1. Assignment 5: Part (A)
The Dine Corporation is both a producer and a user of brass couplings. The firm operates 220 days a year and uses the couplings at a steady rate of 50 per day. u = 50. Coupling can be produced at rate of 200 per day. p=200. Annual storage cost is $1 per coupling, H=1, and machine setup cost is $35 per run, S=35. Since there are 220 days per year and u is equal to 50, therefore yearly demand is 50(220), D= 11000
Determine the economic run size.
Approximately how many runs per year will there be?
Compute the maximum inventory level.
Determine the cycle time, run time and pure consumption time.
Compute the total cost

2. Production, Consumption, Inventory

3. EPQ: One period and One year
p-u u
One Period
One Year

4. Imax, Ordering Cost and Carrying Cost

5. EPQ : Optimal Q

6. The Problem
The Dine Corporation is both a producer and a user of brass couplings. The firm operates 220 days a year and uses the couplings at a steady rate of 50 per day. u = 50. Coupling can be produced at rate of 200 per day. p=200. Annual storage cost is $1 per coupling, H=1, and machine setup cost is $35 per run, S=35. Since there are 220 days per year and u is equal to 50, therefore yearly demand is 50(220), D= 11000
Determine the economic run size.
Approximately how many runs per year will there be?
Compute the maximum inventory level.
Determine the cycle time, run time and pure consumption time.
Compute the total cost

7. EPQ and # of Runs
u =50, p=200, H=1, S=35, D= 11000
Determine the economic run size.
Approximately how many runs per year will there be?
D/Q = 11000/1013 = 10.85

8. Run Time How much do we produce each time? Q
How long does it take to produce Q? d1
What is our production rate per day? p
Run Time?
Q = pd1  = 200d1
d1= 1013/200 = 5.0
p-u u
I max d1 d2

9. Maximum Inventory Maximum Inventory ?
At what rate inventory is accumulated?
p-u
For how long? d1
What is the maximum inventory?
IMax
IMax = (p-u)d1
IMax = (200-50)(5.06) = 759
p-u u
I max d1 d2

10. IMax Pure Consumption Time What is the Maximum inventory?
How long does it take to consume IMax ? d2
At which rate Imax is consumed for d2 days u
Imax = ud2
759 = 50d2
d2 = 15.18
p-u u
I max d1 d2

11. Cycle Time, Total Cost Cycle Time ? d1 + d2 = 5.06 + 15.18 = 20.24
TC = H(IMax/2) + S(D/Q)
TC = 1(759/2) + 35 (11000/1013)
TC = = 759

12. Assignment4: Part (B)
(a) : If average demand per day is 20 units and lead time is 10 days. Assuming zero safety stock. ss=0, compute ROP?
(b): If average demand per day is 20 units and lead time is 10 days. Assuming 70 units of safety stock. SS=70, compute ROP?
(c) : If average demand during the lead time is 200 and standard deviation of demand during lead time is 25.
Compute ROP at 90% service level. Compute ss
(d) : If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days.
(e) : If demand per day is 20 units and lead time is 16 days and standard deviation of lead time is 4 days.

13. Demand Forecast, and Lead Times
Re-Order Point (ROP) in periodic inventory control is the beginning of each period. (ROP) in perpetual inventory control is the inventory level equal to the demand during the lead time plus a safety stock to cover demand variability.
Lead Time is the time interval from placing an order until receiving the order.
If there were no variations in demand, and demand was really constant; the ss is zero and ROP is when inventory on hand is equal to the average demand during lead time.

14. ROP; Fixed d, Fixed LT, Zero SS
(a) : If average demand per day is 20 units and lead time is 10 days. Assuming zero safety stock.
ss = 0
What is ROP?
ROP = demand during lead time + ss
ROP = demand during lead time
Demand during lead time = (lead time) × (demand per unit of time)
Demand during lead time = 20 × 10 = 200

15. ROP; Fixed d, Fixed LT, Positive SS
Problem 4(b): If average demand per day is 20 units and lead time is 10 days. Assuming 70 units of safety stock.
ss = 70
What is ROP
ROP = demand during lead time + ss
Demand during lead time = 200
ss = 70
ROP = = 270

16. Safety Stock and ROP
Service level
Risk of a
stockout
Probability of
no stockout
ROP
Quantity
Average
demand
Safety
stock
z-scale z
Each Normal variable x is associated with a standard Normal Variable z
x is Normal (Average x , Standard Deviation x)  z is Normal (0,1)
There is a table for z which tells us
Given any probability of not exceeding z. What is the value of z
Given any value for z. What is the probability of not exceeding z

17. Common z Values Risk Service level z value .1 .9 1.28 .05 .95 1.65
ROP
Risk of a
stockout
Service level
Probability of
no stockout
Safety
stock z
Quantity
z-scale
Average
demand
Risk Service level z value

18. Relationship between z and Normal Variable x
ROP
Risk of a
stockout
Service level
Probability of
no stockout
Safety
stock z
Quantity
z-scale
Average
demand
z = (x-Average x)/(Standard Deviation of x)
x = Average x +z (Standard Deviation of x)
μ = Average x
σ = Standard Deviation of x
Risk Service z value
level
 x = μ +z σ

19. Relationship between z and Normal Variable ROP
Risk of a
stockout
Service level
Probability of
no stockout
Safety
stock z
Quantity
z-scale
Average
demand
LTD = Lead Time Demand
ROP = Average LTD +z (Standard Deviation of LTD)
ROP = Average LTD +ss
ss = z (Standard Deviation of LTD)

20. Safety Stock and ROP Risk Service level z value .1 .9 1.28
Risk of a
stockout
Service level
Probability of
no stockout
Safety
stock z
Quantity
z-scale
Risk Service level z value
Average
demand
ss = z × (standard deviation of demand during lead time)

21. ROP; total demand during lead time is variable
(c) : If average demand during the lead time is 200 and standard deviation of demand during lead time is 25.
Compute ROP at 90% service level.
Compute SS
ROP
Risk of
a stockout
Service level
Probability of
no stockout
Expected
demand
Safety
stock x

22. ROP; total demand during lead time is variable
Service level
Risk of
a stockout
Probability of
no stockout z
Risk Service level z value

23. ROP; total demand during lead time is variable
z = 1.28
z = (X- μ )/σ
X= μ +z σ
μ = 200
σ = 25
X= × 25 X=
ROP = 232
SS = 32

24. ROP; Variable d, Fixed LT
(d) : If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days.
Compute ROP at 90% service level.
Compute SS

25. ROP; Variable d, Fixed LT
Previous Problem
If average demand during the lead time is 200 and standard deviation of demand during lead time is 25.
Compute ROP at 90% service level.
Compute SS
This Problem
If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days.
Compute ROP at 90% service level.
Compute SS
If we can transform this problem into the previous problem, then we are done, because we already know how to solve the previous problem.

26. ROP; Variable d, Fixed LT
If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days.
Compute ROP at 90% service level.
Compute SS
What is the average demand during the lead time
What is standard deviation of demand during lead time

27. Demand During Lead Time
If demand is variable and Lead time is fixed

28. ROP; Variable d, Fixed LT

29. Now it is transformed into our previous problem where
total demand during lead time is variable
The average demand during the lead time is 320 and the standard deviation of demand during the lead time is 20.
Compute ROP at 90% service level.
Compute SS
Service level
Risk of
a stockout
Probability of
no stockout
ROP
Expected
demand
Safety
stock x

30. ROP; Variable d, Fixed LT
z = 1.28
z = (X- μ )/σ
X= μ +z σ
μ = 320
σ = 20
X= × 20 X= X=
ROP = 346
SS = 26

31. Demand Fixed, Lead Time variable
If lead time is variable and demand is fixed

32. Lead Time Variable, Demand fixed
e) Demand of sand is fixed and is 50 tons per week.
The average lead time is 2 weeks.
Standard deviation of lead time is .5 week.
Assuming that the management is willing to accept a risk no more that 10%. Compute ROP and SS.
Acceptable risk; 10%  z = 1.28