GATE Overflow Test Series | Mixed Subjects | Test 4 | Question: 27

Question

A desktop, a printer and a radio receptor are connected by a single wire to form a network capable of transmitting up to $4\;\text{KBps}.$ For a $5$ second period, all three devices were backlogged (i.e. had data to send). In these $5$ seconds, the desktop and the printer sent $8\;\text{KB}$ each, while the radio receptor did not manage to send anything. If $U$ is the utilization $(U)$ and $F$ the fairness $(F)$ in the network during this period, the value of $U + F = $ _______

Answer 1

$\text{Fairness And Utilization As Numbers:}$

$\textbf{Node throughputs:}$ $x_{i}$ - data bits transmitted by the $i$-th node in $T$ units of time $(i = 0, 1,\dots, n)$

$\textbf{Channel capacity:}$ $C$ - maximal number of bits transmittable in a unit of time.

$\textbf{Utilization:}$ $U = \dfrac{x_{0} + x_{1} + \dots + x_{n}}{C \times T}\quad (0 \leq U \leq 1)$

$\textbf{Fairness:}$ $F = \dfrac{(x_{0} + x_{1} + \dots + x_{n})^{2}}{n(x_{0}^{2} + x_{1}^{2} + \dots + x_{n}^{2})} - \dfrac{1}{n}\quad (0 \leq F \leq 1)$

Now, utilization $U = \dfrac{8 + 8}{4 \cdot 5} = \dfrac{16}{20} = 0.8$

And, fairness $F = \dfrac{(8 + 8)^{2}}{2\cdot (64 + 64)} -\dfrac{1}{2}= \dfrac{256}{256} - \dfrac{1}{2} = 1 - \dfrac{1}{2} = 0.5$

$\therefore$ The value of $U + F = 0.8 + 0.5 = 1.3$

So, the correct answer is $1.3$