Burning a rope to count time

New answer - 116 rupees

So, apparently we can get even cheaper than 120 rupees, this time by buying only a single 16-minute rope, and six 7-minute ropes (cost is $1 \times 32 + 6 \times 14 = 116$). This is how you do it:

1. Create a 2-minute rope with the 16-minute rope and two 7-minute ropes (remaining ropes: $4 \times 7, 1 \times 2$)
2. Use the 2-minute rope to turn two 7-minute ropes into 5-minute ropes (remaining ropes: $2 \times 7, 2 \times 5$)
3. Use one of the 5-minute ropes to turn the remaining 7-minute ropes into 2-minute ropes (remaining ropes: $1 \times 5, 2 \times 2$)
4. Use the two 2-minute ropes to reduce the 5-minute rope to the 1-minute rope you really wanted all along

I no longer dare to say this might be optimal.

Old answer - 120 rupees

The cheapest I have been able to work out so far is 120 rupees: 2 16-minute ropes and 4 7-minute ropes (cost is $2 \times 32 + 4 \times 14 = 120$). You do this as follows:

1. Light the two 16-minute ropes and a 7-minute rope to create two 9-minute ropes (remaining ropes: $2 \times 9$, $3 \times 7$)
2. Use a 9-minute rope and a 7-minute rope to create a 2-minute rope (remaining ropes: $1 \times 9$, $2 \times 7$, $1 \times 2$)
3. Light the two 7-minute ropes and the 2-minute rope to create two 5-minute ropes (remaining ropes: $1 \times 9$, $2 \times 5$)
4. Use the 9-minute rope and one of the 5-minute ropes to create a 4-minute rope (remaining ropes: $1 \times 5$, $1 \times 4$)
5. Use your 5-minute rope and 4-minute rope to create the desired 1-minute rope

I'm reasonably confident this is optimal, but I have no idea how to conclusively prove it.

Start by creating a 2 minute rope as you described. Then start burning three 7 minutes rope at the same time, for 2 min. You are left with three 5 min ropes. Start burning a 16 minute ropes for three times five minutes.

THIS IS NOT AN ANSWER.

Just some thoughts that might help - please let me know if I've misunderstood any questions/answers.

The costs for the 2 kinds of rope are just double the ropes' burn times, so really we're looking to minimise the number of rope-minutes we buy.

The asker's solution involves a [7 minute] rope and 3 [2 minute] ropes. We can make 3 [2 minute] ropes from 3 [16 minute] ropes lit simultaneously with a [7 minute], and then another [7 minute] we light after the first [7 minute] burns. This means we buy 3 [7 minute] ropes, and 3 [16 minute] ropes, for a total of 69 minutes purchased, 138 rupees.

Andrei's solution involves creating 3 [5 minute] ropes, and burning them alongside a [16 minute]. We can create 3 [5 minute] ropes from 5 [7 minute ropes] and a [16 minute] rope - first the [16] burns with the first [7], then with the second [7], then we light the last 3 [7]s and they end up as [5 minute] ropes. This means we buy 2 [16 minute] ropes and 5 [7 minute] ropes, for a total of 67 minutes purchases, 134 rupees.

If we skipped any tricks with creating [x minute] ropes (x!=7,x!=16), we would be burning 7 [7 minute ropes] for 49 minutes, and 3 [16 minute] ropes for 48 minutes, giving us a [1 minute] rope made from the 7th [7 minute] rope. This costs 194 rupees.

I written up these answers in their optimal forms - to create Y [X minute] ropes I assume we can do the process that creates 1 [X minute] rope, but when we light up the rope that ends up being [X minutes] we buy Y of them instead of 1. This is how we can create 1 [5 minute] rope by purchasing 1 [16], 3[7]s, the end rope having started off as a [7], but we can create 101 [5 minute] ropes by purchasing 1 [16] and 103 [7]s, 2 of which are burnt in the process. This is obviously cheaper than buying 1[16], 3[7]s, 101 times.

We have to begin by burning at least 1 [16] and at least 1 [7] - otherwise we're wasting rope, since can't make any guesses or burn at both ends. We must burn at least 1 other [16], because otherwise we're stuck with just a ridiculous number of [7]s (9, by making 4 [5]s (uses up 6) and then burning those as a [20] alongside 3 more [7]s).

So we have to use at least 2 [16]s. Andrei's solution uses that many, and only 5 [7]s - to use only 4 [7]s I think is impossible, given that the 5 is already an optimisation. If we used 3 [16]s we're on par with the asker's solution, so we'd have to use only 2 [7]s - this also seems pretty impossible, not least because we'd be stuck on even numbers.