Design allowances: five added ten per cent, nobody added sixty one

A flow figure crosses five desks and each one puts an undeclared allowance on the number it was handed. Predict what reaches the purchase order, 10%, 50%, or 61%. Allowances applied in sequence multiply rather than add, so 3 of them give 1.331 and 5 give 1.61051, and the handover at which the running figure first passes the size you typed as adequate is named. No allowance is named for any real item and no equipment is selected.

Your prediction before the experiment

Five successive ten per cent allowances add either ten per cent or fifty per cent. The running values stay hidden until the handover count moves once.

What stays the same

Underlying quantity: The same declared basis at every handover
Information state: Unchanged apart from the typed allowance
Size rule: Choose the first typed size greater than or equal to the running figure

The handover trace

Try a scenario

Allowance bearing handovers0 handovers

Only this whole count moves the reveal. The visible rate fields are used in order.

First figure

Typed sizes

Enter 2 to 40 increasing positive values separated by commas. No size is interpolated.

Eight typed allowance rates

Handover 1%

Handover 2%

Handover 3%

Handover 4%

Handover 5%

Handover 6%

Handover 7%

Handover 8%

Boundary

It reveals multiplication across one declared basis and a visitor typed size list. It does not recommend an allowance, select equipment, separate code requirements from discretionary margin, or approve a design.

Here is the maths

The formula, in words

Begin at one. For every active handover, multiply by one plus that handover's typed percentage divided by one hundred. Multiply the first figure by the product. The first typed size at or above the result contains it. Equality remains contained, and no value is inserted between typed sizes.

M(n)=\prod_{i=1}^{n}\left(1+\frac{r_i}{100}\right)

where $M(n)$ is the multiplier, $i$ the handover index, $n$ the number of handovers, and $r_i$ the handover rate.

A worked example

With a first figure of 100 and ten per cent at every handover, one handover gives 110. Two give 121, which first exceeds size 120 and moves to size 135. Five give multiplier 1.61051 and figure 161.051, so size 170 is the smallest typed size that contains it. The increase is 61.051 per cent, not 50 per cent.

100 \times 1.1 = 110

What this does not model

The product does not decide whether any allowance is appropriate, distinguish a code requirement from discretionary margin, select equipment, approve a design, or reconcile a changed scope or information basis.

Check it yourself

At three handovers of ten per cent, the product is $1.1 \times 1.1 \times 1.1 = 1.331$ . At five it is 1.61051. A strict crossing means a running value equal to the initial typed size remains contained.

Source basis

The mechanism is ordinary public multiplication. It reproduces no design rule, code factor, equipment method or wording from sold guidance.

Choosing a real allowance or equipment size is outside this lesson, and those requests are always refused.