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SOLUTIONS

5.1. Campbell Corporation uses Baumol model to manage cash. The cost of transferring money from a money-market fund, which pays 6% interest on balances, to a checking account is $32 per transaction. Campbell needs $13 million annually to pay its bills. Find the annual cost of interest forgone.

$3533 ♥

Solution:

Annual requirement of cash (A) = $ 13 million

Transaction cost (T) = $ 32 per transaction

Opportunity cost of holding cash (R) = 6%

According to Baumol model to manage cash

Total cost = Transaction cost + Opportunity cost

Let C* be the optimum cash balance that minimizes the cost of holding cash

Transaction cost = A x T

C*

Opportunity cost = C* x R

2

Total cost = A x T + C* x R

C* 2

Differentiating with respect to C* we get

0 = - A x T + 1 x R

C*2 2

- A x T = 1 x R

C*2 2

C* = √(2AT) / R

Therefore

C* = √(2*13,000,000* 32) / 0.06

C* = √13866666666.67

C* = $ 117756.81

The annual cost of interest foregone is the opportunity cost

Opportunity cost = C* x R

2

=($ 117756.81/2) *0.06

= $ 3533

The annual cost of interest foregone is $ 3533.

141

5.2. Genentech Corporation, by analyzing its weekly balances in its checking account, has determined that the variance of cash flows is $3,000,000. Further, the cost of transferring money from the checking account to a money market account is $65 per transfer. The interest on the checking account is 1%, while that on the market account is 6%. Genentech wants to keep $5,000 as a minimum balance in the checking account. Find the annual cost of interest forgone.

$606 ♥

Solution:

Variance of cash flows per week (σ2) = $ 3,000,000

Transaction cost (T) = $ 65 per transfer

Interest on checking account = 1%

Interest on market account = 6%

Opportunity cost = 6% - 1% = 5%

Opportunity cost per week (R) = 5%/52 weeks = 0.000962 or 0.0962%

Minimum balance in checking account (L) = $ 5,000

Let C* be the optimum cash balance. As per Miller-Orr model

C* = L + (3/4 x T x σ2/R) (1/3)

C* = 5000 + (3/4 x 65 x (3,000,000/0.000962)) (1/3)

C* = $ 10337.12

Average cash balance = (4x C* - L)/3

= (4 x 10337.12) / 3

= $ 12116.16

Annual cost of interest foregone = Average cash balance * Annual opportunity cost

= $ 12116.16 *0.05

= $ 606

The annual cost of interest foregone is $ 606.

5.3. Miller-Orr, Excel: New Jersey Company has recorded the following balances in its checking account for 15 consecutive Wednesdays:

Week no.	Balance	Week no.	Balance	Week no.	Balance
1	$11,347	6	$23,343	11	$13,907
2	$12,525	7	$17,673	12	$13,623
3	$16,003	8	$15,985	13	$15,249
4	$17,056	9	$12,078	14	$18,466
5	$21,732	10	$10,049	15	$19,567

What is σ, the standard deviation of the net cash flows?

$3010 ♥

Solution:

Week No.	Balance	Net cash Flows	Difference	(Difference)2
1	$11,347
2	$12,525	1178.00	$590.86	349112.16
3	$16,003	3478.00	$2,890.86	8357055
4	$17,056	1053.00	$465.86	217022.88
5	$21,732	4676.00	$4,088.86	16718753
6	$23,343	1611.00	$1,023.86	1048283.4
7	$17,673	-5670.00	-$6,257.14	39151837
8	$15,985	-1688.00	-$2,275.14	5176275
9	$12,078	-3907.00	-$4,494.14	20197320
10	$10,049	-2029.00	-$2,616.14	6844203.4
11	$13,907	3858.00	$3,270.86	10698506
12	$13,623	-284.00	-$871.14	758889.88
13	$15,249	1626.00	$1,038.86	1079224.2
14	$18,466	3217.00	$2,629.86	6916148.6
15	$19,567	1101.00	$513.86	264049.16
Total		8220.00		117776680

The net cash flows for each week have been calculated as the difference between the cash balance of two weeks.

Net cash flow for week 2 = Cash balance for week 2 – Cash balance for week 1

= $ 12,525 - $ 11,347

= $ 1,178

Mean = ΣX/N

Where,

X = Net cash flows

N = Number of observations

Mean = $8,220 /14

=$ 587.14

Difference = Net cash flows - Mean

Standard deviation (σ) = √Σ (Difference) 2/N-1

=√ (117776680)/13

=$ 3010

The Standard Deviation of the net cash flows is $ 3010

5.4. Treasury Securities: Nevada Company has bought T-bills with face amount $5.45 million, with a discount of 4.73%, and time to maturity 73 days. Find its bond equivalent yield, and its yield as a zero coupon bond.

4.84%, 4.94% ♥

Solution:

Face value of T-bills (F) = $5.45million

Discount (d) = 4.73%

Time to maturity (n) = 73 days

Discounted price (B) = Face value of T-bills – Discount

= $5,450,000 – ($ 5,450,000 *0.0473*(73/360))

= $ 5,397,727

Bond Equivalent yield:

Bond equivalent yield = 365d

360 - nd

= (365*0.0473)/ (360 – 73*0.0473)

=0.0484

=4.84%

Yield as a Zero coupon bond:

Present Value (B) = $ 5,397,727

Face Value (F) = $ 5,450,000

Yield = r

Time to maturity in years (T) = n/365 = 73/365

F = B ((1+r) T

$ 5,450,000= $ 5,397,727 (1+r) (73/365)

(1+r) (73/365) = $ 5,450,000/$ 5,397,727

1+ r = (1.009684) (365/73)

1+r = 1.0494

r = 0.0494

= 4.94%

Bond equivalent yield is 4.84% and yield as a Zero coupon bond is 4.94%