Homework 5: Evaluating Net Value and
ROI, Due Wednesday 10/7 at 12pm (noon)
CSCI 510, Fall 2015
30 points
In the lecture charts EC-7 MedFRS Econ Analysis I, the comparative EffectivenessCost Difference E-C for Option A (Accept Available OS) and Option B (Build New OS)
for the MedFRS Transaction Processing System (TPS) was maximized at 2400
TR/sec - $650K = 1750 for Option A and 4400 TR/sec -$1207K = 3103 for Option B
(See the below image from slide 25).
These comparisons are across different quantities. Better comparisons would be to
compute the monetary benefit - cost (Net Value) of Options A and B under different
assumptions of the value of each TR/sec, and also to compare the Return on
Investment (ROI) = (Benefit-Cost)/Cost for Options A and B under different
assumptions of the value of each TR/sec.
Perform three such comparisons (Net Value vs ROI), assuming that the value of each
TR/sec is $0.5K, $1K, and $2K. Does the preferred option for Net Value vary with
the values of each TR/sec? Is the preferred option for ROI the same as the preferred
option for Net Value for the values of each TR/sec?
There are 2 interpretations of this problem. To take the given maximum number of
processors (N) from the slides and book, or to calculate Nmax. First is the scenario of
the student using the given values, called Method 1.
Method 1
Option A
(Note: Nmax is 5 – student might not state it explicitly if they take the values given on
the slides)
TR/sec = $0.5K.
Net Value = $1200K - $650K = $550K. (2 for benefit, 1 for correct answer)
ROI = $550K/$650K = 0.846 (1 for correct answer)
TR/sec = $1K.
Net Value = $2400K - $650K = $1750K. (2 for benefit, 1 for correct answer)
ROI = $1750K/$650K = 2.69 (1 for correct answer)
TR/sec = $2K.
Net Value = $4800K - $650K = $4150K. (2 for benefit, 1 for correct answer)
ROI = $4150K/$650K = 6.38 (1 for correct answer)
Option B
(Note: Nmax is 10 – student might not state it explicitly if they take the values given
on the slides)
TR/sec = $0.5K.
Net Value=$2200K-$1207K = $993K. (2 for benefit, 1 for correct answer)
ROI = $993K/$1207K = 0.82 (1 for correct answer)
TR/sec = $1K.
Net Value=$4400K-$1207K = $3193K . (2 for benefit, 1 for correct answer)
ROI = $3193K/$1207K = 2.64 (1 for correct answer)
TR/sec = $2K.
Net Value=$8800K-$1207K = $7593K. (2 for benefit, 1 for correct answer)
ROI = $7593K/$1207K = 6.29 (1 for correct answer)
Option B is consistently better on Net Value (3 points), but Option A is consistently
better on ROI (3 points).
Method 2
(Nmax is 5 for Option A and 10 for Option B for all scenarios)
Give 10 points for correctly calculating Nmax including all of the steps to calculate the
correct Nmax values. If they get the wrong values, take off 2 points then continue the
grading as though it is correct. If the student does not use a whole number for Nmax
(N is the number of processors, and you cannot use or buy a portion of a processor),
then take off 3 points.
Distribute the points for calculating Net Value and ROI, and doing the final
comparisons as below:
Option A
(Note: Nmax is 5 – student might not state it explicitly if they take the values given on
the slides)
TR/sec = $0.5K.
Net Value = $1200K - $650K = $550K. (1 for benefit, 1 for correct answer)
ROI = $550K/$650K = 0.846 (1 for correct answer)
TR/sec = $1K.
Net Value = $2400K - $650K = $1750K. (1 for benefit, 1 for correct answer)
ROI = $1750K/$650K = 2.69 (1 for correct answer)
TR/sec = $2K.
Net Value = $4800K - $650K = $4150K. (1 for benefit, 1 for correct answer)
ROI = $4150K/$650K = 6.38 (1 for correct answer)
Option B
(Note: Nmax is 10 – student might not state it explicitly if they take the values given
on the slides)
TR/sec = $0.5K.
Net Value=$2200K-$1207K = $993K. (1 for benefit, 1 for correct answer)
ROI = $993K/$1207K = 0.82 (1 for correct answer)
TR/sec = $1K.
Net Value=$4400K-$1207K = $3193K . (1 for benefit, 1 for correct answer)
ROI = $3193K/$1207K = 2.64 (1 for correct answer)
TR/sec = $2K.
Net Value=$8800K-$1207K = $7593K. (1 for benefit, 1 for correct answer)
ROI = $7593K/$1207K = 6.29 (1 for correct answer)
Option B is consistently better on Net Value (1 point), but Option A is consistently
better on ROI (1 point).
Below are details of how to calculate Nmax and towards the end includes calculating
the Net Value and ROI (which are slightly different from above and can be used to
cross-reference students’ work).
Below, “TR” means “transaction”.
The analysis for determining Nmax, the optimal number of processors, based on the
value of VT, the $K value of one TR/sec, follows section 13.3 (book page 210) from
EP-5.
The performance equation is given on slide 11 (of EC-7):
E(N) = (N * (S – P – M * (N - 1))) / T, where:
N = number of processors
S = processor speed (MOPS/sec)
P = processor overhead (MOPS/sec)
M = multiprocessor overhead factor (MOPS/sec/processor)
T = transaction processing time (MOPS/TR)
The values of these parameters (other than N), the resulting E(N) expressions, the
corresponding total value expressions, and the cost expressions (from slide 14) for
the two options are as follows:
Option
Source slide in EC-7
S
P
M
T
E(N)
TV(N)
C(N)
A
B
11
1000
200
80
1.0
80N(11-N)
VT*80N(11-N)
450+40N
14
1000
200
40
1.0
40N(21-N)
VT*40N(21-N)
1007+20N
The total value, TV(N), is the value of one transaction per second, VT, multiplied by
the number of transactions per second, E(N).
The next step is to find the value of N, NMax, that maximizes the net value NV(N) =
TV(N) – C(N). The usual way of doing this is to differentiate NV(N), set the
derivative equal to zero, and solve that for N. The table below uses the
mathematically equivalent approach (from section 13.3) of setting dC/dN equal to
dTV/dN and solving that.
Option
C(N)
dC/dN
TV(N)
dTV/dN
Equation to solve
Resulting formula for NMax
Simplified formula for NMax
A
450+40N
40
VT*(880N-80N2)
VT*(880-160N)
40 = VT*(880-160N)
(880VT-40)/160VT
5.5 – 0.25 / VT
B
1007+20N
20
VT*(840N-40N2)
VT*(840-80N)
20 = VT*(840-80N)
(840VT-20)/80VT
10.5 – 0.25 / VT
With the formulas for NMax, the final results can be computed. Fractions of a
processor should not be allowed (you can’t buy a partial processor).