1) Reorder level = Average daily usage rate x Lead time in days
AVERAGE DAILY USAGE =
Reorder level = Average daily usage rate x Lead time in days
=

*14 = 191.78

From the Standard normal distribution, which z value should be used in the calculation of the quantity of sheets to be ordered? (Round to two decimal places)
Z = 1.645
SINCE THERE ARE 150 SHEETS IN HAND NUMBER OF
SHEETS TO BE ORDERED = 42
2)
3) (a)
Total cost (TC) = quantity (Q)/2 * storage cost (SC) + demand (D)/Q*ordering cost (OC)
OR
TC= Q/2*SC + D/Q*OC
Differentiating TC by Q we get (to minimize TC) d(TC)/dQ = SC/2 – (D/Q^2)*OC = 0 also second derivative is positive ensuring that TC is being minimized or Q= √(2*(D)*(OC)/(SC))
Therefore:
Economic Order Quantity (Q) =
√(2*annual units required*cost of ordering each timeannual cost per average on hand inventory unit)
=
√(2*2000*105)
= 89.44
Rounded off to next higher integer = 90
Therefore at a time 90 HD-60 components should be ordered
(b)
Average Annual cost of ordering HD-60 components is D/Q*OC
=2000/89.44*10 = $223.61
(c)
Annual cost of storing HD-60 components is Q/2 * SC
= 90/2*5 = $225
4) This concept is discussed in the text on pages 692-694. The drum or control of the process is at the bottleneck, resource C. A small buffer inventory should be held between resource B and resource C. A rope (communication) should be directed back to the start of the process, resource A to prevent inventory from building up. Depending upon the situation, a second rope can extend from the finished goods inventory location (another buffer) to resource C as described in the text. An excellent response to this question will include a diagram similar to exhibit 20.9 and/or 20.11.

5) The reference is on page 696 of the text. The authors are distinguishing between a process batch and a transfer batch. Point of view is important here: the answer "one" is from the perspective of each station on the line. The answer "infinity" (or a very large number) is from the perspective of the process.

6)
7) The optimal order will result when the two processes require least time to process the job.
Since the jobs has to go from D to E, so the most optimal order will be when , the process D is not kept idle because of the time taken by process E , as in case with job 5,6,7 as the time required in E is more ,