Question

A company assembles and sells skateboards. One popular model is the "ICE". The final assembly plan for April to September, which also represents 50% of a full year’s demand:

Month	April	May	June	July	August	September	October
Forecast (units)	5000	7000	10000	2000	4000	3000	?
Demand (units)	4500	8000	10000	1900	4000	3200	?

The company is using MRP. The forecast for Skateboard ICE for the next coming six weeks:

Week	w36	w37	w38	w39	w40	w41
Demand (forecast, units)	750	750	750	900	950	1000

The company are buying all components from different suppliers. They are only making the wheel assembly and the assembly of the final skateboard, see also the diagram above.

Item	Stock level	Value/item	Lot size	Lead time (weeks)
Skateboard ICE	1800	$250.00	2000	1
Decks	4000	$20.00	1500	2
Wheel assemblies	1200	$80.00	Even hundreds	1
Bolts and Nuts, M6	25000	$0.50	50000	1
Risers	30000	$1.00	50000	1
Wheels	20000	$8.00	40000	2
Bearings	40000	$4.00	50000	2
Trucks	2000	$19.00	?	2
Nuts, M8	50000	$0.25	100000	1

The company has an ordering cost of $150/order and the inventory carrying cost is estimated to 10%.

3a. If we use moving average with n=5, what is the forecast for Skateboard ICE for October?

3b. If we use exponential smoothing with α=0.3, what is the forecast for Skateboard ICE for October?

3c. Suppose it's now week 35. In what week should production of wheel assemblies start?

3d. In relation to question 3c above, what quantity of wheel assemblies will be needed?

3e. The company have had some problems with the supplier of the truck parts and therefore want to review the setup. But first they want to check how many truck parts they should order each time by calculating the Economic Order Quantity (EOQ).

What quantity of truck parts should the company order?

3f. The supplier of the truck parts replies to the company that if they order in lots of 10 000 each time, they will get a 5 % discount.

What is the total cost for the truck parts if the company order 10 000 each time?

Answer 1

Answer to question 1 :

For n = 5 , forecast for October basis moving average

= Sum of forecasts( for period May to September) /5

= ( 7000 + 10000 + 2000 + 4000 + 3000) / 5

= 26000/5

= 5200

Answer to question 2:

Basis exponential smoothing, relevant equation as follows :

Ft = alpha x At + ( 1- alpha) x Ft-1

Ft = Forecast for period t

Ft-1 = Forecast for period t-1

At = Actual demand for period t

Alpha = 0.3

Also , F0 = A0 ( Forecast for period 0 is assumed to be same as actual demand for period 0)

The below table highlights the forecast values arrived basis exponential smoothing method ( for period April till September)

Forecast for October will be same as Forecast value for September arrived through exponential smoothing

Accordingly, forecast for skateboard ICE for October =   4408    units

Month	Actual demand , A	Forecast, F
April	5000	5000
May	7000	5600
June	10000	6920
July	2000	5444
August	4000	5011
September	3000	4408
October		4408

Answer to question 3:

Lead time of components going into production of wheel assemblies as follows :

Wheels : 2 weeks

Bearing : 2 weeks

Trucks : 2 weeks

Nuts : 1 week

Hence maximum lead time of component = 2 weeks

Thus if wheel assembly is required on 35th week. Production of assembly of wheel assembly should start at = 35 week – 2 week = 33th week

Supply of wheel assembly has a lead time of 1 week for production of Skateboard ICE ,

In other words for skateboards which are required on week 36, production of wheel assembly should start at week 33

Below table presents when Skateboard ICE is required and corresponding production of wheel assembly must start :

Week when skateboard required	W36	W37	W38	W39	W40	W41
Week when wheel assembly production shall start	W33	W34	W35	W36	W37	W38

Answer to question 4:

Annual demand for skateboard ICE

= 2 x ( Cumulative demand for period April to September)

= 2 x ( 4500 + 8000 + 10000 + 1900 + 4000 + 3200 )

= 2 x 31600

= 63200

Since 1 Skateboard ICE requires 2 wheel assembly, annual requirement of wheel assemblies = 2 x 63200 = 126400

However 1200 wheel assemblies are already in stock

Hence number of wheel assemblies which will be required annually = 126400 – 1200 = 125200

Answer to question 5 :

Annual demand for skateboard ICE

= 2 x ( Cumulative demand for period April to September)

= 2 x ( 4500 + 8000 + 10000 + 1900 + 4000 + 3200 )

= 2 x 31600

= 63200

Since 1 Skateboard ICE requires 2 wheel assembly and each wheel assembly requires 1 truck,

1 skateboard ICE requires 2 truck parts

Therefore , annual demand of truck parts = D = 2 X 63200 = 126400

Ordering cost = Co = $150

Inventory carrying cost = Cc = 10% of value of each truck part = 10% of $19 = $1.9

Therefore, Economic Order Quantity ( EOQ)

= Square root ( 2 x Co x D/ Cc)

= Square root ( 2 x 150 x 126400 /1.9)

= 4467.42 ( 4467 rounding to nearest whole number)

COMPANY SHOULD ORDER 4467 NUMBER OF TRUCK PARTS

Answer to question #6:

Order quantity = 10,000 units

Total annual ordering cost, $

= Ordering cost x number of orders

= Ordering cost x Annual demand of truck parts/ Order quantity

= 150 x 126400/10000

= $1896

The discounted price ( after 5% discount) for each truck = 95% of $19 = $18.05

Total annual inventory holding cost , $

= Holding cost per unit per year x average inventory

= ( 10% of discounted price) x Order quantity /2

= ( 10% of $18.05) x 10000/2

= 1.805 x 5000

= $9025

Total annual material cost of truck parts = 126400 ( annual requirement) x $18.05 ( value per truck part) = $2281520

Total cost for truck parts

= Annual ordering cost + annual inventory holding cost + annual material cost

= $1896 + $9025 + $2281520

= $2292441

TOTAL COST OF TRUCK PARTS = $ 2292441