Shortages and service levels. This type of problem can be recognized when the problem asks for the number of units short per order cycle, or per year. Use Formula 13-16 for the expected number of units short per cycle, and formula 13-17 to find the number of units short per year. Use Table 13.3 to

Question

Shortages and service levels. This type of problem can be recognized when the problem asks for the number of units short per order cycle, or per year . Use Formula 13 - 16

[latex]E(n) = E( z)\sigma _{dLT}[/latex] (13–16)

for the expected number of units short per cycle , and formula 13 - 17

[latex]E(N) = E( n)\frac{D}{Q}[latex] (13–17)

to find the number of units short per year . Use Table 13.3

TABLE 13.3

Normal distribution service levels and unit normal loss function

Lead Time Service Level	z	E (z)	Lead Time Service Level	z	E (z)	Lead Time Service Level	z	E (z)	Lead Time Service Level	z	E (z)
0.0082	-2.4	2.403	0.2119	-0.8	0.92	0.7881	0.8	0.12	9918	2.4	0.003
0.0091	-2.36	2.363	0.2236	-0.76	0.889	0.7995	0.84	,112	0.9927	2.44	0.002
0.0102	-2.32	2.323	0.2358	-0.72	0.858	0.8106	0.88	0.104	0.9934	2.48	0.002
0.0113	-2.28	2.284	0.2483	-0.68	0.828	0.8212	0.92	0.097	0.9941	2.52	0.002
0.0125	-2.24	2.244	0.2611	-0.64	0.798	0.8315	0.96	0.089	0.9948	2.56	0.002
0.0139	-2.2	.2.205	0.2743	-0.6	0.769	0.8413	1·00	0.083	0.9953	2.6	0.001
0.0154	-2.16	2.165	0.2877	-0.56	0.74	0.8508	1.04	0.077	0.9959	2.64	0.001
0.017	-2.12	2.126	0.3015	-0.52	0.712	0.8599	1.08	0.071	0.9963	2.68	0.001
0.0188	-2.08	2.087	0.3156	-0.48	0.684	0.8686	1.12	0.066	0.9967	2.72	0.001
0.207	-2.04	2.048	0.33	-0.44	0.657	0.877	1.16	0.61	0.9971	2.76	0.001
0.228	-2	2.008	0.3446	-0.4	0.63	0.8849	1.2	0.056	0.9974	2.8	0.0008
0.025	-1.96	1.969	0.3594	-0.36	0.597	0.8925	1.24	0.052	0.9977	2.84	0.0007
0.0274	-1.92	1.93	0.3745	-3.2	0.576	0.8997	1.28	0.048	0.998	2.88	0.0006
0.0301	-1.88	1.892	0.3897	-0.28	0.555	0.9066	1.32	0.044	0.9982	2.92	0.0005
0.0329	-1.84	1.853	0.4052	-0.24	0.53	0.9131	1.36	0.04	0.9985	2.96	0.0004
0.0359	-1.8	1.814	0.4207	-0.2	0.507	0.9192	1.4	0.037	0.9987	3	0.0004
0.0392	-1.76	1.776	0.4264	-0.16	0.484	0.9251	1.44	0.034	0.9988	3.04	0.0003
0.0427	-1.72	1.737	0.4522	-0.12	0.462	0.9306	1.48	0.031	0.999	3.08	0.0003
0.0465	-1.68	1.699	0.4681	-0.08	0.44	0.9357	1.52	0.028	0.9991	3.12	0.0002
0.0505	-0.164	1.661	0.484	-0.4	0.419	0.9406	1.56	0.026	0.9992	3.16	0.0002
0.0548	-1.6	1.623	0.5	0	0.399	0.9452	1.6	0.023	0.9993	3.2	0.0002
0.0594	-1.56	1.588	0.516	0.04	0.379	0.9495	1.64	0.021	0.9994	3.24	0.0001
0.0643	-1.52	1.548	0.5319	0.08	0.36	0.9535	1.68	0.019	0.9995	3.28	0.0001
0.0694	-1.48	1.511	0.5478	0.12	0.342	0.9573	1.72	0.017	0.9995	3.32	0.0001
0.749	-1.44	1.474	0.5636	0.16	0.324	0.9608	1.76	0.016	0.9996	3.36	0.0001
0.0808	-1.4	1.437	0.5793	0.2	0.307	0.9641	1.8	0.014	0.9997	3.4	0.0001
0.869	-1.36	1.4	0.5948	0.24	0.29	0.9671	1.84	0.013
0.0934	-1.32	1.364	0.6103	0.28	0.275	0.9699	1.88	0.012
0.1003	-1.28	1.328	0.6255	0.32	0.256	0.9726	1.92	0.01
0.1075	-1.24	1.292	0.6406	0.36	0.237	0.975	1.96	0.009
0.1151	-1.2	1.256	0.6554	0.4	0.23	0.9772	2	0.008
0.123	-1.16	1.221	0.67	0.44	0.217	0.9793	2.04	0.008
0.1314	-1.12	1.186	0.6844	0.48	0.204	0.9812	2.08	0.007
0.1401	-1.08	1.151	0.6985	0.52	0.192	0.983	2.12	0.006
0.1492	-1.04	1.117	0.7123	0.56	0.18	0.9856	2.16	0.005
0.1587	-1	1.083	0.7257	0.6	0.169	0.9861	2.2	0.005
0.1685	-0.96	1.049	0.7389	0.64	0.158	0.9875	2.24	0.004
0.1788	-0.92	1.017	0.7517	0.68	0.148	0.9887	2.28	0.004
0.1894	-0.88	0.984	0.7642	0.72	0.138	0.9898	2.32	0.003
0.2005	-0.84	0.952	0.7764	0.76	0.129	0.9909	2.36	0.003

to obtain the value of E(z).

The manager of a store that sells office supplies has decided to set an annual service level of 96 percent for a certain model of telephone answering equipment. The store sells approximately 300 of this model ayear. Holding cost is $5 per unit annually, ordering cost is $25 , and [latex]σd_{LT}=7[/latex].

a. What average number of units short per year will the specified annual service level?

b. What average number of units short per cycle will provide the desired annual service level ?

c. What lead time service level is necessary for the 96 percent annual service level?

Accepted Answer

[latex]SL_{ annual} = 96[/latex] percent D = 300 units a year
H = $5 S = $25 [latex]\sigma _{dLT}=7[/latex]

a. [latex]E(N) = (1 - SL_{amnual} )D = (1 - .96)(300) =12[/latex] units

b. [latex]E(N)= E(n)\frac{D}{Q}[/latex]. Solving for E(n), you have

[latex]E(n)= E(N)\div \left\lgroup\frac{D}{Q}\right\rgroup = 12 \div \left\lgroup\frac{300}{Q} \right\rgroup [/latex]

[latex]Q = \sqrt{\frac{2DS}{H} }=\sqrt{\frac{2(300)(25)}{5} }=54.77 [/latex] (round to 55)

Then [latex]E(n)= 12 \div \left\lgroup\frac{300}{55} \right\rgroup =2.2[/latex].

c. In order to find the lead time service level, you need the value of E(z). Because the value of E(n) is 2.2 and [latex]E(n) =E(z)σ_{dLT}[/latex], you have 2.2 = E(z)(7). Solving gives [latex]E(z) = 2.2\div 7=.314[/latex] . Interpolating in Table 13.3 gives the approximate lead time level. Thus,

[latex]\frac{.307 - .314}{.307 - .324}= \frac{.5793 - x}{.5793 - .5636} [/latex]

Solving,

x = .5728

[ To interpolate , find the two vaues between which the computed number falls in the E(z) column. Then find the difference between the computed value and one end of the range, and divide by the difference between the two ends of the range . Perform the corresponding calculation on the two service levels using x for the unknown value, and solve for x. Often , simply " eyeballing " the unknown value will surffice]

Question 13.P.7: Shortages and service levels. This type of problem can be re...

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