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Second level
Third level
Fourth level
Fifth level
Core problem:
Sometimes a realistic problem seems to have an easily identified core that fits a recognised problem. You can start by taking the core first and working outwards. We decided easily, at the outset, the the core Package Router problem was a required behaviour control problem
Ancillary problems
There are usually sub problems surrounding the core (typically information subproblems). Info subproblems usually easy to separate out from the core; it treats its real world as autonomous
Standard decompositions
Some problem frames are composites and have standard decompositions. When we recognised the information display problem we saw it would require a dynamic model and also needed a static model.
Concerns and difficulties
Examine sub problem for common or characteristic concerns and difficulties – often they will reveal further sub problems.
Session of tennis takes an hour or two; a member’s subscription is valid for a season or for a year, or even longer. This shows there are at least two different sub problems here.
Managing subscriptions is a dynamic domain: members join and leave, pay or don’t pay subscriptions.
Managing court and its lights: treat membership as static
Billing for lights is a third subproblem because it straddles two tempi. Items to be billed are games of tennis, recorded in the fast problem. Billing and payment proceeds at a slower pace, perhaps monthly. Late payers and defaulters may find their membership suspended.
Indicative mood
An assertion of fact. If the motor polarity is set to up and the motor switch setting is changed from  off to on the lift starts to rise within 250 msecs (properties of the motor and winding gear).
Optative mood
Expresses a wish. If an up call button is pressed when the corresponding light is off the light comes on and remains on until the call is serviced by the lift stopping at that floor and leaving in the upwards direction.
Subproblem 1: based on (a) and (b). (a) is given and the requirement is (b)
Subproblem 2: detect failures to satisfy (b)
Subproblem 3: based on (a) the negation of (b) and on (c): (a) is given, NOT (b) is given, and the requirement is (c).
A is indicative – what’s true anyway
B is optative – what customer ideally wants
C is super-optative – what customer must have if B is impossible