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CHAPTER 3: TOLERANCE ALLOCATION 

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3.0 Tolerance Analysis vs. Tolerance Allocation
In the previous chapter it was shown that analytical modeling of assemblies provides a quantitative basis for the evaluation of design variations and specification of tolerances. An important distinction in tolerance specification is that engineers are more commonly faced with the problem of tolerance allocation rather than tolerance analysis.
The difference between these two problems is illustrated in Figure 3.1. In tolerance analysis the component tolerances are all known or specified and the resulting assembly variation is calculated. In tolerance allocation, on the other hand, the assembly tolerance is known from design requirements, whereas the component tolerances to meet these requirements are unknown. The available assembly tolerance must be distributed or allocated among the components in some rational way. The influence of the tolerance accumulation model and the allocation rule chosen by the designer on the resulting tolerance allocation will be demonstrated.
Another difference in the two problems is the yield or acceptance fraction of the assembly process. The assembly yield is the quality level. It is the percent of assemblies which meet the engineering tolerance requirements. It may be expressed as the percent of acceptable assemblies or the percent rejects. For high quality levels, the rejects may be expressed in partspermillion (ppm), that is, the rejects per million assemblies .
In tolerance analysis the assembly yield is unknown. It is calculated by summing the component tolerances to determine the assembly variation, then applying the upper and lower spec limits to the calculated assembly distribution. In tolerance allocation, on the other hand, the assembly yield is specified as a design requirement. The component tolerances must then be set to assure that the resulting assembly yield meets the spec.
The rational allocation of component tolerances requires the establishment of some rule for distributing the assembly tolerance among the components. Sections 3.13.6 discuss the rules available in CATS.
Figure 3.1. Tolerance Analysis vs. Tolerance Allocation.
3.1 Allocation By Proportional Scaling
The designer begins by assigning reasonable component tolerances based on process or design guidelines. The component tolerances are summed to see if they meet the specified assembly tolerance. If not, the component tolerances are scaled by a constant proportionality factor. In this way the relative magnitudes of the component tolerances are preserved.
This algorithm is demonstrated graphically in Figure 3.2 for an assembly tolerance Tasm, which is the sum of two component tolerances, T1 and T2. The straight line labeled as the Worst Case Limit is the locus of all possible combinations of T1 and T2 which, added linearly, equal Tasm. The parabola labeled Statistical Limit is the locus of root sum squares of T1 and T2 which equal Tasm.
Suppose the designer chooses initial values for T1 and T2 based on typical process tolerances for the two component parts. This combination is the point labeled Original Tolerances in the figure. By drawing a line from the origin through this point, then extending it until it intersects the Worst Case Limit, the largest possible tolerances for T1 and T2 are obtained, which satisfy the worst case condition and which still have the original ratio of T1 to T2. Extending this line further, until it intersects the Statistical Limit curve, new values for T1 and T2 are obtained which satisfy the assembly tolerance limit by root sum squares.
Although Figure 3.2 illustrates 1D tolerance accumulation models, the algorithm may be applied to 2D or 3D tolerance accumulation by premultiplying each component tolerance by its tolerance sensitivity.
Figure 3.2. Graphical interpretation of tolerance allocation by proportional scaling.
3.2 1D Example: Worst Case Tolerance Allocation by Proportional Scaling
The following example is based on the shaft and housing assembly shown in Figure 3.3.
Figure 3.3. Shaft and housing assembly.
Initial tolerances for parts B, D, E, and F are selected from tolerance guidelines such as those illustrated in Figure 3.4 [1]. Tolerances are chosen from the middle of the range for each part size. This table is available in the AutoCATS 2D Analyzer's online Tolerance Reference Handbook.
Figure 3.4. Tolerance range of machining processes [1].
The retaining ring (A) and the two bearings (C and G) supporting the shaft are vendorsupplied, hence their tolerances are fixed and must not be altered by the allocation process. The critical clearance is the shaft endplay, which is determined by tolerance accumulation in the assembly. The vector diagram overlaid on the figure is the assembly loop that models the endplay. The average clearance is the vector sum of the average part dimensions in the loop:
Initial Tolerance Specifications
Required Clearance = .020+/.015
Average Clearance = A + B  C + D  E + F  G
= .0505 + 8.000  .5093 + .400  7.711 + .400  .5093
= .020
Dimension A B C D E F G
Nominals .0505 8.000 .5093 .400 7.711 .400 .5093
Tolerances(+/)
Design .008 .002 .006 .002
Fixed .0015 .0025 .0025
The clearance tolerance is obtained by summing the component tolerances by worst case:
TSUM = + TA + TB + TC + TD + TE + TF + TG
= + .0015 + .008 + .0025 + .002 + .006 + .002 + .0025
= .0245 (too large)
Now, solving for the proportionality factor:
TASM = .015 = .0015 +.0025 +.0025 + P (.008 + .002 + .006 + .002)
P = .47222
Note that the three fixed tolerances were subtracted from the assembly tolerance before computing the scale factor. Thus only the four design tolerances are reallocated:
TB = .47222 (.008) = .00378 TE = .47222 (.006) = .00283
TD = .47222 (.002) = .00094 TF = .47222 (.002) = .00094
Each of the design tolerances has been scaled down to meet assembly requirements as shown in Figure 3.5. This procedure could also be followed assuming a statistical sum for the assembly tolerance, in which case the tolerances would be scaled up. Results are summarized in Table 3.1.
Figure 3.5. Tolerance allocation by proportional scaling.
3.3 Allocation by Weight Factors
A more versatile method of assigning tolerances is by means of weight factors. Using this algorithm, the designer assigns weight factors to each tolerance in the chain and the system distributes a corresponding fraction of the tolerance pool to each component. A larger weight factor for a given component means a larger fraction of the tolerance pool will be allocated to it. In this way, more tolerance can be given to those dimensions which are the more costly or difficult to hold, thus improving the producibility of the design.
Figure 3.6 illustrates this algorithm graphically for a two component assembly. The original values for component tolerances T1 and T2 are selected from process considerations and are represented as a point in the figure, as before. The tolerances are scaled, similar to proportional scaling, only the scale factor is weighted for each component tolerance so the greater scale factors yield the least reduction in tolerance.
Figure 3.6. Graphical interpretation of tolerance allocation by weight factors.
Again, although the figure illustrates 1D tolerance accumulation models, the algorithm may be applied equally well to 2 D and 3D stacks. It may also be applied to worst case, statistical, or six sigma tolerance sums.
Note that any components which are vendorsupplied, or subject to other design considerations, can be excluded from the allocation process by declaring them to be "fixed".
3.4 1D Example: Worst Case (WC) Allocation by Weight Factors
The shaft and housing assembly of section 3.2 will be revisited, using weight factors to allocate tolerances. The three tolerances A, C and G are fixed. Design tolerances B, D, E and F are assigned weight factors of 10, 20, 10 and 20, respectively. Weights were determined on the basis of machining difficulty.
The clearance tolerance is obtained by summing the component tolerances by worst limits:
TSUM = + TA + TB + TC + TD + TE + TF + TG
= + .0015 + .008 + .0025 + .002 + .006 + .002 + .0025
= .0245 (too large)
Solving for the proportionality factor with weight factors:
Ti' = PWiTi
TASM = .015 = .0015 +.0025 +.0025 +
P [(10/60)(.008) + (20/60)(.002) + (10/60)(.006) + (20/60)(.002)]
P = 2.31818
Note that only the four design tolerances are reallocated:
TB = 2.31818(10/60)(.008) = .00309 TE = 2.31818(10/60)(.006) = .00232
TD = 2.31818(20/60)(.002) = .00155 TF = 2.31818(20/60)(.002) = .00155
3.5 1D Example: Statistical (RSS) Allocation by Weight Factors
The clearance tolerance is obtained by summing the component tolerances by root sum squares:
TSUM = R(TA^{2} + TB^{2} + TC^{2} + TD^{2} + TE^{2} + TF^{2} + TG^{2})
= R(.0015^{2} + .008^{2} + .0025^{2} + .002^{2} + .006^{2} + .002^{2} + .0025^{2})
= .01108 (too small)
Solving for the proportionality factor with weight factors:
Ti' = PWiTi
TASM^{ = }R(TA^{2}+ P^{2}WB^{2}TB^{2}+ TC^{2}+ P^{2}WD^{2}TD^{2}+ P^{2}WE^{2}TE^{2}+ P^{2}WF^{2}TF^{2}+ TG^{2})
P = 7.57238
Note that the three fixed tolerances are not reallocated:
TB = (7.57238)(10/60)(.008) = .01010 TE = (7.57238)(10/60)(.006) = .00757
TD = (7.57238)(20/60)(.002) = .00505 TF = (7.57238)(20/60)(.002) = .00505
All of the preceeding examples of allocation are compared in Table 3.1. A graphical comparison is shown in Figure 3.7.
Table 3.1 Comparison of Allocation Methods
Proportional  Weight Factors  
Original  Worst Case 
Stat ± 3 sigma 
Weight Factor 
Worst Case 
Stat ± 3 sigma 

Dim  Tolerance  
A  .0015*  .0015  .0015  0  .0015  .0015 
B  .008  .00378  .01116  10  .00309  .0101 
C  .0025*  .0025  .0025  0  .0025  .0025 
D  .002  .00094  .00279  20  .00155  .00505 
E  .006  .00283  .00837  10  .00232  .00757 
F  .002  .00094  .00279  20  .00155  .00505 
G  .0025*  .0025  .0025  0  .0025  .0025 
Assembly Tolerance Scale Factor (P) 
.0150  .0150  .0150  .0150  
.47222  1.39526  2.31818  7.57238  
*Fixed tolerances 
Figure 3.7. Tolerance allocation by proportional scaling and weight factors
3.6 Tolerance Allocation Using Least Cost Optimization
A promising method of tolerance allocation uses optimization techniques to assign component tolerances such that the cost of production of an assembly is minimized. This is accomplished by defining a costvs.tolerance curve for each component part in the assembly. The optimization algorithm varies the tolerance for each component and searches systematically for the combination of tolerances which minimizes the cost.
Figure 3.8 illustrates the concept simply for a three component assembly. Three costvs.tolerance curves are shown. Three tolerances (T1, T2, T3 ) are initially selected. The corresponding cost of production is C1 + C2 + C3. The optimization algorithm tries to increase the tolerances to reduce cost; however, the specified assembly tolerance limits the tolerance size. If tolerance T1 is increased, then tolerance T2 or T3 must decrease to keep from violating the assembly tolerance constraint. It is difficult to tell by inspection which combination will be optimum. The optimization algorithm is designed to find it with a minimum of iteration. Note that the values of the set of optimum tolerances will be different when the tolerances are summed statistically than when they are summed by worst case.
Figure 3.8. Optimal tolerance allocation for minimum cost.
A necessary factor in optimum tolerance allocation is the specification of costvs.tolerance functions. Several algebraic functions have been proposed, as summarized in Table 3.2. CATS uses the Reciprocal Power function: C = A + B/tol^{k}. The constant coefficient A may include setup cost, tooling, material, prior operations, etc. The B term determines the cost of producing a single component dimension to a specificed tolerance and includes the charge rate of the machine. When tighter tolerances are called for, speeds and feeds may be reduced and the number of passes increased, requiring more time and higher costs. The exponent k describes how sensitive the process cost is to changes in tolerance specifications.
Table 3.2 Proposed CostofTolerance Models
Cost Model  Function  Author  Ref 
Reciprocal Squared  A + B/tol^{2}  Spotts  [2] 
Reciprocal  A + B/tol  Chase&Greenwood  [3] 
Reciprocal Power  A + B/tol^{ k}  Chase et. al.  [4] 
Exponential  A e^{B(tol)}  Speckhart  [5] 
Little has been done to verify the form of these curves. Manufacturing cost data are not published since they are so sitedependent. Even companies using the same machines would have different costs for labor, materials, tooling and overhead.
A closedform solution for the leastcost component tolerances was developed by Spotts [2]. He used the method of Lagrange Multipliers, assuming a cost function of the form C=A+B/tol^{2}. Chase et. al. [4] extended this to cost functions of the form C=A+B/tol^{k} as follows:
F([[partialdiff]],[[partialdiff]]Ti) (Cost function) + [[lambda]] F([[partialdiff]],[[partialdiff]]Ti) (Constraint) = 0 (i = 1, . . n)
F([[partialdiff]],[[partialdiff]]Ti) ([[Sigma]] (Aj + Bj/Tj^{k}j)) + [[lambda]] F([[partialdiff]],[[partialdiff]]Ti) ([[Sigma]]Tj^{2 }Tasm^{2}) = 0 (i = 1, . . n)
[[lambda]] = F(kiBi,2Ti^{(k}i^{+2)}) (i = 1, . . n)
Eliminating [[lambda]] by expressing it in terms of TSDO2(1):
Ti = BBC((F(kiBi,k1B1))^{1/(k}i^{+2)} .T1^{(k}1^{+2)/(k}i^{+2)} (8)
Substituting for each of the Ti into the assembly tolerance sum:
T^{2}ASM = T1^{2} + [[Sigma]] BBC((F(kiBi,k1B1))^{2/(ki+2)}.T1^{2(k1+2)/(ki+2)} (9)
The only unknown in equation 9 is T1. CATS iterates the value of T1 until both sides of equation 9 are equal to obtain the minimum cost tolerances. A similar derivation for Ti based on a worst case assembly tolerance sum yields:
TASM = T1 + [[Sigma]] BBC((F(kiBi,k1B1))^{1/(ki+1)}.T1^{(k1+1)/(ki+1)} (10)
A graphical interpretation of this method is shown in Figure 3.9. Note that as the method of Lagrange Multipliers assumes, the constant cost curves are tangent to the tolerance limit curves for the minimum cost tolerance values.
Figure 3.9. Graphical interpretation of minimum cost tolerance allocation.
Numerical results for the 1D shaft and housing example problem are shown in Table 3.3:
Table 3.3 Minimum Cost Tolerance Allocation
Tolerance Cost Data  Allocated Tolerances  
Dimension  Setup A 
Coefficient B 
Exponent k 
Original Tolerance 
Worst Case 
Stat. 
A  *  *  .0015*  .0015*  .0015*  
B  $1.00  0.15997  0.43899  .008  .00254  .0081 
C  *  *  .0025*  .0025*  .0025*  
D  1.00  0.7202  0.46823  .002  .001736  .00637 
E  1.00  0.12576  0.46537  .006  .002498  .00792 
F  1.00  0.7202  0.46823  .002  .001736  .00637 
G  *  *  .0025*  .0025*  .0025*  
Assembly Variation  .0245(WC) .0111(RSS) 
.0150(WC)  .0150(RSS)  
Assembly Cost  $9.34  $11.07  $8.06  
Acceptance Fraction  1.000  .9973  
"True Cost"  $11.07  $8.08 
The Setup Cost is coefficient A in the cost function. Setup cost does not affect the optimization. For this example, the setup costs were all chosen as equal, so they would not mask the affect of the tolerance allocation. In this case, they merely added $4.00 to the assembly cost for each case.
Parts A, C and G are vendorsupplied. Since their tolerances are fixed, their cost cannot be changed by reallocation, so no cost data was included.
Note that in this example the assembly cost increased when worst case allocation was performed. This is due to the fact that the original tolerances, when summed by worst case, gives an assembly variation of 0.0245 in. This exceeded the specified assembly tolerance limit of 0.015 in. Thus, the component tolerances had to be tightened, driving the cost up. When summed statistically, however, the assembly variation was only .00111 in., which was less than the spec limit, allowing the allocation algorithm to increase the component tolerances, driving the cost down. A graphical comparison is shown in Figure 3.10. It is clear from the graph that tolerances for B and E were reduced the most in the Worst Case model, while D and F were increased more in the Statistical model.
Figure 3.10. Comparison of minimum cost allocation results
The advantages of the Lagrange multiplier method are:
The limitations are:
Problems which exhibit these characteristics may be optimized using nonlinear programming techniques or by optimizing tolerances for one assembly spec at a time, then choosing the lowest set of component tolerance values required to satisfy all assembly specs simultaneously.
3.7 True Cost and Optimum Acceptance Fraction
The "True Cost" in Table 3.3 is defined as the total cost of an assembly divided by the acceptance fraction or yield. Thus, the "true cost" is adjusted to include a share of the cost of the rejected assemblies. It does not include, however, any parts which might be saved by rework or the cost of rejecting individual component parts.
An interesting exercise is to calculate the optimum acceptance fraction, that is, the rejection rate which would result in the minimum True Cost. This requires an iterative solution. For the example problem, the results are shown in Table 3.4:
Table 3.4. Minimum True Cost
Cost Model  A  Zasm  Opt. Accept Fract  True Cost 
A + B/tol^{k}  ^{$4.00}  ^{2.03}  ^{.9576}  ^{$7.67} 
A + B/tol^{k}  $8.00  2.25  .9756  $11.82 
The interpretation of these results is that loosening up the tolerances will save money on production costs, but will increase the cost of rejects. By iterating on the acceptance fraction, it is possible to find the value which minimizes the combined cost of production and rejects. Note, however, that the setup costs were set very low. If setup costs were doubled, as shown in the second row of the table, the cost of rejects would be higher, requiring a higher acceptance level.
In the very probable case where individual process cost vs. tolerance curves are not available, an optimum acceptance fraction for the assembly could be based instead on moreavailable cost per reject data. This optimum acceptance fraction could then be used in conjunction with allocation by proportional scaling or weight factors to provide a meaningful costrelated alternative to allocation by least cost optimization.
3.8 Advanced Allocation FeaturesNominal Allocation
Centering the Assembly Distribution Mean
A powerful design tool built into CATS is the ability to adjust nominal dimensions in order to move the mean of the assembly distribution to the center of the tolerance zone. Centering the mean puts the greatest distance between the tolerance limits and the mean, thus minimizing the number of rejects for a given assembly tolerance distribution. Looking at Figure 3.11, we see that the original distribution has its right hand tail extended way beyond the upper spec limit (USL), indicating a large number of rejects. In addition, this design specification is such that its limits are nonsymmetric with respect to the nominal assembly spec dimension. By adjusting the component nominals, the assembly mean may be moved to the center, halfway between the USL and LSL. Clearly, the centered distribution will have fewer rejects.
Figure 3.11. Effects of nominal allocation options
Deciding which component nominals to change is a design decision. Some components may be vendorsupplied and not allowed to change. Others may be too costly to change due to tooling modifications or other requirements. Of course, if the assembly is in production, the fewer parts you have to change, the better.
This problem is similar to tolerance allocation, only here it involves small changes in the mean or nominal values of component dimensions instead of changes in the specified component tolerances. The center of the distribution is shifted without changing the spread of the distribution.
A set of weight factors is provided in CATS for allocation of component nominals, similar to the weight factors used for tolerance allocation. The designer sets the value of the weight factors corresponding to those components he wishes to change. The relative weights determine how much change is assigned to each component.
USL or LSL Justification of the Mean
Small changes may also be made to individual nominal dimensions to shift the nominal spec towards either the upper or lower spec limit. This may be desirable if one of the two spec limits is more important than the other. As with the centering allocation, the distribution is simply shifted without changing its spread. For example, a minimum clearance (LSL) may be critical to an assembly, but the maximum clearance may not be important.
In USL justification, the center of the distribution is shifted such that the +Zasm[[sigma]] limit matches with the USL. Similarly, in LSL justification, the distribution is shifted such that the Zasm[[sigma]] limit matches with the LSL.
All three nominal allocation methods may be used in conjunction with any tolerance allocation method to simultaneously shift the mean and change the spread of the distribution.
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