ISE313 COMPUTER INTEGRATED MANUFACTURING AND AUTOMATION I
FINAL EXAM
15.01.2008
1. Name the three cases of part or product variety in manufacturing systems. Briefly define each of
the three cases. (10 points)
Answer: The three cases of part or product variety in manufacturing systems are
(1) single model,
(2) batch model, and
(3) mixed model.
In the single-model case, all parts or products made by the manufacturing system are identical. In
the batch-model case, different parts or products are made by the system, but they are made in
batches because the physical setup and/or equipment programming must be changed over between
models. In the mixed-model case, different parts or products are made by the system, but the
differences are not significant, so the system is able to handle them without the need for timeconsuming changeovers in setup or program.
2. What is a machine cluster? (5 points)
Answer: As defined in the text, a machine cluster is a collection of two or more machines producing
parts or products with identical cycle times and serviced (usually loaded and unloaded) by one
worker.
3. What are the five enablers that are required for unattended operation of a single-model or batchmodel automated production cell? (10 points)
Answer: As given in the text, the five enablers of unattended operation for a single-model or batchmodel automated production cell are the following:
(1) a programmed cycle that allows the machine to perform every step of the processing or
assembly cycle automatically;
(2) a parts storage subsystem and a supply of parts that permit continuous operation beyond one
machine cycle;
(3) automatic transfer of workparts between the storage system and the machine (automatic
unloading of finished parts from the machine and loading of raw workparts to the machine);
(4) periodic attention of a worker who performs the necessary machine tending functions (e.g.,
parts loading and unloading of the storage subsystem) for the particular processing or assembly
operation; and
(5) built-in safeguards that protect the system against operating under conditions that may be
unsafe or self-destructive to itself or destructive to the work units being processed or assembled.
4. A worker is currently responsible for tending two machines in a machine cluster. The service time
per machine is 0.35 min and the time to walk between machines is 0.15 min. The machine automatic
cycle time is 1.90 min. If the worker's hourly rate = $12/hr and the hourly rate for each machine =
$18/hr, determine (a) the current hourly rate for the cluster, and (b) the current cost per unit of
product, given that two units are produced by each machine during each machine cycle. (c) What is
the % idle time of the worker? (d) What is the optimum number of machines that should be used in
the machine cluster, if minimum cost per unit of product is the decision criterion? (20 points)
Solution: (a) Co = $12 + 2($18) = $48.00/hr
(b) Tc = Tm + Ts = 1.90 + 0.35 = 2.25 min/cycle
 60 
 = 106.67 pc/hr
 2.25
Rc = 2(2) 
Cpc =
$48 / hr
= $0.45/pc
106.67pc / hr
(c) Worker engagement time/cycle = 2(Ts + Tr) = 2(0.35 + 0.15) = 1.0 min
Idle time IT =
(d) n =
2.25  1.0
= 0.555 = 55.5%
2.25
1.90  0.35
= 2.25/0.5 = 4.5 machines
0.35  0.15
n1 = 4 machines: Cpc(4) = 0.5(12/4 + 18(2.25/60) = $0.394/pc
n2 = 5 machines: Cpc(5) = 0.5(12 + 18x5)(0.50/60) = $0.425/pc
Use n1 = 4 machines
5. Name four of the six measures used to assess the performance of a storage system? (10 points)
Answer: The six performance measures discussed in the text are the following:
(1) storage capacity, which is defined and measured either as the total volumetric space available
or as the total number of storage compartments in the system available for items or loads;
(2) storage density, defined as the volumetric space available for actual storage relative to the total
volumetric space in the storage facility;
(3) accessibility, which refers to the capability to access any desired item or load stored in the
system;
(4) system throughput, defined as the hourly rate at which the storage system receives and puts
loads into storage and/or retrieves and delivers loads to the output station;
(5) utilization, which is defined as the proportion of time that the system is actually being used for
performing storage and retrieval operations compared with the time it is available, and
(6) availability, defined as the proportion of time that the system is capable of operating (not
broken down) compared with the normally scheduled shift hours.
6. An AS/RS with four aisles is 80 m long and 18 m high. The S/R machine has a maximum speed
of 1.6 m/s in the horizontal direction. It accelerates from zero to 1.6 m/s in a distance of 2.0 m. On
approaching its target position (where the S/R machine will transfer a load onto or off of its
platform), it decelerates from 1.6 m/s to a full stop in 2.0 m. The maximum vertical speed is 0.5 m/s,
and the acceleration and deceleration distances are each 0.3 m. Rates of acceleration and
deceleration are constant in both directions. Pick and deposit time = 12 s. Utilization of the AS/RS is
assumed to be 90%, and the number of dual command cycles = the number of single command
cycles. (a) Calculate the single command and dual command cycle times, including considerations
for acceleration and deceleration. (b) Determine the throughput rate for the system. (25 points)
t
Solution: (a) Horizontal travel: v =  adt = at = 1.6 m/s,
a = 1.6/t
0
t
y =  atdt = at2/2 = 0.5(1.6/t) t2 = 0.8 t = 2 m
t = 2/0.8 = 2.5 s.
0
t
Vertical travel: v =  adt = at = 0.5 m/s,
a = 0.5/t
0
t
z =  atdt = at2/2 = 0.5(0.5/t) t2 = 0.25 t = 0.3 m
t = 0.3/0.25 = 1.2 s.
0
0.5(18)  2(0.3)
 0.5(80)  2(2)

 2(2.5),
 2(12
. ) + 2(12) = 79 s/cycle = 1.3167 min/cycle


16
.
0.5
Tcs = 2 Max 
0.75(18)  3(0.3)
 0.75(80)  3(2)

 3(2.5),
 3(12
. ) + 4(12) = 130.5 s/cycle = 2.175


16
.
0.5
Tcd = 2 Max 
min/cycle
(b) Rcs Tcs + Rcd Tcd = 60(0.90) = 54 min
Given Rcs = Rcd, 1.3167 Rcs + 2.175 Rcs = 3.4917 Rcs = 54.0
Rcs = 15.465 cycles/hr
Rcd = Rcs = 15.465 cycles/hr
Rt = Rcs + 2 Rcd = 15.465 + 2(15.465) = 46.396 transactions/hr per aisle
With 4 aisles, Rt = 4(46.396) = 185.6 transactions/hr
7. A recirculating conveyor has a total length of 700 ft and a speed of 90 ft/min. Spacing of
part carriers = 14 ft. Each carrier can hold one part. Automatic machines load and unload
the conveyor at the load and unload stations. Time to load a part is 0.10 min and unload
time is the same. To satisfy production requirements, the loading and unloading rates are
each 2.0 parts per min. Evaluate the conveyor system design with respect to the (i) speed
rule, (ii) capacity constraint, and (iii) uniformity principle. (20 points)
Solution: (1) Speed rule: Lower limit:
n p vc
sc
 Max{RL, Ru}
(1part / carrier )( 90ft / min)
= 6.428 parts/min  Max{2, 2}  (OK)
14ft / carrier
v
1 1
Upper limit: c  Min{ , }
sc
TL Tu
90ft / min.
1 1
= 6.428 carrier/min  Min{ , } = Min{10, 10}  (OK)
14ft / carrier
01
. 01
.
(2) Capacity constraint: In this case, the flow rate is interpreted to be the specified load and
unload rates,
that is, Rf = RL = Ru = 2 parts/min
n p vc
sc
= 6.428 parts/min  2  (OK)
(3) Uniformity principle: The conveyor is assumed to be uniformly loaded throughout its
length since
RL = Ru and the flow rate capacity is significantly greater than RL and Ru.  (OK)
Conclusion: The system is feasible.