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At my work we have $2$ machines that are fitted end to end so that an array tape goes in machine $1$ from one end and it gets read $8 \cdot 144 \, \mathrm{mm}$ in $15$ minutes. When comes out of the machine $1$, the tape then can be fed to the machine $2$ which reads the same length$(8 \cdot 144 \, \mathrm{mm})$ in $7$ minutes. Before starting the process we will know how many array sections we are gonna read, thus what is the total length of the tape (e.g., $100 \text{ arrays} \cdot 144 \, \mathrm{mm} = 14400 \, \mathrm{mm}$).

So the question is, is there a way we can derive a mathametical formula to know after how many array being read by the machine $1$ if we start reading machine $2$ we will never have a sceraio when machine $2$ run out of array and try to pull the tape out of machine $1$? Lets say a safe gap of minimum $5$ arrays always should be there between these $2$ machines.

Update: So, thanks to Tobi's awesome answer I was able to use app inventor and develop a very minimalistic app, that can tell me the number of tape to wait for before i can start reading it with machine 2.

Though I'm not good at maths but always love to share things in return, so, I'm more than happy to add the image of the code that I have written, so that whoever is looking for a solution like this can get a fair idea of not only how to work it out but also how to implement it as end to end.

Once more: Thanks Tobi.

Image of the solution implemented in Ai2.appinventor

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  • $\begingroup$ The issue is that the "speed" of machine 2 (one 8*144 in 7 miniues) is (more than) double the speed of machine 1 (one in 15 minutes). Thus, you have to start feeding m-2 at least after m-1 has processed half of the total wotk. In alternative, n°2 machines 1 in parallel. $\endgroup$ – Mauro ALLEGRANZA Nov 25 '16 at 10:52
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    $\begingroup$ I vote to close, because in my humble opinion this task - if it can't be solved by yourself - should at least be paid for. Therefore I recommend educating yourself (and asking question on this platform, should you need help with that) or hiring someone to do it for you. PS: OP's profile is spot on... $\endgroup$ – Stefan Mesken Nov 25 '16 at 12:17
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    $\begingroup$ I've solved it. He shouldn't need to pay for it. The question was an interesting puzzle for me. Why the calculus tag? I didn't need it. Stack Exchange is a site where we get answers for free. Only payments should be donations(if they exist) $\endgroup$ – Tobi Alafin Nov 25 '16 at 12:47
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Let the length of the tape be $D_T$

Let the speed of the first tape(slower one) be $S_1$

Let the speed of the second tape(faster one) be $S_2$

Let the length of the tape for which your request is satisfied be $D_x$

Let the total time for which tape2 will run be $t$

Distance = speed * time

$$D_T = S_2 t \tag{1} $$

$$ D_x+ S_1 \cdot t\gt S_2 \cdot t + 5 \tag{2} $$

$$t= D_T / S_2 \tag{3} $$

Collect like-terms in (2)

$$D_x \gt S_2 \cdot t - S_1 \cdot t + 5, $$

$$D_x \gt t(S_2 - S_1) + 5\tag{4}$$

From (1) $$ t = D_T /S_2 \tag{5} $$

Sub (5) into (4)

$$ D_x \gt {D_T} / {S_2} *(S_2 - S_1) + 5 \tag{6} $$

Plug the values from your question into (6)

$$D_x \gt D_T \cdot \frac{7} {8} \cdot (\frac{8}{7} - \frac{8}{15}) + 5, $$

$$ D_x \gt (\frac{8}{15})D_T + 5 \tag{7} $$

That's your final answer.

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  • $\begingroup$ Can a kind sould please come and help improve my formatting. The answer looks ugly as is. $\endgroup$ – Tobi Alafin Nov 25 '16 at 12:49
  • $\begingroup$ If still any error in my edit, please check once. $\endgroup$ – Narasimham Nov 25 '16 at 19:15

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