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Before we start, idea blatantly stolen from this puzzle

Tom has found another job, with another Boss who always arrives at the same time on weekdays. This boss, however, has a different policy. If Tom is noticed to be tardy by no more than an hour, he must complete a full eight hour shift and earns two Tardiness Points. If he is noticed to be tardy by more than an hour, he must additionally come in and work a 4 hour shift on the weekend, when the Boss comes in at a different time for a 5 hour shift (possibly before Tom is supposed to arrive), and earns 3 Tardiness points. If he earns 5 Tardiness points in a rolling two week period, he's fired on the spot.

Tom is scheduled to work an 8 hour day, from 9 A.M. to 5 P.M, with potential supplemental work on weekends, from 10 A.M to 2 P.M. He is allowed to use the break room once a day, for no more than 15 minutes, and can see whether his boss is in the office (or arriving) on his way both to and from the break room.

Tom wants to minimize the amount of time he works, while keeping his job. What strategy will average the least amount of time working, while never getting him fired.

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  • $\begingroup$ Could you clarify "when the Boss comes in at a different time for a 5 hour shift (possibly before Tom is supposed to arrive), and earns 3 Tardiness points" Is this saying the boss could come in anytime and Tom won't know it? $\endgroup$ – gtwebb Sep 15 '16 at 17:00
  • $\begingroup$ @gtwebb The Boss works for 5 hours on the weekends, always at the same time, but potentially different from the time he comes in on weekdays. Tom will have no reference for when the Boss works on weekends until he works at least one weekend. This is stated because the initial puzzle is trivial if The Boss comes in before Tom on a regular day, because Tom is completely unable to shave any time off his work. $\endgroup$ – Sconibulus Sep 15 '16 at 17:06
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For simplicity, we will assume the day starts at the zero minute and ends at the 480th minute.

To give us discrete intervals, we will assume the following:
1. If Tom is in transit (arriving, going to/from the break room) during the same minute as the Boss arrives, then Tom will witness his arrival. 2. Tom's movement to/room the break room takes one minute. His break while in the break room can be at most 15 minutes.

From the previous question, we know that Tom can probably figure out the Boss' arrival time in 7 days never getting caught. In the worst case, the Boss will arrive close to 480 and Tom will have worked a little over 15 hours more than necessary. The best case is the Boss arrives close to 0 and Tom works very little extra time. The average would then be under 8 hours for Tom to work out the exact time of the Boss's arrival.

Terminology:

$T_a$ - The arrival of Tom. Can be $0 <= T_a <= 480$.
$T_B$ - The arrival of the Boss. Can be $0 <= T_B <= 480$.
$T_b$ - The time Tom leaves to go on break.
$T_r$ - The time Tom returns to his desk.

Safe Strategy

If Tom never gets caught, then our strategy is as follows:

Arrive at $T_a=0$ the first day. Leave to take a break at $T_b=232$, arrive in the break room in the 233nd minute, leave the break room at $T_r=248$, and arrive back at his desk in the 249th minute. There are 6 possibilities.

  1. $T_B=T_a$ - Tom arrives at the same time as the Boss. Tom has determined the Boss' arrival time with no (additional) time worked.
  2. $T_a < T_B < T_b$ - The Boss arrives before Tom's break.
  3. $T_B=T_b$ - Tom witnesses the Boss arriving as Tom goes to take his break
  4. $T_b < T_B < T_r$ - The boss arrives while Tom is on his break.
  5. $T_B=T_r$ - Tom witnesses the Boss arriving as Tom returns from his break.
  6. $T_r < T_B$ - The Boss arrives after Tom's break.

In the 1st, 3rd, or 5th, Tom has determined the Boss' exact arrival time. The wasted time that Tom worked is simply $T_B-T_a$.

For the forth, Tom has narrowed the arrival time of the Boss down to 15 minutes.

For the 2nd and 6th, Tom has narrowed the arrival time of the Boss down to 232 possible minutes. So, Tom can then adjust his arrival and break times to the given interval, always arriving before the Boss.

For all cases, Tom has wasted exactly $T_B-T_a$. On average, this is 4 hours with a $\frac{3}{480}$ chance of no further wasted time.

In the next day, Tom will be able to again take a 15 minute break, this time getting an interval of 108 minutes. Tom's average excess time worked will be 116 minutes with a $\frac{3}{232}$ chance of no further wasted time.

On the third day, Tom will waste an average of 54 minutes and get the interval down to 46 minutes with a $\frac{3}{108}$ chance of being done.

On the forth day, Tom will waste an average of 23 minutes, and get the interval down to 15 minutes. There is a $\frac{3}{46}$ chance of completion.

On the 5th day, Tom will adjust his break to be only 5 minutes. He will get the interval down to 5 minutes, waste on average 7.5 minutes, and have a $\frac{3}{15}$ chance of being completed.

On the 6th and final day, Tom will be able to determine the exact time of the Boss' arrival and waste on average 2.5 minutes.

Thus, the total amount of time wasted is:

$$240 + \frac{477}{480} \times \left( 116 + \frac{229}{232} \times \left( 54 + \frac{105}{108} \times \left( 23 + \frac{43}{46} \left( 7.5 + \frac{12}{15} \times 2.5 \right) \right) \right) \right) = 438.65$$

This is 7:18:38 of wasted time, which is 90% of the original interval.

Risky Strategy

TBD... this is a little more difficult.

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Note: I will use for my convenience a 24h notation for the hours. 1 PM is 13.00, 2 PM is 14.00... 12 PM (midnight) is 24.00

A sure strategy could be

Tom arrives the first day at 9 AM and uses immediately his 15 minutes in the break room. He can determine if Boss's arrival is between 9.00 and 9.15 with his walk to and from the room. If it happens, Tom will start this routine in the next days arriving always one minute later and reducing his pause to check if the Boss arrives at 9.01 or at 9.14 and so on. If the Boss arrives after 9.15 Tom repeates his routine arriving 15 minute later ad have a 15 minutes long pause until he determines the 15 minutes window of arrival, then repeat the "refining" step

If Tom has balls of steel

he can starts his new job with 4,5 or more hours late (for example at 16.00). He argues with the Boss, receives 3 points and a 4 hours duty in the weekend. He now have 4 hours in the weekend which are less than 4+ hours of rest in the first day. Now he can apply the "sure strategy". If after the two firs week he does not found the Boss arrival (15 mins x 5 weekdays x 2 weeks = 2,5 hours, so if the Boss arrives after 11.30 ) at the beginning of the third week he can make the trick again.

The two strategies have in common to

arrive always at work 1 min before the Boss (and do the pause whenever he wants). If Tom is brave, once he determines the optimal time of arrival, he could repeat the "balls of steel trick" every 3 weeks to gain some addictional hour of rest (or until the Boss becomes angry).

and Tom's strategy works if

none of Tom's collegues is a spy and Tom has a good watch.

After the last consideration Tom can set up the

team-work strategy, in which every collegues can arrives every 15 minutes and the team reduces the amount of days needed fore the "sure" strategy

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  • $\begingroup$ This strategy will get Tom down to the minimum amount of work-time per week eventually, but the average number of hours he works getting there will be fairly high, and it will take him longer than the strategies proposed in the answer to the base question. $\endgroup$ – Sconibulus Sep 15 '16 at 17:32
  • $\begingroup$ he must additionally come in and work a 4 hour shift on the weekend I took this to mean that he works another 8 hour shift and then a 4 hour weekend shift for a total of 12 hours. $\endgroup$ – Trenin Sep 15 '16 at 17:33
  • $\begingroup$ @Trenin this is also correct, sorry I glazed over that portion of the answer. 'Teamwork' is also prohibited, you want as few people knowing about your tardiness as possible, in case someone is a snitch. $\endgroup$ – Sconibulus Sep 15 '16 at 17:36
  • $\begingroup$ @Trenin and Sconibulus sorry I misunderstanding the meaning of the "punishment". I will think about it (and yes, teamwork is not a proper solution!). $\endgroup$ – marcoresk Sep 15 '16 at 17:38

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