Tutorial IV
• Problem 32 in Assignment I
• Examples
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Problem 32 in Assignment I
• a. Compute the total inventory
• b. 20000 inventory at the end of quarter 4
• c. 3 requirements for the solution
– Minimum constant workforce
– Total inventory is less than that of part a
– Stock-out occurs in quarter 2
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Examples
• Textbook
– Page 204: Problem 17
– Page 211: Problem 22
– page 212: Problem 25
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Problem 17
•
•
•
•
The Wod Chemical Company produces a chemical compound that is
used as a lawn fertilizer. The compound can be produced at a rate of
10000 pounds per day. Annual demand for the compound is 0.6
million pounds per year. The fixed cost of setting up for a production
run of the chemical is $1500, and the variable cost of production is
$3.5 per pound. The company uses an interest rate of 22 percent to
account for the cost of capital, and the costs of storage and handling
of the chemical amount to 12 percent of the value. Assume that there
are 250 working days in a year.
a. What is the optimal size of the production run for this particular
compound?
b. What proportion of each production cycle consists of uptime and
what proportion consists of downtime?
c. What is the average annual cost of holding and setup attributed to
this item? If the compound sells for $3.9 per pound, what is the annual
profit the company is realizing from this item?
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Solution Sheet
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Solution Sheet
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Problem 22
•
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 $2.5 per wafer, independently
of the number of wafers ordered. Source B will sell the wafers for $2.4
each but will not consider an order for few than 3000 wafers, and
Source C will sell the wafers for $2.3 each but will not accept an order
for fewer than 4000 wafers. Assume an order setup cost of $100 and
an annual requirement of 20000 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.
•
•
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Solution Sheet
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Solution Sheet
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Problem 25
• 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?
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Solution Sheet
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Solution Sheet
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Solution to Problem 17
• D = 0.6x106 per year,
P=10000x250=2.5x106 per year, K =$1500,
h =3.5x(0.22+0.12)=$1.19,
h’=h(1−D/P)=0.9044
• a. Q* = (2x1500x0.6x106/0.9044)1/2 = 44612
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Solution to Problem 17 (cont)
• b. T=Q*/D =44612 / (600000/250)=18.59 days
T1 = Q*/P =44612/(2.5x106/250)=4.46 days
T2 =T−T1 =14.13 days
• c. Average holding cost = h’Q*/2 =20173.55
Average setup cost =KD/Q =20173.94
Total cost =40347.49
Annual profit =(3.9−3.5)x0.6x106 − 40347.49
=240000 −40347.49
=$199652.51
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Solution to Problem 22
• D=20000 per year, K=$100, h=0.2C(Q)
⎧2.5Q
⎪
C (Q) = ⎨2.4Q
⎪2.3Q
⎩
for
0 ≤ Q < 3000
for 3000 ≤ Q < 4000
for
4000 ≤ Q
When c = 2.5, Q*=2829;
when c = 2.4, Q*=2887;
when c = 2.3, Q*=2949.
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Solution to Problem 22 (cont)
• Comparison of Costs
G(2829) =2829x2.5x0.2/2+100x20000/2829
+ 2.5x20000=51414.21
G(3000) =3000x2.4x0.2/2+100x20000/3000
+2.4x20000=49386.67
G(4000) =4000x2.4x0.2/2+100x20000/4000
+2.3x20000=47420
• Q*=4000
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Solution to Problem 22 (cont)
• c. L=0.25 year =3 months
• T*=4000/20000=0.2 year =2.4 months
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Leadtime and Reorder Point
Q
R=D(L−T)
=1000
Reorder
point
R
Place
order
T
Time
Receive
order
Leadtime
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Solution to Problem 25
• D = 140 per year, K =$30,
• h =0.18 C(Q)
350Q
⎧
⎪
C (Q) = ⎨ 350 × 25 + 315(Q − 25)
⎪25(350 + 315) + 285(Q − 50)
⎩
⎧ 350Q
⎪
C (Q) = ⎨ 315Q + 875
⎪285Q + 2375
⎩
for
0 ≤ Q < 26
for
26 ≤ Q < 51
for
51 ≤ Q
for
0 ≤ Q < 26
for
26 ≤ Q < 51
for
51 ≤ Q
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Solution to Problem 25 (cont)
G (Q) = DC (Q) / Q + KD / Q + iC (Q) / 2
When 0<=Q< 26, Q*=12;
when 26<=Q< 51, Q*=66;
when 51<= Q,
Q*=114.
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Comparison of Costs
• G(12) =350x12+30x140/12+350x0.18x12/2
=49728
• G(114) = (285x114+2375)x140/114
+30x140/114
+0.18x(285x114+2375)/2
=45991.36
• Q*=114
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Important Formulas
• EPQ
Q = 2 KD /(1 − D / P )h
*
T * = Q * / D = 2 K /(1 − D / P) Dh
G (Q * ) = h(1 − D / P)Q * = 2 KD / Q *
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Quantity Discounts
• All-units discounts
Gj(Q)=cjD + KD/Q + icjQ/2
• Incremental discounts
G(Q)=C(Q)D/Q + KD/Q +i C(Q)/2
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