PROGRAM : B.
ENG TECHNOLOGY
ENGINEERING: INDUSTRIAL
SUBJECT : FACILITIES LAYOUT DESIGN and
MATERIALS HANDLING
CODE : FACMIB2
TEST : ONE
DATE : 30 August 2021
DURATION : 08:00 – 10:00
TOTAL MARKS : 50
MEMORANDUM
ASSESSOR : Ms B.N MUSHWANA
MODERATOR : Dr R.P MUTYAVAVIRE
NUMBER OF PAGES : 4 PAGES
INSTRUCTIONS : PLEASE ANSWER ALL QUESTIONS
1
QUESTION ONE
1.1. Manufacturing facilities frequently undergo layout design changes. Briefly discuss the
factors that may necessitate such changes. (6)
ANSWER
To build-in flexibility
To reduce or eliminate excessive inventory
To provide for employee convenience, excessive inventory
To promote effective use of people, equipment and space
To optimize quality
To minimize project and unit cost
ACCEPT ANY OTHER RELATED ANSWER
1.2. Mzi’s assembly plant has confirmed orders of 9000 units per month. The plant operates
two (2) shifts per day, five (5) days per week. The average non-productive time per shift is 60
minutes. Plant efficiency is 88% and overall defective rate is 2%. Determine the takt time
necessary to satisfy the customer orders. (6)
ANSWER
𝑞𝑜
𝑞= (1−𝑑1 )
9000
= (1−0.02)
= 9148 units per month
Available time = (2x8x5x4) x 60= 19200mins
Non-productive time = (2x5x4) x 60 = 2400 mins
Productive time = 19200 – 2400 = 16800 mins
Effective time = 16800 x 0.88 = 14784 minutes per month
2
14784
Takt time = 9184
= 1.610 minutes per unit
[12]
QUESTION TWO
Given the figure below, operations D and E represent rework operations on parts that fail
inspection upon completion of operations A and B.
A B C 80 000
7% 9% 5%
D E
40% 60%
2.1.1 Determine the number of units that the process must start with in order to meet
the required output of 80 000 units (7)
ANSWER
𝐼𝐵 = 𝑂𝐴 + 𝑂𝐷
= (1-0.07)q + (1-0.04)(0.07)q
= 0.93q + 0.042q
= 0.972q
𝐼𝐶 = 𝑂𝐵 + 𝑂𝐸
= (1-0.09) 0972q + (1-0.6)(0.09)0.972q
= 0.885q + 0.035q
= 0.920q
𝑂𝐶 = (1-0.05) 0.920q
3
but 𝑞𝑜 = 80 000
and 𝑞𝑜 = (1-0.05) 0.920q
𝑞𝑜
therefore; 𝑞 = 0.920 (1−0.95)
80 000
= 0.920 (1−0.95)
= 91533 or 91534 units
2.1.2 How many defectives are produced at operation C? (2)
ANSWER
Defectives at C = 0.920 x 91534 x 0.05
= 4211 units
2.2 Calculate the cost of defects at operation C; if each defect costs R45. (1)
ANSWER
Cost of defectives at C = 4211 x R45
= R189 495
[10]
QUESTION THREE
3.1 Briefly describe the purpose of a route sheet in process design. (2)
ANSWER
The route sheet lists the operations required to make that part in the proper sequence. The route
sheet would accompany the material from operation to operation telling the operators what to
do. The route sheet will tell the plant personnel about the part number, part name, quantity to
produce (left blank until needed), operation number, operation description, machine number
(if available), machine name, tooling needed, and time standard.
ACCEPT ANY OTHER RELATED ANSWERS
4
3.2 Nine (9) component parts of an assembly are processed through two (2) machines in a
production factory. Table Q3 provides the time standards in pieces per hour at each machine.
Given a takt time of 0.158 minutes/unit, determine the total machine requirements. (14)
[16]
Table Q3
Part 001 002 003 004 005 006 007 008 009
Operations
Parts per 1 1 3 2 2 3 1 2 1
assembly
Punch 1550 1850 1450 2100 1600 1420 1750 1100 -
Drill 1300 980 1100 1750 1050 1175 - 1220 1020
ANSWER
Takt time = 0.158 mins
Standard times in minutes
Table Q3
Part 001 002 003 004 005 006 007 008 009
Operations
Parts per 1 1 3 2 2 3 1 2 1
assembly
Punch 0.039 0.032 0.041 0,029 0.038 0.042 0.034 0.055 -
Drill 0.046 0.061 0,055 0.034 0.057 0.051 - 0.049 0.059
5
Machine requirements
Part 001 002 003 004 005 006 007 008 009 Total
machines
Operation
Parts per 1 1 3 2 2 3 1 2 1
assembly
Punch 0.247 0.20 0.77 0.36 0.48 0.79 0.21 0.69 - 3.78 4
3 8 7 1 7 5 6 4
Drill 0.291 0.38 1.04 0.43 0.72 0.96 - 0.62 0.37 4.83 5
6 4 0 2 8 0 3 4
QUESTION FOUR
Sasekani accessories factory in Doornfontein produces six (6) different molded plastic parts.
After molding, the parts are taken by a conveyor through a spray painting booth and a drying
oven. The company operates a single shift, five days per week. The production schedule for a
particular week is illustrated in Table Q4. Given that the employee personal time is 35 minutes,
plant efficiency is 92%, and hook spacing is 0.4m.
4.2.1 Determine the appropriate conveyor speed to meet production demand. (7)
ANSWER:
Table Q4
Part name Parts per hook Weekly demand Needed Hooks per Hooks per minute
shift
Wiper arm 5 1500 300 0.147
Wheel cover 8 2600 325 0.159
Sun visor 2 1350 675 0.330
Ind. lamp 12 9600 800 0.391
6
Tail lamp 4 1800 450 0.220
Side mirror 6 1200 200 0.098
1.345
Available time = 8x5x60 = 2400 mins
Non-productive time = 35x5 = 175 mins
Productive time = 2400 – 175 = 2225 mins
Effective time = 2225 x 0.92 = 2047 mins
Conveyor speed = Hooks per min x Hook spacing
= 1.345 x 0.4
= 0.538 m/min
4.2.2 From the spray booth, the components pass through a backing oven. The paint needs 15
minutes to cure. Calculate the required length of the oven. (2)
ANSWER:
Length of the oven = 0.538 x 15
= 8.07 m
4.2.3 It is realized that there is insufficient space to accommodate the oven as originally
designed. How can we reduce the size of the oven and still achieve the desired curing time for
the paint.
ANSWER: (3)
Make use of volumetric space. Design multi-later or U-shaped oven.
7
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TOTAL MARKS: 50
