CHAPTER 15
PRODUCTIVITY MEASUREMENT AND CONTROL
QUESTIONS FOR WRITING AND DISCUSSION
1. Total productive efficiency is the point where
technical and allocative efficiency are
achieved. It is the point where the optimal
quantity of inputs is used to produce a given
output.
when trade-offs among inputs exist. No value is attached to productivity changes.
9. Profit-linked productivity measurement and
analysis is an assessment of the amount of
profit change—from the base period to the
current period—attributable to productivity
changes.
2. Technical efficiency means that for any mix
of inputs, no more of any one input is used
than necessary. Allocative efficiency means
that the least costly and most technically efficient mix is chosen.
10.
Profit-linked productivity measurement allows managers to assess the economic effects of productivity improvement programs.
It also allows valuation of input trade-offs —a
critical element in planning productivity
changes.
11.
The price-recovery component is the difference between the total profit change and the
change attributable to productivity effects.
12.
Activity productivity analysis measures
changes in the productivity of individual activities. It does this by measuring the inputs
and outputs of each activity and comparing
the results against a base period. Both profile and profit-linked measures are possible.
The approach is limited because it can be
applied only to value-added activities.
13.
Process productivity analysis measures
process productivity by calculating two components and then summing these two components. The first component is simply activity productivity. The second component is
activity output efficiency. For the second
component, process output is defined and
individual activity output is viewed as process input. Both profile and profit-linked
measurement and analyses are possible.
This approach can be used to assess
changes in efficiency for both value-added
and non-value-added activities.
3. Productivity measurement is a quantitative
assessment of productivity changes.
4. If the productivity ratio (output/input) has only
one input, then it is a partial measure. If all
inputs are included, then it is a total measure
of productivity.
5. An operational productivity measure is expressed in physical terms, whereas a financial productivity measure is expressed in dollars.
6. Partial measures can be misleading since
they do not consider possible trade-offs
among inputs. They do, however, allow
some assessment of how well individual factors are being used and, additionally, often
serve as input to total measures. Total
measures are preferred because they provide a measure of the overall change in
productivity, and they allow managers to assess trade-offs among inputs.
7. A base period serves as a standard or
benchmark for assessing changes in productive efficiency.
8. Profile measurement and analysis computes
a series of operational partial productivity
measures and compares this series with the
corresponding series of the base period to
assess the nature of the productivity changes. Profile analysis does not indicate whether productivity changes are good or bad
348
EXERCISES
15–1
1.
Combinations B and C are technically efficient. Combination B can produce
the same output for less of each input than Combination A. Similarly, Combination C can produce the same output for less of each input than Combination D. Comparing B and C shows that trade-offs exist among the inputs, and
so it is not possible to say that B is more technically efficient than C (or vice
versa).
2.
Once the technically efficient input combinations are identified, then the least
costly combination should be chosen. Input prices are used to value the
trade-offs (B uses more materials but less labor and energy than C):
Combination B: ($8  110) + ($10  180) + ($2  540) = $3,760
Combination C: ($8  92) + ($10  190) + ($2  570) = $3,776
Combination B is the best choice based on allocative efficiency.
15–2
1.
Output-input ratios (Combination C1):
Materials: 4,000/14,000 = 0.29
Labor:
4,000/7,000 = 0.57
Yes, there is improvement. Current productivity is:
Materials: 4,000/16,000 = 0.25
Labor:
4,000/8,000 = 0.50
Since 0.29 > 0.25 and 0.57 > 0.50, Combination C1 dominates the current input combination, and productivity would definitely improve.
Cost comparison:
Current combination ($5  16,000) + ($10  8,000)
Combination C1
($5  14,000) + ($10  7,000)
Value of productivity
$160,000
140,000
$ 20,000
This improvement is all attributable to technical efficiency. The same output
is produced with proportionately less inputs. (Note that the inputs are in the
same ratio 2:1, and that Combination C1 reduces each input in the same proportion).
349
15–2
2.
Concluded
Output-input ratios (Combination C2):
Materials: 4,000/15,000 = 0.27
Labor:
4,000/6,000 = 0.67
Compared to the current use, productivity is better for both materials and labor (0.27 > 0.25 and 0.67 > 0.50).
Compared to Combination C1, however, C2 has lower productivity for materials (0.27 < 0.29) and higher productivity for labor (0.67 > 0.57). Trade-offs
must be considered.
3.
Cost of Combination C2 ($5  15,000) + ($10  6,000)
Cost of Combination C1 (See Req. 1 above)
Difference
$135,000
140,000
$ (5,000)
Combination C2 is a less costly input combination than C1. (It saves $5,000 of
input cost per quarter.) Thus, less resources are used by C2 than C1, and
moving from C1 to C2 would be a productivity improvement. (The same output is produced at a lower cost.) This is an example of improving allocative
efficiency.
15–3
Productivity profiles:
2009: Power:
Materials:
96,000/12,000 = 8
96,000/24,000 = 4
2010: Power:
Materials:
120,000/6,000 = 20
120,000/27,000 = 4.44
The profile reveals that productivity did improve. Each output-input ratio in 2010
is greater than its 2009 counterpart.
15–4
1.
Profit-linked measurement:
Power
Materials
PQ*
15,000
30,000
PQ  P
$ 30,000
300,000
$330,000
AQ
6,000
27,000
AQ  P
$ 12,000
270,000
$282,000
*120,000/8 = 15,000; 120,000/4 = 30,000
Profits increased by $48,000 due to productivity changes.
350
(PQ  P) – (AQ  P)
$18,000
30,000
$48,000
15–4
2.
Concluded
Price recovery = Total profit change – Productivity-induced change
Total profit change:
2010:
[($4  120,000) – ($2  6,000) – ($10  27,000)] = $198,000
2009:
[($3  96,000) – ($1  12,000) – ($8  24,000)] =
84,000
$114,000
Price recovery = $114,000 – $48,000
= $66,000
Price recovery is the profit change that would have been realized without any
changes in productivity. Thus, without the productivity increase, the company
would have shown increased profits of $66,000.
15–5
1.
Productivity profiles:
Materials .....
Labor...........
Capital.........
Year Before
100,000/25,000 = 4
100,000/5,000 = 20
100,000/10,000 = 10
Year After
120,000/20,000 = 6.0
120,000/2,000 = 60.0
120,000/300,000 = 0.4
Based on the profiles, we see that materials and labor productivity increased
as expected and that capital productivity decreased. The increase in materials and labor productivity was caused by reducing materials and labor usage
by using more capital input. Since a trade-off exists, it is difficult to say
whether the outcome is good or bad using profile analysis. Valuation of the
trade-off is needed.
2.
Profit-linked measurement (P = $5, $10, and 10%, respectively):
Materials ...
Labor.........
Capital.......
PQ*
30,000
6,000
12,000
PQ  P
$ 150,000
60,000
1,200
$ 211,200
AQ
20,000
2,000
300,000
AQ  P
$ 100,000
20,000
30,000
$ 150,000
(PQ  P) – (AQ  P)
$ 50,000
40,000
(28,800)
$ 61,200
*120,000/4; 120,000/20; 120,000/10.
Profits increased by $61,200 due to productivity changes. Assuming that this
outcome will persist, the trade-off is favorable, and automating was a good
decision.
351
15–6
1.
Productivity profile analysis:
Materials .....
Labor...........
Energy ........
Status Quoa
0.50
1.25
2.50
Project Ib
0.60
1.50
3.00
Project IIc
0.55
2.00
3.00
a100,000/200,000;
100,000/80,000; 100,000/40,000.
120,000/80,000; 120,000/40,000.
c120,000/220,000; 120,000/60,000; 120,000/40,000.
b120,000/200,000;
Both projects improve technical efficiency because more output is produced
per unit of input for all inputs. A recommendation cannot be made without
valuing trade-offs.
2.
Profit-linked productivity measurement (P = $8, $10, and $2):
PQ*
Materials .. 240,000
Labor........ 96,000
Energy ..... 48,000
Project I
PQ  P
AQ
$ 1,920,000 200,000
960,000
80,000
96,000
40,000
$ 2,976,000
AQ  P (PQ  P) – (AQ  P)
$1,600,000
$ 320,000
800,000
160,000
80,000
16,000
$2,480,000
$ 496,000
PQ*
Materials .. 240,000
Labor........ 96,000
Energy ..... 48,000
Project II
PQ  P
AQ
$ 1,920,000 220,000
960,000
60,000
96,000
40,000
$ 2,976,000
AQ  P (PQ  P) – (AQ  P)
$1,760,000
$ 160,000
600,000
360,000
80,000
16,000
$2,440,000
$ 536,000
*120,000/0.50; 120,000/1.25; 120,000/2.5.
The analysis favors Project II. Price efficiency is concerned with valuing
trade-offs of technically efficient combinations. The objective is to choose the
least costly and most technically efficient combination. In this example, tradeoffs were valued that included both improvements in technical and allocative
efficiency. The objective is to choose the project that offers the most improvement—whether it be from technical or allocative efficiency or both.
Once these trade-offs were valued, it became clear that Project II was better
than Project I.
352
15–7
1.
Productivity profiles:
Materials
Labor
Base Yeara
0.75
3.00
a600,000/800,000;
b720,000/720,000;
2.
Current Yearb
1.00
2.00
600,000/200,000
720,000/360,000
Income statements:
Sales
Materials
Labor
Gross profit
Base Year
$ 9,000,000
(4,000,000)
(1,600,000)
$ 3,400,000
Current Year
$10,800,000
(4,320,000)
(2,880,000)
$ 3,600,000
Change in income = $3,600,000 – $3,400,000
= $200,000
3.
Profit-linked measurement (P = $6 and $8, respectively):
Materials
Labor
PQ*
960,000
240,000
PQ  P
$5,760,000
1,920,000
$7,680,000
AQ
AQ  P (PQ  P) – (AQ  P)
720,000 $4,320,000
$1,440,000
360,000
2,880,000
(960,000)
$7,200,000
$ 480,000
*720,000/0.75; 720,000/3.00
Change attributable to productivity = $480,000
4.
Price-recovery component = $200,000 – $480,000
= ($280,000)
In the absence of productivity changes, input costs would have increased by
$2,080,000 ($7,680,000 – $5,600,000). This increase would not have been offset by the $1,800,000 increase in revenues, producing a $280,000 drop in
profits. This is the price-recovery component, the amount by which profits
will change without considering any productivity changes. The productivity
improvement adds an additional $480,000 increase in profits so that total
profits increased by $200,000 ($480,000 – $280,000). Thus, the productivity
improvement helped offset the profit drop due to input price increases.
353
15–8
1.
Productivity profiles:
Paying-bills activity:
Base Year
Clerks ................
20,000
PCs ....................
20,000
Supplies.............
2
Most Recent Year
64,000
64,000
8
Moving-materials activity:
Base Year
Labor..................
2
Forklifts .............
4,000
Supplies.............
5
Most Recent Year
1.67
2,500
2.50
The productivity profile for the paying-bills activity indicates an overall improvement in activity efficiency (all partial ratios improved). Since this is a
value-added activity, this means that activity performance has improved. The
moving-materials activity indicates a drop in activity productivity. Less output
and less inputs were used to perform the activity. This is a non-value-added
positive effect. The productivity ratios fail to reveal this, though—unless it
can be shown that reduction of non-value-added activities always leads to a
decline in productivity ratios (which seems unlikely). Thus, changes in profiles for non-value-added activities are difficult to interpret. The best way to
deal with non-value-added activities is to include them as part of the productivity analysis of the process to which they belong.
2.
Profit-linked measure (P = $25,000, $5,000, and $1, respectively):
PQ*
Clerks ......
16
PCs ..........
16
Supplies... 160,000
PQ  P
$ 400,000
80,000
160,000
$ 640,000
AQ
5
5
40,000
AQ  P (PQ  P) – (AQ  P)
$ 125,000
$ 275,000
25,000
55,000
40,000
120,000
$ 190,000
$ 450,000
*320,000/20,000; 320,000/20,000; 320,000/2.
Significant savings have come from process innovation.
354
PROBLEMS
15–9
1.
Profit-linked measure, purchasing (expressed in thousands, where P = $2,
$40,000, and 10%, respectively):
PQ*
Supplies....
60
Clerks ....... 0.03
Capital....... 1,200
PQ  P
$ 120
1,200
120
$ 1,440
AQ
60
0.025
1,200
AQ  P
$ 120
1,000
120
$ 1,240
(PQ  P) – (AQ  P)
$ 0
200
0
$200
*120/2; 120/4,000; 120/0.10 (where the partial ratios for purchasing for 2008
are 2 = 100,000/50,000; 4,000 = 100,000/25; and 0.10 = 100,000/1,000,000).
Profit-linked measure, receiving (in thousands, where P = $2, $40,000, and
10%, respectively):
Supplies....
Clerks .......
Capital.......
PQ*
48
0.06
960
PQ  P
$ 96
2,400
96
$ 2,592
AQ
30
0.05
3,000
AQ  P
$ 60
2,000
300
$2,360
(PQ  P) – (AQ  P)
$ 36
400
(204)
$ 232
*180/3.75; 180/3,000; 180/0.1875 (where the 2008 ratios are 3.75 = 150,000/
40,000; 3,000 = 150,000/50; and 0.1875 = 150,000/800,000).
Profit-linked measure, paying bills (in thousands, where P = $2, $40,000, and
10%, respectively):
Supplies....
Clerks .......
Capital.......
PQ*
90
0.12
600
PQ  P
$ 180
4,800
60
$ 5,040
AQ
5
0.01
1,000
AQ  P
$ 10
400
100
$510
(PQ  P) – (AQ  P)
$ 170
4,400
(40)
$ 4,530
*180/2; 180/1,500; 180/0.30 (where the 2008 partial ratios are 2 = 150,000/
75,000; 1,500 = 150,000/100; and 0.30 = 150,000/500,000).
355
15–9
2.
Concluded
Activity output efficiency (in thousands, where P = $12.00, $14.40, and $28.00,
respectively):
Purchasing ....
Receiving.......
Paying bills ....
PQ*
120
180
180
PQ  P
$1,440
2,592
5,040
$9,072
AQ
120
180
180
AQ  P
$1,440
2,592
5,040
$9,072
(PQ  P) – (AQ  P)
$0
0
0
$0
*3,600,000/30; 3,600,000/20; 3,600,000/20 (where the partial ratios are 30 =
3,000,000/100,000 and 20 = 3,000,000/150,000).
3.
Sum of the two components:
Resource efficiency component:
Purchasing..........................................
Receiving ............................................
Paying bills .........................................
Activity output efficiency ...................
Total process productivity change ........
$ 200,000
232,000
4,530,000
$4,962,000
0
$4,962,000
The sum of the two components gives the change in profits attributable to
changes in process productivity. This shows that even though the procurement output’s demands for the output of the three activities remains unchanged, significant improvements in activity resource efficiency have increased the overall efficiency of the procurement process. Notice that the
biggest increase came from the paying bills subprocess. This can be explained in large part by the elimination of the non-value-added activity of resolving discrepancies.
356
15–10
1.
Productivity profile:
Current system:
Materials: 50,000/200,000 = 0.25
Labor:
50,000/100,000 = 0.50
Computerized system:
Materials: 50,000/175,000 = 0.29
Labor:
50,000/75,000 = 0.67
Materials and labor productivity increase with the acquisition (as claimed by
the production manager).
2.
To compare the alternatives, all inputs must be considered:
Productivity profiles:
Materials
Labor
Capital
Energy
Current
0.25
0.50
0.50
1.00
Computerized
0.29
0.67
0.10
0.40
The productivity profiles indicate a mixed outcome—some ratios improve
and some do not. Trade-offs, therefore, must be valued.
3.
Profit-linked measurement (where P = $4, $9, 10%, and $2.50 respectively):
Materials
Labor
Capital
Energy
PQ*
200,000
100,000
100,000
50,000
PQ  P
$ 800,000
900,000
10,000
125,000
$1,835,000
AQ
175,000
75,000
500,000
125,000
AQ  P
(PQ  P) – (AQ  P)
$ 700,000
$ 100,000
675,000
225,000
50,000
(40,000)
312,500
(187,500)
$1,737,500
$ 97,500
*Since output is the same, PQ equals the inputs for the current system.
The trade-offs are favorable. The computerized system will increase profits by
$97,500.
357
15–11
1.
Productivity profile:
2009 ...........
2010 ...........
Materialsa
0.50
0.60
Laborb
2.0
2.4
a18,000/36,000;
b18,000/9,000;
24,000/40,000.
24,000/10,000.
The profiles indicate an overall productivity increase and thus support the effectiveness of the new process.
2.
Profit-linked measurement (where P = $4.50 and $10, respectively):
PQ*
Materials .... 48,000
Labor.......... 12,000
PQ  P
$ 216,000
120,000
$ 336,000
AQ  P
$180,000
100,000
$280,000
AQ
40,000
10,000
(PQ  P) – (AQ  P)
$36,000
20,000
$56,000
*24,000/0.5; 24,000/2.
Increase in profits due to productivity = $56,000
3.
Price-recovery component:
Total profit change:
2010:
Revenues ($16  24,000) .........
Materials ($4.50  40,000) .......
Labor ($10  10,000) ................
Profit ...................................
$ 384,000
(180,000)
(100,000)
$ 104,000
Revenues ($16  18,000) .........
Materials ($4  36,000) ............
Labor ($9  9,000) ....................
Profit ...................................
$ 288,000
(144,000)
(81,000)
$ 63,000
2009:
Total profit change = $104,000 – $63,000
= $41,000
Price-recovery component = $41,000 – $56,000
= ($15,000)
Without the productivity improvement, profits would have dropped $15,000.
The increase in sales would not have recovered the increase in the cost of
inputs.
358
15–12
1.
Productivity profile:
2009 ...........................
Change I ....................
Change II ...................
Optimal ......................
Materials
1.67
1.43
2.00
2.50
Labor
0.83
1.25
1.00
1.25
Change I shows an improvement in labor productivity and a decline in material productivity. For Change II, both material and labor productivity ratios improved. Which change should be implemented depends on the value of the
productivity trade-offs of Change I versus the value of the productivity improvements of Change II. Profile analysis does not reveal these values.
2.
Cost
2009:
(33,000  $60) + (66,000  $15) = $2,970,000
Change I: (38,500  $60) + (44,000  $15) = $2,970,000
Change II: (27,500  $60) + (55,000  $15) = $2,475,000
Optimal:
(22,000  $60) + (44,000  $15) = $1,980,000
Cost of productive inefficiency:
2009:
$2,970,000 – $1,980,000
Change I: $2,970,000 – $1,980,000
Change II: $2,475,000 – $1,980,000
=
=
=
$990,000
$990,000
$495,000
Potential improvement for 2007:
Change I: $2,970,000 – $2,970,0000
Change II: $2,970,000 – $2,475,000
=
=
$0
$495,000
Change I improves the technical use of labor but does so by trading off materials for labor inputs. The optimal combination defines the relative mix ratio
as 1:2. The mix ratio for Change I is 7:8. Notice that the 2009 mix ratio is 1:2.
Thus, moving to Change I decreases allocative efficiency while improving
technical efficiency (by using less labor—in fact, the amount of labor used
corresponds to the optimal level). Change II, on the other hand, has a mix ratio of 1:2, which is the same as the 2009 and optimal input combination.
Therefore, there is no reduction in trade-off efficiency, and reducing the
amount of each input required to produce the same output is all attributable
to improving technical efficiency.
3.
The profit-linked measurement approach will provide the same outcome
without requiring knowledge of the optimal input combination. Since the output is the same for both years, the inputs that would have been used in 2010
without any productivity change are the same as 2009 usage. Thus, the profitlinked measure is simply the difference between the cost of the inputs for the
two years:
Change I: $2,970,000 – $2,970,000 = $0
Change II: $2,970,000 – $2,475,000 = $495,000
359
15–13
1.
Productivity profiles:
Handling goods ............
Entering data.................
Detecting errors ............
2008
0.10
3.00
1.00
2009
0.20
4.00
2.00
2010
0.50
5.00
4.00
With respect to the demand for activity output, process productivity has improved for 2009 and 2010 for every activity. More process output is being
produced per unit of activity input. The changes definitely improved process
productivity. Comparing multiyear profiles allows managers to see whether
improvement is ongoing and to evaluate whether prior-year gains are being
maintained. Trends in productivity data are useful for measuring continuous
improvement efforts.
2.
Profit-linked measurement (where P = $1, $7, and $2, respectively):
2009:
PQ*
Handling .. 1,650,000
Entry ........
55,000
Errors ....... 165,000
PQ  P
$ 1,650,000
385,000
330,000
$ 2,365,000
AQ
AQ  P
(PQ  P) – (AQ  P)
825,000 $ 825,000
$ 825,000
41,250
288,750
96,250
82,500
165,000
165,000
$ 1,278,750
$1,086,250
*165,000/0.1; 165,000/3; 165,000/1.
2010 (where P = $1.25, $8.00, and $2.00, respectively):
PQ*
Handling .. 1,000,000
Entry ........
50,000
Errors ....... 100,000
PQ  P
$ 1,250,000
400,000
200,000
$ 1,850,000
AQ
400,000
40,000
50,000
AQ  P (PQ  P) – (AQ  P)
$500,000
$ 750,000
320,000
80,000
100,000
100,000
$920,000
$ 930,000
*200,000/0.2; 200,000/4; 200,000/2.
Using a moving base year provides more insight regarding continuous improvement. We should see improvement from one year to the next—not just
with respect to some year in the past. In fact, it is possible to have improvement signaled with respect to a base year in the past for the current year, yet
this improvement took place in the year after the base year. This would then
mean that we have maintained the gains but have not realized any further
gains. Thus, better evaluation of continuous improvement occurs by using a
dynamic base year.
360
15–14
1.
Productivity profile:
Materials ............
Labor..................
Capital................
Energy ...............
a50,000/50,000;
b60,000/40,000;
2009a
1.000
0.250
0.025
1.000
2010b
1.500
0.600
0.012
0.400
50,000/200,000; 50,000/2,000,000; 50,000/50,000.
60,000/100,000; 60,000/5,000,000; 60,000/150,000.
Since the ratio changes are mixed, no statement on overall productivity improvement can be made. Valuation of the trade-offs is needed.
2.
Profit change:
2010: Revenues ($65  60,000) .................
Materials ($10  40,000) ..................
Labor ($12  100,000) ......................
Capital (0.10  $5,000,000) ..............
Energy ($2  150,000)......................
Profit ...........................................
$ 3,900,000
(400,000)
(1,200,000)
(500,000)
(300,000)
$ 1,500,000
2009: Revenues ($70  50,000) .................
Materials ($8  50,000) ....................
Labor ($10  200,000) ......................
Capital (0.15  $2,000,000) ..............
Energy ($2  50,000)........................
Profit ...........................................
$ 3,500,000
(400,000)
(2,000,000)
(300,000)
(100,000)
$ 700,000
Total change = $1,500,000 – $700,000
= $800,000
361
15–14 Concluded
Profit-linked measurement (expressed in thousands, where P = $10, $12, 10%,
and $2, respectively):
Materials ....
Labor..........
Capital........
Energy .......
PQ*
60
240
2,400
60
PQ  P
$ 600
2,880
240
120
$3,840
AQ
40
100
5,000
150
AQ  P
$ 400
1,200
500
300
$ 2,400
(PQ  P) – (AQ  P)
$ 200
1,680
(260)
(180)
$1,440
*60,000/1.000; 60,000/0.250; 60,000/0.025; 60,000/1.000.
Productivity-induced profit change: $1,440,000 increase
Price recovery = $800,000 – $1,440,000 = ($640,000)
This means that without the productivity improvement, a loss of $640,000
would have been realized.
3.
2009 per-unit input cost = $2,800,000/50,000 = $56
2010 per-unit input cost = $2,400,000/60,000 = $40
Yes, the per-unit cost has been reduced by $16, much more than the needed
$5. The division’s continued existence was brought about by improved
productivity. Productivity improvements can maintain competitive ability by
reducing the cost of producing. They also may permit a company to achieve a
sustainable competitive advantage and ensure its long-term survival. Thus,
productivity is an important competitive tool. In this case, because the cost
reduction per unit is $16, the company could take a more aggressive stance
and decrease prices even more than $5—with the possibility of increasing
market share and profitability.
362
15–15
1. Units reworked are a measure of rework activity output.
2.
Profile:
Materials .........
Labor...............
2009
7,500/15,000 = 0.500
7,500/12,000 = 0.625
2010
3,600/7,200 = 0.500
3,600/6,000 = 0.600
Productivity has remained the same for materials but has declined for labor.
Profit-linked analysis:
PQ*
PQ  P
Materials .... 7,200
$ 144,000
Labor.......... 5,760
86,400
$ 230,400
AQ
7,200
6,000
AQ  P
$ 144,000
90,000
$ 234,000
(PQ  P) – (AQ  P)
$
0
(3,600)
$(3,600)
*3,600/0.50; 3,600/0.625.
The decline in activity productivity reduces profits by $3,600.
Reducing the demand for a non-value-added activity is the correct action; this
is true even if the productivity of the non-value-added activity drops. After all,
why worry about improving the efficiency of an activity that should not be
done at all?
363
15–16
1.
Process output measure: Units assembled
Profile:
Rework .......
2009
300,000/7,500 = 40
2010
300,000/3,600 = 83.33
Activity output efficiency has improved. Process output is placing fewer demands on the activity output.
Rework .......
PQ
7,500
PQ  P
$480,000
AQ
3,600
AQ  P
$230,400
(PQ  P) – (AQ  P)
$249,600
By reducing the demand for rework activity, profits have increased by
$249,600. This provides a positive measure of reducing the demand for a nonvalue-added activity—but it only shows up in the analysis of productivity at
the process level.
2.
The total process effect is the sum of the profit-linked measures (resource
efficiency plus activity output efficiency, calculated in Exercise 15–15 and
Requirement 1):
Resource efficiency component ........
Activity output efficiency ....................
Process efficiency..........................
$ (3,600)
249,600
$ 246,000
Actions taken to reduce the demand for the non-value-added activity of rework decreased activity productivity but increased activity output efficiency.
The trade-off is favorable, and the overall effect is to increase profits by
$82,000. This simply emphasizes the fact that changes in non-value-added activities must be evaluated in a process productivity context.
364
15–17
1.
Shop-floor workers relate well to operational measures because they are easily understood. Also, these measures are easily tracked. Charts can be visibly
posted that track the performance of the measures over time. Finally, the
measures are usually available on a more timely basis—they can be reported
daily.
2.
Profit-linked measurement, materials and labor (where P = $5 and $10, respectively):
Materials ....
Labor..........
aBatch
PQa
4,000
2,000
PQ  P
$ 20,000
20,000
$ 40,000
AQb
3,333
2,500
AQ  P
$16,665
25,000
$41,665
(PQ  P) – (AQ  P)
$ 3,335
(5,000)
$(1,665)
size is constant, so PQ consists of the inputs used in the first batch.
(rounded); 10,000/4.
b10,000/3
Profits will drop by $1,665.
3.
Profit-linked measurement, three inputs respectively (where P = $5, $10, and
$1,000):
Materials ....
Labor..........
Quality .......
PQ
4,000
2,000
20
PQ  P
$ 20,000
20,000
20,000
$ 60,000
AQ
3,333
2,500
5
AQ  P
$16,665
25,000
5,000
$46,665
(PQ  P) – (AQ  P)
$ 3,335
(5,000)
15,000
$13,335
Profits now increase by $13,335. This illustrates that operational measures
are limited. A firm also needs a comprehensive, integrated productivity system. Viewing quality as an input as measured by defects is useful.
365
COLLABORATIVE LEARNING EXERCISE
15–18
1.
Productivity profile:
Current Year
Materials ........
0.40
Equipment .....
1.67
Setting A
1.00
1.00
Setting B
0.50
2.00
Setting B signals a productivity improvement for both inputs.
2.
Income statements for the coming year:
Sales revenues ($40 75,000) .......
Cost of inputs:
Materials:
($12  75,000) ......................
($12  150,000) ....................
Machine hours:
($12  75,000) ......................
($12  37,500) ......................
Gross profit ....................................
Setting A
$3,000,000
Setting B
$ 3,000,000
(900,000)
(1,800,000)
(900,000)
(450,000)
$ 750,000
$1,200,000
Setting A provides the greatest increase.
3.
Profit-linked measurement (where P = $12, and $12, respectively):
Setting A
PQa
PQ  P
AQ
Materials ..... 187,500 $ 2,250,000 75,000
Equipment .. 45,000b
540,000 75,000
$ 2,790,000
AQ  P
(PQ  P) – (AQ  P)
$ 900,000
$1,350,000
900,000
(360,000)
$ 1,800,000
$ 990,000
Setting B
PQ  P
AQ
$ 2,250,000 150,000
540,000 37,500
$ 2,790,000
AQ  P
(PQ  P) – (AQ  P)
$ 1,800,000
$ 450,000
450,000
90,000
$ 2,250,000
$ 540,000
PQa
Materials ..... 187,500
Equipment .. 45,000b
a75,000/0.40;
75,000/1.6666.
bRounded.
Setting A is the better of the two by $450,000. Setting A uses more machine
hours but less materials; both inputs are priced the same but Combination A
uses less total inputs.
CYBER RESEARCH CASE
15–19 Answers will vary.
366