APS502 Financial Engineering Individual Assignment

PROBLEM-I
BOND PORTFOLIO OPTIMIZATION:

Problem Statement:
To Formulate a linear programming model to find the lowest-cost dedicated bond portfolio that
covers the stream of liabilities given.

Let,
Xn = Number of bonds
Pn = Price of nth bond
Zk = Excess amount at the end of kth date
Si = Spot rate
fi,j = Forward rate

Portfolio Details:
Spot rates
s1 = 1%; s2 = 1.5%; s3 = 2.0%; s4 = 2.5%; s5 =3.0%; s6 = 3.5%

Date 1 2 3 4 5 6
Obligations
Required 500 200 800 400 700 900

Bond 1 2 3 4 5 6 Price Rating
1 10 10 10 10 10 110 108 B
2 7 7 7 7 7 107 94 B
3 8 8 8 8 8 108 99 B
4 6 6 6 6 106 92.7 B
5 7 7 7 7 107 96.6 B
6 6 6 6 106 95.9 B
7 5 5 5 105 92.9 A
8 10 10 110 110 A
9 8 8 108 104 A
10 6 6 106 101 A
11 10 110 107 A
12 7 107 102 A
13 100 95.2 A

Table 1: Bonds & Obligation Details

,Cost function:

F(X) = P1X1 + P2X2 + P3X3 + P4X4 + P5X5 + P6X6 + P7X7 + P8X8 + P9X9 + P10X10 + P11X11 +
P12X12 + P13X13

Substituting the price of bonds, we get,

F(X) = 108 X1 + 94 X2 + 99 X3 + 92.7 X4 + 96.6 X5 + 95.9 X6 + 92.9 X7 + 110 X8 + 104 X9 +
101 X10 + 107 X11 + 102 X12 + 95.2 X13

Constraints:

Objective: To Minimize the Cost Function based on the following constraints.

1. 1st year liability is 500, equating 1st year cash inflow to match the liability and Z1 extra
amount is carried over to the next year.

10 X1 + 7 X2 + 8 X3 + 6 X4 + 7 X5 + 6 X6 + 5 X7 + 10 X8 + 8 X9 + 6 X10 + 10 X11 + 7 X12
+ 100 X13 – Z1 ≥ 500

2. 2nd year liability is 200, equating 2nd year cash inflow and Z1 amount along with forward
rate to match the liability and Z2 extra amount is carried over to the next year.

10 X1 + 7 X2 + 8 X3 + 6 X4 + 7 X5 + 6 X6 + 5 X7 + 10 X8 + 8 X9 + 6 X10 + 110 X11 + 107
X12 + Z1(1+f1,2) - Z2 ≥ 200

3. 3rd year liability is 800, equating 3rd year cash inflow along with Z2 amount with forward
rate to match liability and Z3 extra amount is carried over to next year.

10 X1 + 7 X2 + 8 X3 + 6 X4 + 7 X5 + 6 X6 + 5 X7 + 110 X8 + 108 X9 + 106 X10 + Z2
(1+f2,3) – Z3 ≥ 800

4. 4th year liability is 400, equating 4th year cash inflow along with Z3 amount with forward
rate to match liability and Z4 extra amount is carried over to next date.

10 X1 + 7 X2 + 8 X3 + 6 X4 + 7 X5 + 106 X6 + 105 X7 + Z3(1+ f3,4) – Z4 ≥ 400

5. 5th year liability is 700, equating 5th year cash inflow along with Z4 amount with forward
rate to match liability and Z5 extra amount is carried over to next date.

10 X1 + 7 X2 + 8 X3 + 106 X4 + 107 X5 + Z4 (1+ f4,5) – Z5 ≥ 700

6. 6th year liability is 900, equating 6th year cash inflow along with Z5 amount with forward
rate to match the liability in that year.

110 X1 + 107 X2 + 108 X3 + Z5 (1+ f5,6) ≥ 900

, Forward rate calculation:

(𝑗−𝑖) 𝑗
(1 + 𝑠𝑖 )(𝑖) (1 + 𝑓(𝑖,𝑗) ) = (1 + 𝑠𝑗 )

Spot rates for the dates are,

S1 = 1%, S2 = 1.5%, S3 = 2%, S4 = 2.5%, S5 = 3%, S6 = 3.5%

(1+f 1, 2) = (1+s2)2 / (1+s1)1 = (1+0.015)2 / (1+0.01)1 = 1.02

(1+f 2,3) = (1+s3)3 / (1+s2)2 = (1+0. 02)3 / (1+0.015)2 = 1. 03

(1+f 3,4) = (1+s4)4 / (1+s3)3 = (1+0. 025)4 / (1+0.02)3 = 1.04

(1+f 4,5) = (1+s5)5 / (1+s4)4 = (1+0. 03)5 / (1+0.025)4 = 1.05

(1+f 5,6) = (1+s6)6 / (1+s5)5 = (1+0. 035)6 / (1+0.03)5 = 1.06

Substituting the forward rates on the above constraints. We get,

1. 10 X1 + 7 X2 + 8 X3 + 6 X4 + 7 X5 + 6 X6 + 5 X7 + 10 X8 + 8 X9 + 6 X10 + 10 X11 +
7 X12 + 100 X13 – Z1 ≥ 500

2. 10 X1 + 7 X2 + 8 X3 + 6 X4 + 7 X5 + 6 X6 +5 X7 +10 X8 +8 X9 +6 X10 +110 X11 +107 X12
+ 1.02 Z1 - Z2 ≥ 200

3. 10 X1 + 7 X2 + 8 X3 + 6 X4 + 7 X5 + 6 X6 + 5 X7 + 110 X8 + 108 X9 + 106 X10 +1.03Z2
– Z3 ≥ 800

4. 10 X1 + 7 X2 + 8 X3 + 6 X4 + 7 X5 + 106 X6 + 105 X7 + 1.04 Z3– Z4 ≥ 400

5. 10 X1 + 7 X2 + 8 X3 + 106 X4 + 107 X5 + 1.05 Z4 – Z5 ≥ 700

6. 110 X1 + 107 X2 + 108 X3 + 1.06 Z5 ≥ 900

PART-I:
Objective:

Minimize Cost function:
i.e., Min. 108 X1 + 94 X2 + 99 X3 + 92.7 X4 + 96.6 X5 + 95.9 X6 + 92.9 X7 + 110 X8 +
104 X9 + 101 X10 + 107 X11 + 102 X12 + 95.2 X13

Constraints: (No restrictions based on asset rating)