In each issue of the Automation Notebook we feature a section of brainteasers. These are the brainteaser answers from Issue 21, 2011 of the Automation Notebook. The brainteaser questions are repeated in black. The answers to the brainteaser questions are highlighted in red with explanations. You can view the brainteasers from Issue 21, 2011 without the answers here:http://library.automationdirect.com/brainteasers-issue-21-2011/
1. Queue See
The non-automated assembly line at Widgets”R”Us is notorious for its uneven quality – and its eccentric QC inspectors. After a full day’s production the three person Quality Control team evaluated the finished parts. Lisa initially flagged half of all the parts as defective, but on further inspection she cleared 10 of the suspect parts. Tim reviewed all the parts passed by Lisa and flagged one third of them as defective, but then he was able to fix two of them. The last inspector – Sally – flagged one fifth of the remaining parts, but then changed her mind about one of them and cleared it. After these inspections the factory was only able to ship 17 parts for the day. What was the total number of parts (good and bad) for the day’s production run?
Answer: There were 34 total parts produced that day:
Sally started with 5/4(17-1) or 20 parts
Tim started with 3/2(20-2) or 27 parts
Lisa started with 2(27-10) or 34 parts
2. Roto-Router
A rotary indexing table has nine workstations positioned around its perimeter. The workstations are located at 15°, 40°, 60°, 110°, 135°, 200°, 275°, 325°, & 350° respectively. The motion control system driving the table can be programmed for only two move distances (in degrees), but those moves can each be made in either direction. For accuracy all the moves must originate at the home position (0°). What are the two move distances that will allow the table to reach all the workstation positions with the shortest number of total moves? (The moves required to return to the home sensor don’t count.) For example: moves of 125° and 75° would work and could reach all the workstations with a total of 28 moves – but there is a better solution.
This motion control system could certainly benefit from the flexibility of a SureServo™ system from AutomationDirect!
Answer: Move distances of 100 degrees and 125 degrees will allow the table to reach all the workstations with only 26 total moves.
Let’s call the 125 degree move the ‘major’ move and 100 degrees the ‘minor’ move:
the station at 15° can be reached with three major moves forward,
the station at 40° can be reached with four minor moves forward,
the station at 60° can be reached with three minor moves backward,
the station at 110° can be reached with two major moves backward,
the station at 135° with one major and one minor – both backward,
the station at 200° with two minors forward,
the station at 275° with three majors forward, and one minor backwards,
the station at 325° with one major and two minors,
the station at 350° with two majors and one minor – both forward, a total of 26 moves.
3. One-Lane Highway
You have N cars that are all traveling the same direction on an infinitely long one-lane highway. Unfortunately – but typically – the drivers would all prefer to go different speeds, but they cannot pass each other. So eventually the cars will group up in one or more clumps.
In terms of N, what is the expected number of clumps of cars?
Answer: The answer is the sum of the harmonic series: 1 + 1/2 + 1/3 + 1/4 + … + 1/n
One good way to visualize this answer is to consider what happens if you were to place the cars onto the highway (in a random position) in order of their speed, from slowest to fastest. The slowest car, placed first, is a ‘clump’. The next slowest car might be placed ahead of or behind it, but it forms a new clump only if placed in front of the slower car – with probability 1/2. And an expected value of 1 * ½ = ½. The next slowest car has three possible positions (in front, in the middle, or last), and forms a new clump in only one of those positions (the front), and that probability is 1/3. And so on — car N only forms a new clump if it’s placed at the front of the pack, with probability 1/N.