 # solutions to homework-1

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1845 Words Grammar Plagiarism  Writing  Score solutions to homework-1 ISyE 6201: Manufacturing Systems
Instructor: Spyros Reveliotis
Solutions to Homework 1
A.
Chapter 2, Problem 4.
(a)
D = 60 units/wk × 52 wk/yr = 3120 units/yr h = ic = 0.25/yr ×\$0.02 = \$0.005/ yr
A = \$12
2 × 12 × 3120
Q∗ =
=
= 3869.88 ≈ 3870 h 0.005
The time between orders is given by
3870
Q∗
=
= 1.24 yr = 14.88 mo
T∗ =
D
3120
(b)
D
3120 units/yr
= \$12
= \$9.67/yr
Q
3870 units
Q
3870 units
Holding cost is h =
× \$0.005/yr = \$9.675/yr.
2
2
The costs are essentially the same. This is always true in the case of the EOQ model.
Set up cost is A

(c) The problem could be where to store all the items. If we were to order 1.24 years worth of styrofoam ice chests at a time it could take up a lot of room.
Chapter 2, Problem 5.
(a) Total (holding plus setup) cost would be
T C = hQ/2+DA/Q = (\$0.005/yr)(3870units)/2+(6240units/yr)(\$12)/(3870units) = \$29.02/yr
(b) The optimum cost would be

2(12)(6240)(0.005) = \$27.36/yr.

(c) Using the wrong value for the demand (100% forecast error) in the EOQ formula results in an increase in cost of only 6%. EOQ is quite robust with respect to parameter values.
Chapter 2, Problem 6.
(a) The EOQ with 60 per week was computed to be 3,870 and the optimal reorder period was
1.24 years or 14.88 months. The closest power of two is 16 months or 1.33 years with a cost of
T C(1.33) = T Dh/2 + A/T = (1.33yr)(3120/yr)(\$0.005/yr)/2+\$12/1.33yr=\$19.37/yr
The power of two on the other side of 14.88 mo is 8 mo or 0.67 yr with a cost of
T C(0.67) = T Dh/2 + A/T =(0.67yr)(3120/yr)(\$0.005/yr)/2+\$12/0.67yr=\$23.20/yr

1

(b) The minimum cost without the power of two restriction is

2ADh = 2(12)(3120)(0.005) = \$19.35/yr. so 16 months has a cost that is only 0.1% over the optimal while the 8 month solution is around
20% over optimal. The total cost in the EOQ model is relatively insensitive to the order quantity used. Since the order period is directly proportional to the order

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