SECTION IV REQUIREMENTS FOR 1000 RATIONS PER DAY

SB 10-495-1/NAVSUP PUB 436/MCO P10110.26B/AFMAN 41-121

SECTION VI

REQUIREMENTS FOR 1000 RATIONS PER DAY

x 5 (the staff fed in 1000s) x 5 (number of days) = 180

The ration factor is a subsistence supply factor which

cans.

denotes the average daily number of "units of issue " or

"pounds of issue" of each food item required to support

If requirements are to be computed for any portion of

1000 persons per day. For example, 5.85 cans of

the 10-day menu other than days 1-5 or 6-10, ration

Bacon, precooked, sliced, No. 10 can, are required for

factors for each item must be computed based upon

100 staff for the 10-day menu (See page 6-3). For 1000

total menu requirements for the specific days desired.

staff 58.5 cans are required for the 10-day period. The

To compute ration factors for days other than 1-5 or 6-

average daily requirement for 1000 staff for 10 days is

10, the Recapitulation of Menu Issues, Section IV-N,

5.85 cans (58.5 cans divided by 10 days). The total

page 4-87 must be used. For example, to compute a

units of bacon required for 5000 staff for the 10-day

ration factor for days 1-3 for bacon for staff, determine

menu is 292.5 cans [5.85 multiplied by 5, (staff fed in

the total quantity required for days 1-3 from the

1000s) multiplied by 10 days]. The weight of each can

Recapitulation of Menu Issues (2.60 cans is the total

is 4.500 pounds (conversion factor); therefore, the

requirement for days 1-3.) Multiply 2.60 by 10 (to

average number of pounds required for 1000 staff for

increase the total quantity for days 1-3 per 100 persons

each day is 26.33 (5.85 cans multiply by 4.500).

to that for 1000), and then divide by the number of days.

The result is the quantity required per day for 1000 to be

To facilitate the computation of requirements for 5-day

fed (8.67 units).

increments, ration factors in units are provided for days

1 through 5, and 6 through 10. The ration factor should

To compute the daily requirement for 5000 persons to

be multiplied by the number of persons subsisting

be fed, multiply 8.67 (1000 fed) x 5 (staff fed in 1000s),

(adjusted to thousands) and the resulting figure

which equals 43.35 cans of bacon per day. Multiply by 3

multiplied by the required number of days.

to get the total requirements for staff for days 1-3 days

(130.1 cans).

For example, the total units of bacon required for 5000

staff for days 1-5 is 7.20 (the ration factor for days 1-5),

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