Yes. I have started a cost-benefits analysis. Cost varies as a function of total production volume (V), the rate of output (x), the date of first delivery (T), and the date of completion of the full production run (m), where x(t) denotes the rate of output at moment t. Moreover, as the total quantity of units produced increases, the cost of future output tends to decline because production-knowledge increases as a result of production experience (this proposition is known as the ‘learning’ or ‘progress’ curve"). That is: dC / dT | x = x0 , V = V0 , < 0. This relationship probably holds for most products produced in large batches using traditional mass-production methods. Moreover it is usually assumed that:
dC / dx(t) | T = T0, V = V0 > 0 d2C / dx(t)2 | T = T0, V = V0 > 0 dC / dV | x = x0, T = T0 > 0 d2C / dV2 | x = x0, T = T0 < 0
5 comments:
perfect! someone should design this into a durable foam structure and sell it in the kiosks on St. Mark's Place.
Yes. I have started a cost-benefits analysis. Cost varies as a function of total production volume (V), the rate of output (x), the date of first delivery (T), and the date of completion of the full production run (m), where x(t) denotes the rate of output at moment t. Moreover, as the total quantity of units produced increases, the cost of future output tends to decline because production-knowledge increases as a result of production experience (this proposition is known as the ‘learning’ or ‘progress’ curve"). That is: dC / dT | x = x0 , V = V0 , < 0. This relationship probably holds for most products produced in large batches using traditional mass-production methods. Moreover it is usually assumed that:
dC / dx(t) | T = T0, V = V0 > 0
d2C / dx(t)2 | T = T0, V = V0 > 0
dC / dV | x = x0, T = T0 > 0
d2C / dV2 | x = x0, T = T0 < 0
So, uh, I'll get back to you on that.
id like to hedge on its derivatives
used to be a set of works!
so true
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