Chapter 25 Problem Set pages 558–561
Problem 5
a. What is the EOQ for a firm that sells 5,000 units when the cost of placing an order is $5 and the carrying cost are $3.50 per unit?
EOQ = [(2SO)/C] .5 = [(2)(5,000)($5)/$3.50] .5 =
EOQ = 119.5 units
b) How long will the EOQ last? How many orders are placed annually?
Annual number of orders:
Sales per day: 5,000/365 = 13.7 units (i.e., 14 units)
Duration of the EOQ: 120/13.7 = 8.8 days (i.e., 9 days)
Annual number of orders: 365/8.8 = 41.4 = 42 orders placed annually
c) As a result of lower interest, the financial manager determines the carrying costs are now $1.80 per unit. What are the new EOQ and annual number of objects?
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EOQ = [(2SO)/C] .5 = [(2)(5,000)($5)/$1.80] .5 = 166.7 units
Annual number of orders: 5000/166.7 = 30
Problem 6
a) What is the EOQ?
EOQ = [2(30,000)($60)/$1.50].5= 1,549 units
b) What is the average inventory based on the EOQ and the existing safety stock?
Average Inventory = EOQ/2 + Safety Stock
Avereage Inventory = 1549/2 + 300
Average Inventory = 1074.5; 1,075 units
c. What is the maximum level of inventory?
Maximum level of inventory = EOQ + Safety Stock
Maximum level of inventory = 1549 + 300
Maximum level of inventory = 1845 units
d. How many orders are placed each year?
Annual no of orders placed = 1550*365/30000 = 18.858 ~ 19 orders
Problem 14
FV= PV* (1 + i) n
FV =$77,345;
PV= $76,789; n= 30/365
77345= 76789 * (1 + i) (30/365)
1.00724= (1 + i) (.08219)
.08219 √1.00724 = 08219 √(1 + i) (.08219)
1.09174= 1 + i
Answer: I= .09174 = 9.174%
Chapter 26 Problem Set Pages 585–586
Problem 3
Interest paid: $65,000 x 0.1 x 120/360 = $2,167
Origination fee: $65,000 x 0.015 = $975
Funds the firm gets to use: $65,000 - $2,167 - $975 = $61,858
The simple rate of interest: i = $3,142/$61,858x 360/120= 15.24%
Problem 6
a. Which terms are better if the firm intends to borrow the $10 million for the entire year?
Bank A loan:
Interest payment (.08)($10,000,000) = $800,000
i = $800,000 x 12 / $10,000,000 x 12
= 0.8 (8%)
Bank B loan:
Interest payment (0.66)($10,000,000) = $660,000
Origination fee: (0.02)($10,000,000) = $200,000
i = ($660,000 x 12) / (10,000,000 - 200,000 x 12) = .0673 (6.73%)
With these rates it would be better for the firm to borrow from Bank A. Bank B may have a lower interest rate however, with the additional fees it comes out to about $60,000 more thank Bank A.
b. If the firm plans to use the funds for only three months, which terms are better?
Bank A loan:
Interest payment: (.08)(10,000,000)(3/12) = $200,000
Fee on the unused balance: (0.05)(10,000,000)(.75) = $37,500
i = ($237,500 x 12) / ($10,000,000 x 3)= 0.095 (9.5%)
Bank B loan:
Interest payment: (0.66)(10,000,000)(.25) = $165,000
Commitment fee: (0.02)(10,000,000) = $200,000
i = ($165,000 x 12)/((10,000,000 - 200,000) x 3= 0.0673 (6.73%)
Even if the firm only intends on borrowing the funds for three months Bank A’s terms are still better. Again, with Bank A it is a legit 8% rather than a 6.6% and additional fees with Bank B.
Problem 8
a) I TC = [Percentage Discount / (100 – Percentage Discount)] * [360 / (Payment Period – Discount Period)]
I TC = [3 / (100 – 3)] * [360 / (45 – 15)]
I TC = .37113 = 37.11
b. I TC = [Percentage Discount / (100 – Percentage Discount)] * [360 / (Payment Period – Discount Period)]
I TC = [2 / (100 – 2)] * [360 / (30 – 10)]
I TC = .36735 = 36.73%
A is more expensive with a higher interest rate.
Problem 11
a)
Amount to be Borrowed = [Amount Required / (1 – i)]
Amount to be Borrowed = [$10,000 / (1 – .10)]
Amount to be Borrowed = $11,111.11
Interest Rate = (Interest / Amount of Loan Used)
Interest Rate = ($11,111.11/ $10,000) = 11.11%
b) straight loan therefore:
Interest Rate = 11%
Answer: A is more expensive as it has a higher effective interest rate
Problem 12
Simple Rate:
I CP = [Interest / Proceeds Used] * [365 / # of days outstanding]
I CP = [($1,000,000 - $982,500)/ $982,500] * [365 / 270]
I CP = .02408 = 2.408%
Compound Annual Rate:
Accumulated Amount = Principal* (1 + i) n
$1,000,000 = $982,500 * (1 + i) 270/3
i = .02415 = 2.415%
