CHAPTER 11
Solution
1. MICHAEL TRACI
a. Optimal order quantity using the EOQ model:
EOQ 
2C p A
CH
Where:
CP = $30 (preparation cost per order)
A = 30 units demand period
CH = $2.00 carrying cost per period
EOQ = optimal order lot size
EOQ 
2$3030units 
 30 units
$2.00
b. Average cycle stock = 30 units /2 = 15 units
c. 12 periods x 30 units = 360 units (annual unit demand)
360 units /30 units per order = 12 orders
12 orders x $30 = $360 ordering cost per year
15 units x $2.00 x 12 periods = $360 carrying cost per year
Total = $720 per year
2. LOUIE’S LOBSTER POTS
a. Here is one spreadsheet layout. The formulae for the expected number of units short differ
for each value of R.
Probs:
.05
.15
Demand During LT
20
21
Cum. Prob.
.05
.20
Pr D > DDLT
.95
.80
Expected No. Short
2.90 1.95
Shortage cost for R = 20 is $72.50
.20
22
.40
.60
1.15
.30
23
.70
.30
.55
.10
24
.80
.20
.25
.15
25
.95
.05
.05
.05
26
1
0
0
Probs:
.2
.4
Demand During LT
20
21
Cum. Prob.
.20
.60
Pr D > DDLT
.80
.40
Expected No. Short
1.50
.70
Shortage cost for R = 20 is $37.50
.2
22
.80
.20
.30
.1
23
.90
.10
.10
.1
24
1
0
0
.0
25
1
0
0
.0
26
1
0
0
b.
11-1
3. THE ICU OPTICAL CLINIC
a.
Demand / Month
Probability
12
.10
13
.15
14
.15
15
.20
16
.20
17
.10
18
.05
19
.05
Average Monthly Demand:
EOQ 
Expected Demand / Month
1.20
1.95
2.10
3.00
3.20
1.70
.90
.95
15.00
2$251215  35units
$.25$30
b.
ROP
15
*16
17
18
19
Buffer Stock = ROP - 15 = 1
Expected
Numbers of
Stockouts
.75
.35
.15
.05
.00
Service
Level
.979
.990
.996
.999
1.000
*Selected ROP
c. The annual cost of carrying buffer stock is $7.50 (one unit).
5. THE SELDOM SEEN RANCH
a. The Z value for a .10 probability of stocking out is 1.282:
Safety Stock = (1.282)(1)(1.25)(40) = 64 bales
b. Order Point = 1,000 + 64 = 1,064 bales
c. SL = 1-0.1= 99%
d. Horace should place an order for 2,500 bales whenever the number of bales on hand goes
below 1,064.
11-2