Reliability Allocation

The first step in the design process is to translate the overall system reliability requirement into reliability requirements for each of the subsystems. This process is known as reliability allocation. The allocation of system reliability involves solving the basic inequality:

where

is the is the allocation reliability parameter for the ith subsystem

R* is the system reliability requirement parameter

f is the functional relationship between subsystem and system
reliability

For a simple series system in which the R’s represent probability of
survival for t hours, the above equation becomes

Theoretically, the above equation has an infinite number of solutions, assuming no restrictions on the allocation. The problem is to establish a procedure that yields a unique or limited number of solutions by which consistent and reasonable reliabilities may be allocated. For example, the allocated reliability for a simple subsystem of demonstrated high reliability should be greater than for a complex subsystem whose observed reliability has always been low.

The allocation process is approximate. The reliability parameters apportioned to the subsystems are used as guidelines to determine design feasibility. If the allocated reliability for a specific subsystem cannot be achieved at the current state of technology, then the system design must be modified and the allocations reassigned. This procedure is repeated until an allocation is achieved that satisfies the system level requirement and all constraints and results in subsystems that can be designed within the state of the art.

In the event that it is found that even with reallocation some of the individual subsystem requirements cannot be met within the current state of the art, then the designer must use one or any number of the following approaches (assuming that they are not mutually exclusive) in order to achieve the desired reliability:

1. Find more reliable component parts to use
2. Simplify the design by using fewer component parts, if this is possible without degrading performance
3. Apply component derating techniques to reduce the failure rates below the averages
4. Use redundancy for those cases where 1, 2 and 3 above do not apply

It should be noted that the allocation process can, in turn, be performed at each of the lower levels of the system hierarchy, e.g., equipment, module, component.

Equal Apportionment Technique

In the absence of definitive information on the system, other than the fact that n subsystems are to be used in series, equal apportionment to each subsystem would seem reasonable. In this case, the nth root of the system reliability requirement would be apportioned to each of the n subsystems. The equal apportionment technique assumes a series of n subsystems, each of which is to be assigned the same reliability goal. A prime weakness of the method is that the subsystem goals are not assigned in accordance with the degree of difficulty associated with achievement of these goals. For this technique, the model is:

or

where

R* is the required system reliability

R*i is the reliability requirement apportioned to subsystem i

Example

Consider a proposed communication system which consists of three subsystems (transmitter, receiver, and coder), each of which must function if the system is to function. Each of these subsystems is to be developed independently. Assuming each to be equally expensive to develop, what reliability requirement should be assigned to each subsystem in order to meet a system requirement of 0.729?

The apportioned subsystem requirements are found as:

R*T = R*R = R*C = (R*)l/n = (0.729)1/3 = 0.90

Then a reliability requirement of 0.90 should be assigned to each subsystem.

AGREE Apportionment Technique

A method of apportionment for electronics systems is outlined by the Advisory Group on the Reliability of Electronic Equipment (AGREE) takes into consideration both the
complexity and importance of each subsystem. The importance, wi, is the probability that the system fails given that a module, i, is critical and fails.
and fails

Let:

i = a counter representing each module, i = 1, 2, 3 …, n

t = system operating time

R*(t) = system reliability requirement at time t

ti = operating time of module i

λi = failure rate of module i

wi = probability that the system fails given that module i is critical
and fails

For n total modules in the system, the contribution of each module containing m components, to the overall system reliability is:

For example,

Each module’s unreliability is:

Equation 1

If an exponential failure is assumed, then the unreliability of a module is also given by

and the probability that the module is critical and fails is

Equation 2

The AGREE allocation process dictates that equation 1 must be equal to equation 2, therefore

Equation 3

Solving equation 3 for the module failure rate

AGREE Example

A system has four subsystems, each with 20 modules, as shown in the picture below (range A8:B11).  The required system reliability is 0.9 for a four hour mission (range B3:B4) and the probability that the system fails when a subsystem fails is 1.0 (range C8:C11). What should the allocated module reliability be if:

1. All modules are equally important?
2. Module 3 becomes twice as complex as the other modules?
3. Module 3 is only 10% as important as the other modules?

Question 1
For the stated inputs, each subsystem must have an MTBF of 152 hours (range F8:F11).   The reliability of each subsystem must be 0.974 (range H8:H11), which when multiplied together results in an overall system reliability of 0.90 (cell H4).

Question 2
If module 3 has 40 components instead of 20 (cell B10), this module now has an allocated MTBF of 95 hours and the remaining three modules must have an MTBF of 190 hours (range F8:F11) to achieve the overall system reliability goal of 0.9 for a 4 hour mission.

Question 3
If the quantity of components is put back to 20, but module 3 now has an importance of only 0.1, meaning that 90% of the failures will not cause the system to fail, the allocated MTBF for this module is only 13 hours instead of 152 hours. Note, the product of the module reliability values, 0.684, does not equal the requirement of 0.9 because not all failures of module 3 will cause a system failure.

AGREE Excel allocation example

Feasibility of Objectives Technique

This technique was developed primarily as a method of allocating reliability without repair for mechanical-electrical systems. In this method, subsystem allocation factors are computed as a function of numerical ratings of system intricacy, state of the art, performance time, and environmental conditions. These ratings are estimated by the engineer on the basis of his experience. Each rating is on a scale from 1 to 10, with values assigned as discussed:

System Intricacy. Intricacy is evaluated by considering the probable number of parts or components making up the system and also is judged by the assembled intricacy of these parts or components. The least intricate system is rated at 1, and a highly intricate system is rated at 10.

State of the Art. The state of present engineering progress in all fields is considered. The least developed design or method is a value of 10, and the most highly developed is assigned a value of 1.

Performance Time. The element that operates for the entire mission time is rated 10, and the element that operates the least time during the mission is rated at 1.

Environment. Environmental conditions are also rated from 10 through 1.  Elements expected to experience harsh and very severe environments during their operation are rated as 10, and those expected to encounter the least severe environments are rated as 1.

The ratings are assigned by the design engineer based upon his engineering know-how and experience. They may also be determined by a group of engineers using a voting method such as the Delphi technique. An estimate is made of the types of parts and components likely to be used in the new system and what effect their expected use has on their reliability. If particular components had proven to be unreliable in a particular environment, the environmental rating is raised. The four ratings for each subsystem are multiplied together to give a rating for the subsystem. Each subsystem rating will be between 1 and 100,000. The subsystem ratings are then normalized so that their sum is 1.

The basic equations are:

and the failure rate allocated to each subsystem is

where

C’k = complexity of subsystem k

C’k = C’k / W’k

w’k = r1k r2k r3k r4k

λS = system failure rate

T = misson duration

N = number of subsystems

w’k = rating for subsystem k

rik = rating for each of the four factors for each subsystem

Feasibility of Objectives Example

A system consists of the following six subsystems

1. propulsion
2. ordnance
3. guidance
4. flight control
5. structures
6. auxiliary power

A system reliability of 0.90 for a 120 hour mission is required.

Solution

Assuming an exponential failure distribution and solving the following equation for the overall system reliability requirement

results in a system failure rate, λ, of 878 failures per million hours (FPMH).

Engineering estimates of intricacy, state of the art, performance time, and environments are made based on input from subject matter experts (cells C2:F7). The overall rating for each subsystem is the product of the individual ratings and is computed in column G. The relative complexity of each subsystem is the overall rating for the subsystem divided by the total for all subsystems (e.g., 750/7720 = 0.097, shown in cell H2).  The allocated failure rate for each subsystem is then the the overall required system failure rate, 878 FPMH computed above, multiplied by the complexity associated with each subsystem, as shown in column I.

Feasibility of Objectives Excel allocation example

Dynamic Programming Approach to Reliability Allocation

The dynamic programming technique is applicable to multistage (or sequential) decision problems. The technique converts such a problem to a series of single-stage optimization problems.

In addition to defining the stages of such a process, four attributes of the problem must be identified if the technique is to be applied:

1. The set of all possible states at each stage.
2. The set of all possible decision alternatives available at each stage.
3. The function transforming from one state to a previous state, which depends on the current state.
4. A function defining the return realized at a specific stage resulting from the state and the alternative chosen.

A multistage decision problem may then be converted to a series of single-stage decision problems as reflected by a set of recursion equations, for which the total return from the optimal set of decision alternatives may be evaluated for maximization or minimization..

For subsystems operationg in series (i.e., no redundant paths), the dynamic programming formulation pertains to apportionment of system reliability requirements among series subsystems in such a manner as to minimize the total expenditure of development effort (e.g., cost). Some basic assumptions which are fundamental to the formulation are discussed below:

1. At any particular stage of the development program (at time of apportionment), the system can be partitioned into n subsystems and the present reliability level can be estimated for each subsystem. Failure of any subsystem will cause system failure. In addition, it is assumed that the subsystem goal cannot be less than its estimated present level.
2. The n subsystems function independently so that expected system reliability resulting from the subsystem goals can be expressed as the product of these subsystem goals, where y is the system reliability goal and y1, y2, etc. is the reliability goal for subsystem 1, subsystem 2, etc.
3. An effort function can be identified for each subsystem, defining the number of units of development effort (e.g., cost) required to raise its reliability level from the present value to any potential reliability goal. The effort may represent a single important resource or a combination of resources, if these can be expressed by a common unit. The effort function may be either continuous or discrete. A continuous mathematical function allows the reliability goal to assume any value between the estimated present level and one. A discrete function limits potential subsystem goals to particular values.

Consider a proposed system comprised of n subsystems, each of which are to be developed independently. These subsystems are to function independently and in series. What reliability goal should be assigned to each subsystem in order that the system goal be satisfied at a minimum expenditure of development effort?

The problem may be converted to a dynamic programming problem as follows:

1. Identify each of the n subsystems as a stage such that an apportionment goal must be determined at each stage. A specific numbering sequence for the stages (subsystems) is not necessary, but each subystem must maintain its assigned identity throughout the entire procedure.
2. Define the set of all possible states at each stage.
3. Define the set of all possible alternatives at each stage.
4. Define the return realized at each stage as a function of the decisions made.

Example Using Dynamic Programming

To exemplify the use of the technique, consider a proposed system which is to be developed as three independent subsystems. The system can be functionally successful if, and only if, each of the three subsystems function properly. In order that the system fulfill its intended role, it should have an overall system reliability of 0.90.. Based on engineering analysis and historical information of similar type equipment, estimates of the state-of-the-art reliability levels of the subsystems are 0.95, 0.95, and 0.97.

What reliability goal should be assigned to each subsystem in order to minimize the total expenditure of development funds? The estimated effort (funds) functions for the three subsystems are as follows, expressed in \$1000 units.

Potential apportioned goals are limited to those contained in these tabled functions above. First, (0.95) (0.95) (0.97) = 0.875425 < 0.90, which  indicates that further development/improvement in subsystem reliability is necessary to meet the system reliability goal of 0.90.

The following table shows the cost functions to increase the reliability for each of the three subsystems. For example, to increase the reliability of subsystem 2 from 0.95 to 0.98 would cost \$81.2K, as shown in cell E21. The reliability at stage 3 is, by definition 1.0, the starting reliability for no subsystems.  If subsystem 3 is considered, the reliability at this stage is the product of the starting reliability and the reliability options for subsystem 3.  These are listed across row 28. Enumeration of all combination’s of options for subsystem 3 reliability (y3) and subsystem 2 reliability (y2) is shown in the table starting at row 30.

Continuing the process described above, but now listing all possible combination’s previously calculated for subsystems 3 and 2 down the column B, starting in row 41 (this is the data calculated in range C33:H36, ranked from the lowest reliability option to the highest reliability option) and considering the reliability of subsystem 1 results in the possible reliability values enumerated in cell range C411:H64. The options that meet the overall system reliability requirement are below the red line.

The following tables consider the cost of various reliability options enumerated in the above table.  These costs were given as inputs for incrementally improving the reliability of subsystem 1. For example, one option is the leave subsystem 2 with a reliability of 0.95, and subsystem 3 with a reliability of 0.97, giving a combined reliability of 0.9215, and increasing the reliability of subsystem 1 from 0.95 to 0.98, at a cost of \$16.5K, as highlighted in yellow, cell F70. This will result in an overall system reliability of 0.9031, as shown in cell F41 of the above table. While the table starting in row 67 enumerates cost options available from varying the reliability of subsystem 1, the table starting in row 96 enumerates cumulative options available by varying both y1 and y2.

The table below, starting in row 108 enumerates cumulative cost options available by varying y1, y2 and y3.  The minimum cost is obtained leaving the reliability of subsystem 3 at 0.97, subsystem 2 at 0.95, which costs nothing, and increasing the reliability of subsystem 1 from 0.95 to 0.98, at a cost of \$16.5K. This result is shown in cell C111. The resulting system reliability for  this option will be 0.9031, as calculated in cell F41, two figures above.

References:

1. MIL-HDBK-338, Electronic Reliability Design Handbook, 15 Oct 84
2. Bazovsky, Igor, Reliability Theory and Practice
3. O’Connor, Patrick, D. T., Practical Reliability Engineering
4. Birolini, Alessandro, Reliability Engineering: Theory and Practice
5. Reliability of Military Electronic Equipment, Advisory Group on the Reliability of Electronic Equipment (AGREE), Office of the Assistant Secretary of Defense, Research and Engineering, Washington, D.C., 4 June 1957; U.S. Government Printing Office.