Production Engineering
PIB360S
Operations & Productivity Tutorial
Memorandum
Question 1
Johan Brink, the Production Manager at Ralts Mills, can currently expect his operation
to produce 1000 square metres of fabric for each ton of raw cotton. Each ton of raw
cotton requires 5 labour hours to process. He believes that he can buy better quality raw
cotton, which will enable him to produce 1200 square metres per ton of raw cotton with
the same labor hours.
What will be the impact on productivity (measured in square metres per labour-hour) if
he purchases the higher quality raw cotton?
Answer:
1000m 2
= 200m 2 per hour
Current labour productivity =
1 ton ∗ 5hours
New labour productivity =
1200 m 2
= 240m 2 per hour
1 ton ∗ 5hours
Productivity improves 20% = (240 - 200) / 200 = 0.20
1
Question 2
Thabo Masonto, the local auto mechanic, finds that it usually takes him 2 hours to
diagnose and fix a typical problem. What is his daily productivity (assume an 8 hour
day)?
Thabo believes he can purchase a small computer trouble-shooting device, which will
allow him to find and fix a problem in the incredible (at least to his customers!) time of
1 hour. He will, however, have to spend an extra hour each morning adjusting the
computerized diagnostic device. What will be the impact on his productivity if he
purchases the device?
Answer:
Productivity with computer =
7 hours per day
= 7 problems per day
1 hour per problem
7−4 3
Productivity improves 75% = 
 = = 0.75
 4  4
2
Question 3
Sabrina Garcia is currently working a total of 12 hours per day to produce 240 dolls.
She thinks that by changing the paint used for the facial features and fingernails that
she can increase her rate to 360 dolls per day. Total material cost for each doll is
approximately R52.50; she has to invest R300 in the necessary supplies (expendables)
per day; energy costs are assumed to be only R60.00 per day; and she thinks she
should be making R150 per hour for her time. Viewing this from a total (multifactor)
productivity perspective, what is her productivity at present and with the new paint?
Answer:
Currently
Using the new paint
Labour
12hrs * R150 = R 1800
12hrs * R150 = R 1800
Material
240 * R52.50 = R 12600
360 * R52.50 = R 18900
Supplies
= R 300
= R300
Energy
= R 60
= R 60
Total Inputs
= R 14760
= R 21060
Productivity =
240/14760
= .016
360/21060 = .017
3
Question 4
How would total (multifactor) productivity change if using the new paint raised
Ms Garcia’s material costs by R7.50 per doll?
Answer:
If the material costs increase by R7.50 per doll:
Using the new paint
Labour
12 hrs * R150 = R 1800
Material
360 * R60.00 = R 21600
Supplies
= R 3000
Energy
= R 60
Total Inputs
= R 23 760
Productivity
360/23760 = 0.015
4
Question 5
If she uses the new paint, by what amount could Ms Garcia’s material costs increase
without reducing total (multifactor) productivity?
Answer:
From the answer to Question 3 we know the following:
Currently
Using the new paint
Labour
12hrs * R150 = R 1800
12hrs * R150 = R 1800
Material
240 * R52.50 = R 12600
360 * R52.50 = R 18900
Supplies
= R 300
= R300
Energy
= R 60
= R 60
Total
Inputs
= R 14760
= R 21060
Productivity =
240/14760
= .016
360/21060 = .017
We want to know how high the material cost could go, using the new paint, before the
productivity drops to the current level of 0.016. In mathematical terms we make the
material cost a variable (X), set the new multifactor productivity value to the current
level, 0.016, and solve for X.
360/[(R180×10) + 360 R(X) + R300 + R60] = 0.016
R(X) = R56.50
It follows then that the new paint could raise Materials cost by no more than
approximately R4.00 (the difference between R56.50 and R52.50) before Ms Garcia
would experience a decrease in multifactor productivity.
5