Business Logistics Assignment Help

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Problem 1 :
Consider a single-server queueing system with infinite queue capacity. Arrivals of customers to this system occur in a Poisson manner at
the rate of 36 per hour. The service times, S, of customers are mutually independent and their duration is uniformly distributed between
1 and 2 minutes, in other words, f S (t) = U[1, 2] in units of minutes.
(1) Compute the expected waiting time in queue, Wq , and the expected number of customers in the queue, Lq , when this queueing
system is in steady state. [Numerical answers are expected.]
(2)Assume now that service times have a negative exponential probability density function
with for t ³ 0 , in units of minutes. Repeat part (1), i.e., compute numerically the values of Wq and of Lq for this case.
(3) Assume now that service times are constant and all have duration equal to 1.5 minutes. Repeat part (1), i.e., compute numerically the
values of Wq and of Lq for this case.
(4) Using the G/G/1 bounds discussed in class, compute numerically upper and lower bounds for Wq and for Lq for the case of Part (1).
(5) Return to the original case (uniform service times) from here until the end of the problem. Suppose that immediately after the
completion of a service, exactly 2 customers are left in the system. Carefully write an expression for the probability that after the end of
the next service, exactly 4 customers will be left in the system. Do NOT evaluate numerically this expression.
(6) Repeat part (4) under the supposition that after the completion of a service, exactly 0 customers are left in the system. Carefully write
an expression for the probability that after the end of the next service, exactly 4 customers will be left in the system. Do NOT evaluate
numerically this expression.
(7) Consider again part (5) where after the completion of a service, exactly 2 customers are left in the system. Carefully write an
expression for the probability that TWO service completions later, exactly 4 customers will be left in the system. Do NOT evaluate
numerically this expression.
(8) Suppose that George begins observing the queueing system at a random time (assume the system is in steady state). [George is not a
customer – he is just an observer.] What is the expected time until George observes the first completion of a service? A numerical answer
is expected in this case.

(9) In this part assume that the system’s customers are of two types: Type 1, who constitute 40% of the customers,
and Type 2, who constitute 60%. Assume that Type 1 customers enjoy non-preemptive priority over the ones of Type
2. Assume also that the arrivals of Type 1 and Type 2 customers are randomly intermingled. Finally note that the
service times of both types of customers are uniformly distributed between 1 and 2 minutes. Compute numerically
the expected waiting time, Wq , for each type of customer and for all the customers
using the system.
(10)Assume we now separate the customers of the system into two classes, A and B. Class A customers constitute
40% of the total and have f SA (t) = U[1, 1.4]. Class B customers constitute 60% of the total and have f SB (t) = U[1.4,
2]. Suppose Class A customers have non-preemptive priority over Class B customers. Compute numerically the value
of Wq for each class of customer. Compute, as
well the value of Wq for all the customers using the system and compare with the answer to Parts (1) and (9). Why
are the answers different or the same?
(11)Repeat Part (10) but with Class B having non-preemptive priority over Class A.
Solution:
We have a single-server queueing system with infinite queue capacity. Arrivals of customers to the system occur in
a Poisson manner at the rate of 36 per hour, that is N= gper minute. The service times, S, of customers are
mutually independent and their duration is uniformly distributed between 1 and 2 minutes.
1. This is an M/G/1 system, and we can use the results derived in class for Lq and Wq. Recall that for a uniform
distribution we have

2. When the service times have negative exponential probability density, our M/G/1 queue is a M/M/1 queue. We
can therefore either use the results for M/M/1 or M/G/1 queueing systems. We choose to confine to use the
formulas from (a). Note that for a negative exponential distribution as = E[S] = 4 and therefore C(S) = 1, p= is as
before.

3. Now a9 = 0 and we therefore Cs = 0.
4. We use the bounds discussed in class:
Thus, the upper and lower bounds are
By Little’s Law,

5. For exactly 4 customers to be in the system at the end next service, we need to have exactly three arrivals during
the service of the next customer. Call this next service time t1. Then the probability of three arrivals given ti is given
by:
We now need to evaluate this expression for all possible service times. The service times are Uniformly distributed
between 1 and 2 and therefore f s(t) = 1 1 < s < 2. Therefore:
6. For there to be exactly four customers in the system after the next completion of service we need exactly four
arrivals during the next service. Our expression therefore becomes:

7. Now there are two customers in the system. The event that after their service there are exactly four customers
in the system is the event that we have four carnivals during the two service times.
We first need to find the distribution of the two service times. Recall from Problem set one, that the convolution
of two uniforms has a triangular shape density. Call Sun = St + S2 then the density for S is shown in the figure
below:

Either George arrives during a busy period or during an idle period. Call the event that George arrives during a busy
period A and the event that he arrives during a idle period At. We can then express George's expected waiting time
using the total probability theorem:
Now we know from a) that P(A) = and P(A`) = lb. If George arrives during a busy period we haw to take random
incidence into account. Using our random incidence formula for expected remainder (2.66 in Larson and Odoni) of
an service period we have:
If George arrives in an idle period his expected waiting time is the expected time until the next arrival (which is 1/A)
plus the expected service time, that is
His total waiting time is therefore
9. Using results in ch 4.9 in Larson and Odoni we can calculate Wo:
Equation 4.107 (Larson & Odoni) gives us the expected wait in queue for differnt priority groups:

10.
We now have two classes of customers.
For Class A customers: l 1 =14.4 per hour and thus
For Class B customers: l 2 =21.6 per hour and therefore

In this new case, Wq is lower than in part 1 and part 9 because we are now giving the priority to the customers
with the shortest expected time.
A corollary takes advantage of this phenomenon ( p239 of the textbook):
To minimize the average waiting time for all users in the system, assign priorities according to the expected
service times for each user class: the shorter the expected service time, the higher the priority of the class.
11.
This time, Class B customers have the priority. They have the greatest expected service time.
Therefore, the overall q W increases.
Numerical answer: q W = 8.28

Problem 2
Problem 4.13 in L + O, with the following important change: assume the queue capacity in front of Q.S. 3 is 0 (zero),
not 1. This will significantly reduce the number of states you need to worry about. To do this problem, try first to
identify all the possible states and then draw the state transition diagram. You will need some extra paper because the
first time you’ll try to draw the diagram, a rather messy picture will emerge.
Solution:
We define the states (i, j, k)
where:
- i describes System 1 and i = 0,1 or b ( for blocked)
- j describes System 2 and j = 0,1 or b
-k describes Sytem 3 and k = 0 or 1
There are 13 states.
The corresponding state transition diagram is:

Problem 3
Airline DeeDee has only three airplanes. (There are many small airlines of this size around the world.) DeeDee has
signed a contract with the ABC Corporation under which ABC will provide unscheduled, non-routine repair services
to DeeDee when aircraft suffer mechanical breakdowns. The service times to aircraft at ABC’s repair facility have a
negative exponential pdf with a mean value of 12 hours. However, it has been observed that, with probability 0.6,
an aircraft will need a second, supplementary repair operation after the first repair is completed. (With probability
0.4, the first repair is satisfactory and the aircraft goes back to flying.) The service times for the second repair, if
needed, also have a negative exponential pdf with a mean of 12 hours. After the second repair
operation, the aircraft is always in a “satisfactory” state for flying. The repair facility services only one aircraft at a
time and will perform all repair operations on an aircraft (one or two), before processing the next aircraft on a first
come, first served basis. Finally, it is known that the time between the instant when repair service to an aircraft is
completed and the instant when that aircraft will break down again for the first time has a negative exponential pdf
with a mean of 3 days. For the sake of simplicity, assume, as well, that both DeeDee and ABC operate continuously,
on a 24-hours a day basis.
(a) Draw a state-transition diagram for this queueing system, making sure to define carefully its states.
(b) What is the expected number of DeeDee aircraft in ABC’s repair facility in steady state? This includes aircraft
being repaired and aircraft waiting for repairs.
Solution:
(a) We define the states (a, b. c) Where:
a=number of planes in first stage of service
b=number of planes in second stage of service
c=number of planes in queue

(b) We need to compute the steady state probabilities. These can be found by solving the following set of equations:

Problem 4
Consider a queueing system with two parallel servers and two spaces for queueing (in addition to the two servers).
This facility serves two types of customers. Type 1 customers are of the conventional type. They arrive in a Poisson
manner at a rate of l1 per minute. The service time to these customers has a negative exponential pdf with a rate of
m1 per minute for each server. Any arriving Type 1 customers who find the system full (i.e., 4 customers in the
system) are lost.
Type 2 customers are unusual. They, too, arrive in a Poisson manner at a rate of l2 per minute, but they arrive IN
PAIRS. [Think of a restaurant, where some of the customers (Type 1) arrive individually, and others (Type 2) arrive in
pairs.] Moreover, when service to each one of these pairs begins, the pair occupies simultaneously TWO servers
(i.e., both of the servers). The servers work together on each of these Type 2 pairs. The service time to the pair has
a negative exponential pdf with a rate of m2 per minute. (Note that this means that the two servers begin and end
service to each Type 2 pair simultaneously and, together, can serve m2 Type 2 pairs per minute if working
continuously on Type 2 pairs.) Type 2 customers who do not find at least TWO available spaces upon arrival are lost.
Please draw carefully a state transition diagram that describes this queueing system. Please make sure to define
clearly the states of the system.
Suppose there are currently four Type 1 customers in the system. Write an expression for the probability that
exactly 3 state transitions later, there will be exactly two Type 1 customers and one Type 2 pair in the system. Your
expression should be solely in terms of l1, m1, l2 and m2. [A state transition takes place whenever there is a new
actual arrival (not a lost arrival) to the system or whenever there is a service completion. The arrival of a new Type
2 pair counts as one transition and so does the completion of a service to a Type 2 pair.]

Solution:
(a) A state is defined by (a,b,c) where:
a = the type of the customer(s) (0, 1, or 2) currently in service (you cannot have
one server occupied by a Type 1 customer and the other server occupied by a
Type 2 customer at the same time),
b = the number of Type 2 customers (0, 1, or 2) in the system,
c = the number of Type 1 customers (0, 1, 2, 3 or 4) in the system.
The corresponding state transition diagram for the system is:

(b) The event of interest can happen only by having a set of three consecutive transitions from state (1,0,4) to
(1,0,3) to (1,0,2) to (1,1,2). We have:

Problem 5
We have solved in class Example 3 of pages 246 and 247 in Larson and Odoni, assuming distinct service rates μ1 and
μ2 at the two queueing facilities. We shall now look at the case in which the first queueing system (the one on the
left) has two independent and identical servers, each with service rate equal to μ1. There is still no waiting space in
front of either of the two facilities.
(a)Draw carefully the state-transition diagram (i.e., analogous to Figure 4.20) for this case. Define clearly the states
of the system. Do NOT do any further calculations.
(b) Assume that you have computed the steady-state probabilities associated with the states in the state-transition
diagram you drew in part (a). Write an expression for L, the expected number of customers in this “network” (i.e.,
this two-queueingfacility system) in terms of the steady-state probabilities.
Solution:
A state is defined by (a,b) where a describes the first facility and b the second one.
a = 0 if there are no customers
1 if there is one customer
2 if there are 2 customers
b1 if there is one customer and he/she is blocked
b2 if there are two customers and one is blocked

b3 if there are two customers and both are blocked.
b = 0 if there are no customers
1 if there is one customer.
b = 0 if there are no customers
1 if there is one customer.
The corresponding state transition diagram for the system is:

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