Relationship between signal rate, data rate, minimum bandwidth
Here's an attempt at an answer, by comparing your equations with the more commonly used ones.

We can write the "pulse rate" or the "signaling rate" $R_p$, measured in baud, or equivalently in pulses per second or symbols per second, as: $$R_p = \frac{1}{k} R_b,$$ where $R_b$ is the bit rate, and $k$ is the number of bits carried by a pulse.

To elaborate: a digital communications signal can be written as a sequence of pulses of different amplitudes, each pulse time-shifted an amount $T_p = 1/R_p$, and with the information carried in the pulses' amplitude: $$s(t) = \sum_j a_j p(t-jT_p).$$

Note that:


For each value of $j$, the pulse $a_j p(t-T_j)$ carries the information in its amplitude $a_j$. The amount of information depends on how many distinct values $a_j$ can take. Usually this number is represented by $M = 2^k$ for integer $k$, and we say that each pulse carries $k$ bits.
Since one pulse is transmitted every $T_p$ seconds, the pulse rate is $R_p = 1/T_p$ baud.


Now, (ignoring some more advanced considerations) the bandwidth of $s(t)$ is the same as the bandwidth of the pulse $p(t)$, denoted $B_p$. The pulse that has the minimum possible bandwidth for a given $R_p$ is a sinc pulse, and in that case the bandwidth of $s(t)$ is given by $$B_{min} = \frac{R_p}{2} = \frac{1}{2}\frac{1}{k}R_b.$$ Practical pulses achieve a fraction larger than $1/2$, or in other words, they require a larger bandwidth. If we define this fraction by $c$, we have $$B = c\frac{1}{k}R_b.$$

In conclusion, it seems to me that:


In your equations, $N$ is the bit rate, $r$ represents the number of bits transmitted per pulse, and $c$ represents the bandwidth efficiency of the chosen pulse. The variable name $k$ is more commonly used than $r$.
The first equation in your book is wrong; it should be $S = N \times \frac{1}{r}$.
The equation for the bandwidth is correct.