= Solution
The number and spacing of primary slip planes remain fixed, so the length projected on their normal is invariant: $l_0\cos\phi_0=l_1\cos\phi_1$. The initial and boundary loading directions are $[514]$ and $[211]$, respectively. With normal $[1\bar11]$,
$$
\cos\phi_0=\frac8{\sqrt{42}\sqrt3},\qquad
\cos\phi_1=\frac2{\sqrt6\sqrt3}.
$$
Therefore
$$
\boxed{\epsilon_p=\frac{\cos\phi_0}{\cos\phi_1}-1=\frac4{\sqrt7}-1=0.5119\simeq51.2\%.}
$$
This is engineering plastic <strain> in the ideal geometric model; its corresponding logarithmic <strain> is $\ln(4/\sqrt7)$. Elastic extension and changes in slip resistance are neglected, as required by this construction.
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