
OPERATIONS COURSE  DIVISION
Exercise 1
On the basis of Army's initial instruction (attached) determine the time
required for one regiment with 2 artillery battalions to move to the border
area and deploy there when:
 depth of regiment column  30 km
 depth of the area of deployment  5 km
 average speed of march  20 km
 distance to the border area  120 km
Answer Use the formula:
t=time the regiment needs to deploy in new area
D=distance to border area
V=average speed of march
G_{K}=depth of regiment column
G_{R}=depth of the area of deployment
t_{p}=time spent for halts
Exercise 2
On the basis of the same instructions determine the time for the division
to deploy and occupy the designated departure area when:
 depth of division columns on 3 routes  80 km
 depth of the area of deployment  35 km
 average speed of march  25 km
 distance to departure area from start line  125 km
 time to complete engineer work  8 hours
Answer
To solve the problem a combination of calculations and application of
norms can be used. Calculations are done both on general terms (time to
complete the deployment for the entire division) and calculations to determine
the time of deployment of different echelons such as first and second echelons,
rear service troops, air defense troops, attack helicopters etc. The method is
the same for all categories. In each case normatives are applied to determine
the time required for specific tasks such as engineer work, establishment of
fire system, delivery of supplies, loading and unloading, establishment of
security, etc.
A. Calculations in general terms:
T=5 + 1 + 45÷15 + 8=17 hrs'.
B. Calculations in specific terms
1. For the first echelon regiments (depth of column 30 km, depth of area
of deployment 10 km)
T=150÷25 + 1 + (30  10) ÷ (0.6 x 25) + 8 hrs .....(1)
T=6 + 1 + 1:20 + 8 hrs=16:20 hrs
2. For second echelon regiments
(use same formula)
T=170÷25 + 1 + (30  10) ÷ (0.6 x 25) + 8 ......(2)
T=6:48 + 1 + 1:20 + 8=17:08 hrs
3. For the rear service
(use the same formula)
T=195÷25 + 1 + (10  5) ÷ (0.6 x 25) + 8=17:08 hrs
Notes
(1) Movement distance for first echelon regiments is assumed to be 150 km on
the basis that the regiments should move another 25 km from the division
deployment line to reach their designated position in the first echelon (125 +
25=150).
(2) The distance for the second echelon is assumed to be 170 km on the basis
that the regiment moves 35 km from the head of the column (depth of first
echelon regiment plus 5 km interval) and to reach its designated area it should
move another 10 km from the divisions to reach its area of deployment in the
second echelon (125 + 35 + 10=170 km).
(3) The distance for the rear service is assumed to be 195 km to include its
distance from the head of the column (65 km) and the 5 km interval (125 + 65 +
5=195 km).
Exercise 3
As the commander of 16 MRD read the instructions of the army commander
(attached), clarify your mission and calculate the following.
 depth and width of the division missions
 width of the area of penetration
 required rate of advance
 number of regiments required in first and second echelon
Answer
1. The depth and width of the division mission is measured on the map in the
following manner
a. Depth of the immediate mission 17 km with a width of 8 km
b. Depth of long range mission is 21 km and a width of 12 km
c. The width of penetration area is 3.5 km, requiring the forces and means of
two regiments (2 km and 1.5 km of penetration area assigned to them), the
remaining 4.5 km of the front should be covered by part of one of the regiments
of the penetration area (the right flank regiment) and forces and means of
another regiments.
2. The immediate mission to be accomplished in  hours (assume 7
hours) therefore the average rate of advance is 17 ÷ 7=2.5 km/h. The long
range mission is to be accomplished in  hours (assume 5 hours), therefore
the average rate of advance should be 21 ÷ 5=4 km/h.
3. On the basis of the width of penetration area which is 3.5 km (the norm
is 2 km per regiment) and the overall width of the division sector i.e. 8 km
there can be two alternatives for the echelonment of the troops:
a. Two regiments in the first echelon, one BMP regiment covering the
front and one regiment in the second echelon to be committed after the
penetration of the enemy's brigade defensive position, while the BMP regiment
will then constitute the second echelon to be committed after the penetration
of enemy's division defenses.
b. Three regiments in first echelon, with the BMP regiment coming to the second
echelon after the attack is begun.
4. There the commander can tentatively determine the following:
 rate of advance
 direction of the main attack and the width of penetration area
 combat formation of the division for the attack
5. Issues mentioned in point 4 are further examined , elaborated and
confirmed during the estimate of the situation and
"recognaisrovka".
Exercise 4
Given the available time to prepare and plan the divisions offensive
battle (from _____________ to _____________) schedule the measures in a table
to include time to organize, prepare and plan the battle.
Answer
(Issue prepared table)
Exercise 5
On the basis of the mission assigned to the division by the army and the
deductions of the clarification of the mission prepare initial instructions to
regiments and a separate combat instructions to BMP regiment to cover the
border.
Answer
(Issue prepared texts)
Exercise 6
As the chief of operation section prepare for the commander calculations
to determine the required correlation of forces and means to support the
assigned rate of advance.
Answer
To solve the problem use the rate of advance nomogram. To use the nomogram
first you have to find the F factor.
F factor=D ÷ KTV_{max}
D=distance (depth) of the mission;
k=terrain coefficient:
1.25 level
1.00 roughlevel
.75 rugged hills
.75 urban sprawl
.50 mountainous
T=time required for action in days and fraction of days;
V_{max}=theoretical speed in km/day;
F=38 ÷ (1.25 x 1 x 60)=0.5
Now see in the nomogram what correlation of forces and means is required
when the F factor is 0.5. The answer is 4.3:1
Exercise 7
Determine the width of the main sector on the basis of the following
facts:
 width of the overall area of the division is 8 km;
 overall correlation of forces and means is 3: 1;
 required correlation of forces in the main sector is 4.3:1;
 correlation of forces and means below which we can not drop in the rest of
the division area is 2:1;
Answer
Use the size of the sectors formula which is as follows:
W_{m}=width of the main sector
W_{o}=width of the overall area
C_{o}=overall correlation of force
C_{m}=required correlation of force
C_{s}=correlation of force below which one can not drop in the rest of
the action area.
W_{m}=8 (3  2) ÷ (4.3  2)=8 ÷ 2.3=3.5 km.
Note: You can increase the width of the main sector area
by accepting a lower correlation of forces and means in the rest of the area of
division attack. For instance use 1.3:1 for secondary sector, then:
W_{m}=8 (3  1.3) ÷ (4.3  1.3)=8 x 1.7 ÷ 3=4.5 km
Exercise 8
Prepare the table of correlation of forces and means and summarize
deductions on the basis of the correlation of forces and means. For calculation
purposes the enemy forces and means are filled in the table.
Answer
Issue prepared table. The following issues are to be clarified.
 does overall correlation match the requirement of the rate of advance?
 can the required correlation of forces and means be established in the main
direction?
 based on the enemy's commitment of his second echelon, determine when second
echelon of our troops should be committed into battle, (the soonest and the
latest time of the commitment).
Exercise 9
Calculate the time for 16 MRD to advance from the assembly area (departure
area) and deploy for shift in to attack from the line of march when:
 distance of attack line from enemy forward line=1 km:
 distance of line to deploy into company column=4 km:
 distance of line to deploy into battalion column=12 km:
 distance of regulating line to line of deployment into battalion column=20
km:
 distance of regulating line from start line=40 km:
 distance of start line from assembly area=5 km:
 depth of first echelon regiments=30 km:
 interval between 1st and 2nd echelon regiments=10 km:
 movement speed into attack=8 km:
 average speed during march=24 km/h:
Answer
This calculation can be done either by using several formulas to calculate
each section of the advance and finally to combine them together or by filling
in the prepared tables.
a. Using formulas:
t_{a}=time for crossing final deployment into line of attack minus
(H) in minutes:
D_{a}=distance of line for going into attack formation from the forward
edge of the enemy position in km:
V_{a}=rate of movement in attack formation in km/h:
t_{r}=time for crossing the line of deployment into company column (H
minutes):
D_{r}=distance of line of deployment into company columns from the line
of deployment into attack formation in km:
v=average rate of movement of units in mounted formation during march:
t_{b}=time for crossing line of deployment into battalion column (H
minutes):
D_{b}=distance of line of deployment into battalion columns from line
of deployment into company columns in km:
t_{rr}=time for crossing the last regulation line (lrl) prior to
deployment into battalion columns (bc):
D_{rr}=distance of lrl to bc:
t_{i}=time for passing the start line (sl):
D_{i}=distance of sl from lrl:
t_{vit}=time to begin movement from assembly area
D=distance of sl from assembly area:
t_{i}=time for crossing the sl by second echelon:
G_{k}=depth of the mounted column of first echelon in km:
D_{k}=distance between the tail of the first echelon and the head of
the second echelon in km:
60 and 90=coefficient for conversion of time in minutes for the average speed
during the deployment on each line:
t_{a}=(D_{a} x 60) ÷ V_{a}=(1 x 60) ÷ 8=7.5 [
H  7.5 min or H  00:07]
t_{r}=t_{a} + (D_{r} x 90) ÷ V=7.5 + (4 x 90)
÷ 24=7.5 + 15=H  22.5 min
t_{b}=t_{r} + (D_{b} x 60) ÷ V=22.5 + (12 x 60)
÷24=22.5 + 30=52.5 min
t_{b}=H  00:53
t_{rr}=t_{b} + (D_{rr} x 60) V=52.5 + (20 x 60) ÷
24=52.5 + 50
t_{rr}=H  01:43
t_{i}=t_{rr} + (D_{i} x 60) ÷ V=01:43 + (40 x 60)
÷ 24=01:43 + 01:40
t_{i}=H  03:23 hrs
t_{vit}=t_{i} + (D x 90) ÷ V=03:23 + (5 x 90)
÷24=03:23 + 00:19
t_{vit}=H  03:42
t_{i}'=t_{i} + (G_{K}
b. Using the tables:
The same calculations can be done by using the tables and filling in the
numbers in each line. Such tables are preprepared in advance in blanks and the
operation staff can use them to do calculations taking different options into
consideration.
Exercise 10
Determine the time and distance to the line of meeting with a counterattacking
enemy reserve when:
 the enemy reserve (up to 2 mech and 2 tank battalions) is sighted 28 km from
the forward line of division's attacking troops:
 enemy's speed of advance is about 15 km/hr:
 a delay of 30 minutes is expected in the enemy movement due to a narrow area
along the road:
 planned air strikes and artillery fire's are expected to delay the enemy for
another 40 minutes:
 the speed of own attacking forces in the first echelon is 4 km/hr due to
isolated enemy's strong points across the front:
 the attack on enemy's position 4 km further in the depth (the troops are
expected to reach there within an hour) is expected to delay 45 minutes for
minor regroupment.
Answer  Use the following formula:
:
t_{v}=expected time of meeting the enemy in hours:
D=distance between opposing forces in km
t_{n}=total delay time for own forces in hours
V_{n}=speed of movement of own forces
t_{p}=total delay time for enemy forces
V_{p}=speed of movement of enemy forces
Solution:
t_{v}={28 + [(0.75 x 4) + (1.15 x 15)]} ÷ (4 + 15)
t_{v}=[28 + (3 + 17.35)} ÷ 19
t_{v}=(28 + 20.25) ÷ 19 : t_{v}=2.54 or 2 hrs and 33
minutes
Now to determine the distance of meeting with the enemy use this formula:
l_{p}=V_{n} (t_{v}  t_{n} )
l_{p}=4 (2.54  0.75)
l_{p}=7.16 km
This means that the first echelon forces will be able to destroy the enemy in
this intermediate defensive position before it can launch its counterattack,
provided the enemy's reserve is delayed by air strikes and artillery fire for
not less than 40 minutes and the enemy does have to slow down 30 minutes delay
to cross the narrow pass and own forces do not take more than 45 minutes to
regroup in order to continue the attack.
If the division commander determines that the line of meeting with the enemy is
not convenient, he can chose to repel the counterattack from a line further in
back or he might want to further delay the enemy so that the first echelon
troops can move further than 7.16 km before the enemy launches his
counterattack. This calculation can also be conducted by filling in the
preprepared form.
Exercise 11
On the basis of the assumptions which were mentioned in exercise 10 determine
the number of antitank weapons (ATGM and AT guns) to repel the enemy tanks
when:
the number of enemy tanks are estimated to be 80:
no less than 50% of enemy tanks must be destroyed:
the probability of destruction of a single tank by one weapon with one shot is
0.2:
up to 8 rounds may be fired by each weapon in the time the tanks are located in
the effective zone of fire:
Answer
To determine the number of antitank weapons you can use the formulas or the
nomogram.
1. Formulas:
M_{n}=degree of destruction of artillery tanks (in percentage);
P_{1}=probability of destruction by one weapon in one shot;
N=required number of anti tank weapons to accomplish the mission;
m=weapon rate of fire or number of shots one weapon can shoot during the time
the target is in within range;
M=expected number of attacking tanks.
This method is a lengthy one and difficult to use under field conditions.
2. The nomogram can be used for up to 40 tanks. However for over 40 tanks
calculation can be made on the basis of what can be calculated by the
nomograms: 60=40 + 20, or 77=40 + 37 etc.
In our example the following calculation can be done on the nomogram:
draw a perpendicular line from 0.5 make mark on the "Required amount of
destruction of targets" scale to intersect with the probability of target
destruction by one round 0.2 curve. From this point draw a horizontal line to
the intersection with the "Number of attacking ground targets  40"
line and then a vertical line up to the "Number of firing by one weapon 
8" line. From this point go along the horizontal to "Required number
of antitank weapons" scale and read 17 then multiply it by 2 to get the
required number of AT weapons for 80 tanks: 17 x 2=34 AT weapons.
Exercise 12
The division commander decided to use the divisional AT reserve at the line of
repulsion of the enemy's counterattack, which per exercise 10 will be 7 km up
from the current forward line of the first echelon troops. The AT reserve is
now 8 km from the forward line. The enemy is 21 km away from the line of
repulsion of counterattack with an expected delay of 1 hour and 10 minutes on
the way (due to planned air strikes and artillery's fire). The speed of advance
of the enemy is 15 km/hr the effective range of AT weapons is 3 km. The AT
reserve will need 30 minutes to deploy on the fire line and prepare for action.
It's speed of movement 12 km/hr.
Determine how much time is available for the division commander and staff to
assign mission to the AT reserve.
Answer Use the following formula:
t=time available for the commander and his staff to assign mission
D=distance of the enemy to the line of contact
t_{p}=total delay time for enemy forces
V_{p}=speed of movement of enemy forces
d=effective range of AT weapons
t_{n}=total time required for AT reserve to move to the line of the
repulsion of the enemy's counterattack and time to prepare for action.
1 First determine the t_{n}/
t_{n}=time for AT reserve to move and prepare for action
V_{n}=speed of movement of AT reserve
t_{r}=time to prepare for action
t_{n}=(8 + 7) ÷ 12 + 0.5=1.75
2. Now determine the t:
t={[D + (t_{p} x V_{p})]  d} ÷ V_{p} 
t_{n}
t={[21 + (1.15 x 15)] 2} ÷ 15= 1.75
t=(21 + 17.5) 2 ÷ 15= 1.75
t=38  2 ÷ 15= 1.75
t=36 ÷ 15= 1.75
t=2.25  1.75=0.5
t=30 minutes.
This means that the division commander and staff should assign mission to the
AT reserve within 30 minutes (not later) so that the AT reserve will arrive and
get prepared on the line of repulsion before the enemy tanks reach the
effective range of AT weapons at the line.
Exercise 13
In planning the commitment of divisions's second echelon the area within 4 km
of the enemy intermediate defensive position, which is to be attacked by the
secondechelon, is open and when the regiment moves and deploys to company and
platoon columns and assumes the combat formation in this area, it should be
covered by artillery strikes conducted on enemy strong points at the line of
commitment and on the flanks. The line of attack is 1 km and the line of fire
safety is 400 m from enemy position. The speed of movement is 20 km/hr and
speed of attack is 6 km/hr. Determine the duration of artillery strike to cover
the deployment and attack of the secondechelon regiment.
Answer
To determine the duration of artillery strike we have to find out the time it
takes the regiment to deploy and move to the line of attack and then to the
safety line of fire in front of enemy position (400m).
Use equations:
t=t_{a} + t_{r} ...........(1)
t=time of artillery strike;
t_{a}=time for crossing the attack line up to the fire safety line;
t_{r}=time spent from deployment in the company columns up to the line
of attack;
D_{a}=distance of attack line from enemy position;
d=distance of fire safety line to the enemy position;
V_{a}=speed of movement in attack;
D_{r}=distance of deployment into company columns;
V=speed of movement of troops;
Solution:
t_{a}=(1 0.4) x 60 ÷ 6=6 minutes
t_{r}=3 x 90 ÷ 20=27 ÷ 2=13.5 minutes
t=t_{a} + t_{r}=6 + 13.5 approximately 20 minutes
This means that the artillery strike should be conducted for 20 minutes to
cover the commitment of the second echelon regiment.
Exercise 14
The division has accomplished its immediate mission and continues the attack in
depth to complete the destruction of the enemy forces in its tactical zone and
accomplish the long range mission by the end of the day.
The first echelon regiments are fighting with the enemy forces, which conduct
delaying action and cover the withdrawal of its main forces across the Schmalin
River. Two enemy battalion size columns 15 km from the river are withdrawing to
the river apparently to establish defense on the river.
The division commander has decided to assign a forward detachment to prevent
the arrival of these enemy battalions on the river. The distance to the enemy
columns from the head of the assigned forward detachment is 15 km. A 20 minute
delay is expected in the movement of enemy columns due to planned friendly air
strikes. The speed of movement of enemy's columns is 15km/hr.
Determine the expected time and rate of overtaking of the withdrawing enemy by
division's forward detachment.
Answer
Here the crucial issue is to overtake the enemy before he is able to cross the
river and establish defense there.
1. First determine how long does it take the enemy to cross the river:
t=D + (t_{p} x V_{p}) ÷ V_{p}
t=time it takes the enemy to reach the river;
D=distance of the enemy to the river;
t_{p}=expected delay in enemy's movement;
V_{p}=speed of enemy's movement;
t=15 + (0.5 x 15) ÷ 15
t=15 + 7.5 ÷ 15=22.5 ÷ 15 1.5
t=1 hr and 30 minutes
Therefore the enemy's columns must be overtaken within less than 1.5 hours.
2. Assume that the speed of movement of the forward detachment is 20 km/hr:
Now  t_{o}=[D  (t_{p} x V_{p} )] ÷
(V_{n}  V_{p} )
t_{o}=time to overtake the enemy (hours);
D=distance to the enemy;
t_{p} and V_{p}=same as in 1;
V_{n}=speed of movement of own forces (forward detachment);
t_{o}=15  (0.5 x 15) ÷ 20  15
t_{o}=15  7.5 ÷ 5
t_{o}=7.5 ÷ 5=1.5
t_{o}=1 hour and 30 minutes
This means that at a speed of 20 km/hr the forward detachment can not catch the
enemy columns before they reach the river. In order to overcome this, either
the speed of movement should be increased or the enemy should be further
delayed by air strikes, airborne assault troop, artillery fires, mines etc.
3. In order to find the required speed of movement of the forward detachment to
overtake the enemy in one hour the following calculation can be done, using the
formulas:
V_{n}={[D  t_{p} x V_{p})] + (t_{o} x
V_{p})} ÷ t_{o}
V_{n}=[15  (0.5 x 15)] + [(1 x 15)] ÷ 1
V_{n}=7.5 + 15 ÷ 1=22.5 ÷ 1=22.5
V_{n}=22.5 km/hr.
Therefore the speed of movement of the forward detachment should be at least
22.5 km per hour to ensure the interception of the enemy within an hour. At
that time the enemy will be 7.5 km from the river.
D=(t x V_{p})  (t_{p} x V_{p})
D=(1 + 15)  (0.5 x 15)
D=15  7.5=7.5 km
Exercise 15
Issue combat instructions to first echelon requirements and AT reserve on
repulsion of enemy's counterattack.
Answer
Issue prepared text
Exercise 16
Prepare combat instructions to the second echelon regiment on its commitment
into combat.
Answer
Issue prepared text.
Exercise 17
Prepare combat situation report to the Army staff on repulsion of enemy's
counterattack and commitment of the second echelon.
Answer
Issue prepared text.
Exercise 18
Prepare combat instructions to the forward detachment to move to the river,
intercept the retreating enemy and to establish bridgehead on the river.

