# Supply Chain Management: Study Notes

Pages: 4 (675 words) Published: June 20, 2014
﻿SYST 4050 Supply Chain Management – Homework 5

1. Harley Davidson has its engine plant in Milwaukee and its motorcycle assembly plant in Pennsylvania. Engines are transported between the two plants using trucks, with each trip costing \$1,000. The motorcycle plant assembles and sells 3000 motorcycles a year. Each engine costs \$500, and Harley incurs uses holding cost of 20 percent.

a) How many engines should Harley load onto each truck (i.e. what is the optimal order quantity)?

D = 3,000
S = 1,000
C = 500
h = 0.2

Q* = SQRT((2DS)/(hC)) = SQRT((2*3000*1000)/(0.2*500)) = SQRT(6000000/100) = 244.979

b) What is the corresponding optimal order frequency?

n* = D/Q* = 3000/244.979 = 12.247

c) What is the cycle inventory of engines at Harley?

Cycle inventory = Q*/2 = 244.979/2 = 122.474

d) What is the corresponding total cost (holding and ordering cost only – do not include material cost)?

Total Cost (TC) = Ordering Cost (OC) + Holding Cost (HC)
= (D/Q*)S + (Q/2)hC = (3000/244.979)*1000 + (244.979/2)*0.2*500 = 12,474.45 + 12,474.45 = 24,494.90

e) If order quantities must be in units of 150, how many engines should Harley ship at a time to minimize total cost (compare the two quantities in units of 150 that are closest to the optimal order quantity)?

Q* = 244.979, hence try Q = 150 and Q = 300.

TC(Q=150) = (D/Q*)S + (Q/2)hC = (3000/150)*1000 + (150/2)*0.2*500
= 20,000 + 7,500 = 27,500

TC(Q=300) = (D/Q*)S + (Q/2)hC = (3000/300)*1000 + (300/2)*0.2*500
= 10,000 + 15,000 = 25,000

Taking Q= 300 minimizes cost.

f) What should the order cost be if a load of 300 engines is to be optimal for Harley? Given that Q = SQRT((2DS)/(hC)). In order to find S, we bring it to the left hand side

S = (hC(Q*2))/(2D) = ((0.2*500*(3002))/2*3000) = \$1,500.00

g) If demand, and thus production, for Harley motorcycle grows, but all other input data remain unchanged. Would you expect cycle inventory of engines at Harley to increase or...

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