computing the expected end-product service time using stochastic item delays

Computing the Expected End-Product Service Time Using Stochastic Item DelaysAuthor(s): Marshall RoseSource: Operations Research, Vol. 19, No. 2 (Mar. - Apr., 1971), pp. 524-540Published by: INFORMSStable URL: http://www.jstor.org/stable/169285 .

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COMPUTING THE EXPECTED END-PRODUCT

SERVICE TIME USING STOCHASTIC

ITEM DELAYS

Marshall Rose

Xerox Corporation, Rochester, New York

(Received September 20, 1969)

This paper derives an expression for the expected completion time of a repair project, such as the servicing of end products, when the servicing is composed of a sequence of repair activities on parts of the end product. These parts are subject to repair with a specified probability and, at first, it is assumed that the completion time of each activity is constant. Subse- quently, this assumption is relaxed so that an arbitrary probability distribution can be specified for the activity completion times. The concluding parts of the paper show how to compute the expected end-product service time for a particular class of activity-time probability distributions.

IIHIS PAPER IS concerned with determining the times required to Repair end products, such as military weapon systems and commercial

transportation vehicles. These end products are repaired periodically, and the repairs often involve sequences of activities or tasks to be performed. These tasks consume time, and the end-product service times are therefore a function of these activity times. This paper derives an expression for the expected end-product service time in terms of the activity times, when these times are constant and when they are stochastic. It then computes the service time for activity times that are distributed in a specified fashion.

CONSTANT DELAY TIMES

A FIRM periodically repairs end products, which must be serviced owing to the failure or the scheduled preventive maintenance of their reparable parts. On the first day an end product arrives for servicing, all the recoverable items to be repaired are removed from this end product, and an order is generated for a repair action. Subsequently, the repaired items will be reinstalled.

The reparable items must be reinstalled in a specific sequence, determined by technological constraints. The reinstallation sequence is indi- cated by Zi, called the (fixed) scheduled item resupply time, and is equal to the minimum number of days after removal for repair that item i can be reinstalled. Furthermore, the item is scheduled to be resupplied and re-

524



End-Product Service Time 525

installed on this date. (A list of symbols is provided at the end of the paper. Throughout, unless specified to the contrary, small letters represent random variables, while capital letters represent quantities that do not exhibit randomness.)

Without loss in generality, it is assumed that the time required to re- install the items after they are resupplied is negligible. If the reinstallation time cannot be safely ignored, then the interval between the resupply times of items scheduled to be reinstalled consecutively can indicate the replacement time of an item.

In the present model, this interval is fixed, and represents the time to complete other end-product servicing tasks. Accordingly, owing to this fixed interval, and to the specific reinstallation sequence, it follows that the actual reinstallation date of one item cannot precede the reinstallation date of another repaired item having the next lower Zi, plus the difference in the scheduled resupply times (reinstallation dates) of these items.

Suppose, for example, that there are three items to be repaired, with Z1 =10 (days), Z2= 5, and Z3 = 12. The reinstallation sequence is, therefore, Z2-Z1-Z3; the minimum time between reinstallation of item two and item one is 5 days, and, between items one and three, 2 days.

If all the items are resupplied by their scheduled reinstallation dates, the end- product service time will be equal to Z3, or 12 days. That is, item two will be reinstalled on day 5, item one on day 10, and item three on day 12.

In general, if all items repaired on an end product are resupplied in their scheduled times, the end product service time is equal to z = maxi Zi.

Suppose that at least one item is not resupplied in its scheduled time. Let Xi be the actual item resupply time, defined as the interval in days from the time a repair action is generated until the time the item is returned to the end product for reinstallation. An item delay is then defined by Si=max(O, Xi-Zi).

In the example, assume X=11, X2=14, and X3=12. Although item three is resupplied on schedule, the others are not.

Item two, having the smallest Zj, must be reinstalled first. It is scheduled for reinstallation on day 5, but incurs a 9-day delay, and is therefore not replaced until day 14. Item one is next scheduled to be replaced, but no earlier than 5 days after item two, which is on day 19. Since item one is available on day 11, it can then be reinstalled on day 19.

Item three is last to be replaced. It is scheduled for reinstallation 2 days after item one, or, in this case, on day 21. Since it will be available on day 12, this item can be reinstalled on day 21. The end-product service time is therefore 21 days.

In effect, a delay of one item extends the earliest reinstallation date of items scheduled to be replaced subsequent to the delayed item by an amount equal to the delay. But the maximal extension is equal to the



526 Marshall Rose

maximal delay incurred by an item. Consequently, because of the nature of this reinstallation procedure, an end-product delay occurs when one or more items are delayed, and is given by d=maxiSi. An end-product service time equal to k=z+d can then be written as k=maxiZi+maxiSi.

To simplify calculating the expected end-product service time in this section, the following notational subscripts are used:

h=the { -h+n+1 }th item scheduled to be reinstalled on the end product, where n is the number of reparable items per end product (not a random variable); and,

h = the item having the {h } th longest delay. With the subscript h, Z becomes the length of the {Ii }th largest scheduled item-resupply time. Hence, Z1 represents the item with the maximal scheduled resupply time, so if it is to be repaired, it will subsequently be reinstalled last. Similarly, using the subscript h, Sh becomes the length of the {h }th longest item delay time, where Si represents the item having the maximal delay.

Using this notation, we may express an end-product's service time by k= maXZh+maxSh. This equation implies that an end-product delay is independent of the number and magnitude of the item delays that are equal to or less than the maximal delay occurring on a specific end product. The expected end-product service time becomes E(k) =E[maxZh]+E[maxSh], where E[maxShl=E(d) is the expected end-product delay. Of course, if it is certain that the same items are repaired when an end-product is re- worked, then the service time can be stated as, simply, K=maxZh+ maxSh=Zl+Sl.

In the more general case, each item has a probability Pi of being repaired (they are statistically independent). If the delays are arranged in descending order of magnitude such that Si is equal to the maximal item delay, then the probability that the end-product delay is SI days is equal to Pi, the probability that the first item is repaired. The probability that the end-product delay is S2 days, where S2 is the next delay time in the sequence, is equal to P2(1 -P1), the probability that the second item is repaired multiplied by the probability that the first item is not repaired. Hence, it follows that the probability that the end-product delay is S3 days is equal to P3(1-Pl)(1-P2), the probability that the third item is repaired multiplied by the probability that both the first and second items are not repaired. (See Note 1.) This argument can be continued to cover all possible end-product delays, from Si down to zero days. As a result, the following expression is obtained for the expected end-product delay:

E(d) = S1P1+S2P2(1 - P1) +S3P3(1 -P1) (1 - P2) + * * .




or

E(d)Ah-1 hpt[ o -j)X(l)

where Pj=O for j=O, and fl=0 (1-Pj) = 1 for h = 1. Using a similar argument to find E[maxZh], we can write the expected

end-product service time as:

E(k) = h- ZhPhjl (1-Pi)+Zh1 ShPhUi=? (1- Pj). (2)

Up to this point it has been assumed that there is one unit of an item per end product. If there are Ni units of an item per end product, the expected end-product delay becomes

E(d) h= h1 Sh[l(1 -Ph) ]j-lp i=0 (3)

and, from equation (2) it follows that the expected end-product service time is

E(kC) =E(d ) + h=a1 2h[1- 1 -Ph)^^- ( = pN ( 4)

STOCHASTIC DELAY TIMES

THE PRINCIPAL finding of the previous section was an expression for the expected end-product service time in terms of the item resupply times. It was demonstrated that this expected value is equal to the sum of (1) the expected value of the maximal item delay per end product, and (2) the expected value of the maximal scheduled item resupply time per end product. This section develops, among other things, this same relation for the case of stochastic item delays.

We first calculate the probability that none of the units of a specific item are delayed on an end product. Let F'(s>O), called the conditional probability of a delay, denote the probability that a unit of an item has a delay, given that the unit is repaired. Then the probability that a unit of this item is delayed, without regard to whether or not the unit is repaired, referred to as the unconditional probability of a delay, can be expressed by Fu(s> 0) = P[F(s > 0)], where s is a random variable representing the delay time of a unit, x - Z, and x is a random variable denoting the actual item resupply time. Hence,

1 - Fu(s > 0) =: 1 - P[Fg(s > 0)] (5)

represents the unconditional probability that a unit of this item is not delayed. Now, if there are N units of this item per end product, the unconditional probability that no unit of this item is delayed on an end product is

[1 -Fu(s>0)]N = [1 -P[Fg(s >)]]. (6)

This equation also represents the probability that the maximal value of a unit delay of the item on an end product is zero.



528 Marshall Rose

Next, we calculate the probability that at least one unit of this item incurs a delay exceeding S days on an end product. Of course, when this occurs, the end-product delay will exceed S days. Let the random variable s have a density f(S), where S in the function denotes specific values that the random variable s can assume. The distribution function for s is then F(s < -f f(S) dS, since s must be equal to or greater than zero. Thus, if there is one unit of the item per end product, the conditional probability that the item's delay is equal to or less than t days can be written as

rt Fg(s<)=1-F(s>O) + ff(S) dS,

where f~gfS~dSJO, for <?W,

IJf (S) d (Fg(s >O), for V, and

1- F(s >O) = the conditional probability that the item's delay equals zero,

fg(S) dS =the conditional probability that an item delay between S and S+dS occurs,

W =the smallest length of an item's delay, and V = the largest length of an item's delay.

The expression for the unconditional probability that the item delay is less than or equal to t days takes into account the fact that the item may not always be repaired. Therefore, this expression is given by

rt Fu(s?!)=1-Fu(s>O)+f fu(S) dS, (7)

where f~4~urSdS JOfor ?W,

AJw (S)1dS{Fu(s>O), for t>V, and

1- Fu(s >0) =the unconditional probability that an item's delay equals zero, and

fu(S) dS= the unconditional probability that an item delay between S and S+dS occurs.

Furthermore, because tf fu(S) dS goes to zero when 0=O, and since S is always equal to or greater than zero, it follows that fu(S) can be replaced by P[fU(S)] in equation (7). In addition, equation (5) can be substituted into equation (7), so that

Fu(s<) = -P[Fg(s>O)]+Pf fg(S) dS. (8)

If there are N units of this item per end product, the unconditional probability that all these units have delay times equal to or less than t days




becomes, from equation (8),

F (s?_ )N= [1-P[Fg(s>O)I+Pf fg(S) dS]N.

Therefore, the unconditional probability that the length of at least one unit's delay time is in excess of t days, for this item, is given by (see Note 2):

1- Fu(s < t)N = 1_[1-P[Fg(s>O)]+P| fg( S) dS Nf. (9)

Of course, equation (9) can also be interpreted as the minimal probability that an end product incurs a delay exceeding t days.

The expected length of the maximal unit delay of this item per end product can now be determined. The distribution function corresponding to (9) has the form 1 -Fu(s ? S)N, and the value of its density function can be found from equations (6) and (9) to be 1-P[F(s>O)]N for S=O, and O[Fu(s< S)N]/OS for W_ S_ V, and zero for any other values of S. Hence, the expected length of the maximal unit delay of this item per end product is

K S{a[Fu(s< S)N]h3S } M.

At this point, we want to determine the expected end-product service time. Observe that the maximal unit delay of a specific item on an end product corresponds to the minimal delay that the end product can incur. Therefore, the probability that an end-product delay is equal to or less than t days, Fe(d ), is equivalent to the probability that all units of the items on the end product also have delays equal to or less than t days. This probability can be expressed by

Fe(d?t)- H Fju(s<t)Ni (10)

From this equation, the probability that the end-product delay exceeds t

days is found to be

1-Fe(d < ) = 1-_ Ii1 Fiu (s < t) Ni.

Thus, the value of the density function for the length of an end-product delay, fe(S), is

fe(S) O[Fe(d_ S)I/OiS for W? S<VI,

where W= mini (Wi), and V = maxim (Vi). As a result, the expected end- product delay becomes

E(d) Sfe( S) dS. (11) w



530 Marshall Rose

Of course, it follows from this equation and from equation (4) that the expected end-product service time is then equal to

E(k) Sfe(S) dS+Z EZZh[l -(1 -Ph)Nh]fl (1jpi)Ni W _=

=

This section has so far derived the relation between the expected end- product service time and the stochastic item delays. Nevertheless, calculation of this relation can be difficult, owing to the problems inherent in ob- taining the required data and to the lengthy computational procedure that is needed. The remainder of this paper is, therefore, concerned with de- veloping procedures to overcome these difficulties. The approach employed involves simplifying the density functions of the item delays.

SIMPLIFYING CALCULATIONAL DIFFICULTIES

Simplifying the Density Function for the Maximal Unit Delay of a Specific Item Per End Product

The expression for the expected end-product delay was shown to be re- lated to the conditional distribution functions JFi(s> S), for all i. These functions consist of two parts. Considering a specific item, we first have 1-P(s>0), the conditional probability that a (repaired) unit is not in delay. Next, we have f'(S) dS, the conditional probability that a (repaired) unit incurs a delay of between S and S+dS days.

It is the calculation of the part of f2(S) in the interval W?< s ? V that causes computational difficulties in determining E(d). This part of f'(S) will be referred to as the 'partial density' (see Note 3). The problem can be simplified considerably if a distribution that is analytically tractable is substituted for this part of f(S). At the same time, in order not to in- troduce significant errors between the actual and estimated relations, we want to retain the actual values of N, PF V, W and F'(s>O) when calculating 1-Fu(s < S)N and E(d).

The characteristics of the substituted distribution should be these: that it is continuous; that it has two nonnegative abscissa intercepts, for the smallest (W) and the largest (V) feasible item delay; and, that it is con- venient to deal with mathematically.

The uniform distribution has these properties. Hence, it is assumed that the conditional item delays are uniformly distributed over the interval W < s < V when these delays are greater than zero. Therefore,

f fg(S) dS=Fg(s >0), and w

f fg(S) dS=1 -Fg(s >0) =Fg(s=O).

Based on these specifications, we first want to calculate the density




function for the maximal length of a unit delay of a specific item per end product (a random variable). If there is one unit of a specific item per end product, the conditional probability that this item incurs a delay exceeding S days, F(s > S), has the form (see also Note 4):

F8(s>S) = -Fg(s>O)(S-W)/(V-W)+Fg(s>O), where

-Fg(s>O)(S-W)1(V-W)={F(>) for S?W,

= -Fg(s>O), for S8? V.

The unconditional probability that an item incurs a delay exceeding S days, denoted by Fu(s>S), is then equal to

Fu(s>S) =-P[F(s>O)](S-W)/(V-W)+[F (s >O) (12)

The unconditional probability that an item incurs a delay equal to or less than S days is therefore given by

Fu(s<_!~S) =P[Fg(s>O)](S-W)1(V-W) +I-P[Fg(s>0)].

If there are N units of this item per end product, Fu(s8 S)N represents the unconditional probability that all units of this item have delays equal to or less than S days. Consequently, the expression for the unconditional probability that at least one unit of this item incurs a delay exceeding S days can be written

1-Fu(s? S)N= 1- P[Fg(s>O) (S-W)]/( V-W)

+1 p[Fg(>0) ]]N.(13)

Equation (13) also represents the distribution function of the maximal length of a unit delay, for this item, per end product. The corresponding partial density for an item's delay, denoted by g(S), is found by taking the derivative of this equation with respect to S and multiplying by minus one. The resulting solution has the following form:

g(S) =N[P[Fg(s>O)(S-W)]/(V-W)+1 (14) -P[Fg(s >O)]]IN1 [P[Fg(s>O)]1(V-W)].

Throughout the remainder of the paper it is assumed that these (partial) densities of the maximal length of a unit delay, for each item, per end product, are uniformly distributed. Formulas for determining the expected end-product delay will then be determined.

Calculating the Expected End-Product Delay When the Maximal De- lay Lengths of the Items Have the Same Range and Uniform Density Functions

In this section the expected end-product delay is calculated for the case in which the probability that at least one unit of a specific item is delayed



532 Marshall Rose

on an end product, denoted by RX, is equal to unity for all items. That is, Ri= 1.0 for i= 1, 2, ***, n. This probability can be expressed by

R = 1 - [1 -P[Fg(s >O)]]N (15)

as seen in equation (13). Furthermore, from equation (5), Fg(s>O)= Fu(s>O)/P, and since Fu(s>0)=1-Fu(s=O), it follows that equation (15) is equivalent to R=1-FFu(s=O)N .

Further, assume that s', for all i, are uniformly distributed over the common interval W < 8 V, where X is a random variable representing the maximal unit delay of a specific item per end product. From these criteria, the density function for the length of any item's (maximal) delay is

g(S) = 1/(V-W), and the distribution function ( < S)-) = fg(S) dS can be written

G~s< SS) =(S-W)/(V W). (16)

Equation (16) represents the probability that the maximal value of a unit delay of this item is equal to or less than S days. If there are n items, G(g S) SW)I(VW)], is the probability that all units of these n items incur a delay equal to or less than S days on an end product. The distribution function

1-G(9_< S)n= 1[(SW)I(VW)]n (17)

then represents the probability that the maximal length of any item's delay on an end product exceeds S days. Because the expected end-product delay is equal to the expected value of the maximal item delay per end product, then

E(d)f St[G_ S) ]SI di. (18)

From equation (17), we find that

a[G( SS)8= [n/(V-W)]( -W)8-1. (19)

Substituting equation (19) into equation (18), we obtain

E(d) = [n/( V-W) S W ds. (20)

To solve for E(d) requires integration by parts. Equation (20) can then be shown to be equal to

E(d) =-[n/(V VW)n] [V(V VW) n/n- (V- W) n+ /n(n+l1)].

Simplifying, the expected end-product delay becomes

E(d) = (Vn+W)/(n+1). (21)




As a final point, it can be noted that, if V is reduced by AV and n reduced by An, the reduction in the expected end-product delay is equal to

AE(d) = [An(V- AV--AVn--W) + AV(n+n2)]/

*[(n-An+ 1) (n+ 1)].

To illustrate this, assume that two items are certain to have at least one unit delayed per end product, and the density function for the maximal length of a unit delay, for each item, is uniform. Also suppose that the range of values that these random variables can take is between 1 and 5 days for both. Therefore, we have R1=R2=1.O, and 1_< 5, i=1, 2. The expected end-product delay is, from equation (21), equal to

E(d)-[5(2)+1]/(2+1)-=11/3. (23)

If V is reduced by 1 day for the first item and by 5 days for the second item, i.e., AV = 1 and An = 1, the reduction in the end-product delay is found from equation (22) to be

zAE(d) = [1 (5-1-2-11)+1(2+4)]/[ (2-1+1) (2+1)]=7/6. (24)

Calculating the Expected End-Product Delay When the Maximal Delay Lengths of the Items Have the Same Range and Can Have Partial Uniform Densities

This section assumes that Ri can be unequal among different items. Each item's partial density of maximal delay lengths (when the delays exceed zero) are still assumed to be distributed uniformly over the common interval W5 ?s ?V. Once again, the objective is to calculate the expected end-product delay.

The partial density for the maximal value of an item's delay is, in this case, gi(S) =Ri/(V-W), and the distribution function Gi(s< S) =

fgi(S) dS is given by

Gi(S S) =(S-W)Ri/(V-W)+(1-Ri) for W?S V.

The probability that all n items incur a delay equal to or less than S days is, therefore,

Gi S<S)n Hi-l [(xS-W)Ri1(V-W)+(1-Ri)].

Hence, the probability that the maximal delay of at least one item exceeds S days has the following form:

1-Gis _ S)n= 1_-HJi-1 [S- W)Ril(v VW) + ( 1-Ri)]. (25)

Now, the expected end-product delay, equal to the expected value of the maximal item delay per end product, is

E(d) = St[G( ) S) ]/dSA} dS. (26)



534 Marshall Rose

Taking the partial derivative of equation (25) with respect to S. substituting into equation (26), and rearranging terms, we find (see the Appendix)

E(d)=,- t [SR/VW ] iJ-Z- _f [Rj(Dj+A2)] dS f (27)

where Dj= [V(1-Rj)-W]/R1. (28)

To illustrate the calculation of the expected end-product delay in this case, let 1 A ss ?5, i = 1, 2. From equation (27), the expected end-product delay is

E(d) [SRi/(4)2][R2(D2+S)] dS+f [SR2/(4)2][Ri(Di+S)] dS. (29)

Substituting the values of Dj from equation (28) into equation (29), and then evaluating the definite integrals in equation (29) yields

E(d) = (1/48) [144 (R1+R2)-112R1R2].

Now, if R1 =R2 = 1.0, then the expected end-product delay is found to be E(d)= 11/3, which is also the solution previously obtained in equation (23). Additionally, if the largest delay is reduced by one day for the first item and by 5 days for the second item, the change in the expected end-product delay is

4

AE (d ) = 11 /3 - ISRi / (4 - 1)] dS.

Since Rf is equal to unity, AE(d) =7/6, which is also the solution obtained in equation (24).

Calculating the Expected End-Product Delay When the Maximal Delay Lengths of the Items Can Have Both Unequal Ranges and Partial Uniform Densities

In the last case, the expected end-product delay is calculated when both Ri and the interval for each item's maximal delay can be unequal among different items.

From equation (11), the expected end-product delay may be expressed by

E(d)=f s [Fe(d>S)]/OS} dS. (30)

If the partial density of the maximal delay length of each item is described by a uniform distribution, then equation (30) can be modified:

E(d) = -o S-[AFe(d> S)]/S} (31)

where AS is equal to unity. This transformation can be made because, in




this case, Fe(d> S) is linear and continuous for a range of unit intervals in S, as long as Wi and Vi, for all i, are integers.

Furthermore, it can be calculated from equation (10) that

Fe(d > ) =1H- 1 G (S S). (32)

Hence, if each item's maximal delay length si is uniformly distributed over the interval Wi_ si Vi when the delay exceeds zero, it can be determined from equation (25) that

1 - 1 Gi SS)=1 - t-1 [l(1 - R + sS) dS], (33)

where Si=Wi when OSWi, Si=S when S<Vi, Si=Vi when SVi, and

gi(S) = Ril (Vi - Wi). (34)

Substituting equation (34) into equation (33) yields

1 t1 Gi (S _ S) =1- i-l [(1 - Ri ) + I[Ril(Vi-Wi)] ds] (35)

Replacing equation (32) by equation (35) and simplifying, we obtain

Fe(d> S) = 1 - H 1 [(1-Ri) +S-Wi)Ri1(Vi- Wi)] (36)

The calculation of the expected end-product delay, given in equation (31), is demonstrated next by making use of equation (36). Several numerical examples are presented.

Sample Problems

Example A. Let n = 2,0 _ 5,02 <10,andRl=R2=1.0. From equation (36) the probability that the end-product delay exceeds 9 days

is calculated; the results are shown in the second column of Table I. In the third column, the change in this probability is determined as S changes by one unit. The final column simply represents the product of columns one and three. It can be seen from equation (31) that the summation of the entries in the fourth column yields the expected end-product delay. In this first example, E(d) is 5.9 days. Of course, column two and/or three reveal that 0.38 is the probability that 5.0 _d ' 7.0.

Example B. Let us now assume that the probability that a unit of the first item incurs an end-product delay is reduced by 0.7. The following information is applicable: n=2, 08_<?5, 0<A2?10, R1-0.3, and R2=1.0.

Using the same technique employed in Example A, we calculate E(d) in Table II, and find it to be 5.62 days. Thus, the reduction in R1 of 0.7 reduces E(d) by 0.28 days. Once again, it can be seen that 0.324 is the probability that 5.0 'd _7.0.



536 Marshall Rose

TABLE I

CALCULATING THE EXPECTED END-PRODUCT DELAY: EXAMPLE A

S Fe(d>S) -AF,(d>S)/AS - S[AF6(d>S)//AS]

0 1.00

I o.98 0.02 0.02

2 0.92 o.o6 0.12

3 o. 82 0.10 0.30

4 o. 68 0.14 0.56 5 0.50 o. i8 0.go

6 0.40 0.10 o. 6o 7 0.30 0.10 0.70 8 0. 20 0.10 0. 8o 9 0.10 0.10 0.90

10 0.00 0.10 1.00

Totals 1.00 5.90

Example C. In Example B, column two was recalculated from the applicable information. Alternatively, incremental relations can be used to find E(d) after Rj is reduced by ARh. That is, expanding equation (36), we obtain

F, (d>S) = 1- [(1-R) +R (Si- W)/(V1- W3)] (37 [(1-R2)+R2(S2-W2)/(V2-W2)I (

Now suppose R2 is reduced by AR2, but V2 is not affected. The probability that

TABLE II CALCULATING THE EXPECTED END-PRODUCT DELAY: EXAMPLE B

S Fe(d>S) -AFe(d>S)/AS -S[AF6(d>S)/AS]

0 I. 000 I 0.924 0.076 0.076 2 o.836 o.o88 0.176 3 0.736 0.100 0.300 4 o.624 0.112 0.448

5 0.500 0.124 o.620

6 0.400 0.100 o.6oo 7 0.300 0.100 0.700 8 0.200 0.100 o.8oo 9 0.100 0.100 0.900

10 0.000 0.100 1.000

Totals I.000 5.620




the end-product delay exceeds s days can then be written as 'A A A

Fe (d>S)-AFe (d>S) = 1- [(1-R1) +R1 (Si-W1)/(VI-W1)] (38) [ (1-R2+ AR2 ) + { R2 (S2 -W2) -AR2 (S2 -W2) }/( V2 -W2)].

To find the reduction in Fe(d > S) that results when R2 is reduced by AR2, (38) is subtracted from (37):

AFe (d > S) = -AR2 (Rl-l )-(1-R1 )AR2 (S2-W2) / (V2-W2) -Ri (S- (39 WI)AR2(S2-W2)/[ (VI-WI) (V2- W2)]+,AR2(R S-I- WI) /(VI- WI).

To illustrate the use of the incremental relation in equation (39) to calculate

TABLE III CALCULATING THE EXPECTED END-PRODUCT DELAY: EXAMPLE C

S LwFe(d>S) Fe(d>S) -AF(d>S)/ S[AFe(d>S)/ AS As]

0 0.1400 o. 86oo I 0.1368 0.7872 0.0728 0.0728

2 0.1312 0.7048 o.o824 o.I648

3 0.1232 O.6128 0.0920 0.2760

4 0.1128 0.5112 o.ioi6 0.4064

5 0.1000 0.4000 0.1112 0.5560

6 o.o8oo 0.3200 o.o8oo 0.4800 7 o.o6oo 0.2400 o.o800 0.5600 8 0.0400 o.i6oo o.o8oo O.6400 9 0.0200 o.o8oo o.o8oo 0.7200

T0 0.0000 0.0000 0.o800 o.8ooo

Totals o.86oo 4.6760

the expected end-product delay, let R2 be reduced by 0.2. The following information obtains: n=2, 0<_ sl5,0 ?52<10, R1=0.3, and R2=0.8.

From equation (39), AFe(d > ,) is calculated in the second column of Table III. The third column is derived by subtracting the second column of this table from the second column of Table II. The remaining calculations are similar to those of the preceding examples, and E(d), in this case, equals 4.676 days. The probability that 5.0 <d <7.0 has now been reduced to 0.2712.

SUMMARY

DETERMINATION OF the optimal time needed to complete the repair of an end product is an important aspect of efficient resource allocation for firms engaging in this type of activity. The first step requires that the relation between the tasks to be performed and the time to complete the end-product



538 Marshall Rose

repair be ascertained. This paper has derived this relation for a specific type of repair, namely, one that consists of reworking the recoverable items of the end products, and replacing them in a given sequence. In cases where the end-product repair consists of tasks in addition to recoverable item repair and reinstallation, the methods described here can be used to depict a single stage in the total repair job.

The paper has demonstrated how to calculate the expected end-product service time. The approach involves simplifying the density functions of the item delays. A need for this simplification arises, owing to the effort involved in calculating the actual densities and their effect upon the end- product service time even when there are small numbers of items per end product.

As a result of determining both the relation between the item resupply activities and the end-product service time and the way in which to calculate the expected end-product service time, it is now possible to compute the optimal resupply times for a given service time by methods suggested by BLACK AND PROSCHAN,[11 BLITZ,[2] JACOBSON,[3] AND ROSE.[4] The optimal service time could then be determined using an approach described by Rose. IM

APPENDIX

To DERIVE THE expression on the right-hand side of equation (27), the partial derivative of equation (25) is taken with respect to X, and after rearranging terms, we get

A[S<S)n] n j Ri7 [(1-Rj)+ S-W)Rj/(V-W)1

E S =;S (V-W)[(l-Ri)+ S-W)Ri/(V-W) J (

Substituting this equation into equation (26) yields

I n A vn Ri H [ (1-R j) +(S'-W) R /( V-W) ];d 4

wi., (V-W) [(1-Ri) + (s-W)Ri1(V-W) l

Collecting terms, equation (41) becomes n f v nA

E(d)=E [SRi/(V-W)] H [(l-Ri)+( -W)Rj/(V-W)] (42) i~~~l W j~~=1, nisi

where n

H [(1-Rj)+(S-W)Rj/(V-W)]=1 for n=1. j=1, jii

Simplifying further, we can write equation (42) equivalently as

n f v E(d) {f [ARi/(V-W)n] I [(1-Rj)(V-W)+(S-W)Rj1 dS}. (43)

j=1, isi




Letting Uj = (1 -Rj) (V - W) and Tj = Uj/Rj, and then substituting into equation (43), we obtain n {v nA

E(d)= {f SRil(V-W)n1 II fRj(Tj+S-W)I d (44) s~~l t w j~~=1, j~i

If we now let Dj= T;-W= Uj1R;-W= (1-Rj) (V-W)1Rj-W,

then D =[V(1 -Rj) - W] /Rj. Equation (44) therefore finally becomes

An l TVW in=

A

E (d)= J [Si/VWn [Rj(Dj+S-)] dS?

NOTES

1. These multiplicative probabilities can be used to generate the density function for end-product delays, and therefore indicate the level of confidence that can be placed in the expected delay.

2. Equation (9) represents the distribution function for end-product delays and, as in Note 1, can be used to estimate the variability of the resulting delays.

3. Because the objective is to calculate expected values, we need only concen- trate on these partial densities. This obtains as a result of the product of f9(S=O) and S=0 being zero, and also since f9(S) is equal to zero when S is not equal to zero and not in the interval W<S<V.

4. In the discrete case, when the random variable s can assume only integer values, this becomes

Pg(s> S) --[Pg(s>O) (S- W+1-)/(V- W+1) ]+Pg(s>O), where

-[P8(s>o) (S-W+,)/(V-W+1)]={? for S<W-1, al(-Pg(s>O), for S>V

and Pg(s>S) is the discrete analogue of Fg(s>S).

ACKNOWLEDGMENTS

THIS PAPER was written at the Center for Naval Analyses. The author is indebted to C. VEENDORP of Tulane University for his helpful comments on an earlier draft.

LIST OF SYMBOLS

Z=The fixed scheduled resupply time of a reparable item, defined as the interval in days from the time a demand for a repaired unit is generated until the earliest date that this item can be reinstalled on the end product.

X, x =The actual resupply time of an item, defined as the interval in days from the time a demand for a repaired unit is generated until the date that this demand has been filled.

S s =The delay time of an item, defined as the difference between the actual item resupply time and the scheduled item resupply time.

K, k-=The end-product service time. P =The probability that a unit of an item is repaired.



540 Marshall Rose

N =The number of units of an item per end product. n =The number of items per end product (not a random variable). z =The minimal end-product service time, i.e., the maximal scheduled item

resupply time per end product. d =The end-product delay, i.e., the maximal item delay per end product.

F9 (s >0) =The probability that a unit of an item is delayed, given that it is repaired.

Fu(s > 0) = The unconditional probability that a unit of an item is delayed. f(S) =The density function for the random variable s.

F(s ?t) =The distribution function for the random variable s. Fe(d _< ) =The distribution function for the random variable d.

fe(S) =The density function for the random variable d. V =The largest positive value of an item delay. W =The smallest positive value of an item delay.

s =The maximal unit delay of an item on an end product. g(S) =The density function for the random variable S.

R =The probability that at least one unit of an item causes an end-product delay.

G(s ' 4) = The distribution function for the random variable s.

REFERENCES

1. G. BLACK AND F. PROSCHAN, "On Optimal Redundancy," Opns. Res. 12, 581-588 (1959).

2. M. BLITZ, "Optimum Allocation of a Subsystem Spares Budget," Sylvania Applied Research Memorandum No. 252, June, 1961.

3. H. JACOBSON, "The Allotment of Equipment Spares on the Basis of a System Cost-Performance Analyses," Sixth Annual Meeting of the Opns. Res. Soc. America, Western Division, September, 1959 (unpublished).

4. M. ROSE, "A Decomposed Network Computation ior End-Product Repair Cost Curves," Center for Naval Analyses Professional Paper No. 19, February, 1970.

5. --, "An Investment Model for Reparable Assets: The F-4 Case," Center for Naval Analyses, INS Research Contribution No. 31, July, 1969.



computing the expected end-product service time using stochastic item delays

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