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04-BS-4 · December 2017

Question 7 of 7: Two-Floor Elevator Control Logic

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-4 Electric Circuits and Power — December 2017
National Exams, 3 hours, closed book (one aid sheet permitted). Any five of the seven questions constitute a complete paper; all seven are solved below as a complete study resource.

Reference texts

Question 7: Two-Floor Elevator Control Logic (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Nine binary sensors A–I as defined above (occupancy, floor position ×2, motion, two corridor buttons, two in-car buttons, door status); corridor buttons are valid only when $A=0$ (empty), in-car buttons only when $A=1$ (occupied); all motion commands require doors closed ($I=1$).

Find. Minimal AND/OR/NOT logic for: (a) UP command from a 1st-floor corridor call, (b) the combined UP/DOWN command including in-car button presses, (c) the cabin LIGHT, (d) the ALARM.

Approach. Build each output as one AND term per valid trigger, OR'd together; "disable when X" translates directly to an AND with the literal $X'$ (complement), and every movement term is additionally gated by the elevator being at the departure floor and the doors closed.

  1. a) UP from the 2nd-floor corridor button. Moving 1st→2nd needs: the 2nd-floor corridor button F pressed, no one already inside (corridor controls enabled, $A'$), the car currently at the 1st floor ($B=1$), and doors closed ($I=1$): $$\text{UP}_{corridor}=\boxed{F\cdot A'\cdot B\cdot I}$$
  2. b) Combined command including in-car buttons. A passenger already inside can also request the 2nd floor via in-car button H (needs $A=1$, still requires $B=1,I=1$); symmetrically for DOWN, corridor button E (needs $A'$) or in-car button G (needs $A$) request the 1st floor from the 2nd ($C=1,I=1$):
    FA'HABIUP
    UP = B·I·(F·A’ + H·A) - move 1st→2nd floor
    EA'GACIDOWN
    DOWN = C·I·(E·A’ + G·A) - move 2nd→1st floor
    $$\text{UP}=B\cdot I\cdot(F\cdot A'+H\cdot A)\ ,\qquad \text{DOWN}=C\cdot I\cdot(E\cdot A'+G\cdot A)$$ Each is a single sum-of-products expression: the car only moves from a floor it is actually at, with doors closed, in response to whichever button type is currently enabled by occupancy.
  3. c) Elevator light. On whenever occupied OR doors open ($I'$):
    AILIGHT
    LIGHT = A + I’
    $$\text{LIGHT}=\boxed{A+I'}$$
  4. d) Alarm. Sounds only when stopped ($D'$) AND not at either floor (between floors, $B'\cdot C'$):
    DBCALARM
    ALARM = D’·B’·C’ (stopped between floors)
    $$\text{ALARM}=\boxed{D'\cdot B'\cdot C'}$$
Final results — Question 7
OutputBoolean expression
UP (1st→2nd)$B\cdot I\cdot(F\cdot A'+H\cdot A)$
DOWN (2nd→1st)$C\cdot I\cdot(E\cdot A'+G\cdot A)$
LIGHT$A+I'$
ALARM$D'\cdot B'\cdot C'$
Check: sensors A–I are stated to never activate two-at-once (e.g. B and C mutually exclusive), so no additional interlock beyond the AND terms above is needed to prevent contradictory commands.
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