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# Urchasing Agent for a Particular Type of Silicon Wafer Used in the Production of Semiconductors Must Decide Among Three Sources. Source a Will Sell the Silicon Wafers for \$2.50 Per Wafer, Independently of the Number of

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22 (P226) A purchasing agent for a particular type of silicon wafer used in the production of semiconductors must decide among three sources. Source A will sell the silicon wafers for \$2.50 per wafer, independently of the number of wafers ordered. Source B will sell the wafers for \$2.40 each but will not consider an order for fewer than 3,000 wafers, and Source C will sell the wafers for \$2.30 each but will not accept an order for fewer than 4,000 wafers. Assume an order setup cost of \$100 and an annual requirement of 20,000 wafers. Assume a 20 percent annual interest rate for holding cost calculations.
a. Which source should be used, and what is the size of the standing order?
b. What is the optimal value of the holding and setup costs for wafers when the optimal source is used? c. If the replenishment lead time for wafers is three months, determine the reorder point based on the on-hand level of inventory of wafers.
Ans: 

= 20,000
K = 100
I = .20

All Units Discount

c0 = \$2.50 c1 = \$2.40 c2 = \$2.30
a.
Q
Q
Q

(0)

(1)

(2

=

2K

1c0

=

2K

1c1
2K

1c2

=

(2)(100)(20,000)
(.20)(2.50)

(2)(100)(20,000)
(.2)(2.40)
(2)(100)(20,000)
(.2)(2.30)

= 2828

= 2887
= 2949

0

only Q is realizable. (Figure 11 is accurate with different breakpoints.)

Cost at Q = 4,000
(20,000)(2.30) + (.2)(2.30)(4,000)  (100)(20,000)
2

(4, 000)

= \$47,420

Cost at Q = 3,000
(20,000)(2.40) + (.2)(2.40)(3,000)  (100)(20,000 )
2

0

Cost at Q = Q

=

3,000

2828

(20,000)(2.50) +

2KIC0

> 50,000

It follows that the optimal order size is Q = 4,000
b) \$1420. (47,420 - (20,000)(2.30))

= \$49,386.67

c)

T = Q/ = 4,000/20,000 = .2 years = 2.4 months.

Hence,  > T

2.4 mos.

.6 mos. 3 months r = (T - ) = (20,000)(.6/12) = 1,000.
23. (P226) Assume that two years have passed, and the purchasing agent mentioned in Problem 22 must recomputed the optimal number of wafers to purchase and from which source to purchase them. Source B has decided to accept any size offer, but sells the wafers for \$2.55 each for orders of up to 3,000 wafers and \$2.25 each for incremental amount ordered over 3,000 wafers.
Source A still has the same price schedule, and Source C went out of business. Now which source should be used?
Ans: Incremental schedule for source B.
C(Q) = 2.55Q for Q  3,000
C(Q) = (2.55)(3,000) + 2.25(Q - 3,000) for Q  3,000
= 7650 + 2.25Q - 6750 = 900 + 2.25Q for Q  3,000
C(Q)/Q = 2.55 for Q ≤ 3,000
= 900/Q + 2.25 for Q ≥ 3,000
G(Q) = [20,000]

900
(100)(20, 000)
900
Q
 2.25 
 (.20)
 (2.25)
Q
Q
Q
2

= 18,000,000  2,000,000  (.20)(2.25)Q  \$45,090
Q

Q

2

= 20,000,000  0.225Q  \$45,090
Q

The minimizing Q occurs where Q = cost =

20,000,000 = 9428
.225

20,000,000
+ (.225)(9428) + \$45,090 = \$49,332.64
9428

Since the annual cost for source A exceeds \$50,000, now source B should be used.

24. (P226) In the calculation of an optimal policy for an all-units discount schedule, you first compute the EOQ values for each of the three order costs, and you obtain: Q(0)=800, Q(1)=875, and Q(2)=925. The all-units discount schedule has breakpoints at 750 and 900. Based on this information only, can you determine what the optimal order quantity is? Explain your answer.
Ans: Since Q

(2)

= 925 is realizable, it must be optimal.

25. (P227) Parasol Systems sells motherboards for personal computers. For quantities up through
25, the firm charges \$350 per board; for quantities between 26 and 50, it charges \$315 for each board purchased beyond 25; and it charges \$285 each for the additional quantities over 50. A large communications firm expects to require these motherboards for the next 10 years at a rate of at least 140 per year. Order setup costs are \$30 and holding costs are based on an 18 percent annual interest rate. What should be the size of the standing order?
Incremental discount schedule
Ans:  = 140
K = 30
I = .18

350
(350)(25)  315(Q  25)
(350)(25)  (315)(25)  285(Q  50)

C(Q) =

for Q  25 for 26  Q  50 for 51  Q

350
875/ Q  315
2375/ Q  285

C(Q)/Q =

for Q  25 for 26  Q  50 for 51  Q

 C (Q)  (30)(140)
 C (Q)  Q 
 (.18)

 Q  2 

Q
 Q 

 

G(Q) = (140) 

Q

(0)

(2)(30)(140)
(.18)(350)

=

= 11.55  12.00 (realizable)

 
875
(30)(140)

875
Q
 .18
 315
 Q  315 
Q
 
 Q
 2

G1(Q) = 140

= 123,700  (56.7)Q  44,178.75
Q

Q

(1)

2

(2)(123,700)
 66 not realizable
56.7

=

 2375
 (30)(140)
 2375
 Q 
 285 
 .18
 285 
Q
 Q

 Q
 2 

G2(Q) = 140 
=

333, 700 51.3Q

 40,113.75
Q
2

Q

(2)

=

(2)(333,700)
 114 realizable
51.3

Compare average annual costs at

Q

(0)

(2)

and Q .

(0)
G0(Q ) = (140)(350) + (30)(140)  (.18)(350)(12)  \$49,728

12

(2)

G2(Q ) =

2

333, 700 51.3(114)

 40,113.75  \$45,965
114
2

Hence, the optimal solution is Q

(2)

which requires maintaining a standing order of 114.