--- a/main/queueing/inst/ctmcbd.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/ctmcbd.m	Sat Mar 08 15:55:05 2014 +0000
@@ -32,7 +32,6 @@
 ##
 ## Matrix @math{\bf Q} is therefore defined as:
 ##
-## @iftex
 ## @tex
 ## $$ \pmatrix{ -\lambda_1 & \lambda_1 & & & & \cr
 ##              \mu_1 & -(\mu_1 + \lambda_2) & \lambda_2 & & \cr
@@ -44,10 +43,8 @@
 ##              & & & & \mu_{N-1} & -\mu_{N-1} }
 ## $$
 ## @end tex
-## @noindent where @math{\lambda_i} and @math{\mu_i} are the birth and
-## death rates, respectively.
-## @end iftex
 ## @ifnottex
+##
 ## @example
 ## @group
 ## /                                                          \
@@ -64,6 +61,9 @@
 ## @end example
 ## @end ifnottex
 ##
+## @noindent where @math{\lambda_i} and @math{\mu_i} are the birth and
+## death rates, respectively.
+##
 ## @seealso{dtmcbd}
 ##
 ## @end deftypefn

--- a/main/queueing/inst/dtmcbd.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/dtmcbd.m	Sat Mar 08 15:55:05 2014 +0000
@@ -33,7 +33,6 @@
 ##
 ## Matrix @math{\bf P} is therefore defined as:
 ##
-## @iftex
 ## @tex
 ## $$ \pmatrix{ (1-\lambda_1) & \lambda_1 & & & & \cr
 ##              \mu_1 & (1 - \mu_1 - \lambda_2) & \lambda_2 & & \cr
@@ -45,9 +44,6 @@
 ##              & & & & \mu_{N-1} & (1-\mu_{N-1}) }
 ## $$
 ## @end tex
-## @noindent where @math{\lambda_i} and @math{\mu_i} are the birth and
-## death probabilities, respectively.
-## @end iftex
 ## @ifnottex
 ## @example
 ## @group
@@ -65,6 +61,9 @@
 ## @end example
 ## @end ifnottex
 ##
+## @noindent where @math{\lambda_i} and @math{\mu_i} are the birth and
+## death probabilities, respectively.
+##
 ## @seealso{ctmcbd}
 ##
 ## @end deftypefn

--- a/main/queueing/inst/engset.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/engset.m	Sat Mar 08 15:55:05 2014 +0000
@@ -28,18 +28,15 @@
 ## request service is exponentially distributed with mean @math{1 /
 ## \lambda}), and offered load @math{A = \lambda / \mu}.
 ## 
-## @iftex
-##
+## @tex
 ## @math{P_b(A, m, n)} is defined for @math{n > m} as:
 ##
-## @tex
 ## $$
 ## P_b(A, m, n) = {{\displaystyle{A^m {n \choose m}}} \over {\displaystyle{\sum_{k=0}^m A^k {n \choose k}}}}
 ## $$
-## @end tex
 ##
 ## and is 0 if @math{n @leq{} m}.
-## @end iftex
+## @end tex
 ##
 ## @strong{INPUTS}
 ##

--- a/main/queueing/inst/erlangb.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/erlangb.m	Sat Mar 08 15:55:05 2014 +0000
@@ -26,18 +26,14 @@
 ## arrival rate @math{\lambda}, individual service rate @math{\mu}
 ## and offered load @math{A = \lambda / \mu} has all servers busy.
 ## 
-## @iftex
-##
+## @tex
 ## @math{E_B(A, m)} is defined as:
 ##
-## @tex
 ## $$
 ## E_B(A, m) = \displaystyle{{A^m \over m!} \left( \sum_{k=0}^m {A^k \over k!} \right) ^{-1}}
 ## $$
 ## @end tex
 ##
-## @end iftex
-##
 ## @strong{INPUTS}
 ##
 ## @table @var

--- a/main/queueing/inst/erlangc.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/erlangc.m	Sat Mar 08 15:55:05 2014 +0000
@@ -21,37 +21,32 @@
 ##
 ## @cindex Erlang-C formula
 ##
-## Compute the steady-state probability that an open queueing system
-## with @math{m} identical servers, infinite wating space, arrival rate
-## @math{\lambda}, individual service rate @math{\mu} and offered load
-## @math{A = \lambda / \mu} are busy. This probability is also called
-## Erlang-C formula @math{E_C(A, m)}.
-## 
-## @iftex
+## Compute the steady-state probability @math{E_C(A, m)} that an open
+## queueing system with @math{m} identical servers, infinite wating
+## space, arrival rate @math{\lambda}, individual service rate
+## @math{\mu} and offered load @math{A = \lambda / \mu} has all the
+## servers busy.
 ##
+## @tex
 ## @math{E_C(A, m)} is defined as:
 ##
-## @tex
 ## $$
 ## E_C(A, m) = \displaystyle{ {A^m \over m!} {1 \over 1-\rho} \left( \sum_{k=0}^{m-1} {A^k \over k!} + {A^m \over m!} {1 \over 1 - \rho} \right) ^{-1}}
 ## $$
-## @end tex
 ##
 ## where @math{\rho = A / m = \lambda / (m \mu)}.
-##
-## @end iftex
+## @end tex
 ##
 ## @strong{INPUTS}
 ##
 ## @table @var
 ##
-## @item A
-## Offered load, defined as @math{A = \lambda / \mu} where
+## @item A Offered load. @math{A = \lambda / \mu} where
 ## @math{\lambda} is the mean arrival rate and @math{\mu} the mean
 ## service rate of each individual server (real, @math{0 < A < m}).
 ##
-## @item m
-## Number of identical servers (integer, @math{m @geq{} 1}). Default @math{m = 1}
+## @item m Number of identical servers (integer, @math{m @geq{} 1}).
+## Default @math{m = 1}
 ##
 ## @end table
 ##
@@ -59,8 +54,7 @@
 ##
 ## @table @var
 ##
-## @item B
-## The value @math{E_C(A, m)}
+## @item B The value @math{E_C(A, m)}
 ##
 ## @end table
 ##

--- a/main/queueing/inst/qncmmvaap.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/qncmmvaap.m	Sat Mar 08 15:55:05 2014 +0000
@@ -34,11 +34,9 @@
 ##
 ## This implementation uses Bard and Schweitzer approximation. It is based
 ## on the assumption that
-## @iftex
 ## @tex
 ## $$Q_i({\bf N}-{\bf 1}_c) \approx {n-1 \over n} Q_i({\bf N})$$
 ## @end tex
-## @end iftex
 ## @ifnottex
 ## the queue length at service center @math{k} with population
 ## set @math{{\bf N}-{\bf 1}_c} is approximately equal to the queue length 

--- a/main/queueing/inst/qsmh1.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/qsmh1.m	Sat Mar 08 15:55:05 2014 +0000
@@ -26,11 +26,9 @@
 ## throughput for a @math{M/H_m/1} system. In this system, the customer
 ## service times have hyper-exponential distribution:
 ##
-## @iftex
 ## @tex
 ## $$ B(x) = \sum_{j=1}^m \alpha_j(1-e^{-\mu_j x}),\quad x>0 $$
 ## @end tex
-## @end iftex
 ##
 ## @ifnottex
 ## @example

--- a/main/queueing/inst/qsmm1.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/qsmm1.m	Sat Mar 08 15:55:05 2014 +0000
@@ -23,19 +23,17 @@
 ##
 ## Compute utilization, response time, average number of requests and throughput for a @math{M/M/1} queue.
 ##
-## @iftex
+## @tex
 ## The steady-state probability @math{\pi_k} that there are @math{k}
 ## jobs in the system, @math{k \geq 0}, can be computed as:
 ##
-## @tex
 ## $$
 ## \pi_k = (1-\rho)\rho^k
 ## $$
-## @end tex
 ##
 ## where @math{\rho = \lambda/\mu} is the server utilization.
 ##
-## @end iftex
+## @end tex
 ##
 ## @strong{INPUTS}
 ##

--- a/main/queueing/inst/qsmm1k.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/qsmm1k.m	Sat Mar 08 15:55:05 2014 +0000
@@ -27,17 +27,16 @@
 ## requests in the system is @math{K}, and the maximum queue length is
 ## @math{K-1}.
 ##
-## @iftex
+## @tex
 ## The steady-state probability @math{\pi_k} that there are @math{k}
 ## jobs in the system, @math{0 @leq{} k @leq{} K}, can be computed as:
 ##
-## @tex
 ## $$
 ## \pi_k = {(1-a)a^k \over 1-a^{K+1}}
 ## $$
-## @end tex
+##
 ## where @math{a = \lambda/\mu}.
-## @end iftex
+## @end tex
 ##
 ## @strong{INPUTS}
 ##

--- a/main/queueing/inst/qsmminf.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/qsmminf.m	Sat Mar 08 15:55:05 2014 +0000
@@ -27,16 +27,14 @@
 ##
 ## @cindex @math{M/M/}inf system
 ##
-## @iftex
+## @tex
 ## The steady-state probability @math{\pi_k} that there are @math{k}
 ## requests in the system, @math{k @geq{} 0}, can be computed as:
 ##
-## @tex
 ## $$
 ## \pi_k = {1 \over k!} \left( \lambda \over \mu \right)^k e^{-\lambda / \mu}
 ## $$
-## @end tex 
-## @end iftex
+## @end tex
 ##
 ## @strong{INPUTS}
 ##

--- a/main/queueing/inst/qsmmm.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/qsmmm.m	Sat Mar 08 15:55:05 2014 +0000
@@ -27,30 +27,26 @@
 ## with @math{m} identical servers connected to a single FCFS
 ## queue.
 ##
-## @iftex
+## @tex
 ## The steady-state probability @math{\pi_k} that there are @math{k}
 ## jobs in the system, @math{k \geq 0}, can be computed as:
 ##
-## @tex
 ## $$
 ## \pi_k = \cases{ \displaystyle{\pi_0 { ( m\rho )^k \over k!}} & $0 \leq k \leq m$;\cr
 ##                 \displaystyle{\pi_0 { \rho^k m^m \over m!}} & $k>m$.\cr
 ## }
 ## $$
-## @end tex
 ##
 ## where @math{\rho = \lambda/(m\mu)} is the individual server utilization.
 ## The steady-state probability @math{\pi_0} that there are no jobs in the
 ## system can be computed as:
 ##
-## @tex
 ## $$
 ## \pi_0 = \left[ \sum_{k=0}^{m-1} { (m\rho)^k \over k! } + { (m\rho)^m \over m!} {1 \over 1-\rho} \right]^{-1}
 ## $$
+##
 ## @end tex
 ##
-## @end iftex
-##
 ## @strong{INPUTS}
 ##
 ## @table @var

--- a/main/queueing/inst/qsmmmk.m	Wed Mar 05 22:53:37 2014 +0000
+++ b/main/queueing/inst/qsmmmk.m	Sat Mar 08 15:55:05 2014 +0000
@@ -31,31 +31,27 @@
 ## is @math{K-m}. This function generates and
 ## solves the underlying CTMC.
 ##
-## @iftex
+## @tex
 ##
 ## The steady-state probability @math{\pi_k} that there are @math{k}
 ## jobs in the system, @math{0 @leq{} k @leq{} K} can be expressed as:
 ##
-## @tex
 ## $$
 ## \pi_k = \cases{ \displaystyle{{\rho^k \over k!} \pi_0} & if $0 \leq k \leq m$;\cr
 ##                 \displaystyle{{\rho^m \over m!} \left( \rho \over m \right)^{k-m} \pi_0} & if $m < k \leq K$\cr}
 ## $$
-## @end tex
 ##
 ## where @math{\rho = \lambda/\mu} is the offered load. The probability
 ## @math{\pi_0} that the system is empty can be computed by considering
 ## that all probabilities must sum to one: @math{\sum_{k=0}^K \pi_k = 1},
 ## which gives:
 ##
-## @tex
 ## $$
 ## \pi_0 = \left[ \sum_{k=0}^m {\rho^k \over k!} + {\rho^m \over m!} \sum_{k=m+1}^K \left( {\rho \over m}\right)^{k-m} \right]^{-1}
 ## $$
+##
 ## @end tex
 ##
-## @end iftex
-##
 ## @strong{INPUTS}
 ##
 ## @table @var