Homotopy limit-colimit diagrams in stable model categories It is shown in Remark 7.1.12 of (a newer version of) Mark Hovey's book Model Categories that, in a stable model category, homotopy pullback squares coincide with homotopy pushout squares. The argument goes as follows: given a square $S$ of the form
$$
\begin{matrix}
W & \to & X\\
\downarrow_h & & \downarrow_g\\
Z & \to & Y 
\end{matrix}
$$
one shows that $S$ is a pullback square if and only if the induced map $\mathrm{HoFibre}(h)\to\mathrm{HoFibre}(g)$ is an isomorphism in the homotopy category. Similarly, $S$ is a pushout square if and only if the induced map $\mathrm{HoCofibre}(h)\to\mathrm{HoCofibre}(g)$ is an isomorphism in the homotopy category. One concludes by observing that $\mathrm{HoCofibre}(h)$ is the suspension of $\mathrm{HoFibre}(h)$ and similarly for $g$.
Question: Is it possible to generalize this statement to say something like as follows?
"In a diagram of the form
$$
\begin{matrix}
W & & \to &&  X\\
 &\searrow & \\
\downarrow &&Y&&\downarrow\\
&&&\searrow\\
Z & &\to && V,  
\end{matrix}
$$
$V$ is a homotopy colimit of
$$
\begin{matrix}
W & & \to &&  X\\
 &\searrow & \\
\downarrow &&Y&&\\
&&&\\\
Z & & &&   
\end{matrix}
$$
if and only if 
$W$ is a homotopy limit of
$$
\begin{matrix}
 & &  &&  X\\
 & & \\
 &&Y&&\downarrow\\
&&&\searrow\\
Z & &\to && V."  
\end{matrix}
$$
 A: This is not true. For example, take $W = X = Y = 0$, and the map $Z \rightarrow V$ to be an isomorphism. Then you've got a homotopy colimit diagram, but it is only a homotopy limit diagram if $Z$ is weakly contractible.
A:  Yes, because both statements are equivalent to the existence of an exact triangle of the form
$$W\longrightarrow X\oplus Y\oplus Z\longrightarrow V\longrightarrow\Sigma W.$$
The arrows are the same as in your diagram, except that you have to change of the sign of $W\rightarrow Y$ so that the first two arrows compose to $0$.  
EDIT: The above 'answer' is shamely false. Let me just offer here a brief explanation of Jacob Lurie's correct answer. If the diagram is a homotopy colimit then we have an exact triangle
$$W\oplus W\stackrel{
\left(\begin{array}{cc}
f&0\\
-g&g\\
0&-h
\end{array}\right)
}\longrightarrow X\oplus Y\oplus Z\stackrel{(f',g',h')}\longrightarrow V\longrightarrow\Sigma(W\oplus W).$$
The homotopy limit condition gives rise to an exact triangle of the form
$$W\stackrel{
\left(\begin{array}{c}
f\\
g\\
h
\end{array}\right)
}\longrightarrow X\oplus Y\oplus Z\stackrel{\left(\begin{array}{cc}
f'&-g'&0\\
0&g'&-h'
\end{array}\right)}\longrightarrow V\oplus V\longrightarrow\Sigma W.$$
If $W=X=Y=0$ then you get an exact triangle by putting $V=Z$ and $h'=1_Z$
$$0\longrightarrow Z\stackrel{1}\longrightarrow Z\longrightarrow 0$$
But if you put $X=Y=0$, $V=Z$, and $h'=1_Z$, then $W=\Sigma^{-1}Z$ since the exact triangle is
$$\Sigma^{-1}Z\stackrel{0}\longrightarrow Z\stackrel{\binom{0}{1}}\longrightarrow Z\oplus Z\stackrel{(1,0)}\longrightarrow Z.$$
