Seek Time Model

Seek time is the time that it takes to reposition the mover above the relevant tips when servicing a new non-sequential request. To model the seek time we used a probabilistic approach. We assume that the starting locations of requests are uniformly distributed across the device.

Because the movement in $X$ and $Y$ may be considered independently [10,6], the seek time can be computed as the greater of the seek times in $X$ and $Y\!$. We computed the average seek time when the seek in $X$ is greater ( $t_{seek~x}(\delta_x)$), and the average seek time when the seek in $Y$ is greater ( $t_{seek~y}(\delta_y)$) and estimated these probabilities $a$ and $b$, which add up to 1, to obtain Equation 9. We reason that a higher average seek time in one direction will have a higher impact on the overall seek time, so $a/b = t_{seek~x}(\delta_x)/ t_{seek~y}(\delta_y)$.

$$t_{seek}(\delta_x, \delta_y) = a \times t_{seek~x}(\delta_x) + b \times t_{seek~y}(\delta_y)$$ (9)

While several models exist to estimate the seek time, the one we adopted is based on simple acceleration rules from Newtonian mechanics [13,10] and is given in Equation 10, for a specified distance $\delta$. The mover can seek achieve a much higher velocity than is used to read/write because it can accelerate during the seek. At the end of a seek, the data must be accessed by moving in the $Y$ direction. Therefore, a settle time, $t_{settle}$ (for the mover to position itself accurately) applies only to the calculated seek time in $X\!$, since the mover has to come to a complete stop in that direction. Notice that seeking takes place within one tip area. Therefore, the seeking distance and the seek time are relatively short, and depend on the mover movement range.

$$t(\delta) = \sqrt{\frac{2 \pi \delta}{a_0}}$$ (10)

We calculate the average seek time in $X$ and $Y$ as follows. First, we calculate the probability function $p(d)$ of an incoming request incurring a certain movement $d$ over the tip array either $X$ or $Y$ (we can consider each direction independently). Because the starting sectors are uniformly distributed, the distances in each direction will be the distance between two uniformly distributed starting sectors. This probability will vary linearly with distance and will be at its maximum for a zero displacement, and it will be equal to zero when the displacement is equal to the movement range. The form of the equation is $p(d) = c - b d$, where $c$ and $b$ are constants. Because $p$ is a probability function, the area under it in the range 0 to $\delta$, where $\delta$ is the movement range in $X$ or $Y\!$, which is $c \delta / 2$ will be equal to 1. Additionally, when $d=\delta$, the probability is equal to zero, $p(\delta) = 0$. Using these constraints, we can calculate the values for the constants $c$ and $b$, which we substitute back in the expression for $p(d)$ to obtain Equation 11:

$$p(d)= 2(\delta-d)/\delta^2$$ (11)

Unfortunately, we need to calculate seek time distributions, not distance distributions. We can use Equation 10 to express $t$ as a function of a displacement $d$, converting Equation 11 from the distance domain to the time domain. This gives us the probability of a seek incurring time $t$ in either $X$ or $Y\!$. After substituting 250 m/$\rm {s}^2$ for the physical parameter acceleration, $a_0$, we obtain Equation 12:

$$p(t)=\frac{500t}{\pi \delta}(1-\frac{250t^2}{2\pi \delta })$$ (12)

The movement ranges $\delta_x$ and $\delta_y$ in $X$ and $Y$ from Table 1 specify the maximum distance we can move in each direction. We use them in Equation 10 to find the maximum seek time $t_{max}$ in $X$ or $Y$, shown in Equation 13.

$$t_{max} = \sqrt{\frac{2\pi \delta}{250}}$$ (13)

The actual seek time is the greater of the seek times in $X$ and $Y\!$, so the probability distribution function $P(t)$ of the seek time when it is greater in one direction than the other will be different than $p(t)$. Reasoning that taking the maximum will bias us towards larger seek times, we approximated $P(t)$ with Equation 14.

$$P(t) = \alpha p(t) t$$ (14)

To solve for the constant factor $\alpha$ we integrated $P(t)$ from 0 to $t_{max}$ and normalized it (because it is a probability function). Using our default values for physical parameters, we obtained $\alpha = 75\sqrt{5}/(8\sqrt{\pi \delta})$, or $\alpha = 11.83 / \sqrt{\delta}$.

The average value of the seek time in one direction, taken over all the requests for which its seek time was greater than that in the other direction, can be estimated by integrating its probability function over all possible times (0 to $t_{max}$), shown in Equation 15.

$$t_{average~seek}(\delta) = \int_{0}^{t_{max}}P(t) t dt$$ (15)

When we substitute in Equations 12, 13 and 14 for $P(t)$ we obtain Equation 16:

$$t_{average~seek}(\delta) = \frac{1}{8} \sqrt{\frac{\pi \delta}{5}}$$ (16)

Seeking in $X$ involves a settling time $t_{settle}$ that we add to the prediction for the average seek time in $X\!$. The final formulae for seek times in $X$ and $Y$ are shown in Equations 17 and 18.

$$t_{seek~x}(\delta_x) = \frac{1}{8} \sqrt{\frac{\pi \delta_x}{5}} + t_{settle}$$ (17)

$$t_{seek~y}(\delta_y) = \frac{1}{8} \sqrt{\frac{\pi \delta_y}{5}} ,$$ (18)

We can now substitute Equations 17 and 18 in Equation 9 to obtain an expression for the seek time, which is shown in Equation 19.

$${{t}_{seek}(\delta_x, \delta_y) =}$$

$$\frac{1}{\sqrt{\delta_x} + \sqrt{\delta_y} + 8 t_{settle} \sqrt{\frac{5}{\pi}}}$$

$$\big( (\sqrt{\delta_x} + 8 t_{settle} \sqrt{\frac{5}{\pi}}) (\frac{1}{8} \sqrt{\frac{\pi \delta_{x} }{5}} + t_{settle}) +$$

$$\delta_y \frac{1}{8} \sqrt{\frac{\pi}{5}}\big)$$