Lab 2: Productivity & Break Even Analysis
Exercise 1:
A manufacturer with his current equipment is able 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 a better quality raw cotton which will enable him to
produce 1200 square metres per ton of raw cotton. What will be the
impact on productivity (measured in square metres per labour-hour)
if he purchases the higher quality raw cotton?
Answer:
Productivity improves 20% = (240 - 200) / 200 = 0.2
Exercise 2:
An 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) i.e. problems per day? The mechanic believes
he can purchase a small diagnostic trouble-shooting device which
will allow him to find and fix a problem in a 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:
Exercise 3:
Joanna 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 at present
is approximately 3.50; she has to invest 20 in the necessary tools
(expendables) per day; energy costs are assumed to be only 4.00 per
day; and she feels that she should be making 10 per hour for her
time.
(a) Viewing this from a total factor productivity perspective,
what is her productivity at present and with the new paint? School
of Informatics and Engineering Manufacturing & Quality
Manufacturing & Quality Page 2 of 2
(b) How would productivity change if using the new paint raised
Joanna's material costs by 0.50 per doll?
(c) By what amount could the material cost increase without
reducing productivity? [Hint: let the cost equal X]
Answer:
(a)
Currently
Using the new paint
Labor
12 hrs. * 10
= 120
12 hrs. * 10
= 120
Material
240 * 3.50
= 840
360 * 3.50
= 1260
Supplies
= 20
= 20
Energy
= 4
= 4
Total Inputs
= 984
= 1404
Productivity
240/984
= 0.24
360/1404
= 0.26
(b)
If the material costs increase by $0.50 per doll:
Using the new paint
Labor
12 hrs. * 10
= 120
Material
360 * 4.00
= 1440
Supplies
= 20
Energy
= 4
Total Inputs
= 1584
Productivity
360/1584
= 0.23
(c)
Its unknown how high the material cost could go, using the new
paint, before the productivity drops to the current level of 0.24.
In mathematical terms the material cost is represented by a
variable (X), the new multifactor productivity value is set to the
current level, 0.24, and it is solved for X.
360/ ((12x10) + 360 (X) + 20 + 4) = 0.24
360 = 0.24(120 + 360 (X) + 20 + 4)
360 = 28.8 + 86.4 (X) + 4.8 + 0.96
325.44 = 86.4 (X)
(X) = 325.44/86.4 = 3.7666 3.77
It follows then that the new paint could raise Materials cost by
no more than approximately 0.27 (the difference between 3.77 and
3.50) before experiencing a decrease in multifactor
productivity.
Exercise 4:
A firm cleans chemical tank cars. With standard equipment, the
firm typically cleaned 60 chemical tank cars per month. They
utilized 10 gallons of solvent, and two employees worked 20 days
per month, 6 hours a day. The company decided to switch to a larger
cleaning machine. They cleaned 60 tank cars in only 15 days. They
utilized 12 gallons of solvent, and the two employees worked 6
hours a day.
(a) Calculate a productivity based on the solvent use and
another based on the labour hours, with the standard equipment?
(b) What are their productivities with the larger machine?
(c) What is the % change in productivities?
Answer:
Solution
Standard equipment OLD
Larger Machine NEW
Output Per month
60 tanks per month
60 tanks per month
Input
Solvent
10 gallons per month
12 gallons per month
Solvent cost per month
10*100 = 1000 SR per month
12*100 = 1200 SR per month
Number of Employees
Two
Two
Employee Hours
6 hours per day
6 hours per day
Number days per month
20 days per month
15 days per month
Employees Hour per month
2*6*20 = 240 hours per month
2*6*15 = 180 hours per month
Single factor Productivity Tanks per Gallon
60/(10) = 6 tanks per gallon
60/(12) = 5 tanks per gallon
Single factor Productivity Tanks per labour hour
60/(240) = 0.25 tanks per labour hour
60/(180) = 0.33 tanks per labour hour
(c)
Percentage change Productivity Tanks per Gallon:
Percentage change Productivity Tanks per labour hour
Exercise 5:
A manufacturing company intends to overcome a bottleneck in its
assembly line by through the addition of new equipment. Two
equipment suppliers have presented proposals. The fixed costs for
proposal A are 50,000 and for proposal B are 70,000. The variable
cost for A is estimated at 12.00 per unit and for B at 10.00 per
unit. The revenue generated by each unit is 20.00.
(a) What is the break-even point in units for proposal A and
proposal B?
(b) What is the break-even point in euro for proposal A and
proposal B?
(c) What volume of output (units) would the two alternatives
yield the same profit?
(d) If the expected volume of units is 8,500 which alternative
should be chosen?
Answer:
(a)
BEPx = F / (P V)
(b)
BEPm = P. BEPx
(c)
(SPa VCa) X-FCa = (SPb VCb) X-FCb
(20 12) X-50,000 = (20 10) X-70,000
8X-50,000 = 10X-70,000
70,000 50,000 = 10X-8X
20,000 = 2X
X = 10000
(d)
Profit = Total Revenue - Total Costs - (Total Variable Costs +
Total Fixed Costs) When expected volume is at 8500 units: Proposal
A- Profit= (20 x 8500 units) - 50000 - (12 x 8500 units) =18000
Proposal B- Profit= (20 x 8500 units) -70000 - (10 x 8500 units)
=15000 Since profit for proposal A is higher than profit for
proposal B when expected volume is at 8500 units, A is chosen.
Current pr
oductivity
=
8 hours pe
r day
2 hours pe
r problem
problems
per day
=
4
Productivi
ty with co
mputer
=
7 hours pe
r day
1 hour per
problem
problems
per day
=
7
74 3
Productivity improves 75% .75
44
-
==
1000 sq yds
Current labor productivity = 200 sq yds
per hour
1 ton*5 hours
=
New labor
productivi
ty
=
1200 sq yd
s
1 ton
*
5 hours
240 sq yds
per hour
=
