16. A part used to repair machines has a normally distributed monthly demand a m
ID: 328733 • Letter: 1
Question
16. A part used to repair machines has a normally distributed monthly demand a mean of 65.0 and a standard deviation of 5.2. If lead time is so predictable that it can be considered a constant 0.25 month and the service level is 90 perc a. What is the order point? b. What is the safety stock level? nt 17. If j 30 percent and EDDLT 740 units: a. Compute the safety stock using the percentage of EDDLT method b. Compute the order point using the percentage of EDDLT method. c. Compute the safety stock using the square root of EDDLT method. d. Compute the order point using the square root of EDDLT method.Explanation / Answer
Mean demand over the period shall be denoted by µDL = 65
And the lead time has been denoted = 0.25 month
If we consider 0.25 month as 1 period, hence the whole month is 4 periods .
Hence, Mean demand over the lead time period = µDS = 65/4 = 16.25
And standard deviation over the month is denoted by ?DL = 5.2
Hence standard deviation over 0.25 month (1 period) = ?DS= ?DL/ =5.2/ = 5.2/2= 2.6
Level of service is defined as probability of not stocking out = LOS= 95% =0.95
Hence safety stock = kLOS x ?ds
kLOS is the safety factor =NORMINV(0.95,0,1) =1.65
Hence safety stock = 1.65 x 2.6 = 4.3 ~4 units
Order Pont= Average demand + Safety stock
= 16.25+4.3 = 20.55= 21
Q17
EDDLT is expected demand over lead time
EDDLT = 740 Units
Percentage method:
Safety stock = j x EDDLT = 0.3 X740 = 212 units
Order Point = EDDLT + Safety stock = 740+ 212 = 952 units
Square root method
Safety stock = sqrt (EDDLT)= square root (740) = 27 units
Order Point = 740+ 27 = 767 units
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