "Con's" • The method is based on subjective assumptions, especially regarding the selection of manual VS computer-assisted portions of a function (MT1) and CAG efficiencies (CPF). • Projected savings are based on a headcount reduction - or at least not adding personnel when business expands. Yet some users have not experienced a manpower reduction from the introduction of computer-aided methods. • The question of introducing existing data into the CAG data base was not addressed, e.g., should all, a portion or no existing data be converted? At what cost? Problems of a dual data base, etc. • There could be large unanticipated costs should catastrophic loss of data occur. File management costs of tracing, controlling, storing and retrieving data in a new form may be significantly greater than anticipated.

Computers in Industry' and salvage, faster through-put, etc. • Some projects cannot be handled any other way, e.g., VLSI's. * Analysis can be used to re-allocate existing CAG resources and is not limited to analysis of a new system. • "Other" cost effective uses will evolve thereby adding to the potential savings, e.g., Technical Publications, Plant Layout, etc. • The hypothetical example used costs required for a one-shift operation. Utilizing a two-shift operation would substantially reduce equipment costs. • CAG technology lends itself to the concept of a "paperless factory", a popular topic today.

7. Summary "Scienttfic Management" principles should be used as much as possible to assess and size a new or re-configure an existing CAG system. This paper has presented one method of applying "logical" steps to configure a system. The amount of iterative refinements is left to the user's imagination and needs. The results should only be classed as "budgetary "' since they are based on subjective selection of data inputs. Yet these results should provide a clear indication for potential use and further analysis. It seems appropriate in closing to state the Golden Principle in support of any analysis based in part on subjective inputs.

6.2. "Pro's" • No estimates of intangible savings were included - and these could be substantial and important, e.g., a better design debugged on paper rather than in the factory results in fewer changes, less scrap

Golden principle: Nothing will be attempted if all possible objections must first be overcome.