# Solution

```Student’s Name: …………………………………..
Student’s No.: …………………………………
ISE 313 AUTOMATION AND COMPUTER INTEGRATED MANUFACTURING
FINAL EXAM
1.
14.01.2010
Provide a definition of material handling.
Answer: Material handling is defined as “the movement, storage, protection and
control of materials throughout the manufacturing and distribution process
including their consumption and disposal.”
2.
The CNC grinding section has a large number of machines devoted to grinding shafts for
the automotive industry. The grinding machine cycle takes 3.6 min. At the end of this
cycle an operator must be present to unload and load parts, which takes 40 sec.
(a) Determine how many grinding machines the worker can service if it takes 20 sec to
walk between the machines and no machine idle time is allowed.
(b) How many seconds during the work cycle is the worker idle?
&copy; What is the hourly production rate of this machine cluster?
Solution:
(a) n =
3.6( 60)  40
= 256/60 = 4.27
40  20
Use n1 = 4 grinding machines because no machine idle time is allowed.
(b) Worker idle time IT = 256 – 4(60) = 256 – 240 = 16 sec
&copy; Tc = 256 sec = 4.267 min
Rc = 4 
60 
 = 56.25 pc/hr
 4.267 
3. A certain type of machine will be used to produce three products: A, B, and C. Sales
forecasts for these products are: 52,000, 65,000, and 70,000 units per year,
respectively. Production rates for the three products are, respectively, 12, 15, and 10
units/hr; and scrap rates are, respectively, 5%, 7%, and 9%. The plant will operate 50
weeks per year, 10 shifts per week, and 8 hr per shift. It is anticipated that production
machines of this type will be down for repairs on average 10 percent of the time. How
many machines will be required to meet demand?
4.
Solution:
AT = 50(10)(8)(1 – 0.10) = 3600 hr/yr per machine
WL =
52,000
65,000
70,000


12(1  0.05) 15(1  0.07) 10(1  0.09)
n = 16,913.2/3600 = 4.67
4.
= 4561.4 + 4659.5 + 7692.3 = 16,913.2 hr/yr
use 5 machines
An overhead trolley conveyor is configured as a continuous closed loop. The delivery
loop has a length of 120 m and the return loop of 80 m. All parts loaded at the load station
are unloaded at the unload station. Each hook on the conveyor can hold one part and the
hooks are separated by 4 m. Conveyor speed is vc = 1.25 m/s. Determine
(a) maximum number of parts in the conveyor system,
(b) parts flow rate; and
conveyor system?
Solution:
(a) Number of parts on the conveyor =
(b) Rf =
n p vc
sc
(c) TL = Tu =
=
(1part / carrier )(1.25m / s)
( 4m / carrier )
1
Rf
= 1/0.3125 = 3.2 sec
120m
4m / part
= 30 parts
= 0.3125 parts/s = 1125 parts/hr
```