Solutions for Chapter 18: Questions and Problems
– 138 –
CHAPTER 18
EVALUATION OF PORTFOLIO PERFORMANCE
Answers to Questions
1. The two major factors would be: (1) attempt to derive risk-adjusted returns that exceed a
naive buy-and-hold policy and (2) completely diversify – i.e., eliminate all unsystematic
2. Treynor (1965) divided a fund’s excess return (return less risk-free rate) by its beta. For a
fund not completely diversified, Treynor’s “T” value will understate risk and overstate
performance. Sharpe (1966) divided a fund’s excess return by its standard deviation.
3. For portfolios with R2 values noticeably less than 1.0, it would make sense to compute
both measures. Differences in the rankings generated by the two measures would suggest
less-than-complete diversification by some funds – specifically, those that were ranked
higher by Treynor than by Sharpe.
4. Jensen’s alpha () is found from the equation Rjt – RFRt = j + j[Rmt – RFRt] +ejt. The aj
indicates whether a manager has superior (j > 0) or inferior (j < 0) ability in market
Solutions for Chapter 18: Questions and Problems
– 139 –
5. The Information Ratio (IR) is calculated by dividing the average return on the portfolio
less a benchmark return by the standard deviation of the excess return. The IR can be
6. The difference by which a manager’s overall actual return beats his/her overall
benchmark return is termed the total value-added return and decomposes into an
allocation effect and a selection effect. The former effect measures differences in weights
7. When measuring the performance of an equity portfolio manager, overall returns can be
related to a common total risk or systematic risk. Factors influencing the returns achieved
Solutions for Chapter 18: Questions and Problems
– 140 –
CHAPTER 18
Answers to Problems
1(a).
60.1
.05
.08
0.05
.07.15
S
P
==
−
=
1(b).
0600.
1.00
.06
1.00
.07.13
Market
.0800
1.00
.08
1.00
.07.15
T
P
==
−
=
==
−
=
Sharpe
Treynor
P
Q
R
S
Market
1(c). It is apparent from the rankings above that Portfolio Q was poorly diversified since
Treynor ranked it #2 and Sharpe ranked it #4. Otherwise, the rankings are similar.
Solutions for Chapter 18: Questions and Problems
– 141 –
2(a). Portfolio MNO enjoyed the highest degree of diversification since it had the highest R2
2(b). Note the mean returns are net of the risk-free rate. Doing the calculations we obtain:
Fund Treynor Sharpe Jensen
ABC 0.975(4) 0.857(4) 0.192(4)
DEF 0.715(5) 0.619(5) -0.053(5)
GHI 1.574(1) 1.179(1) 0.463(1)
Only GHI and MNO have significantly positive alphas at a 95% level of confidence.
3(a). (Information ratio) IRj = j/u where u = standard error of the regression
IRA = .058/.533 = 0.1088
IRB = .115/5.884 = 0.0195
3(c). The higher the ratio, the better. Based upon the answers to part a, Manager C would be
rated the highest followed by Managers A and B, respectively. However, once the values
are annualized, the ranking change. Specifically, based upon the annualized IR, Manger
Solutions for Chapter 18: Questions and Problems
– 142 –
4(b). E(Ri) = 6.20 + (15.71 – 6.20)
4(c)(i). Selectivity1 = 20.2% – 12.85% = 7.35%
Selectivity2 = 7.02% – 8.61% = -1.59%
4(c)(ii).Ratio of total risk1 = 1/m = 20.67/13.25 = 1.56
4(c)(iii). Net Selectivity = Selectivity – Diversification
Net Selectivity1 = 7.35% – 1.99% = 5.36%
Net Selectivity2 = -1.59% – 1.57% = -3.16%
4(d). Even accounting for the added cost of incomplete diversification, Fund 1’s performance
was above the market line (best performance), while Fund 2 fall below the line.
5.
a.
Year
Mgr X
Return
Mgr Y Return
1
-1.5
-6.5
2
-1.5
-3.5
3
-1.5
-1.5
4
-1.0
5
6
7
8
9
Average
Solutions for Chapter 18: Questions and Problems
Semi-dev
Semi-deviation considers only the returns that are below the average.
b.
Sharpe ratio: (average return minus risk-free rate) / standard deviation
Mgr X:
0.435
Mgr Y:
0.452
Best performer
6.
6(a)(i). .6(-5) + .3(-3.5) + .1(0.3) = -4.02%
6(b)(i). [.5(-4 + 5) + .2(-2.5 + 3.5) + .3(.3 -.3)] = 0.70%
6(b)(ii). [(.3 – .6) (-5 + 4.02) + (.4 – .3) (-3.5 + 4.02) + (.3 -.1)(.3 + 4.02)] = 1.21%
7 (a). Dollar-Weighted Return
Manager L:
500,000 = -12,000/(1+r) – 7,500/(1+r)2– 13,500/(1+r)3 – 6,500/(1+r)4– 10,000/(1+r)5+
625,000/(1+r)5
Solutions for Chapter 18: Questions and Problems
– 144 –
7(b). Time-weighted return
Manager L:
Periods HPR
1 [(527,000 – 500,000) – 12,000]/500,000 = .03
TWRR = [(1 + .03)(1 – .0085)(1 + .0217)(1 + .0333)(1 + .0603)]1/5 – 1
= (1.143) 1/5 – 1= 1.02712 – 1 = .02712 = 2.71%
Manager M:
Periods HPR
1 [(692,000 – 700,000) + 35,000]/700,000 = .03857