In this method, the most stringent maintainability requirements (that is, the lowest MTTR values) are allocated to the subsystems and components having the lowest reliability; and conversely, the least stringent maintainability requirements are allocated to the subsystems components having the highest reliability. The assumption is that the most complex items will have the highest failure rates. For that reason, the method is referred to as the Failure Rate Complexity Method (FRCM). The procedure for the method is as follows:
Step 1. Determine N_{i}, the number of each item in the product for which the allocation is being made.
Step 2. Identify _{i}, the failure rate for each item (constant failure rate is assumed).
Step 3. Multiply _{i} by N_{i} to find C_{fi}, item is contribution to total failure rate.
Step 4. Express each item's MTTR, M_{i}, as the product of (_{H}/_{i}) and M_{H}, where H is the item having the highest failure rate.
Step 5. Multiply each result from Step 4 by the corresponding _{i}. The result is C_{Mi}.
Step 6. Using equation 6, solve for the MTTR of the item having the highest failure rate.
where C_{Mi} = M_{i} C_{fi}
Step 7. Solve for the MTTR of the other items by multiplying the MTTR found in Step 6 by _{H}/_{i}.
Table XI illustrates an example of maintainability allocation using the FRCM, for the subsystems shown in Figure 17. The same method was used to allocate the MTTRs found for subsystem B to its components.
Table XI  Allocation Using Failure Rate Complexity
Method
Item 
Step 1.
Determine No. of Items per Product
(N_{i}) 
Step 2.
Identify Failure Rate _{i} x 10^{3}f/H 
Step 3.
Calculate Contribution to Total Failure Rate
C_{fi} = Ni _{i }x
10^{3}f/H 
Step 4.
Express each MTTR (M_{i}) as (_{H}/_{i}) x
M_{H} 
Step 5.
Calculate Contribution to System MTTR C_{Mi} =
M_{i} C_{fi} 
A 
1 
5 
5 
M_{a} 
5 M_{a} 
B 
1 
1.111 
1.111 
4.5 M_{a} 
5 M_{a} 
C 
1 
.833 
.833 
6 M_{a} 
5 M_{a} 



C_{fi}=
6.944 

C_{Mi}=
15 M_{a} 
Step 6. Solve for M_{a} MTTR_{Product} = C_{Mi}/ C_{fi }==> 1.44 = 15
M_{a} =
.67 hours 
Step 7. Solve for M_{b} and
M_{c} M_{b} =
4.5 M_{a} = 3 hours ; M_{c} =
6 M_{a} =
4 hours 
Figure 17  Example of Maintainability
Allocation