A convex $ABCDE$ is inscribed in a unit circle, $AE$ being its diameter. If $AB = a$, $BC = b$, $CD = c$, $DE = d$ and $ab = cd =\frac{1}{4}$, compute $AC + CE$ in terms of $a, b, c, d.$
1. Given that $ABCDE$ is a convex pentagon inscribed in a unit circle with $AE$ as its diameter, we know that $AE = 2$ because the diameter of a unit circle is 2.
2. We are given the side lengths $AB = a$, $BC = b$, $CD = c$, $DE = d$, and the conditions $ab = cd = \frac{1}{4}$.
3. Let $AC = x$ and $CE = y$. Since $AE$ is the diameter, we have $x^2 + y^2 = 4$ by the Pythagorean theorem in the unit circle.

4. Using Ptolemy's theorem for the cyclic quadrilateral $ABCE$, we have:
   \[
   AC \cdot BE + AB \cdot CE = AE \cdot BC
   \]
   Substituting $AC = x$, $BE = \sqrt{4 - a^2}$, $AB = a$, $CE = y$, $AE = 2$, and $BC = b$, we get:
   \[
   x \cdot \sqrt{4 - a^2} + a \cdot y = 2b
   \]
   Squaring both sides:
   \[
   (x \cdot \sqrt{4 - a^2} + a \cdot y)^2 = (2b)^2
   \]
   Expanding and simplifying:
   \[
   x^2 (4 - a^2) + 2x \cdot \sqrt{4 - a^2} \cdot a \cdot y + a^2 y^2 = 4b^2
   \]

5. Using Ptolemy's theorem for the cyclic quadrilateral $ACDE$, we have:
   \[
   AC \cdot DE + CD \cdot AE = AD \cdot CE
   \]
   Substituting $AC = x$, $DE = d$, $CD = c$, $AE = 2$, and $CE = y$, we get:
   \[
   x \cdot d + c \cdot 2 = AD \cdot y
   \]
   Squaring both sides:
   \[
   (x \cdot d + 2c)^2 = (AD \cdot y)^2
   \]
   Expanding and simplifying:
   \[
   x^2 d^2 + 4c^2 + 4cdx = y^2 (4 - d^2)
   \]

6. Adding the two equations obtained from Ptolemy's theorem:
   \[
   x^2 (4 - a^2) + a^2 y^2 + 4b^2 + 4aby + x^2 d^2 + 4c^2 + 4cdx = 4x^2 + 4y^2
   \]
   Using $x^2 + y^2 = 4$ and $ab = cd = \frac{1}{4}$, we substitute:
   \[
   4a^2 + 4b^2 + 4c^2 + 4d^2 + 4ab + 4cd = 16
   \]
   Simplifying:
   \[
   4(a^2 + b^2 + c^2 + d^2 + ab + cd) = 16
   \]
   \[
   a^2 + b^2 + c^2 + d^2 + ab + cd = 4
   \]

7. Finally, we need to find $AC + CE$ in terms of $a, b, c, d$. From the above equation, we have:
   \[
   x + y = 4 - (a^2 + b^2 + c^2 + d^2)
   \]

The final answer is $\boxed{4 - (a^2 + b^2 + c^2 + d^2)}$