Add proof.

cost-models.tex

@@ -43,24 +43,38 @@ \chapter{Cost Models}
     Writing $T_{i}$ for the time taken to execute a program on $i$
     processors,
   \[
-    \frac{T_{1}}{p} \leq T_{p} \leq T_{\infty} + \frac{T_{1}-T_{\infty}}{p}
+    \frac{T_{1}}{p} \leq T_{p} \leq T_{\infty} + \frac{T_{1}}{p}
   \]
   where $p$ is the number of available processors.
 
-  % \begin{proof}
-  %   At level $j$ of the computation DAG there are $M_{j}$ independent
-  %   operations, which can clearly be executed by $p$ processors in
-  %   time
-  % \[
-  %   \Bigl\lceil{\frac{M_{j}}{p}}\Bigr\rceil
-  % \]
-  % Now we have
-  % \begin{align*}
-  %   \frac{T_{1}}{p} &= \frac{\sum _{j} {M_{j}}}{p} & \text{(By definition)} \\
-  %   & \leq \sum _{j} \Bigl\lceil{\frac{M_{j}}{p}}\Bigr\rceil \\
-  %   & \leq \sum _{j} \Bigl\lceil{M_{j}}\Bigr\rceil
-  % \end{align*}
-  % \end{proof}
+  \begin{proof}
+    Suppose that the computation DAG has $n$ levels, and at level $j$
+    of the computation DAG there are $M_{j}$ independent operations.
+    Then the total work of our program, and the time taken on a single
+    processor, is
+  \[
+    T_{1} = \sum_{i=1}^{n}M_{i}.
+  \]
+  In the case where we have an infinite number of processors, each of
+  the $M_{j}$ independent operations can be executed in a single step, so we also have
+  \[
+    T_{\infty} = n
+  \]
+  because each level can be executed in one step. For the case where we
+  have $p$ processors, we denote the time taken to execute level $i$
+  of the DAG by $T_{p}^{i}$, where
+  \[
+    T_{i}^{p} = \Bigl\lceil{\frac{M_{i}}{p}}\Bigr\rceil \leq
+    \frac{M_{i}}{p}+1.
+  \]
+  The intuition behind the inequality is that if we need to execute
+  $k$ independent nodes and we have $p$ processors, then we can
+  perform $\lceil \frac{k}{n} \rceil$ sequential iterations, with each
+  chunk executing (up to) $p$ nodes in parallel. Finally,
+  \[
+    T_{p} = \sum_{i=1}^{n}T_{i}^{p} \leq \sum_{i=1}^{n}\Bigl(\frac{M_{i}}{p}+1\Bigr) = \frac{T_{1}}{p} + T_{\infty}
+  \]
+  \end{proof}
 
   \end{theorem}