The network consists of four loops connected in series. Each loop consists of two 1 ohm resistors in series connected in parallel with two ohm resistors in series. Thus the effective resistance R of each loop is given by $\frac{1}{R}=\frac{1}{2}+\frac{1}{4}$ giving $R=\frac{4}{3}$ ohm. Therefore the equivalent resistance of the network = resistance of four loops connected in series  each having an effective resistance of $\frac{4}{3}$ ohm = $4\times \frac{4}{3}=\frac{16}{3}$ ohm