{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework Solutions 02

# Homework Solutions 02 - IEOR E4007 G Iyengar Oct 8th 2008...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: IEOR E4007 G. Iyengar Oct. 8th, 2008 Solution to Homework #2 1. Simple problems on simplex Consider a linear program with n = 4 variables and m = 2 constraints. Suppose the current basis B = { 1 , 2 } and the constraints are given by x 1- 3 x 3 +3 x 4 = 6 x 2- 8 x 3 +4 x 4 = 4 For each of the following set of objective vectors decide whether the current basis is optimal, if not then move to a better basis or declare the problem is unbounded. (a) max c = 1 1- 3- 1 There are two non-basic variables x 3 and x 4 . We will consider them one by one: • x 3 : First check the reduced cost ¯ c 3 = c 3- c B A- 1 B A 3 =- 3 + [1 , 1] 3 8 = 8 > So, x 3 is an improving direction. We can keep increasing x 3 as long as ¯ x B- x 3 A- 1 B A 3 ≥ In this case- A- 1 B A 3 ≥ 0, therefore, we can increase x 3 indefinitely. Conse- quently, the problem is unbounded. (b) max c = 0 0- 3 1 • x 3 : First check the reduced cost ¯ c 3 = c 3- c B A- 1 B A 3 =- 3 + [0 , 0] 3 8 =- 3 < Not an improving direction. • x 4 : First check the reduced cost ¯ c 4 = c 4- c B A- 1 B A 4 = 1- [0 , 0] 3 4 = 1 > So, x 4 is an improving direction. We can keep increasing x 4 as long as ¯ x B- x 4 A- 1 B A 4 = 6 4- x 4 3 4 ≥ Thus, x 4 = min { 2 , 1 } = 1, and the new basis is B = { 1 , 4 } . 1 (c) min c = 1 1- 3 8 • x 3 : First check the reduced cost ¯ c 3 = c 3- c B A- 1 B A 3 =- 3 + [1 , 1] 3 8 = 8 > Not an improving direction. • x 4 : First check the reduced cost ¯ c 4 = c 4- c B A- 1 B A 4 = 8- [1 , 1] 3 4 = 7 > So, x 4 is also an improving direction. Thus, current basis is optimal. 2. Tinyco Cash Flow Management 25pts The cash position of the company is described as follows. • Initial holding \$200,000 in short-term bonds • Demand for cash in months i = 1 , 2 ,..., 5:- 100, 300,- 500,- 10, 30 (in thou- sands) • Cash earns 0.5% per month and short-term bond earns 0.9% per month • Tinyco can borrow up to \$100,000 at 1% interest per month • Transaction cost of 0.2% is charged for cash ↔ bonds The goal of the company is to maximize cash position at the end of five months while meeting its liabilities. Formulate the LP that solves this problem. We use the following variables • b i : bond position at the end of month i • c i : cash position at the end of month i • l i : amount borrowed in month i with b = 200, c = 0 and l = 0, and make the following assumption • the bond coupon (or interest) is paid in cash, i.e. we do not pay any transaction cost for that. Then the constraint at the end of month i is given by (1 . 005 c i- 1 + 1 . 009 b i- 1- 1 . 01 l i- 1 ) | {z } old wealth- ( c i + b i- l i ) | {z } new wealth- . 002 | b i- b i- 1 | | {z } cost ≥ d i , 2 where d i denotes the demand in month i . If we assume that the bond coupon is reinvested then the cost term would change to | b i- 1 . 009 b i- 1 | . Thus, the optimization problem is given by max c n s.t. (1 . 005 c i- 1 + 1 . 009 b i- 1- 1 . 01 l i- 1 )- ( c i + b i- l i )- . 002 | b i- b i- 1 | ≥...
View Full Document

{[ snackBarMessage ]}

### Page1 / 12

Homework Solutions 02 - IEOR E4007 G Iyengar Oct 8th 2008...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online