Goeller on Telecom
Traffic
A First Course in
Telephone Traffic Engineering
Epilogue
Anyone who has worked through this course and
learned how to call up programs and get numerical answers to traffic
problems is a long way down the road to understanding one of the
most interesting aspects of system design. However, at the end of
this first course, it is wise to pause for a moment to survey what
we have not yet covered.
"Limited source gain," mentioned briefly in
Chapter 2, is a topic of considerable interest in some
circumstances. And just as Binomial is related to Poisson, so the
Engset formula is related to Erlang B and an unnamed formula,
sometimes called Delay, is related to Erlang C. These formulas are
well worth study in the future.
In the past, the graded multiple has been the
subject of intensive study. Although modern systems do not require
the graded multiple in the way SXS systems with their ten (or
twenty, in England) outlets per level did, we still find occasion to
use some trunks one way outgoing, others one way incoming, and have
both groups overflow into a two-way group. Graded multiple works
well in this context, but it is more important as a guide to
Wilkinson's Equivalent Random Theory, he basis of automatic
alternate routing.
With regard to queuing, we have so far only
studied delays of calls whose length is described as "exponential,"
or varying greatly around a mean value, with many more short calls
than long. The Crommelin-Pollaczek formula has been worked out for
calls or operations of fixed length, and is of particular interest
in the design of switching systems themselves. Queue order other
than first come first served (FIFO) has also been studied
extensively, with Random and LIFO offering a considerable
literature.
Comment has been made as to how Erlang B has no
delay in queue, Poisson requires calls to remain in the system for
one average holding time whether served partially or not at all,
while Erlang C has calls wait for a trunk and then use the trunk for
a full holding time. There is a certain amount of recent work
investigating queues of fixed length prior to overflow to some other
facility group, which is worthy of attention.
Data system design makes use of many of the
formulas programmed to go with this course, but often looks at
things from a somewhat different point of view that needs special
exposition. Many other disciplines are now involved with queuing and
other traffic ideas, and are busily reinventing triangular wheels to
obfuscate the essential unity and universal applicability of
traffic. Exploring these highways and by-ways is both interesting
and profitable.
But for the moment, this is a good place to
pause and consolidate our gains. Nothing teaches traffic theory as
well as traffic practice, and no final exam is better than comparing
actual results with those predicted before the fact. A good network
may not be a thing of beauty or even a joy forever, but it cause for
satisfaction when it all come together.
Frankly, I hope you have as much fun working
and playing with telephone traffic as I have had for these many
years. Good Luck, and may all your networks be optimal!
Lee Goeller
Haddonfield, NJ
Feb. 3, 1983
Appendix:
Some Simple Derivations of Telephone Traffic Formulas
(Webmaster's Note: The
"Panels" mentioned in this appendix are not included in this web
edition. We hope to add them at a future date.)
The Binomial Distribution
In Chapter 2, we used the Binomial distribution
as a model for finding the number of extensions off hook in a small
PBX, given that we knew the probability of any one extension being
off hook. The actual formula is as follows, where ^ signifies
exponentiation (i.e. p^n is p multiplied by itself n times).
P(n)= C(L,n)*p^n*(1-p)^(L-n)
Just looking at it doesn't help much. However,
if we study it, we will see that it is the most logical thing in the
world. First of all, P(n) is the probability a PBX serving L lines
or extensions has exactly n extensions off hook when we take a look
at the system. This is what we are trying to calculate. C(L,n) is
shorthand for the number of ways we can take n things from a group
of L, and we will study it in a minute. The lower case p is the
probability that one extension is off hook, and (1-p) is the
probability that it is on hook. It must be one or the other, so
p + (1-p) = 1,
or certainty. We multiply p and (1-p) by
themselves a certain number of times (raise them to powers), as will
be explained.
Let's look first at the C(L,n), or the number
of combinations we have when we choose groups of n things from a
supply of L. Let's use a specific problem to make things clear.
Assume we have 10 boxes, and three marbles, one red, one green, and
one yellow (stick with me--we'll get back to telephones in a
minute). We want to find all the ways we can put the three marbles
in the 10 boxes, with only one marble per box, or C(10,3).
We take the red marble first. We have 10 boxes
we can put it in. So there are 10 ways we can deal with it. Then we
take the green marble. One box is full, so we have 9 places left.
The yellow marble has only 8 places to go. This says we have a total
of 10*9*8=720 ways we can place the red, green and yellow marbles in
the 10 boxes, and proves that mathematics sure beats experimenting
in some instances.
But now, suppose we didn't care whether we
chose the red, green or yellow marble first, second or third. We
could select any one of the three to go first, any one of the two
remaining to go second, and have only one left to choose last. The
number of ways we could choose would thus be 3*2*1=6. If all the
marbles were yellow, we could choose them six different ways and not
even know we were making a choice. Thus the ways we could put three
yellow marbles in our 10 boxes would be 720/6=120, a much smaller
number. This is our numerical value of C(10,3)
There are several ways we can express C(10,3)
in terms of mathematical symbols. For instance:
C(10,3) = (10*9*8)/(3*2*1)
But there is a way of writing 3*2*1 that is
simpler. We would write it as 3!, read "three factorial." To write a
"factorial," we start with a given number, multiply it by the next
smaller number, then by the next smaller, all the way down to one.
For instance, 10! = 10*9*8*7*6*5*4*3*2*1.
This immediately suggests something else we
could do. In our formula, we could multiply 10*9*8 by 7! and have
10!. To keep from changing the value of the equation, we would have
to also divide by 7!, and we could write our formula as C(10,3) =
10!/(3!*7!)
Mathematicians love formulas like this, and
write them without numbers to be perfectly general:
C(L,n) = L!/(n!*(L-n)!)
What this tells us is the number of ways we can
choose n things from a group of L, or find the number of different
ways we might find n extensions off hook on a PBX with a total of L
lines or extensions.
We still have a way to go to find out the
probability that exactly n extensions out of L are off hook. Let's
take our "6 CCS per line" and work with it. When we specify 6 CCS
per busy hour per extension, we are assuming that each phone will be
off hook for 600 seconds or 10 minutes during the hour, so the
probability of finding it off hook is 6/36 or 10/60 or one sixth.
Thus p is 0.1666, and (1-p) is 0.8333.
If our PBX has 10 extensions and we want to
find out the probability that only the boss, his assistant and his
secretary are off hook (and everybody else is on hook), the
probability would be given by the following formula:
P(boss, asst, sec) =
p*p*p*(1-p)*(1-p)*...
until we get a total of seven (1-p) terms.
(Remember, the rules of probability say that if a AND b are supposed
to happen, we multiply the probabilities of a and b alone). This
could be written as
P(boss, asst, sec) =
p^3*(1-p)^7.
But what we really want is the probability of
any three (and exactly three) people, being off hook. Well, we know
how to do that. We know how many different ways we can have 10
things taken 3 at a time. We multiply the probability of it
happening one way by all the ways it can happen, and we come out
with our original formula:
P(3) =
C(10,3)*(0.1666)^3*(0.8333)^7
or, in more general terms,
P(n) = C(L,n)*p^n*(1-p)^(L-n)
That's it. That's probably how Molina reasoned,
and how the problem is dealt with in all first courses in
probability and statistics. It is a perfectly general situation, and
extends far beyond telephones.
For the non-mathematical reader who has stuck
with me through all this, it breaks my heart to have to say we are
not yet quite done. What we really want is the probability that the
people on our PBX find all trunks busy. Let's simplify the problem
and assume no intra-PBX calls so that all off-hooks are associated
with trunks. If we have 10 extensions and 3 trunks, is the
probability that exactly 3 extensions are off hook equal to the
probability of blocking?
The answer is no, from the way the problem is
defined. We said the phones were off-hook. We did not say they were
necessarily talking to somebody. The assumption underlying Binomial
(and Poisson as well) is that each call stays in the system for one
average holding time, whether a trunk is available or not. If an ATB
is encountered, the call waits and, if a trunk comes free, it uses
the trunk for the remainder of the holding time. Thus for three
trunks, the Binomial probability of blocking is the probability that
3 or 4 or 5 or 6, etc. calls are present in the system, some of them
waiting their turn. If they don't get a trunk, they vanish at the
end of one holding time. So we have to calculate the probability
that 3, 4, 5, 6, 7, 8, 9, OR 10 phones are off hook, add them up,
and then we have the Binomial blocking probability. (Again, the
rules of probability say that if we are going to have either a OR b,
we add the probability of a to the probability of b). In terms of
actually making the calculations, it is easier to calculate the
probability that none OR 1 OR 2 trunks are busy, add these up, and
subtract the result from 1.
Another point. We have been using the "wire
chief's point of view" here. If we wanted to calculate the
probability that a "particular user," upon coming off hook, would
find all trunks busy, we have to take the remaining L-1 extensions
and examine their behavior, which is what the particular user
observes. Thus we would go through the same calculations with L-1
replacing L.
From Binomial To Poisson
It is possible to modify the Binomial
distribution so that it becomes the Gaussian or bell-shaped curve
often used by statisticians, but it is also possible to shift it
into the Poisson distribution used by traffic engineers. What we do
is let p get small and L get large, but keep the product constant.
Multiplying p times L gives us the average traffic in the system,
because p is the average occupancy of one extension and L is the
number of extensions; p*L is obviously A, the offered traffic. If p
is the probability that an extension is making a call via a given
trunk group, A is the total average traffic offered to the trunk
group.
To go from Binomial to Poisson, we start by
writing out the first several terms of the Binomial distribution for
L extensions, assuming first that none, then one, then two, etc.,
lines are off hook. This is shown in Panel 1 at the end of this
appendix. In Eq. 0, we find the probability that no lines are off
hook, P(0), is given by (1-p)^L, because 0! and p^0 both equal 1,
surprising as this may seem. In P(1), we get a term of the form
(1-p)^(L-1). If we multiply and divide this term by (1-p), we will
get P(0)/(1-p), which will turn out to be helpful. We can regroup to
put the equation in the final form shown.
Going on to P(2), we do the same trick, but
this time we multiply and divide by (1-p)^2. This lets us express
P(2) in terms of P(0). If we keep on, we can express the Binomial
probability of any number of lines being off hook in terms of P(0).
Eq. n shows the general form for P(n). We will play with this form
of the equation to go from Binomial to Poisson. To simplify Eq. n,
we note that p is always less than 1, and we are assuming that p
gets very small. Thus, right away, we see that 1-p approaches 1
without much in the way of error. Further, L is going to get larger
and larger so that each term in the product from L to (L-n+1) will
not be much different from L as long as n, the number of off-hook
lines, is small by comparison. This simplifies the product of the L
terms to L^n and L^n to (A/p)^n. Finally, we get Eq. n', where
P(n) = (approximately)
(A^n/n!)*P(0)
This is what we want, because if we can find
P(0), we can find all the rest of the probabilities.
To find P(0), we have to resort to the trick
shown in Panel 2. Here we take the natural logarithm of both sides
of the equation Eq.0 from Panel 1, expand the right hand side in a
Taylor series, throw away the terms that get very small, and take
the anti-log of what's left. Expanding things in a series is another
trick that the mathematicians do all the time; and don't ask me to
explain it. The idea, however, is to turn something complicated like
a logarithm into something simpler, like an ordinary number.
If p is much less than 1, p^2 is even smaller,
p^3 is smaller yet, etc. So we drop everything except the first
term, keeping only lnP(0)=L*(-p). If we remember that L=A/p, we find
that ln P(0)=-A. Taking the antilogarithm of both sides of the
equation, we finally get P(0)=e^(-A), where e is the base of natural
logarithms. This is what we have been seeking.
We can now run down the equations in Panel 1,
making P(1)=A*e^(-A), P(2)=(A^2/2!)*e^(-A), P(3)=(A^3/3!)*e^(-A),
etc. The general term is P(n)=(A^n/n!)*e^A. This is the Poisson
approximation to the Binomial distribution, and a perfectly good
distribution on its own. What we have now is the probability of
exactly n calls being in progress at any time when the average
number of calls in progress from a large PBX during the hour is A.
As with Binomial, the Poisson blocking
probability requires us to find the probability that N or more calls
will be in the system, where N is the number of trunks available.
Once again, calls stay in the system for an average holding time,
whether served or not.
Erlang B and Statistical Equilibrium
Although, as we have seen, telephone traffic as
a series of random events can be studied by comparison with coin
tosses and dice rolls, there are other approaches as well. Erlang
reasoned that, if calls arrive at random and drop out after they
have been served, it should be possible to calculate the status of a
trunk group at time T if we know its status at time zero.
If we know the average arrival rate, we can
estimate how many phone calls will appear in a given period. If we
know how long it takes to serve each call, we know how fast they
drop out. If we know how many calls we have in the system at the
beginning, how many we add during T, and how many finish talking and
hang up, we can easily tell how many calls are in the system at the
end of T.
As we saw in Chapter 1, the offered traffic in
Erlangs represents the arrival rate, the number of calls appearing
during the average holding time of one call. For example, 14.73
Erlangs of traffic, composed of calls with holding times of 243
seconds, means that 14.73 calls will arrive during the 243 second
interval (4 minutes and 3 seconds), 3.63 calls will arrive per
minute, and the probability of a call arriving in one second is just
about 0.06.
The probability of a call finishing and hanging
up is obviously proportional to the number of trunks in use at any
time, and inversely proportional to the average holding time. If no
trunks are busy, there is obviously no probability of a call leaving
the system. If six trunks are busy, however, the probability of one
of those six calls finishing during the next interval of time is
quite high. If calls are 243 seconds long, they can be served at the
rate of 3600/243=14.81 calls per hour, 0.247 per minute, or 0.0041
per second per trunk. That is, the probability of a call in progress
finishing up during the next second will be 0.0041, and if 5 calls
are in progress at the time, the probability of one of them
finishing will be 0.02.
If we make T, our study period, long, like one
minute, a number of calls can arrive and depart. But if we make T
very short, like one second, the probability of more than one call
arriving or dropping out becomes vanishingly small. By assuming no
more than one call arrives or departs during T, the mathematics
becomes much simpler.
Let's assume we have 3 calls in the system at
the end of our study interval T. How did it get that way? Well, we
might have had 3 calls at the start of T, and no calls arrived or
departed. Or, we might have had 2 calls and one arrived. Or,
finally, we might have had 4 calls, and one departed. These are the
only possibilities if we assume no more than one change per second.
The arrival rate is the offered traffic A
divided by the average holding time. The departure rate is the
number of calls in progress at any time, n, again divided by the
holding time. For everything to work out right, holding time must be
in seconds. For very heavy traffic systems (100 Erlangs, for
instance), the one-second interval may be too long. Then tenths or
hundredths of second might be needed.
Panel 3 shows a series of equations using these
probabilities in a system with 4 trunks. The lower case "a" is the
arrival rate, calls during interval T. If we multiply a by T, we
have the probability of a call arriving during T. With T at one
second, the multiplication is easy. The lower case "r" is used for
service rate, or the reciprocal of holding time (this makes the
algebra easier). Again, we multiply r by T to get the average rate
at which calls are serviced and drop out.
Each equation in Panel 3 gives the probability
that n calls are in the system at the end of T, in terms of
conditions at the start of T. Note that in Eq.0, no calls can arrive
during T to produce 0 calls in the system, and in Eq. 4, for the
last trunk in the group, no calls can drop out to get down to 4,
since 5 calls cannot be present in a 4 trunk system. Eq.n is the
general form the equation takes.
The probability that calls neither arrive nor
depart needs explanation. The probability that no calls arrive is 1
minus the probability that one call arrives, or (1-aT). The
probability that no calls drop out is 1 minus the probability that
one call drops out, or (1-nrT) where n is the number of calls in
progress at the start of the interval T, and r is the service rate,
as defined above. The probability that neither event happens is the
product of the two terms, or
[1-(a+nr)T+anrT^2]
As we have seen, a and r are small, and T is
purposely made very small. Thus the last term is much smaller than
the others, and can be ignored.
In Panel 4, we multiply out the middle term of
the general equation from Panel3, subtract the old P(n) from the new
P(n) on the left side of the equation, and divide the whole thing
through by T. This gives Eq. 1. And Eq. 1 shows the rate of change
in P(n), or P(n)new-P(n)old divided by the time interval T. Now,
here is where statistical equilibrium comes in. We assume that,
whatever the probability of n calls being in the system may be, it
does not change during the busy hour. Thus the whole left side of
the equation is equal to 0. Students of calculus will immediately
recognize a derivative; derivatives, or rates of change, are always
0 when things are not changing. Note that it is the probability of n
calls in the system that does not change--not the number of calls
which is fluctuating with random traffic around a constant average
A.
We can now rearrange Eq. 1 in Panel 4 to get Eq.
2, which shows the probability of (n+1) calls being in the system,
assuming we know P(n) and P(n-1). Then, in Eq. 3, we go back to A,
using the original definitions of a=A/h and h=1/r. This puts the
equation in a form we can use, assuming we can find some way to get
started in evaluating the P(n) terms.
That turns out to be easy. If we set n=0, and
note that we have no probability of having -1 calls in the system,
P(1)=A*P(0) as shown in Panel 5. If we stick this in the equation
for n=1, we get P(2) in terms of P(0). Continuing down Panel 5, we
can get as many terms as we like. But we note that the general term
for P(n) is (A^n/n!)*P(0). All we need now is P(0).
To evaluate P(0), we note that, for a group of
N trunks, the probability that some number of calls, from 0 to N, is
in the system is 1 or certainty. Thus we can write:
1=P(0)+P(1)+P(2)+P(3)+...+P(N)
which we have just shown to be equal to
1=P(0)+A*P(0)+(A^2/2!)*P(0)+(A^3/3!)*P(0)+...+(A^N/N!)*P(0)
Because every term on the right has a P(0) in
it, we can factor it out and solve. Thus
P(0)=1/[1+A+(A^2/2!)+(A^3/3!)+...+(A^N/N!)]
and we can stick it back into our general
equations for P(N). This gives
P(N) =
(A^N/N!)/[1+A+(A^2/2!)+(A^3/3!)+...+(A^N/N!)]
which is the basic Erlang B formula for
blocking when A Erlangs of traffic are offered to N trunks. It is
the probability that exactly N trunks are busy. If fewer are busy,
there is no blocking because a call arriving can find a trunk;
further, if a call finds all trunks busy, it vanishes.
It is interesting to compare Erlang B with
Poisson. As we have seen, the Poisson probability that we have
exactly N calls in the system is
P(N) = (A^N/N!)*e^(-A)
But if we raise e to the -A power, it is the
same as having 1/e^A. So we can write
P(N) = (A^N/N!)/e^A
Students of mathematics know that e^A can be
expressed as an infinite series of terms of the form
1+A+(A^2/2!)+(A^3/3!)+... where the three dots mean the series goes
on forever. Thus Erlang B is like Poisson except that the series
stops after N terms. Keep in mind, also, that the probability of
exactly N calls being in an N trunk system is not the Poisson
blocking probability.
Suggestions for Further Study
Before attempting to study traffic theory
seriously, the reader must have a command of algebra and elementary
calculus; a first course in probability and statistics is also
highly desirable. Then Petr Beckmann's "Elementary Queuing Theory
and Telephone Traffic," from ABC TeleTraining, Geneva, IL, can be
attempted. This is the most readable book I have found on the
subject.
A more advanced book is Donald Bear's
"Principles of Telecommunication Traffic Engineering," published by
Peter Peregrinus, Ltd., and available through the IEEE in the United
States. Leonard Kleinrock's two-volume "Queueing Systems," from
Wiley, can be a whole career in itself.
Other helpful books include Ramses Mina's
"Introduction to Teletraffic Engineering" from Telephony, and the
two classics on probability, "Probability and Its Engineering Uses,"
by T. C. Fry (Van Nostrand) and Feller's "Probability Theory and Its
Applications" (Wiley). Syski's "Introduction to Congestion Theory in
Telephone Systems" (Oliver and Boyd), although mathematically
complex, contains a wealth of historical information. Both Syski and
Frankel (Tables for Traffic Management and Design, ABC TeleTraining)
contain extensive bibliographies.
Unfortunately, the most influential book on
Telephone Traffic has never been published. Called "Probability and
Statistics, Fundamentals - Applications to Traffic and Design
Problems," it was prepared for the Communications Development
Training Program (Kelly Kollege) at Bell Labs, apparently by R. I.
Wilkinson. The first edition came out in 1948, and various editions
followed. The coming of the Xerox machine has permitted generations
of telephone people, working temporarily at Bell Labs, to take
copies back into the real world. It is a great pity that this
excellent volume is not more generally available.
Short paragraphs describing each program.
5/8/83
(Webmaster’s Note: These
programs were on a disk that accompanied the original book. They are
not included in this web publication)
1. EXPO. Using average holding
time, EXPO calculates the probability that a given call will last as
long as or longer than a given interval. Exponential distribution of
holding times is assumed. Program then asks user to input holding
time in decimal minutes for specific duration probabilities.
2. SIMU-1. SIMU-1 simulates user
behavior on a PBX where each extension is assumed to be off hook 10
minutes (6 CCS) during the busy hour. Average and peak numbers of
calls in progress are accumulated on successive runs. User is asked
to put in PBX line size and a random number to initialize the random
number generator.
3. SIMU-2. SIMU-2 simulates user
behavior on a PBX where each extension's off-hook time during the
busy hour can be entered as required. Average and peak numbers of
calls in progress are accumulated on successive runs. User is asked
to input PBX line size and per cent of time average extension is off
hook.
4. BINOM-1. BINOM-1 applies the
binomial formula to calculate the probability that a particular
caller in a finite group will find a given number, and at least that
number, of the remaining callers off hook. Average occupancy per
source (caller) and number of sources are entered. Occupancy will be
a decimal number less than 1, typically on the order of 0.12.
5. BINOM-2. BINOM-2 applies the
binomial formula to calculate the probability that a given number,
and at least that number, of callers are off hook at one time, when
the total number of callers (sources) and the average occupancy per
source are specified. This is called "the wire-chief's point of
view." Occupancy per source is a decimal number less than 1,
typically in the range of 0.12.
6. POISSON. POISSON uses the
Poisson formula to calculate the probability that a given number,
and at least that number, of calls are in the system at a given time
when the offered traffic in erlangs (hours of use per hour) is
specified. Carried traffic and average delay are also presented.
POISSON assumes a call is in the system for one average holding
time, whether served or in queue; it can approximate short term
queuing. User inputs hourly offered traffic in Erlangs and average
call holding time in decimal minutes.
7. SIMU-3. SIMU-3 simulates the
Erlang B behavior of telephone calls offered to a trunk group. Under
Erlang B assumptions, calls arriving at random and finding all
trunks busy simply vanish. These are counted as overflows, and grade
of service is calculated for each hour's simulation by dividing the
number of overflows by the total number of offered calls. User
specifies offered traffic in Erlangs, average call holding time in
decimal minutes, and number of trunks in group. A random number must
also be supplied to initialize the random number generator
8. ERL-B. ERL-B is the Erlang B
traffic distribution which relates traffic offered to a trunk group
to the traffic offered to and carried by each trunk. Total carried
traffic and grade of service are also calculated. User inputs
offered traffic in Erlangs (hours of use per hour).
9. ERL-GS. ERL-GS is the Erlang
B distribution inverted. The user specifies the desired grade of
service, and the program finds the offered, carried and lost traffic
as well as the average occupancy for trunk groups from 1 to N trunks
in size. N should be less than 80. Grade of service is put in as a
decimal number less than 1. Typically, it might be 0.03.
10. ERLGSHC. ERLGSHC is
identical to ERL-GS but uses a printer for output rather than the
computer's CRT terminal.
11. ERL-C1. ERL-C1 is the Erlang
C distribution used for queuing. Probability of entering the queue
and delay in queue, measured in holding times, are resented, along
with number of calls in queue. These averages are taken over the
whole hour and also the part of the hour when all trunks are busy.
Calls arriving at random are assumed to stay in queue until served.
Thus offered equals carried traffic. The user specifies hourly
offered traffic in Erlangs.
12. ERL-C2. ERL-C2 is the Erlang
C distribution used for queuing. Probability of entering the queue
and delay in queue in seconds or minutes are presented, along with
number of calls in queue. These averages are taken over the whole
hour and also the part of the hour when all trunks are busy. Calls
arriving at random are assumed to stay in queue until served. Thus
offered equals carried traffic. The user specifies offered hourly
traffic in Erlangs and holding time in decimal minutes.
13. CARRIED. CARRIED uses Erlang
B to permit the user to specify carried rather than offered hourly
traffic for a trunk group; the user also specifies holding time in
decimal minutes, and percent traffic recalling and overflowing to
another group. The program shows traffic carried per trunk, assuming
the same hunt order is always used; it also summarizes the first
attempt, recalling, offered, carried, overflowing and lost traffic
in terms of hours and calls.
14. TAPRD. TAPRD calculates the
cost of WATS and calls via MCI and SPCC network services using the
carrier-specified tapered rates. The user inputs total hours carried
per month and the number of trunks in the group. The program finds
the cost for that situation, as well as the cost at 15, 40, 80 and
200 hours of average use per trunk.
15. B1-MONTH. B1-MONTH relates
monthly traffic to busy hour traffic and calculates traffic on each
trunk, using Erlang B and assuming all hours are busy hours. User
inputs monthly paid traffic, average holding time per call in
decimal minutes, overhead percent (to allow for dialing, ringing,
etc.), business days per month, busy hours per day, and percent of
traffic recalling and overflowing to toll. The program summarizes
first attempt, recalling and offered traffic, as well as carried,
overflowing and lost traffic in terms of both hours and calls, per
busy hour and per month.
16. B2-MONTH. B2-MONTH is
similar to B1-MONTH but uses a "shaped day" rather than assume all
hours are busy hours. A business day of 9 hours is assumed, with 3
busy hours, 3 hours at 2/3 of busy hour traffic, and 3 hours at 1/3
of busy hour traffic.
17. B3-MONTH. B3-MONTH is
similar to B2-MONTH but uses offered traffic rather than carried
traffic. While B1-MONTH assumes a business day of 9 hours, B2-MONTH
permits a "window" to be selected. The window might by 6 hours, for
instance, representing the time an east coast location can
communiate with the west coast (11 am to 5 pm in the east, 8 am to 2
pm in the west). Similar to B2-MONTH, one third of the window is
assumed to operate at busy hour traffic level, one third at 2/3 of
that level, and the rest at 1/3 of the busy hour traffic level.
18. C-MONTH. C-MONTH takes
monthly offered (or carried, as they are the same under Erlang C),
and finds out the queuing behavior when traffic is at the busy hour
level, 2/3 of that level, and 1/3 of that level. The user inputs
hours of paid traffic per month, and average holding time in decimal
minutes.
19. MULTI-B1. MULTI-B1 uses
Erlang B to estimate traffic carried by circuits in each WATS band
and overflowing to the next higher band, and then prices out the
carried monthly traffic. All hours are assumed to be busy hours. The
user inputs monthly carried traffic, either measured or postulated,
overhead percent factor, business days per month, busy hours er day,
and per cent traffic for each WATS band. Different trunk-group sizes
can then be tried for each service area.
20. MULTI-B2. MULTI-B2 is the
same as MULTI-B1 except for using a shaped day. The business day is
assumed to be 9 hours long, with three busy hours, three hours at
2/3 of busy hour traffic, and three hours at 1/3 of busy hour
traffic.
21. MULTI-B3. MULTI-B3 uses a
shaped day, similar to MULTI-B2, but it uses offered rather than
carried paid monthly traffic. Traffic can run up the WATS bands and
overflow to toll; then the entire cost is priced out. After the user
has input offered monthly paid traffic, overhead percent factor,
business days per month, busy hours per day, and average cost of
toll in dollars per minute, the program can be run for various
configurations of circuits in each group.
22. MINTREE. MINTREE is used to
find minimum distances between nodes in a network. The user inputs
the V and H coordinates of up to 10 nodes, and the program finds the
"minimum spanning tree," or minimum length path that connects all
nodes.
21. MAXTREE. MAXTREE is used to
find the highest community of interest among up to ten nodes. The
user inputs the two-way traffic between each air of nodes in a
network, and a "maximum spanning tree," emphasizing heaviest traffic
routes, is presented.
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